Preface
I.
xiv
The Nature ofThermodynamics 1.1
What is ther1nodyn
1.2
Definitions
I
3
ics?
5
1.3
T he kilomole
1.4
Limits of the continuum
1.5
More definitions
1.6
Units
I.7
Temperature and the zeroth law of thermodynamics
1.8
Temperature scales
•
6 6
7
9
Problems
I0
12
16
v
Contents
•
Yl
2.
19
Equations of State 2.1
Introduction
2.1
Equation of state of an ideal gas
21 22
.
2.3
23
s' equation for a real gas
2.4
P-v-T surfaces for real substances
2.5
Expansivity and compressibility
2.6
An application Problems
25 27
29
31 •
3.
The First Law ofThermodynamics Configuration work
3.2
Dissipative work
3.3
Adiabatic work and internal energy
3.4
Heat
3.5
Units of heat
3.6
The mechanical equivalent of heat
3.7
Summary of the first law
3.8
Some calculations of work
40 41
43
Problems
4.
37
3.1
45 46
46 47
48
Applications of the First Law 4.1
Heat capacity
4.2
Mayer's equation
4.3
Enthalpy and heats of transformation
4.4
Relationships involving enthalpy
4.5
Comparison of
4.6
Work done in an adiabatic process Problems
64
SI
53
u
54
and h
59
61 61
57
35
Contents
5.
••
VII
5.1
The Gay-Lussac-Joule experiment
5.2
The joule-"Thomson experiment
5.3
Heat engines and the Camot cycle Problems
6.
69 72 74
80
The Second Law ofThermodynamics 6.1
Introduction
6.2
The mathematical concept of entropy
6.3
lrreversible processes
6.4
Carnot's theorem
6.5
The Clausius inequality and the second law
6.6
Entropy and available energy
6.7
Absolute temperature
6.8
Combined first and second laws· Problems
7.
67
Consequences of the First Law
85
87 88
89
91 94
97
98 I03
I 04
I 07
Applications of the Second Law I 09
7.I
Entropy changes in reversible processes
7.2
Temperature-entropy diagrams
7.3
Entropy change of the surroundings fc>r a �eversible process
7.4
Entropy change for an ideal gas
7.S
The Tds equations
7.6
Entropy change in irreversible processes
7. 7
Free expansion of an ideal gas
7.8
Entropy change for a liquid or solid Problems
Ill
I I0
Ill
113
I 21
1 22
118
Ill
Contents
VIII •••
8.
Thermodynamic Potentials 81
Introduction
8.1
The Legendre transformation
8.3
Definition of the thersrnodynamic potentials
8.4
The Maxwell relations
8.5
The Helmholtz function
8.6
The Gibbs function
8.7
Application of the Gibbs function to phase
8.8
An application of the Maxwell relations
8.9
Conditions of
.·
Problems
.
9.
. .
.
1 29 1 30 1 31
1 34 1 34
1 16
e equilibrium
s
1 37
142
1 43
1 46
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The Chemical Potential and Open
149
Systems 9. I
The chemical potential
9.2
Phase equilibrium
9.3
The Gibbs phase rule
9.4
Chemical reactions
9.5
Mixing processes Problems
.
I 0.
127
·.
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•,
.
.
.
15I
•
ISS 1 57
160
1 62
166
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.•
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.
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. .
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.
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.
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•
.
.
.
.< <
:
·. ·.·
.
· :
.
.
. . .· ·
The Third Law of Thermodynamics I 0.1
Statements of the third law
I 0.2
Methods of cooling
I 0.3
Equivalence of the statements
175
I 0.4
Consequences of the third law
178
Problems
1 79
I7I
1 74
· .·.·.• .•.
· .. ·
.
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1 69
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Contents
II.
ix
11 1
Basic assumpti ons
11.2
Molecular flux
ll.l
Gas pressure an d the ideal ps law
I1.-4
Equ
.
•
186 188
190
. '
11.5
Specific heat capacity of an ideal gas
I 1.6
Distribution of molecular spee ds
II.7
Mean free path and collision frequency
I 1.8
Effusion
191
193 198
10 I
•.
processes Problems
203
208
213
Statistical Thermodynamics 12.1
Introduction
12.2
Coin-tossing
ll.l
Assembly of distinguishable
11.4
Therpanoclynamic probability and e ntropy
12.5
Quantum states and energy le\'els
12.6
Density of quantum states
.
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:
.
..
..·
.
.
.
.
. .
..
.
.
215 116 es
111 113
215
229
23 I
Problems
13.
183
of enersr
II.9
12.
181
inetic Theory of Gases
The
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Classical and
. .
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:
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uantum Statistics
I ].1
Boltzmann statisti cs
13.1
The method of Laa•an� multipliers
13.3
The Boltzmann
1].4
The felstfli•Dirac
ll.S
The
235
bution
138
ZaiZ 244
136
233
.
x
Contents 13.6
Dilute gases and the Maxwell-Boltzmann distribution
13.7
The connection between classical and statistical thermodynamics
14.
Comparison of the distributions
13.9
Alternative statistical models
253
254
2S7
The Classical Statistical Treatment of an Ideal Gas
261
14.1
Thermodynamic properties from the par tition function
14.2
Partition function for a gas
14.3
Properties of a monatomic ideal gas
14.4
Applicability of the Maxwell-Boltzmann distribution
14.5
Distribution of molecular speeds
14.6
Equipartition of energy
14.7
Entropy ch ange of mixing revisited
14.8
Max well's demon Problems
15.
248
13.8
Problems
246
265 266 268
269
270 271
273
275
The Heat Capacity of a Diatomic Gas 279
15.1
Introduction
15.2
The quantized linear oscillator
I 5.3
Vibrational modes of diatomic molecules
I 5.4
Rotational modes of diatomic molecules
15.5
Electronic excitation
15.6
The total heat capacity Problems
263
289
287 288
279 282 284
277
Contents
16.
XI
•
The Heat C Introduction
16.1
Einstein's theory of the heat
16.3
Debye's theory of the heat capacity of a solid
of a solid
296
305 •
308 agnet
17.3
Properties of a spin·l/2
17.4
Adiabatic demagnetization
17.5
Negative temperature
17.6
Ferromagnetism Problems
agnetism
307
etism
318
321
325
328
33 I
Bose-Einstein Gases ation
333
18.1
Blackbody
18.2
Properties of a photon gas
18.3
Bose-Einstein condensation
18.4
Properties of a boson gas
18.5
Application to liquid helium Problems
293
30 I
Introduction
17.1
19.
193
The Thermodynamics of 17. I
18.
291
16.1
Problems
17.
of a Solid
338 340
345 347
349
Fermi-Dirac Gases 19.1
The Ferwni energy
19.2
The calculation of p.(T)
353
355 357
liS
XII
Contents
••
Free electrons in a metal
1 9.4
Properties of a fermion gas
1 9.5
Application to white dwarf stars
364 367
370
Problems
20.
36 1
19.3
lnformationTheory
373
20. 1
Introduction
20.2
Uncertainty and information
20.3
Unit of information
20.4
Maximum entropy
20.5
The connection to statistical thermodynamics
20.6
Information theory and the laws of thermodynamics
3 75
375
3 79 381 384 386
.
20.7
Maxwell's demon exorcised
387
388
Problems
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Appendices
A.
Review of Partial Differentiation 39 1
A. I
Partial derivatives
A.l
Exact and inexact differentials Problems
39 I
393
3 99
40 I
B.
Stirling's Approximation
C.
Alternative Approach to Finding the Boltzmann Distribution
D.
Various Integrals
407
403
: ::
.
xiii
Contents
Bibliography
41 3
Classical thertnodynamics
Kinetic theory of gases Statistical mechanics Special topics
41 3
4 14 41 4
41 5
Answers to selected problems
Index
424
417
This book is intended as a t ext for a one -sem est e r u nder gr aduat e cour se in t herm al p hysics Its obj ective is to provid e third- o r fo urt h year p hysics stu dent s wit h a solid intr oduct ion to t he cl assical and st at ist ical t heor ies of t her m ody namics. No prep arat ion is ass umed beyond colle ge-level general phy si cs and advanced calcul us. A n acquaintance with prob ab il ity and st at i stics is h elp ... f ul b ut is by no means necessary. The curr ent pr actice in many coll e g e � is to <>ffer a cour se in cl assical therm odyn am ics with littl e or no ment ion of the st atist ical t heory or vic e versa. Th e argument is t hat it is in1p ossible to do j ustice t o b ot h in a one semester cour se. On t he b asis of m y own te ach i ng exp er ience I str ongly dis agree. The standar d treatment of temperature work, heat entropy, etc. oft en seems to t he student l ike an endle ss coll ect i
·
�
..
..
only limited light on the underlying physics and can be abbreviated. The fun
of classical therm(>dynamics can easil y be grasped in little semester. leaving a n1 pl c tin1e t o gain a reasonably thorough
damental concepts more than half -a
of the statistical method. statistical t h erm odynam i cs
understanding Since
n ot
structure the entire
good r e as ons not providing a
The p h ysics •
XIV
course
subsumes the classical results�
why
around the statistical approach? There are
classical t h e < l ry is ge ne ral
simple._ and d irec t kind of visceral, int uit iv e com pr e hen si on of t herm al processes. st udent not confronted wit h thi s rem ark abl e phenom enol ogical t o do so. The
�
'I
Preface
XV
conception is defi ni teiy
deprived. Tt) be sure, the inadequacies of c l assic al
ther
modynamics becorne apparertt upon close scrutiny and invite inquiry about a
more fundamen t al descripti on This, of course, exactly reflects the historical .
development of the subject. If only the statistical picture is presented, however,
it
is my observation that the student
fails to appreciate fully its more abstract
concepts, given no exposure to the related classical ideas first. Not only do clas
si c al and st a tistical the rtn odyn amics in this se11se corr1 plemcnt each oth er, they
also bea u ti fully illustrate the physicist's perpetu al stri vi ng for descriptions of gr eater power , el egance, uni ver s ali ty, and freedon1 from ambiguity.
Chapters 1 th rough
10
r epres e nt a fairly traditional introduction to the
classical theory. Early on emphasis is placed on the advantages of expressing
the fu!tdamental la\vs in terms of state variables, qu anti tie s whose differentials
are exact. Accordingly, the sea rch for i nt egr ati ng factors for the differentials of work and heat is discussed. The elaboration of the first law is follow ed by chapters on app lic ati ons and consequences. Entrop y is presented both as a
useful mathematical variable and as a phenomenological construct necessary to explain why there are processes permitted by the first law tha.t do not occur
in nature. Calculations are then gi ven of the change in entropy for variclJS
reversible and irreversible. processes The therm od ynamic p otentials are .
br oach ed vi a the Legen d re transformation fo llov. in g elucidation of the ratio '
nale for having precisely four such quantities. The conditions for stable
equ i
librium are examined in a section that rarely appears in undergraduate texts.
Modifications of fundarnental relations to deal with open systems are treated
in Chap ter
9 and the third l aw is given its due in Chapter 10.
The kinetic theory of gases, treated in Chapter 1 1 , is
con
cerned with the
molecuJar basis of such thermody n amic prop erties of ga�es as the tempera
ture, pressure, and thermal energy. It represe11ts, both logically and historically, the tran sition between classic a l thermodyn am i cs and the s tat istical t heo ry
.
The unde rlying principles of equilibrium statistical thern1odynamics are
introduced in Chap ter 12 through consideration of a simple c oin tossing -
experiment. The basic concepts are then defined. The statistical interpretation of a system con taining many molecttles is observed to require a knowledge of the properties of the in div idua l molecules making up the system. This infor
mation is furnished by the <.tuantum rnechanical notions of energy levels, quan tum states, and intermolecular ft1rces. In C h apt er 13, the explication of
classical and quantum statistics and the derivation of the particle distribution functions is based on the method of Lagrange multipliers. A di scussio� of tl1e
connection betwe en cla ssic a l an d statistical the rm o dynamics c om pl etes the development of the mathematica1 formulation of the statistical theory.
Chapter 14 is devoted to the statistics of an ideal gas. Chapters 15 througl1 19
present important examples of the application of the statistical method "The .
last chapter introduces the student to the basic ideas of inforn1ation theory and
offers the intriguing thought
that statistical t hermo dynam ics is but a spe-.
cial case of some deeper, more far-reaching set of physical principles.
Preface
xvi
Throughout the book a serious attempt has been made to keep the level of the chapters as uniform as possible. On the other hand. the problems are intended to vary somewhat more widely in difficulty. In preparing the text, my greatest debt is to my students, whose response has provided a practical filter for the refinement of the material presented herein. A.H.C. Drew University
ACKNOWLEDGMENTS In addition to my students at Drew University, I owe .thanks to two colleagues and friends, Professors Robert Fenstermacher and John OJlom, who have encouraged me at every turn during the writing of this book. I am indebted to .
Professor Mark Raizen of the University of Texas at Austin. who reviewed the manuscript and used it as the text in his thermal physics course: his comments were invaluable.
I am especially grateful to Professor Roy S. Rubins of the University of Texas at Arlington for his thoughtful and thorough critique. I also received useful feedback from other reviewers, whose suggestions contributed substan tially to an improved text. Th ey are AnjtJm Ansari_ University of Illinois at Chicago; John Jaszczak, Michigan Technol(>gical University: David Monts,
· Mississippi State University; Hugh Scott, Oklahoma State University: Harold Spector, Illinois Institute of Technology at <:hicago; Zlatko Tesanovic, Johns Hopkins University. I thank my editor Alison Reeves and her assistants, Gillian Buonanno and Christian Botting, for their support, guidance, and patience. Production editors Richard Saunders and Patrick Burt of WestWords Inc. were particu larly helpful. FinaiJy, I am extremely graleful to Heatl1er Ferguson. who turned my lecture notes into a first draft, and to l.,ori Carucci and her daughters Amanda and Brigette. who prepared the final manuscript. Without all of these people, the book would never have seen the light of day. •
•
•
•
I
1. 1
What Is Thermodynamics?
3
1.2
Definitions
5
1.3
The Kilomole
6
1 .4
Limits of the Continuum
6
1 .5
More Definitions
7
1.6
Units
9
I. 7
Temperature and the Zeroth Law
1 .8
of Thermodynamics
10
Temperature Scales
12
I
1.1
WHAT IS THERMODYNAMICS?
Thermodynamics is the study of heat in the field of physics. The central con cept of thermodynamics is temperature. Since temperature is not expressible in terms of basic mechanical quantities such as mass, length, and time, it is evi dently a fundamental notion that sets thermodynamics apart from other branches of physics. The development of ther1nodynamics provides some of the most fasci nating chapters in the history of science. It began at the start of the Industrial Revolution when it became important to understand the conversion of heat into mechanical work. The experin1ents of Joule, Hirn, and others, and the the oretical studies of Helmholtz resulted in the principle of the conservation of energy when applied to therrnal phenotnena. The principle became the first law of thertnodynamics. Mayer postulated the equivalence of heat and work and made an estimate of the mechanical equivalence of heat. The subsequent progress of thermodynamics owes much to the
· Frenchman, Sadi Carnot, whose treatise of 1824 led io one of the most far
reaching principles of physical science: Carnot's theorem, which is, in effect, the second law of thcrrnodynamics. Actually, the principle emerged before the first law. It was the outgrowth of Carnot's interest in the practical question of the efficiency of steam engines. His \vork, put in simple mathematical form by Clapeyron, attacked the more fundamental problem of the efficiency of heat .
engines in general. The concept of entropy began to appear quite early in papers by Clausius and William Thomson (Lord Kelvin), but it was not until
1865
that
Clausius saw fit to give it a name and a full definition. Later, Nernst and Planck added a third law, which is a statement about the behavior of thermo dynamic quantities, including entropy, at the temperature of absolute zero. In this period classical thermodynamics was worked out in essentially its present form. It is a phenomenological theory, describing the macroscopic prop erties of matter, most of which are amenable to direct measurement. No assump tions are made about the fine structure of material substances. No attempt is made to explain
underlying causes or to provide a mechanistic description. As a 3
The Nature ofThermodynamics
Chap. I
4
.
.
.
...
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.
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:
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Figure 1.1 Ludwig Boltzmann, 1844-1906 whose work led to an understanding of the macroscopic world on the basis of molecular dynamics. (Courtesy of American Institute of Physics/Emilio Segre Visual Archives.)
.
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consequence" the theory has the advantages c.>f great s implicit y.. broad gene rality
,
and a close connection between experimental results and familiar concepts. The
noted experimentalist
�W. Bridgman said .. ''The J aws of thermodynamics have a
different feel from most of the other laws of physics. There is something palpably verbal about them
they smeJI of their hum
Toward. the end of the nineteenth century, \vhen the aton1ic nature of .,
matter began to be understood .. ways were found to e x pr ess the pressure. tem
perature, and other macroscopic p roper t ies of a gas in terms <)f a ver a ge values su ch
of the properties of molecules,
as the 1 r kinetic energy. The r es ul ts. with
which Maxwelrs name is closelv asS()Ciated. came to he c
ory of gases. The kinetic theory eventually expanded int o the f ar more comprehen sive statistical mechanics (or statistical thermodynamics) of BoJtztnann and •
Gibbs, and ultimately encompassed the ideas of quantum mechanics. The statistical approa ch takes account of the molecular C()nstitution of matter and
reveals a deeper foundation on which the laws of thermodynamics exist . The
connection between the classical and statistical theories is associated \Vith the fact that macroscopic measurem e nt s arc averages of the hehavior of astro
nomical numbers of particles. Thermodynamics is by no means
a
closed sub.iect. In recent years it has
been possible to ext en d the statistical theory
to
include nonequilibrium
processes and even nonlinear effects. At the present time. the re1ationship
Sec. 1.2
Defrnit!ons
5
between thermodynamics and information theory, and the study of chaotic behavior in thermodynamic systems, attract a great deal of attentiott. Modern
ideas, furthermore, have been shown to be applicable to a wider variety of phenomena than hitherto suspected. Thermodynamics has captured the imagination of many of the greatest minds of science. Einstein, who was captivated by the subject, wrote: "A theory is the
more impressive the greater the simplicity of its premises, and the ntore
extended its area of applicability. Therefore, the deep impression which classi cal thermodynamics made upon n1e. It is the only physical theory of universal content concerning which I am convinced that, within the applicability of its
basic concepts, it will never be overthrown.''That is quite a statement, indeed!
1.2 DEFINITIONS In developing the basic ideas of thermodynamics the originators of the theory were careful not to be so concise as to render definitions sterile. Tne following brief definitions will be illustrated by examples in future sections.
Thermodynamics can be described as the study of equilibrium
in which
temperature is an
properties
important variable. All of the words in italics need
to be defined.
In thermodynamics we are concerned with a S}'Stem, some portion of the physical world. The system could be a container of gas, a piece of metal, a mag
net. The system must not interact chemically with the vessel that contains it.
(The behavior of a liquid must not be influenced by the test tube that holds it.) A system may exchange energy with other systems, which constitute the
roundings
comprtse a •
open
sur
of the given system. The system, together with its surroundings,
untverse. •
We classify S)'Stems as to whether they are open, closed, or isolated. An system can exchange mass and energy with its surrottndings. A
closed
system cannot exchange mass with its surroundings, but can exchange energy in other forms. An isolated system cannot exchange mass or energy in any form with its surroundings; it is completely cut off from other systems. The quantities we use to describe the macroscopic behavior of a system
are called
properties,
observable characteristics of a system. Other names are
thermodynamic variables or thermodynamic coordinates. An extremely impor tant concept is that of a state variable, a property whose differential is exact. Properties are extensive or intensive. An
extensive
property is propor
tional to tl1e mass. An example is the volume V: if the mass is doubled, the vol ume is doubled (assuming that the density remains constant). An
intensive
property is independent of the mass. Temperature Tis an intensive property; its value is not affected by a change of mass. Pressure ther examples of intensive properties.
P and density pare fur
Chap. I
6
The Nature ofThermodynamics
Inherently extensive properties are given by capital letters. Inherently intensive properties are denoted by lowercase letters. There are two important exceptions: The temperature Tis always capitalized to avoid confusion with the timet: and the pressure P is capitalized to distinguish it from the probability p. An extensive property can be converted to an intensive property by dividing by the mass. This is called a specific value: Specific value -
value of the extensive property -
mass of the system
•
In this text, we will go back and forth between extensive properties and their corresponding specific values.
1.3 THE KILOMOLE '
The kilomole is a unit of mass defined as follows: I
kilomole - mass m in kilograms equal to the molecular weight.
Thus one kilomole of oxygen gas (02) is equal to 32 kg. (The mole is a unit of mass more familiar to chemists: one mole is equal to the mass in grams. Thus a mole of oxygen gas is 32 g.)
1.4 LIMITS OF THE CONTINUUM
We tacitly assume that classical thermodynam ics is a continuum theory, that properties vary smoothly from point to point in the system. But if all systems are made up of atoms and molecules (as our definition of the kilomole implies), it is reasonable to ask: how small a volume can we be concerned with and still have confidence that our continuum theory is valid? We can obtain an approximate answer to this question by invoking Avogadro's law, which states that at standard temperature and pressure (0°C 3 and atmospheric pressure), one kilomole of gas occupies 22.4 m and contains 6.02 x 1026 molecules. (The latter is called Avogadro�s number, NA . ) Then .3 22.4 m kilomole -l
m3
•
This molecular density is sometimes called Loschmidt's number. Using it we can easily show that a cube one millimeter on each side contains roughly 1016 mole cules, whereas a cube one nanometer (1 0-9 tn) on a side has a very small prob-
Sec. 1.5
7
More Definitions
ability of containing eve11 one molecule. We can therefore be r�asonably cer tain that classical thermodynamics is applicable down to very small macro scopic (and even microscopic) volumes, but ultimately a limit is rea ched where
our theory will break down .
1.5 MORE DEFINITIONS
The state of a system is defined as a condition uniquely specif i e d by a set of properties. Examples of such properties are pressure, volume, and tempera ture. The question arises: how tnany properties are required to specify the state of a thermodynamic systerrt? By "required number" we mean that every time a system with the gi v en properties is subjected to a particular environ
ment, every feature of its sllbsequent behavior is identical. In classical mechanics, if the displacement and velocity of a system are known at some instant of time, as well as the forces acting on it, the behavior of the system is predicted for all times. Most frequently, the t hermodynamic state of a single component system is also specified by two independent variables.
An equilibrium state is one in wh ic h the properties of the system are un i form throughout and do not change with time unless the system is acted ttpon by ex t ern al influences. A non-equilibriltm state characterizes a system in which gradients exist and whose p r ope rties vary with time (the atmosphere and the
oceans are examples). State variables are properties that describe equilibrium states. An equation of state is a functional relationship among the state vari
ables for a system in equilibrium. A path is a series of states through which a
system passes.
In introductory physics, a ·'change of state" is frequently used to den ote a transition from a liquid to a gas, or from a solid to a liq1Jid, etc. In therlnodynam •
ics, such a change is referred to as a charzge of phase or phase tratzsformation.
If the pressure .P, the volume V, and the t en1 pe ra t u re Tare the state vari
ables of the system, the equation of state takes the form f(P, V,
T)
=
0.
( 1.1 )
This rel a tions hip reduces the number of independent variables of the system from three to two. The function f is assun1ed t o be given as part of the specifi cation of the system. It is customary to represent the state of such a system by a point in three-dimensional p V- T space. The equation of state then defines a surface in this space (Figure 1.2). Any point lying on this surface represen t s a state in eq uilib rium. In thermodynamics a state automatically means a state in eq uilibr iu m unless otherwise specified. The boundary bet\\'een a system and its surroundings th roug h which changes may be imposed is called a system wall. An adiabatic wall is a boundary tha t permits no heat interaction with the sttrrou n dings The word comes from ..
.
The Nature ofThermodynamics
Chap. l
8
T te ta s m iu r ib il u q e --, _ /-
---f(P. V. T) v
Surface in P- V- T space. Points on the surface represent equilibrium states of the system whose equation of 0. state is f ( P. V. T )
Figure 1.2
I'
=
•
-
ia ad is m ste sy ed lat iso An ... h ug o r h g t in go t no .. ng ni ea m s. the Greek adiabato y l ri e ssa c ne t no is ll wa c ati iab ad an th wi t m s e y s a r. ve we Ho d. batically containe
isolated. Mechanical interactions can take place th rough adiabatic walls. For e xample material can be removed or added� the volume can change, a mag ne ti c field can be applied, etc. A diatltert11al �rail is a boundary that freely allows ,
heat to be exchanged� The Greek word diatl1ern-zos me a ns Hheat through.�' If a system s t rong l y interacts \\'tth its walls. the pr
cated. Think of panc ake batter stick ing to the griddlt!. or sulfuric acid 1n an
iron ve ssel !
·
A process is a c h an ge of s t ate expr c ,sed in terms of a path along the
e qu ation of state surface in Figure 1.2. In a c_vclicaiJJrt)cess. the initi al and final
states are the same. A qltasi-static prl)cess is a process in \\'hich. at each instant,
the system departs only infinitesimally fr<)m an equilihrium state. That is. changes of state are described in te rms of differentials. An example is the
gradual compression of a gas. A reversible prtJcess
is
a proces s \\'hose directi<)n can l)e reve rsed hy an
infinitesimal change in some property. It is a quasi-static process in which no di ssipati ve forces such as friction are p rese nt All re versible processes are .
quasi-static but a quasi-static process is not necessarily reversible. For exam ..
ple, a slow lea k in a tire is quasi-static but not reversih1c. A reve rs i ble process
is an idealization: friction is al w a y s present. An irre\'ersihle pr<>cess in v ol ve s a finite change in a prope rty in a given step and includes dissipation (energy ·
J oss ) All natural processes are irreversible. .
In many proce sses some IJT<)perty of the ,
s yst e n1
remains constant. An
isobaric process is a process in \\'hich the pr\!ssure is constant . If the
con stant the process ,
is
v ol ume
is
iS(JCIInric. And if th� temperature d <)e s n 't cl1 an ge the .
process is called isotlrertnal.
Suppose a piston enclosing a gas is immersed in a heat bath� so that the
gas is kept at constant temperature. The gas is slowly co mp resse d (Figure
1.3).
Sec. 1.6
Units
,. p
diathertnal wall .. . . . .. . . . . .
. . . ·. .. . . . . . .. . . . . . .. . . ;t . . , , • . • .
•
.
. .
.
•
•
.
•• •
.
. 0
.
�
.
•
piston
•
. . " ·. . .. . . .
.
•'
.
.
.
.'· .;
..
.: .> :.>. .
.
.
v . . . .
.
.
••:0 ••: . . . .. . . .. ". . . . . . •. . .. .... .• . .. . . • •
. .
.
.
.
..
.
•
.
.
.
(P;, V,) v
/
heat bath
gas
(a)
(b)
Figure 1.3
Example of a quasi-static, reversible and isothermal process. (a) A piston slowly compresses a gas held at constant temperature. (b) A P-V diagram representing the process; (P;, V;) and (P1, V1) denote the initial and final states, respectively. The process is quasi-static, reversible, and isothermal and can be described by a path in a
P- V diagram connecting the initial and final states through a series
of infinitesimal changes.
If, instead of
a
gradual compression, the piston is given a violent push,
sound "'aves or shock waves and turbulence are generated, accompanied by stro11g temperature and pressure gradients. The process is an irreversible, non equilibrium process that cannot be represented by a path in the
P- V plane.
Only the end points can be plotted, representing the initial state and the final state reached after equilibrium has been eventually reestablished.
1.6 UNITS The lntertlational System of Units (SI) will be used, almost exclusively. The system is based Oil the fundamental units of length, mass, and time
the meter
(m), the kilogram (kg) and the second
(s). respectively. In this system, the unit
of pressure is the pascal (Pa), equal to
1
Nm -2•
Other units of pressure in common use ar e 1 ba r -- 105 Pa, 1 atm<>sphere (atm) 1 torr
�=
1.01 X
lOS Pa.
133.3 Pa.
The atmosphere and the torr (named after Torricelli) are derived units. The atmosphere is based on the use of a manometer to measure pressure (Figure 1.4).
Chap. I
10
The Nature ofThermodynamics
vacuum
p h
. :· . . ·\. . . . . . . . > ·. . ·· · ·· . · . · ..-- m crcu rv · · . . . .· . · : .. : . .: . · : . . . . . . . . . .. . ,..,,,· :. .·. .· . · ·. . . . .. . . . . .. . .. ... .:.. . ...... . .· ·. . . . ... . .· . . . . ... . . . .. .. . . . ... ..·. . . . . . .. :
.
.
. •
.
.
.
.
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.
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:• · ·· · · .· :� · . �·
:
.
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•
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.
.
.
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•
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:
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•
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• •
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0
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.,
•
•
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:..•
.
.
.
.
.
•
.
•
.
.
.
.
,
•
•
Figure 1.4 A simple mercury manometer used to measure the atmospheric pressure.
.
.
·
•,
.
.
· '
The he ight of a column of mercury is balanced by the force exerted by
the gas on the liquid surface. A p re ss u re of 1 atmosphere cau� es the mercury to rise to a height of
76 em. That
is. P
hpg. where P is the pressure of the
=
atm osphe re his the height of the mercury column. pis the density of m ercury
.
,
and g is the acceleration of gravity. Thus
1 atm
=
(0.76m) 1.36
The torr is then defined
as
,.
,.
•• ..,
0
••'
__
·
..,, �
•
, , . '"'
o
. ,
1
=
.
7fi0
-
9 .8
\
m·
� s-
J
1.01
=
X
105
Pa.
of 1 mm of mercury:
atm
=
·
133.3 Pa.
�..s,.·,
,
... \� \4� ��...., ,�, �..:-';., , t.:.� W" '. #t::--T �·"C� · . .\IT-"' <('C 'U�•.-.2 ., ,.. I -..-',.•TY" "'E:I •' C\•••� . _,,�-·��� ,.. ._, ,.- .._ ..... � •.... o•""•...,....__.,.--; �• """ :WV �- �"'·� . ·' ...., �_·..- • •_.. .._,. , ��, � . 7.. _ �� •�.... · ..· , • , •�•' . . "� . .. X ... ..._ • -'_ -· � • • • · ;, 'f -.. • · � • .. ' ' ' ' o '• ' � r:m tw.• ,. __. � (J, :: ., , ,.. , _,.. _, ,.. '4"' •• '• • ">o+. ':oo o ., .... , \oo'.., • o • O: 'ot-;-''·•�•- , .. �t oil' _ , ,_ .,.� ";�, : ' - " :o:,;: , 'YT � ;. • •, ; \,- .' ,:.. � ' ".J ' • • o o". • "': ', • .,. o N' ,..,· , � «- - •
..,_•.•c. •r��.-,. �,. • '
l O"
the pressure
1 t(1ff
•,
x
kg
•
.,.,,,.\- ' ,
•
..
...
..
..._
�•
,
;."J> -
'
•
1.7 TEMPERAT UREANDTHE ZEROTH LAW OF THERMODYNAMICS Temperature is
a
mor e suhtlc property than pre s s u re Its ori gin is the so-called .
zeroth Jaw of thermodynamics.* The zeroth law is based on
e xpe ri men ts
(as
are all physical laws) and is concerned with properties of systems in thermal
eq uili briu m .. that is, systems in eq u il i b rium connected by a diathermal wall. The law states: If t�vo s_vstems are separate(r in ther1nal e(Jllilihriul1l •vitlz they
are i11 eqttilibrilll11
a tl1ird syste111.
�-vith each fJTher.
.
Let systems A� B. and ..
tainer. We shall use C
(_�each consist of a 1nass of flutd in an insulated conas a reference. We choose to describe the state of each
. *There are some \\'ho hc1icve that this should n•)t be dignified by ca11ing it a . Jaw." .
.
Temperature and t:he Zeroth Law ofThermodynamics
Sec. 1.7
A
c
B
(a)
II
c
(b)
Figure l.S Systems in therntal contact: (a) systems A and C in equil ibritun ; (b) systems B and C in thermal equilibrium.
thermal
system in terms of the state variables P and V. We place A in contact with C through a diathermal wall as in Figure 1.5(a). The system C can be thought of as a thermometer. Its state is given by the pair of variables ( �·, Yc). From observations, if we choose a particular value for PA, then VA will be uniquely deter111i ned (only one variable is inde pendent). The condition under which A and C are in equilibrium may be expressed by the equation
(1.2) where F1 is some function of the four variables. We assume that this equation can be solved for Pc:
(1 .3) Next we place system B and C in thetttlal contact (Figure 1.5(b)). For equilibrium, (1. 4) or
(1.5) Equating Equations (1.3) and (1.5), \Ve obtain the condition under which A and B are separately in equilibrium \vith C:
(1.6) But, according to the zeroth l aw A and B are then in equilibrium with each other, so that ,
(1.7)
The Nature ofThermodynamics
Chap. I
11
Solving for P_4 we obtain ..
( 1 .8) Now, Equation ( 1 .6) can also be solved for PA in p ri ncipl e : �
PA = g ( V4 , PB .. VH .. Vc ) .
( 1 .9)
as ere wh les, iab var r fou by ed 1in ern P" det s i t tha t es t a s .9) 1 ( on i at u q E Equ atio n ( 1 .8) says that it is a fun ctio n of t)nl y thre e. TI1i s can only mea n tha t the functions f1 and f2 in Equ atio n ( 1 .6) cont ain Vc in such a form that it can cels out on the two sides of the e qua tion . f<)f e x am p l e ,
!t
f2
=
=
<�> 1 < PA ..
vA ) ' ( vc )
4J J ( p8 , VH ) ( ( .\1(· )
+
11 ( vc ) ..
+ 7J ( VC )
•
When the cancellation is performed w e ha ve ..
( 1 . 1 0) •
Extending the argument to additional systems, we get
For any system in thermal equili brium with a given system .. we can choose to w ri t e
( 1. 12) where we define T as the empirical
e qu a t i on
tetnperature and Equation ( 1 . 1 2 ) is the
of state of t he system. Equat ion ( 1 . 1 1 } th e n says that systetns in ther m a l equilibrium with one a n o th e r h ave the s a me temperature. Thus tenzpera ture is a property of a sys t em that determ ines if thermal equili brium e xi sts \\'ith some other syst em . The next section contains a scheme f,)r me asuri n g the temperature. Let C be a •·thermometer, ..' an arbi trari l y sclecttd standard body. Then the relative temperature of the systems A a n d B c a n bt� compared without bringing them into contact \Vith each other.
1 .8 TEMPERATURE SCALES
To assign a numerical value to the tcn1pcrature of a system. we choose as a ther mometer a substance that has a so-called tJzernzometric propert.,v tha t changes with temperature and is easily measured. A n example is the volume of a fluid that expands on heating (think of the familiar liq ui d in glass thermometer). -
-
Sec. 1 .8
Temperature Scales
13
· We choose a thermometric property X that is linearly relate d to the tem
perature over as wide a range as possible: X =
Here
a
and
Prior to
b
(1.13)
aT + b.
are constants. We choose reference points to llefine the scale.
1954, two
fixed points were used, the ice and steam points of water.
The scale was determined by assigning the numerical values ooc Celsius) to the ice point and l00°C to the steam point.
(C
for
An excellent choice for the thermometric property X is the pressure of a
gas. The constant-volume gas them1ometer, shown in Figure
1.6, is a practical
method for measuring the temperature of an object. Several observations were made when this method was used (Fig ure
1 .7):
a.
The P- Tcurve is very nearly linear over a wide range of temperature.
b. The curve is increasingly linear as the pressure decreases. c.
A linear extrapolation of the plot gives
P
=
0 at
T
=
.
- 273.l5°C. This
turns out to be true for all gases although the slope of the curve is differ ent for different gases.
I -
capillary
---.....--
.... :. ,.._, .. . . :. .
.
"
.
. . .
.
.
.
.
. .
.
.. .. . . ' . · .. . . . ... .
. .
h
rigid _....,.__ diathermal wall
. .·.:..��. . : : : .• •': : : .. .. .
.
-. -. :·: . . . ..
.
.
.
.
·.· ·..
.
. ·. ,'• •• • •• 0 • • • .
.
. .. .. .•.... .. ...... .. ·. .. . . :·: .·.·:· . : · •. .. ... . • . ! . .·:. . . . 0
.
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.
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.
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. . . .
. . .
.
.
p gas
·.· < :· : •• • • •:••: .. · .·.• . .. • :. . , . .·. · : . ... •.. • ••• • • • • •• . • ::· :_ · . . :· : : · . ;: :. . -.: .· : . . ·,·.: . ... : ... . . ·... . . . . .. . . ' ... . . ... ·. . - .; ; ·- :: .. :.···..· ..·.·>.' ... . .: . . •. : :-.·-:... : . �.... ..... . .. .:. ...'·...' .. •, ' .: .. . ·•·.·. . :-. -: ·:-.. . . . . .·..... .. .. ... . .. :-..·:·� �-=· · · . .......·...· ... .. •,. . . ... "' . . . -: : \ ':::: � : . . : : : . . . . . . :-:- �·.·.�.·. •, · · . . · ..-:. .. ....·.... ·. · :: .. ,· : . .. . . • ... ...: ··.:· . : •• •• •• .. .. . · ··· . . . . ' . : :.. '-: . . . ·.. . .. . . . . .. .....·.. .. .. .•. . . ·. . .:-�· ··:·: :. · ..:-.:::�::: · . · :::::·:... •-::�• :• • :..·•• .::::�-�::: --��=: : ·.· ·.., ,:.. . .·.. '•' • ... . ..•..·.• •� • •. •. 0 • ·.· :- : ::� :?:· . :_:-::·· :·· : ..
. .
0
. . . • .. . · . . . .
.
0
,
. ..
.
.•
.
. .
.
.
.
.
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.
.
.
· .
0
. ·
mercury manometer
.
..
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. . .
.
. .
.
.
· < .
. . .
. •
.
.
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.
.
.
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.
Figure 1.6 Simplified constant-volume gas thermometer. The pressure is given by P hpg, as in Figure 1 .4. =
Chap. I
14
The N ature ofThermodynamics
.
.
I
I
I _ _ _ _ _ _ _ ...._ __. _ _ _... . __. _ __ � ;...
Figure 1.7
r
Pressure versus temperature for a gas
ther1nometer. The ice and steam points of water are fixed points. The temperature at zero pressure is obtained by linear extrapolation.
For a gas thermometer. then. P
==
aT
+
b.
The zero pressure point gives ()
=
-
273 . 1 5a
+
b.
Thus P
=
af7� ( c'C )
We can now choose an a bsol ut e ice and steam points. We want
.
+
sca l e such
P
==
273. 1 51. �
( 1 . 1 4)
that there are 1 00°C between the
aT ( K. )
�
( 1. 1 5 )
where K is t h e abbreviation for kelvin. the unit of temperature on the absolute scale (no degree sign ). It follows that ( 1 . 1 6) Now we need only one reference poi nt to dc.�finc the slope a in Equ a t i on ( 1 . 1 5 ) . In 1 954 the reference point was taken to be the triple point of water. the pres sure and temperature at which ice and liq uid \Vater coexist in eq uilibrium with saturated vapor. The tri ple point temJ>erature of water is 0.01 :)C, that is� 0.01 degrees above the i cc point at atrilospheri<; pressure. Equations ( 1 . 15 ) and ( 1 . 1 6) give
Sec. 1 .8
15
Temperature Scales
T(K) --
273.16
=
P
( 1 . 17)
-
Prp '
where Prp is the pressure at the triple point. A better definition, taking into account the convergence of th.e P-T curves for gases at low temperature (Figure 1.8), is
( 1 . 18) The curves of the figure are constructed in the following way. An amount of gas in the bulb of a constant-volume gas thermometer is exposed to a cell con t ai nin g water at the trip le point. The pressure is recorded. With the volume held constant, the bulb is surrounded with boiling water at atmospheric pres sure, the gas pressure is measured, and the temperature at the steam point, T5 , is calculated usin g Equation ( 1 . 17). Then some of the gas is removed from the bulb, the lower triple point pressure is measured, and the new values of the pressure and temperature at the steam poi n t are found. The process is continued and the resulting curve is extrapolated to the ordinate. Although the curves for different gases have different slopes, they are discovered to inter sect in the limit as PT P approaches zero. The triple point reference is precisely reproducible and absolute zero is precisely defined.
373.8
t-
373.6
�
� ;;;; ;;; ================= He2 H 37 3.0
...-�--..J..-----"--'--'"'L-.
20
0
•
Figure 1.8
40
60
80
100
120
Prp(torr)
Readings of a constant-volume gas therrnometer for the temperature of condensing steam as the density of the gas (and hence PrP) is lowered. The extrapolated value of T5 is the same for all gases.
Chap. I
16
u se.
Two other temperature scales are in
The Nature ofThermodynamics
The familiar Fahrenheit scale is
related to the Celsius scale by the equation
( 1 .19) On this scale the ice point is 32°F and the steam point is 2 1 2°F. The Rankine scale is derived from the Fahrenheit scale: ( 1 .20) The Rankine scale is widely used in enginc.!ering: it does not use the ""degree'' symbol.
We shall see later that it is possible t o define the absolute scale of tem perature independently of the thermometric properties of gases.
.
·
. ·
.
.
.
.
. ..
.
..
.·
.·
..
.. .
.·
:
. .
:
. '
· .
.
.
'
.
.
.
.
.
.
. .
.
.
:
.
.
.
. .
.
.
-
:
.
.
P RO B L E M S
1-1 Classify the follo\\'ing systems as ope n_ closed. or isolated:
(a) A mass of gas in a con tainer \Vith rigid. impermeable. di athe rm a l walls. (b) A mass of gas in a container \Vith rigid, impermeable. adia b a t i c wa11s.
(c) A sugar solution enclosed hy a men1branc permeahJc only to water that is immersed i n a large container of wa1cr.
1·2 · U si ng the terms de fi n ed in the chapter. characterize the foJiowing processes as
completely as possible: (a) The temp e ra t u re of a gas. enclosed i n a cy l i nd e r provided with a frictionless piston . is slowly increased. The pressur� remains constant . (b) A gas, enclosed in a cyl inder provided with a piston, is slowly ex pande d . The temperature remains constan t . ll1ert is a force of friction bet,veen the cylin der wall and the piston . (c) A gas enclosed in a cylinder provided \Vith a frictionless piston is quickly compressed. (d) A piece of hot metal is thro\vn into cold \\'ater. (Assume that the system is the metal, which n e it h er contracts nor expands. ) (e) A pendulum with a frictionless su pport swings back and forth . (f) A bullet is s topp ed by a targe t . 1-3 On a plot of volume versus temperat ure. dra\\' and label l ines indicating the fol lowing processes.. each pr t)C c c d i n g from t he san1e initial state � • . �).
(a) An isothermal expansion. (b) An isothermal compress i on.
(c) A n isochoric increase in temperature.
Problems
17
1-4 Estimate the pressure you exert when standing on the floor. Express your answer
in atmospheres, in pascals, and in torr. Repeat the calculation for spiked heels.
1-S Let the resistance R of a piece ot· wire be a thermometric property for measur
ing the temperature T. Assume that
R = aT + b,
where a and b are constants. The resistance of the wire is found to be 5 ohms when it is at the temperature of melting ice and 6 ohms when it is at the temper ature at which water boils at atmospheric pressure . If the ice point is taken as 100° and the boiling point as 500° on a particular scale, what is the temperature 5.4 ohms? on this scale when R =
1-6 The following table gives the observed values of the pressure P of a gas in a con stant-volume gas thermometer at an unknown temperature T and at the triple
point of water as the mass of the gas used is reduced. P-n· (torr) P (torr)
1 00 1 27.9
200 256.5
300 385.R
400 5 1 6.0
Determine T in kelvins to two decimal places by considering the limit PrP-+0 ( pI Pr p ) . What is this temperature in °C? 1-7 The resistance R of a doped germanium crystal is related to the temperature T
through the equ��ion
log R
=
4.70
-
3 .92 log T�
when R is in ohms, T is in kelvins, and the logarithm is taken to the base 1 0. In a liquid helium cryostat, the resistance is measured to be 218 ohms. What is the temperature? 1-8 A thern1ocouple consists of two wires made of dissimilar metals that are joined together to form an electrical circuit. A thermal electromotive force (emf) e is
generate(J when the two j unctions are at different temperatures. When one junc tion is held at the ice point and the other is at a Celsius temperature of T, the therntometric function is given by 2 E = aT + bT ,
where e is in millivolts. Calculate the constants a and b for a thertnocouple read ing 60 m V at 200°C and 40 mV at 40oc·c. What temperature corresponds to a reading of 30 m V? 1-9
(a) Consider the linear relationship between the thermometric prop e rty the temperature T given by
X and
X = aT + b.
Suppose that the ice and steam points are used as fixed points with temper atures of oo and 100° respectively. Show that T
=
100
X - XI
X.( _ XI .
Chap. I
18
(b )
The Nature ofThermodynamics
If, instead, the thertnometric function is chosen as
T
=
aln .�Y
+
h,
show that on the new scale
1-10 At what temperature do the Fahrenheit and Kelvin scales give the same reading? 1-11 The temperature of the normal boiling point of nitrogen is
77.35
K. Calculate
the corresponding value of the temperature on the (a) Celsius. (b) Fahrenheit, and (c) Rankine scales.
2. 1
Introduction
21
2.2
Equation of State of an Ideal Gas
22
.
2.3
Van Der Waals' Equation for a Real Gas
23
2.4
P-v-T Surfaces for Real Substances
25
2.5
Expansivity and Compressibility
27
2.6
An Application
29
19
2. 1 INTRODUCTION In thermodynamics, we add to the basic dimensions of mass, length and time the "fourth dimensio n" of temperature. We saw in Chapter 1 that the concept of temperature is intimately associated with the. notion of systems in thermal equilibrium. Perhaps the simplest example of a thermodynamic system is a homogeneous fluid say, a gas or vapor.* Its state, as we have observed , can be described by an equation of the fottn f ( P, V , T)
=
0.
(2. 1)
The equation of state connects the three fundamental state variables, only two of which are independent. It is an expression of the results of experiments. Every system has its own equation of state. We note that the equation of state does not involve time. Classical ther modynamics deals mainly with systems in thertnal, mechanical, and chemical equilibrium that is, thermodynamic equilibrium and does not concern itself with the rate at which a process takes place. The mechanical variables occur in "canonically conjugate" pairs, one an extensive variable, the other intensive. Thus the volume V (extensive) is "con jugate" to the pressure P (intensive). The situation is analogous to that in mechanics, where the generalized position coordinates qk and the generalized momenta Pk form canonically conjugate pairs. In thertnodynamics, the quantity PdV is the differential of work (with units of energy). We are led to ask if there is an extensive state variable canon ically conjugate with the temperature whose product has the unit of energy. This very important question will be addressed in Chapter 6.
* The laws of thennodynamics apply to all forms of matter, but they are most easily formu lated and understood when applied to gases.
21
Chap. 2
22
Equations of State
1.2 EQUATION OF STATE OF AN IDEAL GAS The equation of state of a system composecl of m kg of a gas whose molecular weight is M is given approximately b):
(2.2 ) .
where R is a universal constant� having the same value for a11 gases:
J
Since
n
( 2.3)
mj M is the number of kilomoles of the gas, we can write
PV
=
nRT.
(2.4)
This equation is called the equation of sta1e of an ideal gas or perfect gas. It includes the laws of BoyJe't Gay-Lussac� Charles.. and Avogadro .. which were discovered over a period of 200 years. In 1 8 1 1 Avogadro postulated that at a given temperature and pressure equal volumes of all gases con tain e.quaJ numbers of molecules. The reason for this is that the molecules in a gi ven sample of gas have negligible volumes compared with the volume of the sample itself. In Equation (2.4) we note that the extensive variable V divided by n� the number of kilomoles of the gas.. is the specific volume ·v. Thus the equation of state can be writ ten Pv RT. The projections of the surface f ( P, v, T) 0 on the P-v plane, the P- T plane and the v- T plane are sh ow n i n Figure 2. 1 . We use whatever diagram is most appropriate for the process we are in terested in. =
==
..
p
p
T
Vz
v
(a) Figure l.l
::;
;.__--- -�
(h)
T
p
...._
___.
_ _ _ _
T
(c)
Diagrams for an ide al gas. I n ( a ) the isothern1s are eq uilateral hyperbolae: in ( b ) the isochores are straight l i nes: and in (c) t h e isobars are also straight lines.
Sec. 2.3
23
Van Der Waals' Equation for a Real Gas •
2.3 VAN DERWAALS' EQUATION FOR A REAL GAS The characteristic equation of an ideal gas represents the behavior of real gases fairly well for high temperatures and low pressures. However, when the temperature and pressure are such that the gas is near condensation, important deviations from the ideal gas law are observed. Among the numerous equations of state that have been introduced to represent the behavior of real gases, that of van der Waals is especially inter esting because of its sim.plicity and because it satisfactorily describes the behavior of many substances over a wide range of temperature and pressure. Van der Waals derived his equation from considerations based on kinetic theory, taking into account to a first approximation the size of a molecule and the cohesive forces between molecules. His equation of state is a
P + 2 ( v - b) -= RT, v
(2 .5)
where a and b are characteristic constants for a given substance. For b 0, Equation (2.5) reduces to the equation of state for an ideal gas. a The tettn aj v2 arises from the intertttolecular forces due to the overlap of electron clouds. The constant b takes into accottnt the finite volume occupied by the molecules; its effect is to subtract from the volume term. 2 Multiplication of Equation (2.5) by 1J yields the equation =
=
Pv3 - ( Pb + RT) v2 + av - ab
= 0.
(2.6)
The result is a cubic equation in v with three roots, only one of which needs to be real. In Figure 2.2 some isotherms calculated from the van der Waals equa tion have been drawn. constant. The correction As T increases, the curves approach Pv terms in the van der Waals equation become less important. For T < Tc , there is a local maximum and minimum value of P. The transition between the two types of curves is a curve having an inflection point CP, the so-called critical curve, T = Tc . Imagine that the volume is decreased by loading a piston con fining the gas in a cylinder. Suppose one moves along an isotherttl for which T < Tc . After we reach a maximum of the c•Jrve, the pressure begins to fall. This is an unstable region since the pressure no longer increases as the volume is diminished. Actually, this portion of the isotherm is not really traversed at all because the gas undergoes a change of phase. * Assume that the compression =
* The transition from a gas to a liquid, from change" or a .. phase transformation."
a
liquid to
a
solid, etc. is termed a "phase
Chap. 2
14
Equations of State
p
d Figure 2.2
Isotherms for a van der Waals gas.
b ...... a
c
�--- -----+
·v
begi ns at p oi nt a i n the diagram. Part of the gas begins to liquefy at b and the pressure remains constant as the volume is further decreased as long as the temperature is held constant. Between b and d. liquid and vapor are in equi librium. Finally, at d liquefaction is complete. Thereafter the curve rises steeply because it takes a large increase of pressure to compress a liquid. The curves with no extrema ( T > Tc ) can have no regions of this kind. Above Tc it is impossible to liquefy a gas, n<, matter how large the pressure is. The interface between a liquid and a vapor is not discernible. The critical values Vc , Tc . and Pc of a st1bstance can be expressed in terms of the constants a and b that appear in van der Waals' equation. For T Tc . =
(2.7) The po i n t at which the tangent to the curve is horizontal is given by
iJP av
c
0 =- =
RTc --
� +
( Vc -- b ) --
2a -
(2.8)
Vc� ,
and the point of inflection is the point at which the rate of change ts zero:
of the
slope
•
c
= = 0
2RTc
3 ( vc - b )
_
6a
4. vc
(2.9)
Sec. 2..f
P-v-T Surfaces for Real Substances
15
Solving these equations yields Vc
Tc
=
3b, 8a
=
21Rb ' (2. 10)
Evidently, Pc vc
� RTc for a van der Waals gas at the critical point. The rela
=
tions of Equation (2.10) suggest that the equation of state of a van der Waals gas can be written in terms of the critical variables, eliminating the constants a and b. This is indeed the case (see Problem 2-5).
1.4 P-v-T SURFACES FOR REAL SU BSTANCES Real substances can exist in the gas phase only at sufficiently high tempera tures and pressures. At low temperatures and high pressures transitions occur to the liquid phase and the solid phase. The P-v- T surface for a pure substance includes these phases as well as the gas phase. Figure 2.3 is a schematic diagram of the P-v-T surface for a substance that contracts on freezing. Notice the regions (solid, liquid, gas or vapor) in which the sttbstance can exist in a single phase only. Elsewhere two phases can exist simultaneously in equilibrium, and along the so-called triple line, all three phases can coexist . •
_
._._,
_
,_
_
_
_
_
_
_
_
p
p
CP
-
.--..
-
-
-
-
T
Jiqui
CP gas vapor
-
-
-
�
-
Figure 2.3
freezing.
---
.-
-
---
._
-
--..,
T
triple line _. ----
_ _ _ _ _
P-v- T surface for a substance that contracts on
Chap. 2
16
Equations of State
p
p
S-L
S-L CP
liquid
•
CP
liquid
solid solid vapor
TP
vapor
TP T
�------• T
(a)
(b)
p. T diagrams for (a) a su bs t ance that contracts on freezing: and
Figure 2.4
(b) a substance th at e x p a nds on free zi n g.
The physical distinction between t he various phases is straightforward. A solid has a definite volume and shape. A liquid has a fixed volume but not a fixed shape. A gas has neither
a fixed volume nor shape. A ·vapor is a gas at any tem
perature less than the critical temperature. The projections of the surfaces on the
P- T plane are of special in terest
( Figure 2.4). The L-V curve is the vapor pressure curve, for which the liquid and vapor phases coexist in equilibrium. It is also known as the hsaturated , vapor, or boiling-point curve. The S-L curve is the freezing point curve. and the
S-V curve is the sublimation curve. At T
=
Tc the specific volumes of the saturated liquid and vapor
become equal. For T
>
Tc there is no separa tion into two phases; the interface
between the liquid and the vapor is indiscern i ble. By raising the temperature a bove Tc then increasing the pressure. and finally reducing the temperature, �
one can m ake an "end run" around the critical point. In this way, it is possible to m ove from a vapor to a liquid without crossing the vapor pressure curve. No critical point can exist for soJid-Iiquid equil ibrium (the S-L curve). That is because solids and liquids possess diffe rent symmetry properties. A normal liquid is isotropic, whereas a solid has a cryst alline structure whose ori entation defines a particular set of directions. The transition from one sym metry to another is strictly a discontinuous process. For C02 the critical temperature is
3 l .0°C
. and the critical pressure is
72 atmospheres. The triple point occurs at - 56.6°C a nd about 5 atmospheres. When heat is supplied to solid co� (dry ice) at atmospheric pressure.. it sub
limes and changes directly to the vapor phase.
17
Expansivity and Compressibility
Sec. 2.5 v
I I I
I I I I I t I
I
lOOOC
T
l.S
Sketch of the specific volume of water as a function of temperature.
For H 2 0 the critical point is 374.2°C and 207 atmospheres. The triple 3 point is 0.01 oc and 4.6 torr (6.0 X 10- atm ) . Thus the pressure must be
reduced to a low value to reach the triple point.
Water and a few other substances expand on solidif}·ing. Water expands about 10 percent on freezing. A block of ice floats with nine-tenths of its vol ume below the sutface of the 'Nater and one-tenth above. (An iceberg floats with one-seventh of its volume above the surface; the difference is because sea water is denser than fresh water and common ice is full of air b1Jbbles.) If
water contracted on freezing, the results would be disastrous, but ice floats. During a change of state, there is always a change in volume but the tempera ture and pressure are constant. For example.. at the freezing point of water, the
10 percent, as noted. But at the steam point the expansion is enormous: 1 cm3 of liquid water at l00°C becomes 1600 c.m3 of steam at this change in volume is
temperature. Changes of state are accompanied by changes in molecular forces. In Figure 2.5 the specific volume is sketched as a function of tetnpera
ture for water. Since v is the reciprocal of the density, it follows that the ma:
2.5 EXPANSIVITY AN D COMPRESSIB!LITY Suppose that the equation of state of a given substance is written in the form v
=
v ( T, P ) .
(2. 1 1 )
Chap. 2
18
Equations of State
Taking the differential, we obtain
dv
ilt'
tiT +
-
:.=
iiT
clv
r
\ ri P
r
tl P.
(2. 12)
(The subscript denotes the q ua ntit y held (:onstant . ) Two important measur able quantities are defined in terms {)f the pa rti al derivatives. The expansivity. or coefficient of volume e xp an sion is given by .
•
f3
rlV \ .l . aT ; P
1 =
t'
--
(2. 1 3)
This is the fractional change of volume resulting from a change in temperature, at constant pressure. Simi l a rl y the isothermal compressibility is defined as ,
•
K
-
=
-
1
-
v
iJv
(2.14)
i1 P .,.
.
the fractional change in volume as the pressure change� with the temperature held constant. The negative sign is used since the volume always decreases wi t h increasing pressure (at consta n t tempera t ure ): that is. the partial derivative is
inherently negative, an d the isothermal compressibility is a positive q uantity. For()an ideal gas, v = RTI P. and 1
R
1
f3 = � p = r ·
(2. 1 5 )
The higher the temperature� the I<)WCr the expansivity. AJso, K
=
-
1
--
l'
. RT \ -
, P_
-
1
(2. 1 6 )
-
p
•
The isothern1al compressi b i l i ty of an ideal gas is small at high pressures. . For a l iqu id or a solid .. in contrast to a gas� {3 and K are nearly constant over a fairly wi de range of temperature and pressure. This experimental fact allows us to develop an approximate equation of st ate that is useful in many applica tions. Substituting Equation (2. 1 3 ) and ( 2. 1 4 ) in Equation (2. 1 2 ), we obtain tl v
=-
{3vd7'
-
( 2. 1 7 )
I<" Vd P .
We assume t h a t the volume change is sm a i J when the temperature and pres sure - are changed so that to a first approximation, v ·v0 (a constant) and f3 and K are constants. Then ..
�
..
Sec. 2.6
An Application
29
Integrating, we have v
dv
=
{3v0
T To
dT
-
P
KV0
dP,
•
so that
v
=
v0 [1 + {3 ( T
-
T0 )
-
K ( P - P0 ) ].
(2.18)
This is an approximate equation of state for a liquid or a solid. The volume increases linearly with an increase in temperature and decreases linearly with an tncrease tn pressure. •
•
Suppose that the temperature of a block of copper is increased from 127°C to 1 37°C. We wish to know what change in pressure would be necessary to keep the volume constant. For copper in this temperature range,
Then, Equation
�p
(2.18) gives 10-5 �T = ( 10) 2 K 10-1 7.6 X
{3 =
·
5.2
X
=
6.8
X
107 Pa
��
680 atm !
2.6 AN APPLICATION Suppose that we wish to calculate the decrease in pressure of a fluid when it is
cooled from T1 to T2 • We know that the equilibrium states of the fluid are fixed by specifying two of the state variables that are related by some equa tion of state
f ( P, V, T)
=
0.
(2 .19)
We assume that the process in which the fluid is changed from an equilibrium state ( P1 , T1 ) to another equilibrium state ( P2 , T2 ) is isochoric that is, the
volume is unchanged. We suppose for a moment that the fluid . is cooled reversibly. This could be achieved by placing in contact with the fluid a series
of large bodies (reservoirs) ranging in temperature from T1 to T2 to effect a quasistatic cooling through a sequence of equilibrium states. For any one of these states we may write Equation (2.1 9), or, solving for P,
P
=
P (v� T ) .
(2.20)
Chap. 2
30
Equations of State
Taking the differential , w e ob ta in
aP a t'
c/P =
dv
r
+
iJP
I
,
aT
(2.2 1 )
dT. t'
Since we are assuming that the cool i ng takes pl ace at constant volume, the first term on t he right-hand side is zero. Thus ,.
r,
-
p,
-
L
pl
-
iJ P ,..
-
.
ar
T.
•
' 1.
(2.22)
dT.
Unfo rtuna tely.. the in te gran d is unknown. However.. using the cyclical relation •
given in Appendix A, we have
aP aT
--
aP 1 r
,.
dl'
f
Th us
riP
riT
-
-
��
1
-
ar
.
av
iJ V
-
a,v
-
aP
p
---
aT
riP
T
p
-
�
-
--
-
•
K
r
Substituting t his result in Equat ion (2.22)� we obtain P,-
-
PI
{3 =
K
-
( T*\-
-
Tl
)
( 2. 2 3 )
.
if {3 and K are i ndependent of T This is negative i f T., < T1 indicating that the pressure is reduced on cooling. Note that this result could have been obtai ned immediately from Equation (2. 1 8). In practice, the cooling would not be reversible because one simply heats the fluid and lets it cool. The result wou ld be large temperat ure gradients within the fluid itself and between the fl uid and its surroundings. Thus the intermedi ate states are no t equilibrium states and the equation of state cannot be applied. However_ the initial and fi nal states ar£· equilibrium states, and i t doesn't matter how we go from state 1 to state 2 to determi ne the change in the state P2 P, i s independent of the path, bei n g variable P. This is because a P deterr11i ned only by the end points. We can choose any convenient path to cal culate changes in state functions for processes between a pair of eq ui li briu m states. The most con ve nie nt path is a reversible path. ..
•
=
-
Problems
ll
Ou r fundamental state variables P, v, and T are all exact differe ntia ls whose integrals are independent of the path . Fotltlulating all thertnodynamic relations in terttts of state variables will be a goal of the theoretical develop
ment; this is called the condition ofintegrability. When we encounter a thertno dynanlic quantity whose differential is imperfect, we will seek an integrating factor to give us a quantity that we prefer to work with. Appendix A includes a comprehensive discussio�1 of exact and inexact differen tials.
P RO B L E M S
2-1 How many kilograms of helium gas are contained in a vessel of I liter volume at
50°C if the pressure is one atmosphere? (The atomic weight of helium is 4.)
2-2 A tank of volume 0.5
m3 contains oxygen at a pressure of 1.5
x
106 Pa and a
temperature of 20°C. Assume that oxygen behaves like an ideal gas. (a) How many kilomoles of oxygen are in the tank? (b) How many kilograms? (c) Find the pressure if the temperature is increased to 500°C. (d) At a temperature of 20°C, how many kilomoles can be withdrawn from the tank before the pressure falls to 10 percent of the original pressure?
2·3 A cylinder provided with a movable piston contains an ideal gas at a pressure P1 , specific volume v1 , and temperature T1 • The pressure and volume are simul
taneously increased so that at every instant P and v are related by the equation P
A = constant
= Av,
(a) Express A in terms of the pressure P1 , the temperature T1 , and the gas con
stant R. (b) Make a sketch representing the process in the P-v plane. (c) Find the temperature when the specific volume has doubled, if T1
=
200 K.
2-4 For an ideal gas the slope of an isotherrn is given by
iJP av •
-
r
-
p
-
-
v
and that of an isochore is aP aT
= t•
p
-
T
'
.
Show that these relations give P v = RT, the equation of state.
;
2-S Noting that at the critical point, the three roots of the van der Waals equation .
are equal (i.e., ( v Vc )3 = 0), show that the critical values of the specific vol ume, temperature, and pressure are given by Equation (2. 10). Show that, in -
Chap. 2
ll v'
terms of the reduced quantities
vfv" , T'
=
T/Tc
=
and
Equations of State
P'
=
P/Pc .
the van
der Waals equation becomes
P'
-r .
3-
n
·v
,� ..
,
I -
-
3
2-6 Usi ng the Berthelot equation of state� P
=
.
RT
· -
1J
·-
- -
b
-
�·-
a
.,
-·-
Tv-
.
show that IJ(
=
)b.
r
Tc =
(Hint: As noted in the text iJ PI rl l '
Compare the numerical value of in the fo llowi ng table:
Ba
'J 2 1bR
P c
·
fZtiR T2b 3b t
=
·
..
0 and il� PI riv2
=
RTc_ - / Pc vc. with ·
=
0 at the critical point.)
the experimental values given
Substance He H, Q., co, H ., O -
-
-
..
3.06 J.27 :\.42 J.6) 4.29
2-7 Using the Dieterici equation of state. p
=
RT
-
-
v - b
e a/ RTr
•
show that a
Pr = - 2 2 " 4e b and find the numerical value of RTc u la t ed experimenta] values"?
2-8
(a)
How docs this com p a re with the tab
Making use of the cyc l ica l re lation ( Eq u a t io n ( A .7) in Appendix A ) . fi nd
the expansivity f3 of
(b)
/ Pc Vc .
a
substance obeying the Dieterici e q uation of state i n
Problem 2-7.
At h ighe r temperatures and l arge specific vol utnes ( low densities) all gases a pproxi ma t e an ideal gas. Show that for large values of T and v. the e x pres s ion for {3 obtained in ( a ) goes over to the corresponding equation for an ideal gas.
2-9 Show that in general
ap · -
aP
r
+
dK
-aT
=
r
0.
Problems
]3
1-10 A hypothetical substance has an expansivity /3 2bTfv and an isotbertnal com pressibility K = afvt where a and b ·are constants. Show that the equation of =
state is
v
-
bT2 +
aP
=
constant.
Z-11 Suppose that
f3
=
v-a Tv
K = '
3 ( v - a) 4Pv
•
Show that the equation of state is
4 3' P (v - a) where
=
AT
,
a and A are constants.
2· 12 Show that � and K are infinite at the critical point. 2-13 A glass bottle of nominal capacity 250 cm3 is filled brim full of water at 2QOC. H the bottle and contents are heated to SOOC, how much water spills over? (For water,
f3 ::: 0.21
x
10- 3 K - 1 . Assume that the expansion of the glass is negligible.)
•
I
3. 1
Configuration Work
3.2
Dissipative Work
37 •
4
40
•
. .
.
'
..
· 3 .3
Adiabatic Work and Internal Enetgy
3.4
Heat
43
3.5
Units of Heat
45
3.6
The Mechanical Equivalent of Heat
46
3.7
Summary of the First Law
46
3.8
Some Calculations ofWork
47
•
41
35
3.1 C ONFIGURATION WORK
Configuration work is the work in a reversible process given by the product of some intensive variable (P, for example) and the change in some extensive variable (V, for example). Let y represent such an intensive variable and X the corresponding extensive variable. In the most general case, where more than one pair of variables may be involved, llW
y; d X; , i
=
=
1, 2, . . n.
(3.1)
.
•
'
Here ltW represents the work done by virtue of infinitesimal changes in the extensive variables X;; 71W is not an exact differential, as indicated by the notation (see Appendix A). The variables X1, X2, etc.� are said to determine the configuration of the system, and
•
'
is called configuration work. Consider the configuration work done by the system consisting of a gas enclosed in a cylindrical container by a piston (Figure 3.1). The gas expands against the pressure associated with the external force F (possibly due to the weight of the movable part of the piston).lne work done by the gas is aW
=
F dx
=
PA d.t
=
P dV.
Here dx is the displacement of the piston, A is its cross-sectional area, P is the pressure (the intensive variable), and V is the volume (the extensive variable). Since P and dV are positive, ltW > 0, denoting work done by the sy�tem. This sign convention will be used throughout the text. We assume that the process is reversible and that the system is in equilibrium at all times. No friction (dis sipation) is involved. The integration is nontrivial since P P(V) in general. =
37
The First Law ofThermodynamics
Chap 3
38
.
F=
dx
�
.
PA
----
� --
----
�-
-
-
:
4 ·
'·
final
level
--
original level
--: -------
•
Fipre 3.1
Expansio.n of an enclosed gas against the pressure produced by an external force.
TABLE 3.1
Various examples of configuration work.
Svs tem
Intensive Variable
•
gas
..
Extensive Variab1e
.
V (volume) A (area) q ( c har ge ) M ( ma�nctization) P (polarization)
P (pressure) r (surface te n sio n ) £ (electromotive force) 8 ( ma g n et i c fi e l d ) E (electric field).
liquid. or solid
film electrolvtic cell �
magnetic material dielectric material
...
PdV I'dA Edq Bt/M Ed'P
In thermodynamics configuration work is often called PdV work for obvious reasons. However. there are many examples of configuration work, ....
'�
some of which are listed in Table 3.1. For a surface film of liquid._ the reversible work required to increase isothern1ally the surface area of the film by an
amount
dA
is aW
fd A._ where
f
is the surface tension. In an electrolvtic cell such as a lead storage battery._ the configuration work is the product of the =
electromotive force
£
•
•
of the cell and the amount of charge transferred in the
chemical reaction . Magnetic work is done <)n a magnetizable material by a magnetic field.
I( for
example� the material is placed in a solenoid that gener
ates a field B, the incremental work done is
lldM, where dM is the change in
total magnetization of the material. Analogously. an electrically susceptible material inserted between the plates of a parallel-plate capacitor experiences a change of total electrical polarization
tiP: the
work is·
EdP,
where E is the
electric field strength in the region between the plates. The configuration work done in the change of volume of an ideal gas can
be easily calculated in some special cases. For an isochoric process.. dV aW
=
=
0. so
0. For an isobaric process, \ ..
W
=
PdV
=
P
\
liV I I
=:
P(Vh- Vn).
(3.2)
Sec. 3.1
Configuration Work
39
p
p
----
3.2 Work done Fi in an isobaric process.
v
Here Va is the initial volume and Vb the final volume. The work done is evidently the area under the curve in a P-V diagram. lbis is the shaded rectangle of Figure 3.2 for the case of an isobaric pr oce ss For an iscthert11al process, the temperature is constant, the P-V curve for an ideal gas is a hyperbola, and .
.
W
=
nKT
v, dV Vo
V
V,
=
nRT In V.
(3.3)
,
a
(Figure 3.3). In both the isobaric and the isothermal process, if Vb > Va, the process is an expansion, and work is done by the system (W > 0). If Vb < V,, the process is a compression, and work is done on the system (W < 0). This important sign convention will be followed throughout the text.
p
Figure 3.3
Work done in an isothertnal process involving an ideal gas.
v
Chap. 3
40
The First Law ofThermodynamics
p A
I •
Figure 3.4 Work is path-dependent; the area under curve A is differe nt from the area under curve B.
B
• I '
I I
I I
VII
It must be emphasized that work is not a state variable. This is easily seen in Figure
a
v
property of the system: W is not
3.4. Since f P dV is the area under
the curve, different results are obtained for paths A and B. Thus a.
aWis not exact.
b. W depends on the path of integration. c.
The integral around a closed path is not zero: that is,
:: : :::::::*;:: :?.::::=�:::-:-:�::: �:=:�::s: :/�:: ��: �:: . :::
�aW
* 0.
�):::::::=->.�::�-=.�:: �:;:·
. •'•··..··.·.·.. ·•· .·.·.· ·. .·.·.·.·: . ..·.·. · . . ... .. .... •. . . .. ... . . ...• • .. ...... , . .. . ..... ... . . • · .... ... ·.. ... • · ·. . . ·....·..·.··.·. . ·.·:.·. .· ·.· .· ·. ·�..·: =:.:·:.·.•.·.·. ··· '·I'·'·· ' : .: .::·.·::-::: .. : : :: : :: : : �:.::::·; : :::: :,:: . : ·:. :: :::·::·. ::·:::�:=:=:;:: �� ·:�(: : :::.� :: ·.::.::: :·:.:•.· : : ·:: ::·= : :..· .:·.:·: :, :: =:{"-· ::� � < � :·:·:x::: :: ::: :: ::": . ·.
....
· ...
•
3.2 DISSIPATIVE WORK Dissipative work is work done in an irreversible process. Such work is always done on the system. In the general case� both configuration 'W'ork and dissipa tive work may be done in a process. The total work is the algebraic sum of the two kinds of work. If a process is to be reversible. the dissipative work must be
zero. Since a reversible process is necessarily quasistatic� then to specify that a process is reversible implies that the process is quasistatic. There are two well-known examples of dissipative work. The first is stir ring work. Consider a stirrer immersed in
a
fluid. The stirrer and the fluid
together constitute the system. The stirrer is attached to a shaft projecting through the wall of the container and an external torque is applied at the outer end of the shaft (Figure
3.5).
Regardless of the sense of rotation of Lhe shaft, the external torque is always in the same direction as the angular displacement of the shaft, and the work done by the external torque is always negative (work is always done on
the system). The work is given
by aw
=: -
Td8.
(3.4)
Sec. 3.3
Adiabatic Work and Internal Energy
. ..
41
.
. .
,.. .' -�.; ;:: ....... ... . . ..· . .·... . .... :::� .._.,'-"'-"" .
.
·
;
�. ·
. . .. . ..�
./'..-/'�""'"""' .
...... ·:. ...... .
.
·
..:.>
.
. . ... . .. . . . ' . '. .. .
.. •
.
.
.
.
:
..
. ..
.
.
.
Figure 3.5 Stirring work. A stirrer is inunersed in a fluid and an external torque is
·. .
:�·
. · .
.
:
.
:.
. •
� ·.:. ... ·�
.
..
. ·.··..., .·. . .. . :·
.
· .
. .
::
... . .. .
. . .
... ,•.•
. .. .. . .<. .
.
.
:
··
·
·
:
. •.
-
applied. I •
I R Figure 3.6 Electrical work. A current is passed through a
I
resistor.
Clearly, reversing the sense of the torque
T
reverses the direction of the angu
lar displacement d8. A second example of dissipative work is the work needed to maintain an
electric current I in a resistor of resistance R (Figure 3.6). Quantitatively,
(3.5) Here ttW is the work done in time dt on the system comprising the current and the resistor. Reversing the direction of current flow does not affect the sign of the work.
3.3 ADIABATIC WORK AND IN TERNAL ENERGY
A fundamental result of the greatest importance is arrived at by considering adiabatic processes between two equilibrium states of a system. Having defined configuration work and dissipative work, we can think of different adiabatic paths between two states a and b of the system. In the P- V diagram of Figure 3.7, the path labeled 1, denoted by the solid line, could represent stir ring work (at constant volume), followed by an adiabatic compression of the fluid. (The system is assumed to be thermally insulated from its surroundings.)
Chap. 3
42
p
The First Law ofThermodynamics
b
Figure 3. 7
Three different adiabatic paths representing processes in which an isolated system is carried from state a to state b.
a
�-------+�
Correspondingly, path
2�
v
portrayed by the dashed line, could depict an
adiabatic compression followed by electrical work at constant volume. T he dotted line of path 3 could delineate somt� combination of mechanical and electrical work performed simultaneously. Now, it is an experimental fact that the work done in the three processes is the same. To be sure, experiments have not been performed for all possible adia batic processes between al1 equilibrium states of all possible systems. However, the whole edifice of thermodynamics is consistent with the conclusion that The total work done i11 all adiabatic processes be/ween any t�vo equilibriLlnz states is i1z dependen t
of the path.
"
�tJ
lf"''ad
=
•c
J
is path-independent and
lfWad is an exact differential.
(We can remove the bar
from the differentiaL but only if we restrict ourselves to adiabatic work.) It fol lows that we can define a property of the system, a state variable U, such that
ll
,,
In differential form, (3.6) We note that dU is the negative of the adiabatic work done bj' the system, so it is the adiabatic work done on the system. We define U as the internal energy of the system. Thus Equation (3.6) states that the increase in internal energy is
Sec. 3.4
43
Heat
equal to the work done on the system in any adiabatic process. Alternatively, the work done on the system (with no heat flow) results in an increase in its internal energy. To summarize, experiments show that an energy function of state can be defined by the amount of work done in an adiabatically enclosed system. This
is not to say, however, that any two states of a system can be coupled by an adi abatic path.
3.4 HEAT
We wish now to extend our ideas to processes that are not adiabatic. If the sys tem is not contai11ed within adiabatic walls, it is nevertheless possible to bring about a given change of state in quite different ways. "-\ beaker of water may be brought from 20°C to l00°C by electrical work perforrned on a resistor immersed in it, or, alternatively, by lighting a Bunsen burner under it. The lat ter process inyolves no work at all (Figure 3.8). For any change between given states, d U can always be uniquely deter mined by carrying out an experiment under adiabatic conditions, for which dU = -llWad· If the conditions are not so specialized, then dU =1: - t1Wad' in general. Instead we may write the equation dU
=
7/Q- aW,
or
t!Q
=
ltW - 7/Wad.
'·
(a) Figure 3.8
(b)
Two ways of raising the temperature of water: (a) by electrical work; (b) by applying heat.
Chap. 3
44
The First Law ofThermodynamics
.
In other words, we can define heat flow as follows: Heat flow into the system is equal to the total work done by the system minus the adiabatic work done.
Then aQ
==
dU + dW,
(3.7)
•
which is the first law of therr11odynamics in differential form. In words, the first law states that The heat supplied is equal to the increase in intertzal energy of the system plus the work done by the system. Energy is conserved if heat is taken into account.
·
Note that heat is not a property (state variable) of the system; only the inter nal energy is. It can be shown that the quantity liQ exhibits the properties that are commonly associated with heat. These properties are summarized as follows: 1.
The addition of heat to a body change� its state.
2.
Heat may be conveyed from one body to another by conduction, convec tion_ or radiation.
3.
In a calorimetric experiment by the tnethod of mixtures, heat is con served if the experimental bodies are adiabatically enclosed. We note that since U is a state variable�
dU
=
0.
That is, in a cyclic process, the change in internal energy is zero so that the work performed is equal to the heat absorbed by the system. One useful way of looking at the first law follows from transposing the terms in Equation (3.7):
dU
=
71Q- aW.
The increase in internal energy of a system equals the heat flow into the system minus the work done by the system. Think of a bank: aQis a deposit� itW a withdrawal. What's left in the bank is d U. The sign convention for aQ is 7/Q > 0 for heat flow into a system, and 71Q < 0 for heat flow Ollt of the system.
Sec. 3.5
Units of Heat
45
The temperature of a body alone is what determines whether heat will be transferred from it to another body with which it is in contact, or vice versa.
A large b·tock of ice at ooc has far more internal energy than a cup of hot water;
yet when the water is poured on the ice some of the ice melts and the water becomes cooler, which si
·
tes that energy has passed from the water to the ice.
When the temperature of a body increases, it is customary to say that heat has been added to it; when the temperature decreases� it is customary to say that heat l1as been removed form it. When no work is done, �U
=
Q,
which says that the internal energy change of the body is equal to the heat transferred to it from the s11rroundings. One definition of heat is: Heat is energy transferred across the boundary of a system temperature difference only.
as
a result of a
This definition is not entirely satisfactory, however. Changes of state (e.g., from ice to water or from water to steam) involve the transfer of heat to or from a body witlzout any change in temperature. It should also be kept in 111ind that heat transfer is not the only way to change the temperature of matter: a body that has work done on it may become hotter as a result, and a body that does work on
something else may become cooler. A more precise definition of heat is:
Heat is the change in internal energy of a S)'Stem when no work is done on or by the system.
3.5 UNITS OF HEAT
Since heat is a form of energy, the correct SI unit of heat is the joule (J). However, the kilocalorie is widely used. It is defined as follows: 1 kilocalorie (kcal)
the heat required to r&ise the temperature
1 kg of \Vater from
of
14.5°C to 15.5°C.
The relationship between the joule and tlte kilocalorie is: 1J
=
4 2.39 X 10- kcal,
1 kcal
�
4184 J.
The calorie that dieticians use is the same as the kilocalorie and is some times written "Calorie" to distinguish it from the ordinary, smaller calorie associated with the cgs S)'Stem of units.
Chap. 3
46 · . ·. :.:.:_:.
. .
:.
····:
. ..
..
.
. ·
.
.
:
· . .
. . . .·.
The First Law ofThermodynamics .
·
.
:
'
. .
.
.
.
.
.
.
.
.
'
. .
.
'
.
. ·
.
.
.
.
3.6 THE MECHANICAL EQUIVALEN T OF HEAT
We have noted that if the configuration \York and the dissipative work are both zero.
or the heat flow in increases the internal energy. Suppose. on the other hand. that the configuration work is zero and that work is done with a stirrer immersed in a fluid kept at constant volume and thermally insulated. Then
where Wd is the dissipative work done on the system and is inherently negative. The right-hand sides of the equations in the two examples give rise to the same change in internal energy. Thus work and heat are .. equivalent'' in this sense. However, heat and work can be differentiated from a microscopic view point. When energy is added to a system in the form of heat, the random motion of the constituent molecules is increased. Consider gas in a cylinder. If heat is added through diathe1tt1al walls, this increases the random kinetic energy of the molecules which means a rise irt temperature. Now let the gas be confined in a cylindrical piston with adiabatic walls. When the piston is pushed in� the mole cules striking the piston are accelerated. in the direction of its travel. However. any organized motion initially imparted to these m<>lecules is rapidly random ized by collisions, either with the walls or with other gas molecules. Again� the increase in random kinetic energy appears as an increase in temperature. •
3.7 SUMMARY OF THE FIRST LAW 1.
Energy is conserved. Heat is energy transferred to a system causing a change in its internal energy minus any work done in the process.
2.
The quantity U is a generalized store of energy possessed by a thermody namic system which can be changed by adding or subtracting energy in any form.
3.
The internal energy U is a state variable: it is e x te nsiv e
.
4. The first law can be expressed in differt�ntial form as aQ
=
dV + aw.
5. For a reversible process, aW is so lely 71Q
==
dV +
con figura tj on (•lo PdV'�) work, so that
PdV.
The First Law ofThermodynamics
Chap. 3
48
If the expansion is isobaric� d P = nR
W
( T,,
-
=
0 and
T,,)
P ( V,,
=
-
V.,) .
Again, work is done by the system.
Equation (3.8) is especia lly useful for deter m in ing the work done
compressing a solid. Suppose that
a
.
m
10 g block of copper is isothermally (and
quasistaticaJJy) compressed at a t emperature of 0°C. If the initial pressure is 1 atm and the final pressure is
1000 atm. how much work is done on the block
of copper? We assume that whereas the pressure change is large. the volume cha n ge is small,
V,,
=::;
V.,
=
V. Then aw
=
-
KPVtiP�
and ,
,
'
PdP
KV =
-
-.
. I' .
For copper. K
=
7.5
x
lO
�,
=
-
( P,,- - Po-) . �
(3.10)
...,
Pa- 1 at OOC and the density pis 8.93 g( em)-·'. Thus 10 g
m
Since Pa2 << P,?, and
12
,
�
l
m�
_
_
1
__
6
3
1.01 x 108 Pa we find that
1
a comparatively small amount of work. At this stage of our study� we cannot say much about what happens to the heat and the internal energy in these processes. We need some more basic concepts, which we shaH introduce in subsequent chapters.
PROBLEMS
3-1 Ten kilomoles of an ideal gas are cotnpressed isothermally and reversibly from a pressure of 1 atmosphere to 10 atmospheres at 300 K. How much work is done? 3-2 S t e am at a constant pressure of 30 atn1ospheres is adn1itted to the cylinder of a steam engine. The length of the stroke is 0.5 m and the diameter of the cvlinder is 0.4 m. How much work is done by the steam per str oke? .,
�
Problems
49
3-3 An ideal gas originally at a temperature
and pressure P1 is compressed
7;
reversibly against a piston to a volume equal to one-half its originai volume. The temperature of the gas is varied during the compression so that at each instant the relation P
(a) (b) (c)
=
A V is satisfied, A
=
constant.
Draw a diagram of the process in the P- V plane. Find the final temperature T2 in ternts of T1• Find the \York done on the gas in terms of n, R. and T1•
3-4 Ice at ooc and at a pressure of 1 atmosphere has a density of 916.23 kg m-:l, while water under these conditions has a density of 999.84 kg m-:t. How much work is done against the atmosphere when 10 kg of ice melts into water?
3-5
(a)
Derive the general expression for the work per kilomole of a van der Waals gas in expanding reversibly and at a constant temperature T from a specific volume v1 to a specific volume v2•
(b)
Find the work done when 2 kilomoles of steam expand from a volume of 30m�
a
volume
of
60 m-'
at
a
temperature
of
l00°C. Take
3.19 x 10-2 m3(kilomole) -t. lQ-" Jm3(kilornole) -2 and b Find the work of an ideal gas in the same expansion. a =
(c)
to
5.80
x
==
3-6 The temperature of an ideal gas at an in.itial pressure P1 and volume
Vj
is
increased at constant volume until the pressure is doubled. The gas is then expanded isother1nally until the pressure drops to its original value, where it is compressed at constant pressure until the volume returns to its initial value.
(a) (b)
Sketch these processes in the P- V plane and in the P- T plane. Compute the work in each process and the net work done in the cycle if n
=
2 kilomoles, P1
=
2 atm, and V1
=
4 m�.
3·7 Two kilomoles of a monatomic ideal gas are at a temperature of 300 K. The gas expands reversibly and isothermally to twice its original volume. Calculate (a) the work done by the gas; (b) the heat supplied. (Note that for an ideal gas UrxT, so � U
=
0 in an isothermal process.)
3-8 A gas is contained in a cylinder fitted with a frictionless piston and is taken from the state a to the state b along the path acb shown in figure 3.9. 80J of heat flow into the system and the system does 30J of work.
p
Figure 3.9 P- V diagram for Problem 3·8.
c
b
a
d v
The First law ofThermodynamics
Chap. 3
50
(a) How much heat flows into the systetn along the path adb if the work done by the gas is 10 J?
(b) When the system is returned from b to
along the curved path, the work
a
done on the svstem is 20 J. What is the heat transfer? .,
(c) If Ua
=
0 and Ud
=
40 J, find the heat absorbed in the processes ad and db.
3-9 A volume of 10 m3 contains 8 kg of oxygen at
a
temperature of 300 K. Find the
work necessary to decrease the volume to 5 m)
(a) (b) (c) (d) (e)
At constant pressure.
At constant temperature. What is the temperature at the end of the process in (a)? What is the pressure at the end of the process in (b)? Sketch both processes in the p V plane. ..
3-10 The pressure on 100 g of nickel is increased quasistaticalJy and isothermally from zero pressure to 500 atm. Calculate the work done on the material, assum ing that the density and isothermal compressibility remain constant at the val ues of8.90
x
103kgm-3and6.75
x
10-11Pa-1,respectively.
3-ll During the ascent of a meteorological balloon its volume increases from 1 m3 to 1.8 m3, and the pressure of the ideal gas inside the balloon decreases from 1 bar to
!
bar-(1 bar= 105 Pa). The internal energy of the gas is U =BOOT joules,
where Tis in kelvins.
(a) (b) (c) (d)
Assume that in this process (the ascent) V
=
AP
+
B. Find A and B.
If the initial temperature is 300 K what is the final temperature? ..
How much work is done by the gas in the baJJoon? How much heat does it absorb?
•
'
4.1
Heat Capacity
53
4.2
Mayer's Equation
54
4.3
Enthalpy and Heats ofTransformation
57
4.4
Relationships Involving Enthalpy
59
4.5
Comparison of u and h
61
4.6
Work Done in an Adiabatic Process
61
51
4.1 HEAT CAPACITY
Q is
Consider a process in ·wrhich heat
added to a system, causing it to change
from one equilibrium state to another with an accompanying rise in tempera ture !l T. The heat capacity C of a system is defined .as the limiting ratio of the heat absorbed divided by the temperature increase:
lim
c
�r ....o
Q �T
=
71Q . dT
The heat capacity is not truly a derivative beca.use
aQ
(4.1)
-
is not an exact differential. We interpret
Q is not a state variable and
'llQ
as an infinitesimally small
flow of heat and dT as the resulting change in temperature. Furtherttlore, "heat capacity" is a misnomer because it implies th at
a
system is capable of
holding heat, which it is not; a term connoting the storage of internal energy would be more appropriate. The heat required to produce a given temperature change is propor tional to the mass of the system. We define the specific heat capacity, often abbreviated to "specific heat," as the heat capacity per unit mass:
c=
1 n
tJQ dT
(4. 2)
The SI units are J kilomole -l K -t. The specific heat capacity depends on the conditions under which the heating takes place
that is, on what parameters are held constant during the
heating process. There are two important cases:
(1)
the specific heat c11, where
the heat is supplied at constant volume; and (2) the specific heat
cp,
in which
the heat is added at constant pressure. Thus llq
dT
and
cp
=
(4.3)
v
53
Chap.4
54
Applications of the First Law
•
c,
.l kilomole-1
K -J
-
I •
0
400
200
800
600
1 000
1200
7"( K) Plots of c,. and cp for copper versus temperature at a pressure of one atmosphere. (Adapted from Thermodynamics.. Kinetic
Figure 4.1
Theorv, and Staristica/ Therntt>dvnanlics 3rd edition by F.W. Sears and ..
..
•
•
•
G.L. Salinger.) •
The distinction between the two heat capacities is important chiefly for gases because of the large coefficients of thermal expansion. In general. the specific heats are functions of temperature and can be regarded as constants only over limited ranges of temperature. As an example., consider the specific heats of a solid such as copper� shown in Figure low temperatures,
Cp
::=
cr
and both ten(l toward zero as the temperature
approaches zero. For high temperatures,
25
x
10� J kilomole- 1 K
-
1•
4.1. At
c,.
approaches the constant value of
Many solids exhibit this behavior� an en1pirically
established fact known as the law of Dulong and Petit. to be discussed in Chapter 16.
4.2 MAYER'S EQUATION We wish to find the relationship betweert
c,.
and
cp
for an idea) gas. For a
reversible process. the first law is dU or� in terms of intensive variables
=
..
tiLt Since
u
7/Q- Pd�.�.
=
aq
-
(4.4)
Pdt'.
is a function of the state variables. \Ve can write. say.
.
(4.5)
Sec. 4.2
Mayer's Equation
55
Here we assume that the equation of state is
Pv
=
RT.
(4.6)
From Equation (4.5),
du
au av
=
r
v d
au aT
+
dT.
(4.7)
v
Combining Equations (4.4) and (4.7), we have
tlq
=
au aT
dT
(4.8)
+
v
To obtain cv, we divide this equation by dT and hold the volume constant so that dv =0. The result, which holds for any reversible process, is tlq
au aT
-
-
dT
· v
•
(4.9)
v •
We shall see that for an ideal gas, u u(T) only. This follows from the Gay-Lussac-Joule experiment (see Chapter 5). Thus =
au av
T
=0
(4.10)
for an ideal gas. The substitution of Equations (4.9) and (4.10) in Equation ( 4.8) gives ·
tlq
=
cvtJT· + Pdv.
(4.11)
Differentiating Equation (4.6), we have
Pdv
+
vdP
=
RdT.
(4.12)
Using this in Equation (4.11) leads to
·
tlq
=
(ct.
+
R)dT - vdP.
Now
Cp=
tlq
dT
p'
(4.13)
.
56
dP
Applications of the First Law
Chap.4
=
O. Then llq =
dT
p
c,. +
R
c,. -+·
R.
=
c1'1.,
or Cp
=
(4.14)
This relation is known as Mayer's equation.* It states that over the range of
variables for which the ideal gas law holds, the two specific heat capacities differ by the constant R. We introduce the ratilJ of specific heat capacities Cp
y=
c,.
=
(4.15)
.
For a monatomic gas at room tempera t u re
For a diatomic gas at room temper at ur e
.
values will be derived in Chapters R
=
cp
,
=
cp
�R
==
..
i R.
cr
=
cr
=
�R
..
and y =
� R. andy
14 and J S. The difference cp
-
=
c,.
1.67.
1.40. These is equal to
8.31 x 103 J kilomole ·-I K- 1 for a wide range ()f temperatures. For solids
the difference is about 5 x 1 O·' J kiJomole K- 1 at 1200 K by way of comparison. ..
Finally. from the foregoing analysis we can calculate the internal energy
of an ideal gas. From Equation
( 4.9) we can write
(' v =
since
11
ar
u
depends solely on Tin the case of an ideal gas. Then
or ·T ll
-
-
u0
=
-T
CvdT.
If cv is independent of the temperature.
(4.16) *Julius Robert von Mayer. a German physician. Jiscovercd the equivalence of heat and
work and first enunciated the principle of conservation of energy
( 1842 ).•
Sec. 4.3
Enthalpy and Heats ofTransformation
57
The specific heat capacity at constant volume is nearly constant over a region of temperature extending from approximately 200 K to 1200 K for most gases. Thus Equation (4.16) is applicable for many practical problems.
4.3 ENTHALPY AND HEATS OF TRANSFORMATION •
The heat of transfotmation is the heat transfer accompanying a phase change. A change of phase is an isothetultlal and isobaric process and entails a change
of volume, so work is always done on or by a system in a phase change. Thus
is the specific work involved in the process. Now •
du
=
aq- Pdv� •
or, for a finite change, .
'
so that
I I •
,
I
(4.17)
• •
I . .
I
Here l is the latent heat of transformation per kilomole associated with a
given phase change. Let h
u
+ Pv.
The quantity h is called the specific enth.1/py. The origin of the word is the Greek verb thalpein meaning ''to heat." Since
u,
P, and v are all state variables,
h is also a state variable. (If two variables have exact differentials, their prod
uct and sum also have exact differentials.) With this definition, Equati on
(4.17) becomes
which is to say that the latent heat of transformation is equal to the difference in enthalpies of the two phases. Consider now the three possible phase changes. Let 1 denote a solid, 2 a liquid, and 3 a vapor. Then .
Applications of the First Law
Chap.4
58 •
I 12
/23
/11
=
h'' - /z' � solid to liqttid (fusion).
=
h'"
=
h'" - h'
-
h"--+ liquid to vapor
'
(vaporization),
�solid to vapor (sublimation) . .
Here the first superscript is associated with the .'iecond subscript; that is, dou-
ble prime is associated with the subscript 2 denoting a liquid etc. This is merely a notational convention. Since enthalpy is a state function.. ,
tlh = (.)� the change of enthalpy in a diagram as in Figure 4.2.
cyclical process is zero. This can be depi cted in a p T ..
Consider a cyclical process around the triple point and close enough to it so that only changes in the enthalpy occur during phase transitions. Then solid
:.
vapor (heat flows in)
..
dhl
vapor� liquid (heat flows out) ..
�h 2
liquid � sol id (heat flows out).
�h3
=
=:
�
Here
r
CP
solid liquid Figure 4.2 Pressure· temperature plot for a substance that contracts on freezing. The circle denotes a cycJicaJ process around the triple point for which the total change of enthalpy is zero.
ST.�Rl' vapor
.
.
T
/13 I:.2 /21
=
-
1-r:,
= -lr2
Sec. 4.4
Relationships Involving Enthalpy
TABLE 4.1 substances.
and fusion for various
Latent heats of
Latent Heat of Vaporization at Steam Point,
Mercury Alcohol Gasoline
Latent Heat of Fusion
1 atm
538 kcal/kg 63 207 95
Water
59
at Melting Point,
1 atm
80kcal/kg
Water
3 5 77
Mercury Lead Aluminum
•
so
or
(4.18) (All these latent heats are intrinsically positive.) This is an important result. Some values are given in Table 4.1. The latent heat of sublimation for water is 538 + 80 618 kcal kg-1• =
The SI unit for latent heat is J kg -t. Thus, for example, 8 kcal 53 kg
x
4184
J kcal
=
2.25
X
106
J kg
.
4.4 R�LATIONSHIPS INVOLVING ENTHALPY In Section 4.2 we assumed that
u = u(v, T).
Whereas the selection of
from among the three fundamental stat� variables
P, v,
v, T
and Twas arbitrary,
we will find that this choice is a natural one both from an experimental and mathematical viewpoint. Similarly, the natural choice in the case of the thermodynamic variable
h is h
=
h (T, P).
(4.19)
Then the analysis proceeds in a manner exactly parallel to what was done in the case of the internal energy. Here aq h
=
du + Pdv,
= u + Pv,
(4.20 ) (4.21)
Sec. 4.6
Work Done in an Adiabatic Procass
61
since, from Equation (4.27), h depends on T only. Hence we can write T
h - h0= cpdT = cP ( T
T0)
-
To
if
Cp is a constant. For an ideal gas, then, the specific enthalpy is given by h=h0+cp(T-T0).
::: : ·::;::: ·:::::·:::::-:; : ::··:::: :·:::= · :: .:.:·: : :;·::::.:·:::: : .::·;: :::. · ::: :: : ::::=: �::: :: ::-: =: ::.·: :: :;:·:= -::: ::�:::::· ::�
(4.29)
::*:::·.:=;:::::.:::=:::.:=·\:: .::·=: ;::·:.:.:'::::: ·:.:·:·:::::::::::.: :·:·:.: ::::· :::::·::: ': :: :: .:.::::·:::::·:. ::·;::.;:.:: ::: · ::::· :;:::.: :·.:::· .:.:· :::.: :: : ..:::: ;: :· :·:
4.5 COMPARISON OF
u
AND h •
Table 4.2 summarizes the parallel expressions involving the internal energy and the enthalpy.
TABLE •.2 enthalpy.
Analogous relations involving the internal energy and the
Reversible process:
Internal energy u
Enthalpy h
du
dh
Cv
=
=
tlq- Pdv dU dT
Cp
=
ilq + vdP ah
=
v
llq c"dT au 0 av r
Ideal gas:
=
+
Pdv
=
ar
p
ilq cpdT- vdP iJh 0 iJP T =
=
T
h- h0
c,.dT
=
T.,
0
4.6 WORK DONE IN AN ADIABATIC PROCESS We· have seen that the specific work
P( v2 - v1), where
v1
w
done in an isobaric process is
is the initial specific volume and� is the final volume.
Similarly, for an isotherlttal process in which the substance is an ideal gas,
(4.30) We now wish to find the specific work done in an adiabatic process involving �n ideal gas. Setting llq 0 in Equation (4.28), we obtain =
vdP = cpdT.
•
.-
(4.31)
•
62
Chap.4
Applications of the First Law
We also have 7/q which for 7/q
=
=
c,.dT + Pdv�
0 yields Ptlv
=
-
c,.dT.
(4 32 ) .
Dividing Equation (4.31) by Equation (4 32 ) gi ves .
·vdP
cp -
-
Pdv
-y ..
--
-----
c,.
:__
or dp
(/1) . -y·
- =
( 4.33)
'l'
jt>
This equation can be easily integrated to give ( 4.34)
where K is a constant of in t e g ra t ion . This is the relationship between the pres
sure and volume for an adiabatic process involving an ideal gas. Since y > 1
�
it follows that P falls off more rapidly with v for an adiabatic process than it
does for an isothermal process (for which J:lv
==
constant).
The work done in the adiabatic process is V2
1
•
Ptlv
1C =
=
v-Ydv
K
==-
•
1 (Kv Y)
1)2
.
. .
1-'VI
--
VI
v,
Pv1 at bot h limits: if we use K Now, K P1v{Y at the lower limit �'e ohtain K =
=
P2vl at the upper limit and
=
lV -=
For an expansion� v2
> v1
..
lV >
l
1 -y
t.
0. and the work is done bj� the gas: for
pression the work is done by the surroundings
reve rs i ble adiabatic process.
1�-' = u!
ful expression for an ideal gas.
( 4.35)
---- r P2L2 ), - PI v IJ·1
-
lt2
==
cr(T1
a
com
the gas. Note that for a
T2) .. which is another use
Sec. 4.6
Work Done in
an Adiabatic
Process
As an example of an adiabatic expansion, consider an ideal mona tomic gas enclosed in an insulated chamber with a movab le pis ton The ini. tial values of the state variables are P1 = 8 atm, V1 = 4 m3 and T1 = 400 K. .
The final value of the pressure after the expansion is P2
=
1 atm. We wish to
find \12, T2, W, and d U. For an ideal monatomic gas, the ratio of specific heats 1 is
5/3.
For an adiabatic process,
Then T2 is found from the ideal gas law: 13.9 4
(400)=174K. •
As the gas expands, the temperature and the pressure drop (no heat flows into or out of the system). The work done by the system in the ex pansion is
=
---
= 2.74 Finally, since
Q
=
X
106 J.
0, dU = - W= -2.74
x
106 J. The internal energy
decreases. This result could also be found by
using the expression
AU= Cv(T2 - T1 ): 3R
=
(8
X
=-2.74
PtYt 3R 1.01
X
105
400 X
X
4) 3 - (174- 400) 2;
106 J.
Work is done by the system at the expense of internal energy.
Applications of the First Law
Chap.4
64
PROBLEMS
4-1 The specific heat capacity
c,.
of solids at Jow temperature is given by the Debye
T3 Jaw: c I)
T =A 0
3
.
The quantity A is a constant equal to 1 �.4 x 1 o-� J kilomole -I K- 1 and 6 is the Debye temperature, equal to 320 K for NaCI.
(a) What is the molar specific heat capacity at constant voll:Jme of NaCI at 10 K and at 50 K?
(b) How much heat is required to raise the temperature of 2 kilomoles of NaCl from 10 K to 50 K at constant volume?
(c) What is the mean specific heat capacity at constant volume over this tem perature range?
4-2 The equation of state of a certain gas is { P + b) v aT + ht• + u0• energy is given by u
=
RT and its specific internal
=
(a) Find Cr. (b) Show that c,.
-
c,.
=
R.
4-3 Show that
au (Refer to Equation ( 2.13 ). )
4-4 Show that
dU
K = PK'V- (cp- Cv) -. aP , f3 (Refer to Eq uat i ons (2.13) and ( 2.14 ). )
4-5 When a material is heated .. it cxperiencl!s a very slight increase in mass since E me:.. Estimate the fractional change in mass when the temperature of a block of copper is rai sed from 300 K to 400 K. Take cp for co.pper to be 2.6 x. 104 1 kiJomolc- 1 K- 1• Its atomic weight is 29. =
4-6 Consider oxygen as an ideal gas w i th c,. (5/2)R. Suppose that the tempera· ture of 2 kilomoles of 0" is raised from 27°C to 227°C. (a) VVhat i s the increase in inte rnal energy? (b) What is the increase in enthalpy? =
-
4-7 The specific heat capacity at constant pressure of a gas varies with the tempera ture according to the expression (
•
( · P == tl
+ hT
'
' 2 T -
-
·
Problems
65
where a, b, and c are constants. How much heat is transferred in an isobaric
process in which a kilomole of gas experiences a temperature increase from T
to2T?
4-8 Show that the following relations hold for a reversible adiabatic expansion of an ideal gas :
(a) TV1-l
(b)
T pt-lh
constant.
=
=
consta nt
.
4-9 An ideal gas undergoes an adiabatic reversible expansion from an initial state
(T1, v1) to a final state (a) Show that
(T2,
v2).
In
T, \
_::
=
Tl
( 'Y
-
1 ) In
v1 v2
where y is the ratio of specific heats.
(b)
If TJT1
=
2/5 and
vJv1
=
2. show that the final state is not accessible from
the initial state via any adiabatic reversible process involving any known ideal gas.
,
4-10 The equation of state for radiant energy in equilibriunt with the temperature of the walls of a cavity of volume Vis P
equation is U
(a)
(b)
=
aT4V.
=
aT4/3, where a is a constant. The energy
Show that the heat supplied in an isotherrnal doubling of the volume of the
cavity is (4/3)aT4V.
Sho\\· that in an adiabatic process, VT3 is constant.
4-11 An ideal diatomic gas, for which
cv
=
5R/2
..
occupies a volume of 2m3 at a pres
sure of 4 atm and a temperature of 20° C. The gas is compressed to a final pres
sure of 8 atm. Compute the final volume, the final temperature, the work done. the heat released.. and the change in i_nternal energy for:
(a) A reversible isothermal compression . (b) A reversible adiabatic compression. 4-12 An ideal monatomic gas having an initial pressure P0 and an initial volume V0 undergOC!s an adiabatic expansion in which the volume is doubled.
(a) Show that the fmal pressure is 0.314 Pr}· (b) Show that the ch ange in enthalpy in the process is
-0.93 P0VQ.
4-13 One ki l ogram of ice at -20°C is heated to ooc and melted. The resulting water is heated to l00°C, vaporizes, and the steam is furth er h�ated to 4QO�>C. Assume that all the processes are isobaric. Calculate the c hange in enthalpy in kilocalo ries and in
joules.
(The specific heats at constant pressu;e of ice, liq uid water,
and steam are 0.55, 1.00, and 0.48 kcal kg-1, respectively.)
4-14 An auto mob ile tire is inflated to a pressure of 270 kPa at After three hours of high speed -
driving the
the beginning of a trip.
pressure is 300 kPa. What is the
internal energy change of the air in the tire between pressure measurements? Assume that air is an ideal gas with a constant specific hea t capacity Cv
and that the internal volume of the tire remains constant at 0.057 m3.
=
5R/2
Chap. 4
66 4-15 Shortly after detonation the fireball of
a
Applications of the First law
uranium fission bomb consists of a
sphere of gas of radius 15 m and temperature 3 x lD-c; K. Assuming that the
e xpa n sion is adiabatic and that the fireball remains sphe ri c a L estimate the radius of the ball when the temperature is 3000 K. (Take y =
1.4.)
5.1
The Gay-Lussac oule Experiment
69
5.2
The oule Thomson Experiment
72
5.3
Heat Engines and the Carnot Cycle
74
67
5.1 THE GAY·LUSSAC-JOULE EXPERIMENT
We have postulated that in general u
=
u(T, v ) . The first efforts undertaken
to measure the dependence of the internal energy of a gas on its volume were made by Gay-Lussac and later by Joule in the middle of the nineteenth cen tury. The results of the experiments showed that the internal energy is a func tion of T only and does not depend on the volume v. This, like many other properties of gases, is on ly approximately true for real gases and is assumed to hold exactly for ideal gases. Before describing the experiments, we consider the following arguments. Using the cyclical and reciprocal relations for partial derivatives given in Appendix A, we can write •
au iJ·v
aT -
-
av
-
T •
dlt
II
aT
For a reversible process, we have (Equation
(5.1)
,,
(4.9))
au
Ct
,
=
aT
·
Thus Equation (5.1) gives aT
...
dU
av
= -c 1.
v
av
.
(5.2)
ll
Equation (5.2) implies that if we can measure the change in temperature of a gas as the volume increases while the internal energy is held constant, we can
69
Consequences of the First Law
Chap. S
70
diaphragm I
�as sa ...
mpl�
,
I .-----·
vl - vo
vo. To -
thermal insula t1on" •
Figure 5.1
�
-
-
vacuum
Setup for the Joule experin1cnt prior to rupture of the diaphragm.
determine the dependence of the internal energy on the volume. The question is: how can we keep
11
constant duri ng the expansion? Since
ev i dently we want an adiabatic process in \\'hich no work is done. The pr inci ple used by Joule is shown �chematically in Figure 5.1. A sam ple of the gas unde r investigation is enclosed in one insul�ted vesseL separated by
a
d i a phragm from an initially evacuated cham
ber . The t empe r atu re of the gas is measured w i th
diaphragm is rup t u red
�
the
gas under goes
tial specific volume of the gas.
c1
portion of a thermal1y
a
a
thern1ometer. When t he
.free epl·pansiv11. He re
V0
i s the ini
the final specific volume:�� and T1 are the ini
tial and final temp e r a t u re s respectively. In the .
e xpa n sion
no work is done and
the internal energy is unchanged since no �nergy is lost or gain ed . * The fina] temperature� which can he measured. is u, oT Vo
av
dv. u
from basic mathematics. The subscript on the d e ri va t i ve is appro priat e since
the internal
e ne rgy
coefficient. Joule
\\'as
is held c onsta n t. The· integrand is kno\\'n as TJ, the
unable
to
measure
any
temperature
change
Joule
for
air.
Subsequent experitnents using various gases ind ica t e that the temperature d e cr e as es bv at most a verv sn1all amount. Summarizing these resul ts .
.
�
a v.
0.001 K kilomole m-3.
u
is finite. However. the rupture of the diaphragm results in a rapid expansion during which the gas is so highly nonuniform that we cannot assign a distinct value to the pressure. See Sect ton 7.7 for a further discussion of free expansion. )jl
Although
<
..
no external work is done. Pdv
Sec. 5.1
The Gay-Lussac Joule Experiment
71
It can be shown (Problem 5-3) that for a van der �'aals gas,
Since the constant
a
is zero for an ideal gas, \Ve conclude that 11 is exactly zero
in that case:
iJu
'TJ =
dV
•
= 0
and
u
(5.3)
(T)
=
u
for an ideal gas. This very in1portant experimental result can be proved theoretically. Consider Equation (4.8), which is valid for any reversible process. If \Ve divide the equation by the temperature T, we obtain
·
au T
T aT
dV
T
v
+ P dv.
(5.4)
T
In Appendix A we note that an inexact differential can be made exact if an inte grating factor can be found. We shall see that the integrating factor
1/T makes
the left-hand side of Equation (5.4) an exact differential, a function of the state variables
T and v. * If d z
=
M dx + N dy, then the condition that dz be exact is iJM -
iJN =
ay
-
iJx
•
In this case, the condition is 1
au av TaT
=
a 1 au aT T av
+
P
'
where the subscripts on the partial derivatives have been omitted for conve nience. Carrying out the differentiation, we have +
1
iJ2u
TaTav
1 iJP +-
--
TaT'
or
au av •
__
r
T
iJP
aT
- P.
(5.5)
v
This fact is intimately associated with the introduction of the concept of entropy. as will
be shown in Chapter 6.
Consequences of the First Law
Chap. 5
71 � s a g l a e d i n a r o F
RT
p
=
v
-
..
so that
du dV
=
T
T
_
a
_
-P
--
dT
RT -
=
RT
-�
v
p
v
,
(5.6)
0.
==
v
of ity lid va e th s on d p n e de f oo pr e Th v. on nd pe This proves that u does not de the assumption that aq/T is an exact differential and the applicability of the
analysis to an irreversible process. These matters will be discussed in Chapter 6.
5.2
THE JOULE-THOMSON EXPERIMENT
The Joule-Thom son experimen t also known as the Joule-Kelvin experiment"* gets the same result as the Jou l e e xp e ri m ent in a different way. It is important ._
because the temperature drop in Joule "s experiment is difficult to measure. The experiment involves a throttling process, performed by forcing gas th rough a porous plug to a region of lower pressure in an insulated cylinder (Figure 5.2). A continuous throttl i ng process can be pe rformed by a pump that mai n tai ns a constant high pressure on one side of a porous wall and
a
constant
low pressure on the other side. The objective is to measure the temperature difference T2 llq
=
0
and
-
T1•
S ince the p rocess takes place in an insulated cylinder,
u2 - u1
+
1v =
0'� The specific work done in forcing the gas
t hrough the plug is 0 = w1
P1
dv
=
P1,v1 ..
-
li l
while the work done by the gas in the
ex p a n sio n
P2 ti'v
UJ2 =
==
is
P2v2•
·0
The tota l work is therefore
(5.7) •
WiJJiam Thomson became Lord Kelvin at the end of the nineteenth centun·. •
Sec. 5.2
initial state
The Joule-Thomson Experiment
7l
••• • •• •• • • • • • •• • • • • ••• • •• • ••
,...____,
__..-
porous plug
.. . • •• • •• •• • • • • • •• • • •• • •• • • •• •••
final state
(b) Figure 5.2
Principle of the Joule-Thomson porous plug experiment. Here (a) corresponds to the initial state and (b) to the final state, with the piston in the fully advanced position.
It follows that
or (5.8) Thus a throttling process occurs at constant enthalpy. In the Joule-Thomson experiment, the pressure and temperature of the gas before passing through the plug are kept constant and the pressure on the other side is varied while the corresponding temperature is measured. The results give a locus of points on a T-P diagram representing the isenthalpic state of the gas (Figure 5.3). The numerical value of the slope of the curve at any point is called the Joule-Thomson coefficient �, where
aT ·-
aP
•
(5.9)
h
The gas is cooling when /L is positive and heating when J.L is negative. The point 0 is called the inversion point. where JL =
The experimental result is that for most gases over a reasonably wide range of temperatures and pressures, the T.P curve is approximately flat and
Chap. 5
74
Consequences of the First Law
T
•
I
•
tnvers1on
J
po1nt •
I I
I t I I J
Figure 5.3 A curve of constant enthalpy for the throttling process.
J.L
==
I
p
0. It can be shown in a manner completely parallel to the analysis of the
Joule experiment that
ah
-
. riP
=
-cp
T
iJT aP
•
11
Thus, for an ideal gas� ah aP
r
=
and
0
h
=
h(T).
(5.10)
The results of the two experiments are equivalent: that is. dll
-
iJh
-
iJ·v
r
aP .,.
=
0.
Again, the result can be proved theoretically.
5.3 HEAT ENGINES AND THE CARNOT CYCLE
The heat engine is a highly useful concept in thermodynamics and reminds one of the subject's origins. A heat engine is a system that receives an input of heat at a high temperature.. does mechanical work., and gives off heat at a lower temperature. Before discussing a practical heat engine.. we consider two extreme cases. In Figure 5.4(a) .. work is done
on
the system and is converted to heat. The heat
in turn is used to raise the internal energy <>fa heat reservoir. In this case, the heat out (of the system
M) is equal to the work applied because the state of
the system is unaltered. Examples are stirring heat or the heat generated by the flow of current through a resistor.
Sec. 5.3
Heat Engines and the Carnot Cycle
T
T
heat reservoir
Q w
M
--
(W
=
75
Q ��
machine or sub5tance
Q)
w
( W < Q)
(a)
(b)
Figure 5.4
The concept of a heat engine. In (a), work is done on the system and is converted to heat. In (b), heat is extracted from a reservoir and is converted to mechanical work. This configuration is not possible.
Case (b) is the reverse of c.ase (a). Heat is extracted from a reservoir and is converted to mechanical work by the machine. The question is: can the work done by the system be equal to the heat in? Can the conversion be 100 percent efficient? The answer is no. The second law of thermodynamics states unequiv
ocally that it is impossible to construct a perfect heat engine. (This is sonle times referred to as the Kelvin statement of the second law, which will be discussed in the next chapter.) Thus case (b) must be modified as shown in Figure 5.5. Heat is absorbed from the higher temperature reservoir, work is done, and heat is given up by the engine to a lower temperature reservoir. The engine works in a cycle, so that the state of the system is the same at the end of the cycle as at the beginning.
Fig11re S.S Modification of Figure 5.4b to include a cold temperature reservoir. •
Q2
heatabsorbed
Qt
heat given off
uthermal pollution,
Consequences of the First law
Chap. 5
76
The efficiency of the engine 11 is equal to the work done by the system
(the machine or working s ubst ance ) divided by the heat absorbed Q2:* JW!
W
= 71
Q�
==
-,Q�
output input
=
·
(5.11)
•
, In other words� the efficiency is the ratio of �-\vhat you. get" to ��what you pay. ..
Applyi ng the first law to
the syste m
.
we have
In writing this, we arc mindful of our sign convention in which the heat flow into the system and the work done b.,v the system are p os i tive q uantit ies.. whereas heat flow out of the system and W()fk done on the system are inher ently nega tive We assume that the engine operates in a cycle. Since the state .
of the system is unaltered in a cyclical process, there is no change in its internal
energy (recall that U is a state variable and � U = becomes!
0
in a cycle). The first law
(5.12) Subs t itut in g this
in Equation (5. t 1 )
.
we
obt a in
•
(5.13)
Q�
Q�
We now consider by far the nl<)St important case of an engine operating in a cycle
namely. the Carn<)t engine. We assume a four-step reversible
process in which the ""working substance'' is an ideal gas. The four steps are an
(a to b in a P-l' diagram), an adiabatic expansion (b to c). an isothermal compression (c tel tl). and an adiabatic compression (d to a ) , returning the gas to its original state (Figures 5.6 and 5.7). Referring to Tahle 5.1. we sec that the t<)tal work done in the cvcle is isothermal expansion
.,
•
w Note that
�p dV
=:
"''�
* 0: the
+
W' + w,
work
must
W"
+
==
I
p t/V4
be greater than zero if the engine is to
be useful. We should remind ourselves that because P and ables,
�dP
=
0 an d
* The efficiency
#tiV
=
71 is not to
Some texts writ�
lV
=
{}2
0, hut t hi s does
he mistaken for the -
value signs often leads to confusion.
Q1• n1caning "
-
'
n o t imply that
.Joule coefficient TJ.
::: ;Q�! - Qt/. The
V are �p dV
state vari
om i ssion of
the
=
0.
absolute
Sec. 5.3
Heat Engines and the Camot Cycle
77
•::· ··r�:�: .:·· ::::.� : ... :. : -:: ··�: _ �!� -;: �:-· ·.. : -:· :. :o:: ..:-. ·• :::.�;��-· ·.::•·.:•.:� •··. :;< . :<, . -� · . . ·.·.· . • ..< ,.� · ... .. .·. •.•,·,· .·.·.·.· • · � • • .... - � .- �·: >�·· ;s-. · , : ··� . · <-�;- . ,:. � -· · ·, ..,.? .v� � if.::' X·. ! �,. : :��ro'Y.o " =::;:::� : : :• � � ...-,= : :;,�•; � ; •,; :: � oo-;•'·•• •�•='� :�·f. ' � ':J•J' • • '•.•-:�;:• :.�:�� 'o • ' • · :·:·:�. · =· . <'·:..:· A · · ··:· .·:.- �·. ··: '.:·.-7 ;-� • >:· ·�. ; :�· :.;; · ·'· ·: • • . ·. ·=·. · • , � · .· . . · ·. . ...·.·.··... .·. .•.r�.)'4';; _ ., • ,.". '•· ..·· - ·,v., .,·. . .,, ..· �· .·v · .·., ,.,.· __, ..;·..-.<·Zo\ . " ., : ·.-•'.�>. .,.. ,x � ..;A·� A.::··:·x· ··. (.�·.• ·• .·:J··•·., .·� ·.·; · .·.·., ·. ··;,.,,, v : �.· ,. : . ; � ; , :' ;-: :· �·: :·. .· · . • . •••• . ' . . .. •· x . . • • i• • < . ··� · •· . -.. •.. . •. . , ...· .. ... · . ?·. •• .. . ..;'. ..:-�·.•. ..\:.:: ,�· . .....·. ..· •·· • ·� . ·. ·, · •·.:on ··· ··�·-·,,..... .. -�·., .. � ••. ., ..& ' . . .. .. .. · ,, . . ·.· • . . • �. .. . · . · . · ) , . . · . · . • o....., •·'\· � , • N, ,•• •'••• ' .o,... ••, . .... ..•...•..•.·••,•• ••••• oC- •. . ..·•....-:.. • ....••,.•... ... . .. , ,. , -,•• ,... ..•• ..·.• � ..,-·.. ,· ..•., ...... .. ....-,.� ... . • . . . . . .• .., . · · ,. . · •0 . • ::��=; :� � �: �; =* ! : ��:::���:�:: : £ � i � � :: : ; :: -[: ::� § : : =�� .; :;;; . :} :{ · � • . . . • ·. ·;-,:.·-··. -·•.'.>·:.·.-·:.;;...:. � ·: -:·;.o: ·:.·.·"�"' .>.·. •· ':$;..A·;'(.·.'· • .·· ·..; :.. ··· . .. . ;;� �·• · "·X...,. ; - ..,··,.� ··• .·· .-:....:-;,. · • v . v .·;• .· eo •'_.. �,,� · ...._ :-· . ,;;.•·,.· o .l:- A ·.;, !: ' " r . , • . ,. � � : • <·�· ; � 6· X J' -' .. O. . ; o., • • ' '• . · ..,, .. . �· -Y� o o '( • < ; : · ' il . . .., ,r.· _, : .o : ' ·1':":"1": · � • · • . ' lb :······ ·..o· ·:.� :.,. ..� · : · '�'...1.: ��, :• ·· �A ':'«�•· ·• �·.,!. �. • • ' · . .. · " • " � ' • , . �'! '� , /'.,• _,, �. •_.y_,, �_. � 'A A .;o,.•·•�r • •' • ' • • • • • .• h' •A • . :oro'"" ._ , r,. · A•.t" • • ' ' -· o J ·�' '7 � , · • )•,.. ,• (o ·.••:<·' , •• ,T . )<;o ,•,.• , • , ,•o -'o•' • u••• ,•• •L� ;Jit; o. ;.o • o-� ·, 0 , �,, • � . : 7.;�-:·::.<· :;; .... : � . � ... ;.,>:. 0:.....:.):;; :' -,:.!, : '• •-:.. . .. .,. f',� -" : ..·.·... . o ,..• ·. ·. �..... .... ....· >•·o:�. Y. .� , • . . ·· 1·.·· ,•: · •. • ·•<" · A·•; ·�. ...... ':'-' · o•• .;;,;� ·· N(··..·"'· · 0• '•)0 ·· . .�:'·.•':' · .;,..· .;; · · �·;··.<· · · · '• · · . . • . . • :.; ... " . ..>¥.<, . ;, ·� � �-(·:.( �.> •.� �\; �=�\ X ,-. ..�-= ' ·.: :-0 ·• A' . . :�j.·o.·� ·•. .... - ": ... �-�· · "=-; �. · · •. · . · .·.·• •.:,. ._.x� . .�A. :.:,· . · ·' . .·· ,··, ···. ·..,.;.·) > i'\:· �:0� • ;;� :: c : :c;" :< =. ·.�··.·��··.·'' �� -! � > ..,_>�� :'. " .� � ·� � ; =-·· • · < • • ; ..A , ' ·�· • •.·· • ··· ! " • X � · · · · !· 0 · ' • · ' ·.�!-•'•:. -:>·-,;,..:� ��- : .(< ·' :·�� �: ·�:� '·.h'� � ''-'t. .:.- •. o, · '.v J .;. .• Y .'\ �. .� .•-..' L . ... ,,.,__ ' '·» .. ·•'•�;·,. ·•·•· ''·• "" ' ' � �•'• '
"•••·.�.> ·. •,•..... •.-.·••••·.•:·• .• ·• . <\! ••' o••, ••. .'••0 •'·•. · ,· y,··o·, ·.?...·.-·..... ....·•.••·• •.••.•.••••..;• •:!-.•• • · • • . . • • •.•�.·0 . ·· ·,. . <· • · . · · . . • ·� • ,· .··. •,' • ·�( • • • • • .··.·.,·. . ... -... ·�·-· · , · : .· • . ·,� )• .·;�······· • ,., ..•••.·:·�• ·.-�.·••••·;·• :-:-x. .... .-. · .· ·•· · :··· . �.·:-: .•:· ·: :•··�·: :. .. · ·:·.·: · �·. · · :,:;< •: ,..;;•··: :·,• • � • . . • ; • • · •. · , , , ' ' , -:•:_ . '\. .� ".t ... , • . • . •.I< --� . . · . '" • . .r . •oV'.'''" •... 9f.• •·. , . .. .'v,. " � . ... _, ,.A.. . . · '"" .. , . •".' t.• .·•".-•'.•'·. · '" ""f" V�" • oo oo •" • • • • • • . •• .. .. o • • . • X .. . ..,v • • • . ' · •.•• .. ... , ... . · t ·.; < · •• . • , , , . ,. .· . .) · . .. .-. .. · . . . � . · , . . • . , .. . < •. •• . .,, .. .. . ..t[!tL,' '"• ',., . • .. .. '1 .� o· . �•· •.• •;�. . -: ·: · �?:..: .�.·-·,.:·� ...; .. : ; • :·.-. : ... , ..-._,;. •• '· ' :-:· :-:t.·:•:. <· :l : l . ; . . : . :.'. . .. ·••. . .. . . • • . -..; · ,, v.·•X. .:..t X•'· •. ••., ,--· ·,)'..,·•..•·.,.. .: <.. , ...," ".....•,..... . ,•.· :.:< �·: � :, . . �·..·'< ····)•· •,· -""' · -• V.,• ..,• ·.v •..... x•,.... ... �.A• '� .• ,.·...•.• •' '"' •vv-1+ . \....,.•.,. >·•' • . . . ,,., ...... ..:. ' ', ' '.1'·.0. . .. ...,,... , . &: .., .. , � .. ··· · · · . . . . • . ·. v ·:.o.., .. 6 • .• (...•' ,.x••,,·.� .-� ..•x_ . . ·:o • •••. . ... ....·..•... ,; .-,..,.,· ,,r ; · .· .··..··.··ol':.·, , ·.." ' ;r. <.., .·.,. ·. .·�· •.· ,· ···"'·�.· . . '-�.-.· •• '· · ·. v ·.•·•;·• . . ·.: , : . A ·.{! < � , � • · : > 7 · '· . : . , .. , � "" � ' -"'.,, · . • •• ,• • :··:· ... ,0 •*•• ,... . · "'!' • . ·.« ·A.� :-� • ,•••�. . ·.• '·-· · • 0 ,, . ., .... " . .... • , .. ,... • • • � . ?.i . •• �' )0•• ·· . ••·5-·'>•'.t•.� ,,•,.•.., i'· .- -�...- , .,•. v,· ·,:•.,'«.> ,..:.�Y<'f.. .• · W• , ·:l!'.:.: _,. ., � "· <1.'. ('\ ' • , 'L• . • • ·.· • ·.,.•� ·.· ·. • · - ., ·�....:J,.. , ,. -':' . :• r. • • ;... · ./.. ·. · . .,..•.,,..•,o,..., · .•r....,..... . .� 0 ) � •• . "-·•.•;.� ·Y." . : :: ·• :•:: • �:.;; " �.�;: ; ;: �.,. •.. <•:;_> . -. ":; . . -;,-.. ���••�:=• : .;:: :::: � � : ..•... ...., .·"·:it .< . � •.,0 'V • :...:..•>:. ...'•-:':• : . • .. � • • , . . · x-: 0:• �,. . " . . · • • ,. ' • ' ' ·. � · ·. ' ' • : �•" *" # " A v, , 'li . • J< . , · ' " '", o ·�· ,... · , •" • •., ' . . ·..•..v .y , ••� \' • '. .... ...:.-. ..._ ,• •,_.+ •••=-- . )'• ·$ .. , •� • •�.•r. .•.....'rol'. :... ·� . � ., •. ?S . .... . . "'....·..''....... ..•,..••. .,·.-�·� . . . > . ._:. . ., ....-•�::t, . , · '-"-o-· .. · .·.•. :.0. · . • · · · <'·'"-' • �,"" ..;, '·· • •"• • ·"'.e;.' NV' •. ,·Jo,o:'•.···. �····'···'·'··<'·•·,.· W "·�·#:'···'···� . . · , · ' �·., · . ... -:;•·�Y."!'" "'=•.:,(,:�r-��•<·' '. ··-•'":•'":>:·:Ci;·:··•.·.�; , . . . �::;�;;;:;.:�:;:"'?;�:'�. : ·:·: :- .:�::: :::. :*:: ·-:� =·�,�:::• :=:��; '• •• •• •.·.'•.·.. � ••.·••· ...-;,o•.., • • ·,·•. .··• ···. ·••.· ·:1) :,C:• • •· .•· :·· •·>. ....� , .�.;.. o -,.o,, . ".· •'. , • . • • -... r , . ( · .�. A . . · . · . . ·.· . . . • • · • · •••.. ... . .,'i., •.• ,••·�••. ..,_ . '1 ' ... 0 ..,-. • 0 · . • . • • • • • . . · · •·•••4 •.,..., • •• . • , • • . • . . . • 'If. ·· � · · ·:·• : • • • · • · . . � . ;. : : : v ._•:-!·o_t,;�vo -:� ""0 :o : · · �0 � : : : : .. . : ·; • . ·>� · -;, ·:· · ·' · : ; · · . . . .· . •J '• • • -":" • '\1 • • oo ·•oo •.' ·:.••-• .•••·.·. ..••. .o • •••. ...• ..o...o .. o..... .... .•.,.... """' ...,,0. ·••••• ,.·.·. ·: .• • > ·o · ·.·. ..,,Yo . .. . . 0 ...,•. '. . , � . .· '.. . ... · · • ·.. ·�. •..... ...• •••• • • •• •. • . . •.•...· ·� ·· ··· •••••••••• • .•• . . ... .. ••• ••-' ,,
� � �� r
�: ?.� i�
.. .
�
:
.
-
o
t
..
.. ._
.
.
,
•
. ,_
_,, ,.
..
.
.
�-::.:A -'.!�� � ,_. ,� � v o,
'
����: ����:..:.� �
. �,
�
x
�-.. ..-� .- �"-��-l .
,
•
.._.
•• · o roo
.
oo
oo
,
..
'• '
.
�
' ' -!\
_.
.··o. ..
..
.,.
.-_.,
.. ..
.... �
�� Y··�V ·.�-'.A.� • �"' � ..
� ..:.' .
.,
..
..
0
...
v
......
...
. .
.
..
...
'o' o ... . 0 .. .. o
� ::�;x . .
...
..
'o
....
•
..
o
0
, .. •
o
o
...
0 ..· ..... .
'
....
. .�
,_ ..
.
..
•
•.
.
...
....
..• -� .
....
.1.
..
..
.
•
...
'•' "'
..
�
'.
..
...
.
0
insulator
insulator
isotherrnal expanston
adiabatic expanston
isothermal compresston
adiabatic compresston
(a-+ b)
(b .... c)
(c-+ d)
(d-+ a)
•
Figure 5.6
•
•
•
The four steps of a Carnot cycle.
p Q-.
-
(adiabat)
r2 (adiabat)
5.7 P-V diagram for a Carnot cycle. The slopes of the isotherrns and adiabats are exaggerated.
TABLE 5.1
c
v
Heat transfer and work for the four steps of a Carnot cycle. Heat
Process a
b
�b
Isothermal expansion
.. c
Adiabatic expansion
•d d-+a c
lsothertnal compression Adiabatic compression
Q2
>
O(in)
0 Q1 < 0 (out) 0
Work
0 (done by system) W' > 0 (done by system) W1 < 0 (done on system) W" < 0 (done on system) w2
>
Chap. 5
Consequences of the First Law
aQ -ttw.
(5.14)
78
We can write the first law dV
=
On the isotherms. for an ideal gas d U alone. Thus aQ
Now
b and
=
l!W. Here Q2
=
=
0 since U depends on the temperature
W2 and Q1
are on the same adiabat py-r
c
W1. Hence
=
(>
0 since vb >
(<
0 since V:1
=:
��)�
(5.15)
< �-).
constant. Since P
(5
=
.
1 6)
nRT/V� we
have
so that Tv;--
1
=
constant and (5.17)
Similarly, a and d lie on the same adiabat. so TV �
If we divide Equation
u
(5.17) by
y-
I
=
Tv I ./, ..
I
·
Equation (5.18) and take the
(5.18) 1/(y-
l)th
root. we obtain
V,,
v:. -
(5.19)
-
-
v:, •
Combining Equations (5.15), (5.16), and (5.19), we obtain
Ql -·
Q: Using this expression in Equation
==
-
r1
-
(5.20)
-
T:..
(5.13) gives (5.21) -
for an ideal gas. Since T1
<
T2, it follows that 11
< 1.
Sec. 5.3
Heat Engines and the Carnot Cycle
79
The work perforttted during the adiabatic steps of the cycle can be calcu lated using Equation (4.35). However, we note that the use of Equations (5 .11), {5 .15), and (5.21) yields W, the total work done, which is the quantity of greatest interest. It can be easily shown that this is just the area enveloped by the curves in the P- V diagram delimiting the cyclical process. It is important to emphasize that a Carnot engine operates between
only two reservoirs and that it is reversible. Also, if a working substance other than a11 ideal gas is used, the shape of the cunyes in the P- V diagram will, of course, be different. Carnot theorized that the efficiency given by Equation (5.21) is the maximum efficiency for any engine that one might design. We note that the efficiency would be 100 percent if we were able to obtain a low temperature reservoir at absolute zero. However, this is forbidden by the third law. A Carnot refrigerator is a Carnot engine in reverse (Figure
5.8). Work is
done on the engine with the result that heat is removed from a low tempera ture reservoir {the interior of the refrigerator) and delivered to a high temper ature reservoir {the surrounding room). In this case we define a coefficient of performance
c
as the ratio of the heat Q1 extracted from the low temperature
reservoir to the work done on the system
again, ..'what you get" divided by
"what you pay." Because Q1 is positive (heat tlo\v into the system) and W is negative (work done on the system), we introduce a minus sign in order to make the coefficient of perfor111ance a positive quantity:
(5.22)
The last step follows from Equation (5.20), which applies to the Carnot refrig erator as well as to the Camot engine. Since T1 < T2, the coefficient of perfor mance of a refrigerator, unlike the thermal efficiency of a heat engine, can be made much larger than unity.
T,
...
w
--
Figure 5.8 The concept of a Carnot refrigerator.
M
•
Chap.5
80
Consequences of the First Law
refrigerant (heat out to surroundings)
(e.g., freon)
(
high pressure
�
condenser
high pressure vapor
liquid
compressor throttle --
w
low pressure
low
liquid/
pressure
vapor
vapor
evaporator long capillary tube
Figure 5.9
Ot
(heat in
from refrigerator space)
Schemati c diagram of a typical refrigerator.
A schematic dia gram of a typical refrigerator is shown in Figu re 5.9. The refrigerant is a substance chosen t<> be a saturated liquid at the pressure and temperature of the condenser. The liquid undergoes a throttling process in
which it is cooled and is partially vaporized. The vaporization is completed in the evaporator: the heat of vaporization is heat absorbed by the refrigerant from the low temperature reservoir (the interior refrigerator space). The low pressure vapor is then adiabatically compr(�ssed and isobarically cooled until it becomes a liquid once again.
Refrigerators are designed to extract as much heat as possible from a cold reservoir with small expenditure of work. The coefficient of performance of a household refrigerator is in the range 5 to 10.
PROBLEMS
S-t The Gay-Lussac-Joule experiments demonstrated that
if l'
r
=0
for an ideal gas. Show that this result in1plies that
ilh
-
,iJP
=
0.
T
(Hint: Write t h� differential dh and assunte that h
=
h(P . T)
and u
=
u
( 1'. T).)
Problems
81
5·2 Assume that the specific internal energy of an ideal gas is given by
u (a) (b)
=
u0
+
cv(T - Tv), u0
=
constant.
Show that the Joule coefficient 71 is zero. Show that the Jouie-Thomson coefficient JL is zero.
S-3 The specific internal energy of a van der Waals gas is given by
u
u0
=
+
'
a cvT - -, u(), a constants. v
(a) Find an expression for 71· Sho\v that 11 = 0 if a == 0. (b) Find an expression for the specific enthalpy h as a function of v and T. (c) (d) (e)
Show that JL
=
K
RTv
Cp
( V- b)
-
v Cp
.
Calculate the isothertnal compressibility Show that if
a
=
b
=
0,
K =
v
RT
, and p.
K
=
for the van der Waals gas.
0.
5·4 Show that
(a) (b) (c) 5-S
iJh aT ah
= v
-
-
av
T
ar av
Cp 1-
= h
{3p, •
K
JLCp -
•
VK
J.L --- --
v(
p,/3 - K)
·
Carefully sketch a Carnot cycle for an ideal gas on (a) a u-v diagram; (b) a u-T diagram; (c) a u-h diagram; (d) a P T diagram. -
S-6 A Carnot engine is operated between two heat reservoirs at temperatures of 400 K to 300 K.
(a)
If the engine receives 1200 kilocalories from the reservoir at 400 Kin each cycle, how much heat does it reject to the reservoir at 300 K?
(b)
If the engine is operated as a refrigerator (i.e., in reverse) and receives 1200 kilocalories from the reservoir at 300 K, how much heat does it deliver to the reservoir at 400 K?
(c) (d)
How much work is done by the engine in each case? What is the efficiency of the engine in (a) and the coefficient of perfor mance in (b)?
5-7 Fifty kg of water at ooc must be frozen into ice in a refrigerator. The room tem perature is 20°C. The latent heat of fusion of water is 3.33 x lOS J kg-1• What is the minin1um power required if the freezing is to take place in one hour?
S-8 W hich gives the greater increase in the efficiency of a Carnot engine: increasing the temperature of the hot reservoir or lowering the temperature of the cold reservoir by the same amount?
Consequences of the First Law
Cha p. 5
81
S-9 An idea ) monatomic gas is the w or k ing substance of
Carno t engine. During
a
the isothermal expa n s ion the volume doubles. The ratio of the final volume to the initial volume in the adiabatic expansion is 5. 7. The \\·ork output of the engine is 9 x 1 0" J in each cycle . Compute the temperature of the reservoirs between which the engine ope ra t e s. ( A ssutne
n
=
1 kilomole).
5-10 A h e � t engine ha ving two kilotnolcs of an ideal ntonatomic gas as the working substance under goes a revers ihle cycle of t.>peratio ns n1ade up of the four steps shown in Figure 5.1 0. Here ( T, V, ) '· ( T�. �,) ( 400 K. 10m ). ( T�. V-l) .
=
(a)
=
=
( 300 K. 2 m .� ) . ( T". V� ) ' ( 300 K. I 0 m· ). Find:
=
( 400 K. 2 m .�)
.
the heat flow in each step:
(b) the work done by the gas in the cyc le : (c) th e thermal efficiency of the engine. 5-11 Durin g some integral number of cy cles a r�ve rsible engine works between three .
heat reservoirs. lt absorbs
Q� joules
Qj joules
of heat from a reserv oir at T, .. also absorbs
of heat from a reservoir at T> de l ivers
Q.� j o u) es
at T�. and performs W j oules of n1echanical \\'ork. If T1
T'
=
300 K. Q1
=
1200 J. a n d "·'
-=
of heat to =
a
400 K. T:
rese rvoi r =
200 K.
200 J. find Q� a nd Q�.
5-12 A stude nt clain1s to have co ns tr u cted
a ptototype cyc lical eng ine that operates
between two reservoirs at temperatures of 500 K and 350 K. respectively. In
each cycle the engine e xt racts 6000 J of heat fro m the high temperature reser voir. rejects �900 J of heat to the lo\\: ten1p�rature reservoir and per forms 2100 J
of work on the surround i n gs Would you be prepared to give financial support to .
the development of the engine._,
5-13 A heat pump is a device t h at can he used in winte r to \\'arm a house by refriger a t i ng the ground .. the outdoor a i r or the \\'ater in the mains. A Carnot engine. op er a t i ng in reverse as a heat pump, extracts heat Q: from a cold re s er voir at t e m pe rat u re T� and delivers heat Q� to a reservoir at a higher temperature T: in .
e ach cvcle . •
(a) D e fi ning the coefficient of performance c as the ratio of what y ou get to .. what you pay:' derive a n e x pression for c for a c·arnot heat pump. (b) On a day when the outside· temperature is 32° F. the interi or of a h ouse is maintained at a ten1perature of oX:- F by means of a Carnot heat pump ··
'"
.
Calculate its coefficient of perforn1ance.
p 2 ..... -isotherm
isochore -
•
Figure 5.10
P- V diag ra m for
a four-step cycle.
-·�/
�isochore
isotherm '-----· ---.-.+-
\ .'
Problems
83 p c
adiabat
Q,
-
b
e
Fig11ae S.ll
P-V diagram for
...-.
d
I I I l
adiabat a
.. ..._ __. ..
___
the Otto cycle.
v
v,
-
5-14 The behavior of a four-stroke gasoline engine can be approximated by the so called Otto cycle. shown in Figure 5.11. The processes are as follows: e �a
isobaric intake (at atmospheric pressure)
a
adiabatic compression (compression stroke)
)l
b
b--.c
isochoric increase of temperature during ignition (Gas combustion is an irreversible process. here replaced by a reversible isochoric process in which heat is assumed to flow in from a reservoir.) adiabatic expansion (power stroke) isochoric decrease of temperature (exhaust valve opened)
a--.e
(a)
isobaric exhaust (at atmospheric pressure)
Assume that the working substance is an ideal gas and show that the effi .. ciency is given by
'vhere
(b)
r
=
vnft'h is the compression ratio of the engine.
Calculate ,1 for the realistic values
Note: Since )'
2:
r =
5 andy
=
1.5.
1 "Ne want r as large as possible. The maximum practical obtain
able value is approximately 7. For greater values, the rise of temperature upon compression is large enough to cause an explosion
before
the advent of the
spark. This is called preignition. For diesel engines preignition is not a problem and higher compression ratios are possible. This is partly the reason that diesel engines are inherently more efficient than gasoline engines.
·
Consequences of the First Law
Chap. 5
84
5-15 Consider the Joule cycle, consisting of two i�obars and two adiabats (Figure 5.12).
Assume that the working substance is an ideal gas with constant specific heat capacities Cp and c, Show that the efficiency is ..
1J=l-
pl
()'-I
li)'
p,
•
-
p
2..,_
. _ .. .. ..._
_
3
4
Figure 5.U P- V diagram for the Joule cycle.
6.1
Introduction
87
6.2
The Mathematical Concept of Entropy
88
6.3
Irreversible Processes
89
6.4
Carnot's Theorem
91
6.5
The Clausius Inequality and the Second Law
94
6.6
Entropy and Available Energy
97
6. 7
Absolute Tern perature
6.8
Combined First and Second Laws
•
98 103
85
6.1 INTRODUCTION
The first law of thermodynamics is remarkable in its generality, simplicity, and utility. Yet our statement of it is in some respects less than satisfactory. Furthermore, in our discussion of heat engines in the previous chapter, we hinted that the first law does not constitute a complete theory because certain processes that it perntits do not occur in nature. The problems are as follows. First, classical thermodynamics is concerned with states of equilibrium and various processes connecting them. We have seen that there is a very substantial advantage in dealing with state variables, changes that are expressible as exact differentials. Then the exact process by which a final state is reached from a given initial state is immaterial; the transi
tion is i n de pen dent of the particular path taken. However, two of the three quantities appearing in our statement of the first law the work perfottned and the heat exchanged are inexact differentials. The question arises: Is there any way in which we can write the first law in terms of state variables only? A second concern is that our theory emphasizes reversible processes that is, quasi-static processes in which no dissipative forces such as friction are present. But reversible processes are idealizations whereas real processes are irrev . ersible, and the first law draws no distinction between them. This leads us to ask: Is there some state variable by which we can distinguish between a reversible and an irreversible process? Finally, in Chapter 5 we noted that certain processes are impossible that are by no means prohibited by the first law. It is not possible to construct a machine that convet1s all the heat it extracts from a reservoir into useful mechanical work. It is not possible to transfer heat from a cold body to a hot body without supplying energy in the forrn of work. The first law says nothing about these prohibitions, these "principles of impotency," as they have been called. Is there, then, something missing from our theory? Do we require a sec ond fundamental law for a complete description of our world as it actually is? The answer is yes to all of these questions. We will begin by introducing the concept of entropy as a useful mathematical variable. Phenomenological observations will give the concept its physical meaning. 87
The Second Law ofThermodynamics
Chap. 6
88
6.2 THE MATHEMATICAL CONCEPT OF ENTROPY The first law in its most general forn1 (f
=
7/Q-
( 6.1)
7/W.
Neither aQ nor ztW is an exact differential: only their difference is exact. We saw that for a reversible process the work is configuration work alone� w·hich can be expressed as aW,
PdV.
-=
where the subscript denotes a reversible process. Thus a"'� ---�
p
=
(6.2)
t/V.
Now.. the volume Vis a state variable and tf\l is an exact differential. By multi plying llW, by the integrating factor l/ P. \Ve have replaced an inexact differen tial with an exact one. Is there an i n tegra t i n g factor for liQ or liQ, that \villlead to an t!Xact dif ferential? We note that Pis an intensive state
ia ble while ZIW, is an exten
va r
sive variable. The quantity aW,jP is therefore extensive. Also, two of the three fundamental state variables
P, V,
and T appear in Equation
( 6.2). This leaves
T
as the \'unused'' fundamental state variable and suggests the relation ..
aQ, ..
T
=
( 6.3)
(/ 5.
That is, the reciprocal of the ahsoJute temperature is an integrating factor that permits the replacement of the inexact differential aQ, by the exact differen tial dS. Equation
(6.3) is in fact Clausius's definition of the e11tr(JPY S. The term
is derived from the Greek word en-+ trepei11 for ··turning,, or Hchanging.�· When we substitute Equations (6.2) and (6.3) in Equation (6.1) we obtain
( 6.4) for a reversible process. Thus we have succeeded in \vriting the first Jaw entirely in terms of state variables. There
are
(I) (6.4)
t wo remaining concerns:
exact differential: and (2) Equatio_n
\\'C
have not proved that dS is an
appears to hold for reversible
processes only. This would seem to indicate that we have replaced a com pletely general statement of the first law with one that is restricted to a special case. Both of these problems will be addressed later in this chapter.
Sec. 6.3
Irreversible Processes
89
6.3 IRREVERSIBLE PROCESSES We have already encountered irreversible processes in discussing dissipative work. Consider the following processes and the results observed in nature:
1. A battery will discha:-ge through a resistor, ieleasinog energy (Figure 6.1 ). The reverse will not happen: adding energy to the resistor by heating will not cause the battery to charge itself. 2. Two gases, initially in separate adjoining chambers, will mix uniformly when the partition separating the chambers is removed (Figure 6.2). The gases will not separate spontaneously once mixed. 3. A gas in one portion of a chamber is allowed to undergo a free expan sion into an evacuated section of the chamber (Figure 6.3). The gas will not compress itself back into its original volume. 4. Heat flows from a high temperature body to a low temperature reservoir in the absence of other effects (Figure 6.4). The reverse process does not take place. If T2
>
T1 initially, the body comes to temperature T1•
.. . . . . '.. . .. .. . ... . . .. . . . ... .:·· : : : ;·:
..______,/ I
Figure 6.1
Battery
discharging current through a resistor in a medium to which heat can be supplied.
. . ·. .·' :· :. ·: : �< . · :.:·. .· ::::.·' . .·. ·: · ... :' : :- ;··. ·. . .. .. . .. . . ... . =··: : .: . '·. : ;·· ·: . . . . .. 0 . .. .' : .• . 0 .. . . . . . .. . ·. :-:- ·.. :.0.�:.:: . '..·=·· -. . 0ooo 0 oo• . .. . · ) :�·· · ::: ': .. . .. . .... .. ... . .' . . . ... ;..:-: ' . .·:-; ::.:. ·..:-:· . ..... :(... . .... ·.·.. . ... · :"%" . ...···..··· . .·. .·:... vv . . . • . · . . · · • .. . · . ... ....... • . • - . .. · :( -:•.• .._ . .• . .. •.• . ·... . •. · · �·-·.-. ·:·. ·. >:·.·· .,·.. •... '�-'1' , ·. ·· ·. . ·.. ·.··.· ....... .. .. . . ,.·.·.· ... .·.•... ,· .•.· ·, · ·.· .· · .· . .... . . •..., .... · ,. :. . .• .• . • ... .. . . . ... . . , · · . · .. ·· · . ··.·.···.· ·.·..·..··...· �• · � • · · · · · ' · · · , · · · . • . · · . . • · . . ·.: · . • ·.·. . ...·...· . . .· .. . · • . • . .' .;-'·'• ' •' . � ... · : : . ; · .:. . . . . · � . .. • ... . · . • -... . • , ... ·. · -:· . · · · . · ·; ·.·.• · : · ·; -:-:· · · · · : · . -: . · . . ·. · ·: · · . · · . .. . . · . . . . . , .. .. . . . ...· · ..· .• · ··. .. . .. · .· .....·.·. · ..· ..· · ''""' .. · .· . '. ' . •.. . .. . .. '\' . . . .. ... . . . . . . .. .· . . . . .. .. ' .•'. (... . .. . . . . •. · .·.. ·..·.·..•. .•. . •..··•.··· ·•· ',·.. .�·...... · .·..·. , . . . • · . . .. .... · . · . . . ·. · · · ,. . . . )' .... · . ·• , . .. · . . . .. .· · · . . , ' . • ·. . ... . · · · . • ' . • · . · ··· · . · · )'• ' . . . . • . · · ·· · .· · · . . - · · . . ·. · : :· : : .: �::� �;:j?,��;=� :�; : :::.��:: :�: : : : ::=: : -:��:��=:�::��� �:�:;-�=:�:::�::: : ·: . : :::: ::::�;. :.:�;:.: ? :::.:�.::�:-: :�;::::::::�� :�( �� :_:_; . = � : : � ::
I
; {� ,
.
\
::
:; �; �;.::�:: : :��
'
::: �:!:��-
�
. .. . . . . . ... .. . .. . .. . ·· · · · .. . . . -· '·�··. . .. ... . .. .. . .. .. · ..._;;. : :-:. : -: . ;·. :-. : · :.·. : . : ·.. : � � . . .· ;. ·. ·. :-:-· .·: · . . . .· :·· ·: :-:-:· :- · : : : .· : . •.. ·'. �· >.· · :· .-: ·-:-: . . . · · · · .. · · • • •.• : . •.• •• • :·: ·:· / , • ...:.-: :�:··-: ·.· · :: · . . :·: :· :-:-�:-:: · .• . ::··�=:- � · -: :: :.;:: : : · :. ... � :: -::.:·, ; ; , :;. ;::.··· :. : ·.·:··:: :·: -:··: :> :: ·· . . . 0 . . : o , . •..� . • .. .. . . . . .. . .. . . . . . .. . ... . . ... . .. . .. : ··· �·· · · · · · ····· • ·· • , ··"· • ·-: -:·X ·: ··'� .: ....�. :.' · v ' ·,·: ·:·· .· :- :- : -: � :-:-··: :·: : : -: :·: .· : :' :>:·.-:·. .. ·· • ; . ·: : -- :-.- : -: .·:.·: ·7 ··:< : :• .' · ••. -:-:· .· �� :-:· .' .· -: ·:�· Y -;· ·; :: . : · ·: · · · . · : : . . . . . · · · · . · · . . .· . ·� .• •.· • •.. ·· ·: "·" .•. ., · .-·..:· ·· ···· ·. •. ·.·..• ·.... • ·.·.·.•.· . ·.·.·•.·•·. . · .· ··:: ·:·:.;.: :: • . . >.· . ·: ·:-·: .: · ;... •·• ··: ·: ·· · -- :- .. ·: ·:·. -:· ·. ·.·;.• ·- : ..... � ::_;.· .::::::. .��:� . o:o::::. : : :�0:0 .;.; �··':· :� :. ... . . ·..·.�.. -... ·:· :-:.: -'ooo• . 0 0 .. . · .. o
.
:-�·:.
.
. •
0
'oO .'o0o .·0· _ ·.· . .......· .
�:;»�
. .. . . . . . .·.
·. .· : : . 0 . 0: 0.: 0• . .0:. . o o
X
X
0
.o,o. o:o,·O o. :o O o o·oooo:-oo ,
:·:>: ):
· ·' •· :-:- > .· : '· ...:'-!:-: .. 0 . . 0. . 0 o• 0 0 0o o' ooO o o
0 .
<:.<}0
X
0 X
=-o.:oooo.o:
Q
X
.o
"
�::� )oo>:oooo o
�....:-.: ·.:.·;·.· ,.. ·: :: :�= � ;: : ..... . . :. k-.·< 7 ·: ·.·. · .--: o :·0:;·,.,·. .. , o:··. .; . "< · •.. ·. ::�: . : �·· 0 oo0 -o : . . · 0 ·>. :/ · ;::·:· :. :0::::: 0 0 0 0
0
� :0�.� �.
�
•
.
•
?
X
Q
. . ..
0
���;�:
:S ��::;:
o:' :00 0-:0 .
. :;:::: : ; . ·•· ·.. . ..-..•, .. ..... . .. ·.·.... ..... . . . ..-� . . .. ·.. ... · .. . , .·.� ·.. . .. ·. ·. . ·. -:· ���� �� -·:� :-:·:· . . .... , ........: �.. . ..· ··· ··. . ... · ·. . . ·. . . .. . . · . .. . .. .-�.. .-. .· -. .· •..· · . . .•,·· ·. . . ··.� . . . ....,..·.·:-- .· ;,;. · . .. . . . · · : .· { · . . . . : :: ::; ·=} ..-:�: -�::: �;: .:: :: :� ' .::.: .:::;:: : .;. , :;.::, : : ;. : : :. '·::�: ::::�-:·:·.':'· ;: ::. � ·.: ::· · :: :. :::.. .::;: · -� .· · : -: ::::: :::::::;::} .; . :::-:� :.: � ·:: ·::::< ·:· . . . . . . . . · . : · · · · . · · · · · . , . .. . . . . ·.·. . . : .;. . . : ... •.; ,. .... •• ,. : . . . . ....-..· .. ...... .··. .-:· � : .·. . . ,.;..; .....,. .. . ··•··:.- . •. ·. .·.· ...·..•.·.••.· ...· . ·· . .. . , .··.· ·...·..·:...:....... . .;. ·�-··.; .. ....·.·;·.....·· . ... ·,. ...•,.,..:.. : .... X..:;:.• ·.". . ·: ·· . · . ·. ··,.7• ·.· •··......·,.. . . • . · · · ·..· ' .· . · ::: ' · .. : ;.t,;- •. i : : .; : ;-: - � ; ;- .: : · . : -:- : ::.· � :·: · � ·:·: ·�; - :· ; . . <' ; :- : .•·: :- ·=· :- : : :- :� ..; . .. . -:-:-..: -: �.:-. =· ;� : :� : ·:.::: _,-:- ;- .: : ·: �·:.::: :::� : -::- . : ; :;;.
�
X
..
.
.
(a) Fig111e 6.2
:.•
•
(b)
1\vo different gases, (a) before mixing, and (b) after mixing.
Sec. 6.4
Camot's Theorem
91
T Figure 6.6
Schematic diagram of a device forbidden
Q
by the Kelvin-Planck
statement of the second law. It is impossible to have W
=
M
Q.
w
6.4 CARNOT'STHEOREM
We have said that Carnot argued that efficient engines must operate as closely as possible to a Carnot cycle. Using the Clausius statement of the second law, we can prove Carnot's theorem which states:
No engine operating between m·o reservoirs can be more efficient than a Carnot engine operating between those same two reservoirs. We consider two engines, M (Carnot) and M' (hypothetical), and assume 17'
>
11· That is,
(6.5) The engines are depicted schematically in Figure
6.7.
Since a Carnot engine is reversible, it may be driven backward (as a
refrigerator) by the work from M'. We can arrange the position of the adiabats T,
...
M
M'
r, (a) Fi811re 6.7
(b)
Two engines: (a) a Camot engine M and (b) a
hypothetical engine M' with efficiency assumed to exceed that of the Carnot engine.
92
Chap. 6
The Second Law ofThermodynamics •
Q,
-
M'
"'
M '
Figure 6.8 The two engines of Figure 6.7 harnessed together, with the hypothetical engine driving the Carnot engine configured as a refrigerator.
so that in one cycle of each engine� M uses exactly as much work as M' pro duces. Thus I WI I W' 1. The Hcomposite" engine is shown in Figure 6.8. Paying careful attention to signs� we note that =
W'l1'I, I - ;Q'I-IQ I 2
and
or
The equality
f WI
=
j W' I therefore gives
or
Taking account of the signs in Equation (6.5) and noting that
W
=
; W' i, we
see that jQ2 > IQ2'1 and therefore Q11 > Q1'j. We conclude that the compos' I Q 1 I from the cold reservoir ite engi1;1e does no work� but extracts heat I Q1 and delivers an amount Q2 - �Q2': to the hot reservoir. This conclusion vio, lates the Clausius statement, so our hypothetical engine cannot exist. It fol-
lows
77'. We can show that if the Clausius statem�nt is violated, the Kelvin-Planck statement is also violated. In this sense the two statements are equivalent.
that TJ
>
Sec. 6.4
Carnot's Theorem
93
•
M'
M
w
M
w
I
(a)
(b)
Figure 6.9
(a) A composite engine in violation of the Clausius statement; and (b) the equivalent engine, in violation of the Kelvin-Planck statement.
Referring to Figure
6.9, we consider
a hypothetical machine M' that, in viola
tion of the Clausius statement, extracts heat from a cold reservoir and delivers the same amount of heat to the hot reservoir (no work is done on the machine). The Carnot engine M absorbs heat Q2 from work
W, and rejects heat Q1• The work done is lV
ite engine,
no
=
Q2
the hot reservoir, does +
Q1• For the compos
heat is exchanged at the cold reservoir, but heat Q2 + Q1 is
extracted from the hot reservoir and an equal amount of work is done. This violates the Planck-Kelvin statement.
Continuing the proof of Carnot's theorem, we have shown that 11 > 11'· Suppose that the hypothetical engine were reversible. Then TJc
>
(6.6)
1J,,
where the subscript C denotes the Carnot engine and
r
the reversible hypo
thetical engine. Since both engines are reversible, the Carnot engine could have been used to drive the other engine backward (as a refrigerator), attd we would have the result 1Jc If both Equations
<
TJr
(6.7)
(6.6) and (6.7) are1o be satisfied, TJc
=
'Ylr·
(6.8)
Chap. 6
94
The Second Law ofThermodynamics
That is to say, all reversible engines operating between the same reservoirs have the same efficiency 11 = 1
-
T1/T2• Irreversible engines will have a lesser
efficiency. This is Carnot's theorem.
6.5
THE CLAUSIUS INEQUALITY AND THE SECOND LAW
For a Camot cycle we have seen that (Equation
5.20): (6.9)
The quantity Q/T is known as the Carnot ratio. From this we deduced the effi
ciency 'T1 = 1
-
.
T1/T2 and proved that the efficiency of all reversible cycles has
this value. Applying the Carnot ratio to an infinitely narrow Carnot diagram (finite temperature difference but infinitesimally small quantities
of heat
extracted and rejected), we obtain
( 6.1 0) We consider an arbitrary reversible cycle and represent it as
a
continu
ous contour in a P-V diagram. We replace the process by infinitely narrow Carnot cycles (Figure
6.10).
The fact that the continuous contour is replaced by a sequence of infini tesimal saw-toothed steps is not problematical: the adjacent adiabats cat1cel each other, leaving only the contributions at the boundary. By an extension of Equation
(6.10), we obtain ,...._
l!Q, �
;
l!Q, = 0
_._....�
.
T
.
( 6.11)
p
Figure 6.10
P- V diagram of a
reversible cycle. The process is replaced by infinitesimal Camot cycles.
L....-..--------.
v
Sec. 6.5
The Clausius Inequality and the Second Law
95
where the integration is carried out over the whole corttour. The subscript
r
emphasizes the reversible nature of the cycle. Since the integrand is the differ ential
dS of the entropy, it instailtly follows that dS is an e.tact differential and
S is a state variable. Consider next an irreversible cycle. An irreversible Camot cycle that operates between the same temperatures and produces the same work has a smaller efficiency 11' than the efficiency 71 of a reversible Carnot cycle. That is, 11' < TJ, which implies that
or
(6.12) For an infinitely narrow Carnot diagram we have
instead of Equation
(6.10). Following the
same reasoning as before, it is clear
that for an arbitrary cycle that is partly or wholly irreversible, we must have
' Q l! T
0 .
<
(6.13)
Combining Equations (6.11) and (6.13) we obtain the famous Clausius
71Q
<
T
'
inequality
(6.14)
0'
where the equal sign holds for reversible processes. Equation
(6.14) is some
times taken as a statement of the second law.
1
Finally, we consider the change in entropy in an irreversible process. Let � 2 be an irreversible change and 2-+ 1 be any reversible path connecting
the two states in the P- V diagram of Figure
ttQ -
2aQ \.
1
T
lzt
+
=
T
6.11. Then
J
2
T
.
Equation
(6.14) gives
The Second Law ofThermodynamic�
Chap. 6
96 p
/
_,
,
__
...
I
I I Figure 6.11
An irreversible change (dashed line) and a
l
reversible path connecting states 1 and 2.
�---- ----•
V
Chang ing the order of integration and the sign of the second
27/ Q •
\.•
< -
T
t
term� we h ave
I
F y
( irrc:vcrsihk)
Thus
I
or_ in
differential
T
.
form.
( 6.15) •
Again, the equality
s
i gn
holds for
a
reversible process, and the inequali ty
appropriate for an irreversible p rocess
.
Suppose now that the sys t e m is is()]ated. Then
�5' for
a finite process. We
=
s�
-
..�,
?
is
()
aQ
=
0 an d d5;
c
0 or·
( 6.16)
(isolated system)
conclude that
The entrop)· (Jf an isolated s_vstetn increases in any irre·versible prlJCess
and is unaltere(i i11 any reversible pr()Cess. This is the principle o.f increas-
1ng entropy. •
It is to be noted that this statcn1cnt refers to net entropy changes. It does
not say that the entropy of part of the system cannot decrease. If. for example.
lower
heat flows from body A to body B at a contained in an adiabatic enclosure, �5
ofaQA), but
�5
=
uSA+ �58 will st il l
..
1
temp era t ure with bo t h bodies ,
is then nega ti v e ( becau se of the sign
he posi ti v e ( figu re 6.12).
Sec. 6.6
Entropy and Available Energy
97
A
B
Figure 6.12 Two bodies in thermal contact in an adiabatic enclosure.
adiabatic container
The adiabatic enclosure contains everything that interacts during the process and we can define this assembly as our universe. (This is not to be confused with the real universe, which may or may not constitute an isolated system.) If.. then, our universe consists of a system and its su rr oundings, it fol lows that
•
�Suniverse
=
�Ssystem + LlS�urroundings
>
0.
(6.17)
The fact that the entropy of an isolated system can never decrease in a p rocess provides
a
direction for the sequence of natural events. The laws of
mechanics are second-order equations in time t and are unaltered by the replacement oft with - t. As far as these equations are concerned, all physical
processes can run backward as well as forward. Clearly this is not so; the story •
of Humpty Dumpty illustrates that processes go only in the direction of increasing entropy. It is for this reason that the law of increasing_ entropy is
often described as p roviding "the arrow of time'' for the evolution of natural processes.
6.6 ENTROPY AND AVAILABLE ENERGY Another way of ex p ressi ng the results of the second law is in te1tns of what is known as available energy. It is impossible to utiliZe all the i n terna l energy of a body for the production of mechanical \Vork, since work c;an only be obtained by extracting heat from the body and giving it to an
e n g i ne
whose efficiency is
less than unity. Suppose that the temp er a ture of a body is T2 and that a
reve rsible engine works between T2 a nd T1, which is tt1e tem perature of a large reservoir to \Vhich the engine can give up hea t and is also the lowest ten1pera ture available (Figure 6.13). Then, if the body gives an amount of heat
the engine, only part of it
namely� aQ( 1
-
T1/T2)
aQ
to
can be converted into
mechanical work. If the engine is irreversible, still less work can be obtained. The available energy is defined to be liQ( 1 energy is T1llQ/T2.
-
T1/T2), while the unavailable
Chap. 6
98
The Second Law ofThermodynamics
T�
-
dQ M
dW
Figure 6.13
Diagram showing the amount of heat available to do mechanical work:ltW ltQ(l T1/T2). =
T1
-
Corresponding to the principle that the entropy always increases in a spontaneous pro�ess there is the principle that the available energy always
decreases in an irreversible cvcle. We could state that
·
.,
There exists no process that catz i1zcrease the available energ}' in the universe. The property of increasing entropy is eqttivalent to saying that energy is
always being degraded into forms that are more and more difficult to use for the production of work. The significance of entropy becomes clearer when we consider heat to be the energy of molecular motion. The lack of complete availability of that energy for the production of work is due to the randomness of the m olecular
motion. It is impossible to reduce the motion of each molecule simultaneously to zero by the action _of forces acting on the hody as a
whol e
�
and so it is impos
sible to extract all the heat energy from a body Thus the property of increas .
ing entropy means that the molecular motion of an isolated system always tends to become more random, and the entropy can be thought of as a mea sure of the Hrandomness" of the internal motion of a system. This connection between entropy and randomness will be made quantitative by means of sta tistical thermodynamics.
6.7 ABSOLUTETEMPERATURE The Carnot c ycle together with Carnot�s theorem can serve as the basis for d efin ing an absolute temperature scale. Carnot's theorem shows that the ratio
Q1jQ2 has the same value for
all reversible engines that operate between the
same temperatures. The special fact about a reversible Carnot cycle is that the efficiency is independent of the nature of the working substance. It can there
fore be used to define an absolute scale of temperature as follows.
Sec. 6.7
Absolute Temperature
99
Let 8 denote an empirical temperature, based on the ex pansion of an ideal gas. Divide a Carnot C}'Cie into tw o subcycles, each using the same mater
ial (Figures 6.14 and 6.15). Let the path a -+ b in a 8- V diagram represent an isothern1al expansion during which heat Q2 flows into the system. Similarly, c -+ d is an isothet·1nal
compression during which heat Q1 flows out of the system. For the cycle abcda,
6 •
a v· t
6.14 Representation in the 8- V plane of a Camot cycle divided into two sub-cycles.
Bt
"" -
,._,
-
----
-
-
�------------
L-----� v
M
e.l
M
Figure 6.15 The tw� Ca� ot engines in series dep1cted tn the 8- V diagram of Figure 6.14.
-
The Second Law of Thermodynamics
Chap. 6
100
f is an unknown function of the two temperatures 91 and 82. lenge is to find f. To do th is consider cycle ahefa. For this subcycle,
where •
Our
chal
,
S im ilarly for the subcy cle fecdf, ,
Thus
or
Because the right-hand is a function of fJ1 and 82
only, the
left-hand side must
also depend on these variables alone. This is only possible if
and
Then
independent of the properties of any given substance. Here
is another
unknown function. Since the scale of temperature is arbitrary.. we can introduce a thermodynamic temperature scale using
T
=
Ac/> ( 0)
A
�
constant,
=
( 6. 1 8)
so that
IQ i
I
I :
TI
---
! Q,! ,
- I
T_,
•
-
( 6.19)
This ratio is the same as the ratio of the temJ)eratures obtained in our analysis of the Carnot cycle.. in which the working substance is an ideal �as.
Sec. 6.7
but
.
Absolute Temperature
101
The problem remains to determine the function cP ( (}) . The proof is long,
IS
worth the effort. Using the definition of the entropy, we can write the
first law in the following form (in terms of intensive variables):
ds
l,(du + Pdv).
=
(6.20)
T
This equation is true in general, as wiiJ be discussed in Section
and v to be the fundamental independent variables and let u
dtt
au aT
=
dT + v
Substituting this expression in Eq uation
ds But also,s
=
dU
1 =
dT +
T iJT
v
au av
T
dv.
=
6.8. We take T u(T, v). Then (6.21)
(6.20)� we have 1
au
T
dV
+ T
P dv.
(6.22)
s(T, v), so that
ds It follo,vs from Equations
=
as aT
(6.23)
dT + v
(6.22) and (6.23) that au T aT 1
as
-
iJT
-
v
-
(6.24)
' v
and •·
as
1
au
T
dV
-
av
r
T
(6.25)
+P.
Now
a av
as
--
-
iJT
v
r
-
we can therefore differentiate Equation important relation
a as aT av
'
T
v
(6.24) with resp�ct to v �nd Equation
Combined First and Second Laws
Sec. 6.8
101
where A' is a constant of proportionality. An appropriate choice of A' makes 8
equal to T; we can then use T to represent both the empirical temperature and the equivalent thermodynamic temperature. The definition of the Kelvin scale is completed by assigning to T1 in Equation (6.19) tlte value of 273.16 K .. the temperature of the triple point of water. For a Carnot engine operating between temperatures Tand 71, we have
IQf
(6.30)
The smaller the value of Q, the lower the corresponding value T. The smallest value of
Q
is zero and the corresponding value of T is zero, called absolute
zero. In a Carnot cycle, heat is transferred during the isothermal processes.
Hence, if a system undergoes a reversible isothermal process without heat transfer, the ten1perature at which the process takes place is absolute zero. This is a fundamental definition of absolute zero.
6.8
COM·BINED FIRST AND SECOND LAWS
We saw early in the chapter that for a reversible process, dU
where
llQ,
=
71Q,
=
T dS and 7/W,
ltW,
-
=
T dS
-
(6.31)
P dV,
P dV. In its most general fortn, the first law can
=
be written dU
=
aQ - aW
The second law states that T dS
=
aQ,
>
(general).
aQ
(6.32)
for an irreversible process. We
can write
llQ, where
e
=
ZIQ
+
e
(irreversible),
(6.33)
is a positive quantity. Substitutin.g this in Equation (6.31) gives dU
=
aQ
+
E
-
aW,
(irreversible).
Comparing Equation (6.34) with Equation (6.32) we see that
(6.34)
aW
<
aW,
,
-
or
,
more specifically,
aW
=
aW,
-
e
(irreversible).
(6.35)
This is exactly what we expect: the total work done by the system is the reversible configuration work (the useful mechanical work) plus the (negative)
The Second Law ofThermodynamics
Chap. 6
104
dissipative work associated with frictional forces. The latter appears as heat in
Equation (6.33). Therefore, dU
==
aQ- aw
�
aQ,
€-
·-
(6.36)
(aW,- e).
or dU
(general).
T dS - P dV
=
(6.37)
Evidently Equation (6.37) has the same universality as Equation (6.32). It�s just that
7/Q is
identifiable with TdS
and
t1W w i t h PdV
only for
a
reversible
process. In fact. Equation (6.37) is not restricted to a process at all: it simply expresses a relationship among the state variables of a system and the differ ence between the values of these variables for two neighboring equilibrium states. Consider two irreversible pr<)cesscs. For the free expansion of a gas. al\1
==
0 but PdV is finite. S imil arly
.
ztQ
==
0 but TdS has a nonzero value (the
entropy increases in the process). For adiabatic stirring, aQ 0 but TdS =f. 0 and dS� > 0. Also. dV 0 hut aW =f. 0, since stirring work is done. Equation (6.37) is by far the most important relation in classical thermo =
==
dynamics.
PROBLEMS
6-1 Jt\ (�arnot engine operates to be an ideal gas. Take y rn u m
volun1e (c anJ
on
I kg of rne:thane ( CH-l), \vhich we sha1l consider
1.35. The rat10 of the maximum volume to the rnini th e <�arnot cycle Jiagran1 of Figure 5.7) is 4 and t h e
=
a on
cycle efficiency is 25 percent. Finu the entropy increase of the methane during the isothertnal expansion.
6·2 Find the ch�·n!gc in entropy of the system during the follo\ving processes: (a) i kg of \Vater is heated reversibly by an electric heat1ng coil from 20°C to 80 c ( c p l cal g I c c· : =: 4. 1 8 X ] 0 3 J kg - I K I ) . (
.
==
.
..
(b) 1 kg of ice at ooc and 1 atn1 pressure melts at the same temperature and 5 pressure. (The l aten t heat of fusion is 3.34 x l 0 J kg- 1.) (c) 1 kg of steam at lOOoc� and 1 atm pressure condenses to \Vater at the same tem perature and p r e ss u r e (The latent he a t of vaporization is 2.26 x 106 J kg- 1.) .
6-3 The Jow temperature specific heat
of
a diamond varies with temperature
according to 3
J
kilomole -I
K -l
�
. .
'
Problems
105
where the Debye temperature 8
2230 K. What is the entropy change of 1 g of
=
dia m ond when it is heated at constant volume from 4 K to 300 K?
(The
atomic
weight of carbon is 12.)
6-4 An electric current of 1 A flows far 10 s in a resistor of resistance 25 oh1ns. The resistor is submerged in a large volume of water, the temperature of which is 280 K. What is the change in the entropy of the resistor? Of the water?
6-S A thermally insulated resistor of 20 ohms has a current of 2 A passed through it .
for 1 second. It is initially at 20°C. The mass of the resistor is 5 g; cP for the resistor is 850 J kg -l K -I.
(a) (b)
What is the temperature rise? \\'hat is the entropy change of the resistor and the universe?
(Hint: In the actual process, dissipative \Vork is done on the resistor. I magine a reversible process taking it between the same equilibrium states.)
6-6 An inventor claims to have developeq an engine that takes in 108 J at a temper ature of 400 K, rejects 4 x 107 J at a temperature of 200 K, and delivers 15 kilo watt hours of mechanical work. Would you advise investing money to put this engine on the market?
6-7 Derive an expression for the entropy of an ideal gas
(a)
As a· function of
T and V.
(b) As a function ofT and P.
Assume that the specific heats of the gas are constants.
6-8 An ideal monatomic gas undergoes a reversible expansion from specific volume v1
to specific vo l ume
(a) (b) (c)
v2.
Calculate the change in specific entropy
�s if the expans i on is isobaric.
Calculate as if the process is isothermal. Which is larger? By how much?
6-9 A kilomole of an ideal gas undergoes a reversible isothermal expansion from
a
volume of 5 liters to a volume of 10 liters at a temperature of 20°C.
(a)
What is the change in entropy of the gas? Of the universe?
(b) What are the corresponding changes of entropy if the process is a free
expansion?
6·10
(a)
Show that for reversible changes in temperature at constant volume,
cv
(b)
=
T(iJsfiJT)v.
3 bT for
aT + a metal at low temperatures. Calculate the Assume that C0 variation of the specific entropy with temperature. =
6.11 Consi� er a van der Waals gas. Show that cv is a function ofT on ly. (Hint: use Equation (6.26)).
(a) (b)
Sho w that the specific internal energy is
(c)
Show that the specific entropy is S ==
Cv T
dT
+ R In
(V
-
b)
+ S0•
Chap. 6
106
The Second Law ofThermodynamics
6-12 When there is a heat flow out of a system during a reversible isothermal process� the entropy of the system decreases. Why doesn't this violate the second law?
6·13 One kilomole of
a
monatomic ideal gas is
cycle shown in Figure 6.16. Here PJ
==
around the reversible closed 3 3 10 atm. V1 2 m , and V2 4m •
ca rr ied
=
=
Calculate the change in entropy for each leg of the cycle and hence show that the entropy change for the complete cy cl e is zero.
p I
isochore ---r-
adiabat
---.. -..
-
2
Figure 6.16 P- V diagram for Problem 6.13.
v
isobar
7 .I
Entropy Changes in Reversible Processes
I 09 .
7.2
Temperature-Entropy Diagrams
I I0
7.3
Entropy Change of the S_urroundings for a Reversible Process
II I
7.4
Entropy Change for an Ideal Gas
I 12
7.5
T he Tds Equations
I 13
7.6
Entropy Change in Irreversible Processes
118
7. 7
Free Expansion of an Ideal Gas
121
7.8
Entropy Change for a Liquid or Solid
122
107
7.1 ENTROPY CHANGES IN REVERSIBLE PROCESSES .
To illustrate the calculation of the entropy change, we first consider reversible processes. In terms of specific quantities, the first law for a reversible process is
aq,
=
du
+
(7.1)
Pdv,
or
aq, T
du =
p
T + T
dv
ds.
=
(7.2)
We shall examine some special cases. 1. Adiabatic process: aq,
=
0, ds
=
0, s
=
constant. A reversible adiabatic
process is an isentropic (constant entropy) process. We note in passing that an irreversible adiabatic process is not isentropic. 2. lsothertnal process:
2 ztq, 1
-
q,
(7.3)
-
T
r·
3. Isothertnal (and isobaric) change of phase:
(7.4)
Here lis the latent heat of transformation . .
4. Isochoric process: We assume that u v
= constant in an isochoric process, u
du
=
=
=
u( v, T)
in general. Since
u(T), as for an ideal gas, and
cvdT. Thus
Sz - St
=
2 I
Cv
dT
T
. 109
Applications of the Second Law
Chap. 7
110
If ct, is constant over the temperature range
T2
T1• we have
-
(7.5) S. Isobaric process: It is convenient to use the specific enthalpy in an iso baric process. Since
d/1
du + Pdv + vdP.
=
(7.2) yields
Substitution in Equation
h h(P. T) in general. Since P =constant here, h only and dh cpdT. Then
Assume
==
=
h(T)
=
2
Cp
I
if
.
·
:
· · .
·
· ·
·
. .
. .
.
Cp
· .
·
· .
·
: ·_ .. , > _
· · ·
.
.
.
Cp)n
=:
-
T
T2
(7.6)
is constant.
:
· ·
dT
.
.
.
.
.
·
_
. .
·
·
·
.
. .
.
.
·
·
·
:
·
·
· .
.
.
.
·
: .
· .
:
· ·
·
· .
_
.
:
· .
:
· .
.
,
:
·
·
. .
.
· .
.
.
:
:
.
7.2 TEMPERATURE-ENTROPY DIAGRAMS The total quantity of heat transferred in
a
reversible process from state 1 to
state 2 is given by 2
(/,
Ttl,..
=
I
This equals the area under a curve in a T-s dtagram. For a reversible i�othermal process Tis a constant and the curve is a horizontal line. For batic (and isentropic) process. dJ
=
a
reversible adia
{J if T =I= 0 and the curve is a vertical line.
Consider a T-s diagram for a Carnot cycle (Figure 7.1 ). Recall that a--+ b is an isothermal expansion b ..
compression, and d
�a
�c
an adiabatic expansion. c
�d
an isothermal
an adiabatic compression. lt follows that the T-s dia
gram is a simple rectangle for a Carnot cycle. The area under the curve is
Sec. 7.3
Entropy Change of the Surroundings for a Reversible Process
III
T
b
Figure 7.1
r)
T-s diagram for a
t----
--
d
c
Carnot cycle.
Since f du
s
=
0, it immediately follows that w =
T ds
=
q,2
+
q,,.
7.3 ENTROPY CHANGE OF TH E SURROUNDINGS FORA REVERSIBLE PROCESS
In every process in which there is a reversible flow of heat between a system and its surroundings, the temperatures of the systetn and the surroundi;tgs are essentially equal, differing only by dT. The heat flow Ollt oftlte surroundings at every point is equal in magnitude and opposite in sign to the heat flow into the system (Figure
7.2). Thus dssystem
+
dssurroundings
==
dsuniverse·
But
7lq, =
T + dT
surroundings
surroundings
( lts ) surroundings·
surroundings T+ dT
Figure 7.2
Heat flow into a system from its surroundings in a reversible process.
system T
dq, > 0
Applications of the Second Law
Chap. 7
112 Thus
Ids I surroundings
I tis luniverst!
and
- ! tis �ystcm
=
==
o.
In any reversible process, the entropy change of the universe is always �ero. TI1is means that any change in entropy of the system will be accompanied by an entropy change in the surroundings equal in magnitude but opposite in
sign. Entropy is conserved in
a rev e rs ible
process. Of course, onl y
idealized -
processes are reversible; all natural processes are irreversible and entropy is not conserved in general.
7.4 ENTROPY CHANGE FORAN IDEAL GAS With liu
==
ccdT in Equation
(7.2), we have (7.7)
for a reversible process. For an ide_al gas� P/T
ds
==
c,.
tlT T
+
R
=
dv v
R/v. so .
Integrating, \Ve have
+ R In
if
cr
v'"'
-
•
(7.8)
is constant. In any thermal process. (lnly the cl1ange in entropy is impor
tant. In chemistry, however� it is important to be able to assign an absolute value to the entropy. Statistical thermodynamics will make such an assignment possible.
(7.8) holds only for an ideal gas, it has characteristics typical of practically all solids. liquids, and gases: ( 1) if only the temperature is varied. the higher the temperature rise� the greater the increase in entropy: (2) Although Equation
if only the volume is varied, the l a rg e r the volume expansion. the greater the entropy tncrease. •
The implications of these results are important. Consider the isentropic expansion of a gas. An increase of volume provides a positive contribution to an- entropy change. From a molecular point of view, the available energy levels of the system become more closely spaced as the volume increases and the mol ecules tend to become more randomly distributed among them. The entropy
The Tds Equations
Sec. 7.5
II 3
tends to increase. To keep the entropy constant, there must
be a compensating
negative contribution to the entropy change. which is provided by a decrease in temperature. The drop in temperature that takes place when a gas expands adi abatically (and reversibly) may therefore be regarded as an effect that is needed to offset the volume increase, in order to keep the entropy constant.
7.5
THE Tds EQUATIONS
From the combined first and second laws, expressed as
Tds
du
=
+
Pdv,
(7.9)
we can obtain some powerful results known as the
''Tds
equations." They
involve writing the specific entropy as a function of two independent coordi nates, that is, two of the three fundamental state variables
P, v, and T. The
equations are:
(7.10) av a
aT av The ferred in
a
(7.11)
p
aT aP
P
Tds equations T
have a variety of uses:
and integrating;
(7.12)
�v
v
reversible process (aq,
dividing by
cJ<
cp
=
(3)
(1)
they give the heat trans
Tds); (2) the entropy can be obtained by the equations express the heat flow or
entropy in ter1ns of measurable properties such as cp, {j,
K,
T, etc; (4) they can
be used to deterntine the difference in the specific heat capacities cp and cv; and
(5) the equations can provide relations bet\veen pairs of coordinates in a
reversible adiabatic process in which
ds
=
0.
Th•! derivation of the Tds equations is straightforward. The key is noting that the entropy is
a
state variable whose partial derivatives satisfy the condi
tion for an exact differential. The first
Tds equation was essentially derived in
Section 6.7 in the steps leading to Equation (6.26), from which the final result can easil}'· be found. Here we shall derive the second the derivation of the third as a problem. Let
T and P be the independent variables. Then Tds
=
du + Pdv.
Tds equation and leave
Applications of the Second Law
Chap. 7
114
The enthalpy ish
+ Pv, so that
u
du
dh - Pdt\ - vdP.
=
Thus
Tds
=
dh - vdP
rill =
CIT
p
dT +
d
T
or
ds
With s
=
=
1
ah
T
iJT
1
ah
T
aP
+
dT P
-
.
v
dP.
(7.13)
r
s ( T, P). we have
ds
as =
dT
aT
as
+
P
aP
dP.
(7.14)
r
Since T and Pare independent._ it follows that as
1 -
aT
-
ah
T
P
(7.15)
-
AT
p
and as iJP
1
ah
=
riP
T
T
-
T
(7.16)
v .
The differential ds is exact. Therefore we can equate the mixed second-order partial derivatives of s: a aP
a2 s
as iJT
p T
-
-
-
-
aP aT
Substituting Equations (7 .15) and
--
aT aP
a
as
iJT
aP
� -
•
T p
(7 .16) atld carrying out the differentiation"
we get 1
a2h
1
a2h
-
1
-
-
T iJP aT
av
T aT aP
ah
-v.
--
iJT
P
aP
r
Sec. 7.5
The Tds Equations
liS
Two of the terms cancel, and we have •
ah =
iJP
-T
T
av
+
iJT p
v,
(7.17)
analogous, incidentally, to Equation
(6.26). cp. Using this and Equation (7.17) For a reversible process, (ahjiJT)p =
in Equation (7.13), we obtain the result
dP. Finally, since
(avj iJT) P
=
v{3, we have T ds
=
cptlT - Tv/3 dP.
As an example, suppose that one kilomole of an ideal gas undergoes a
P1 to P2• We wish to find the quantity of heat transferred in the process. For an ideal gas, {3 1/T. Thus
reversible isotherrnal change in pressure from
=
Tds
cP.�T. - vdP
=
RT =
-
p
dP.
0
Then
T ds
q, =
=
P2dP
- RT
P,
J'
=
- RT In
P2
•
}'1
Heat is absorbed if P2 < P1 and evolved if P2 > P1• An important application of the Tds equations is the determination of cp - Cv, the difference in the specific heat capacities of a given substance. Equating the first and second Tds equations, we have
Cp dT - Tvf3 dP Solving for
dT
dT, we get
=
TJ3 dv K(Cp - Cv)
=
cvdT
+
T�
K
dv.
(7.18)
•
+
Tv/3 dP (Cp - Cv)
=
aT av
dv P
+
aT dP. aP v
Applications of the Second Law
Chap. 7
116 Thus
aT
=
T{3 --
·
and
Tv{3 = --( cP c )
ciT
--
•
aP Solving the first equation for
v
l'
-
cp - ell
·
and using the reciprocal relation gives
or
(7.19)
The equation for since
Cp
(aTIaP) l: gives
the same result. This equation is noteworthy
is measured in experiments.. whereas
cv
be calculated from theory� the equation affords
is difficult to measure but can a
comparison of the two. Also..
the quantities on the right-hand side of the equation are all positive, showing that
cP
cP
ct.
-
is always greater than =
It is left to the student to show that
c r·
R for an ideal gas.
Equation (7.19) can be used to calculate the specific heat capacity .
of a solid. As an example, let us calculate ph eric pressure, K =
9.5
x
Cp
=
29
x
c,.
for copper at 1000 K. At atmos
103 J kilomole -l K-
1 ..
{3 = 6.5
x
10
-
5
K
-
t
V n
=
nz
np
=
63.6 kg kilomole- 1 --
-·-
X 103 kg
8.96
m-3
=
7.1
x
l
10-· m-1 kilomole- 1•
Hence
Cl,
=
�
and
2 10-1 Pa-1• The specific volume in m3 kilomole-1 can be found
from the atomic weight of copper and its density:
v =
C1•
Cp
= 29
2 Tv{3 -
x
K
103
=
-
29
3.2
1
X 10· -
x
103
=
( 100()) (7.1
26
X 10-3) (6.5 X 10-5)2
- ------
9.5 x x
I0-12
lW J kilomole-1 K-1•
117
The Tds Equations
Sec. 7.5
This high-ternperatttre value is very nearly equal to 3R, a result that will be explored in detail in Chapter 16.
. Finally, we consider a reversible adiabatic process ds = 0. Equation
(7.12) gives
Cp_dv = - C,J< dP
f3
{3v
(7.20)
'
or ---
1
OV
iJP
v
s
=K
Cv
K
=
Cp
y
.
In analogy with the isothermal compressibility K we introduce the adiabatic compressibility
1
Ks = -V
Since 'Y
>
1,
Ks
<
K
av aP
--
(7.21)
•
K; the adiabatic compressibility is less than tt1e isothermal
compressibility because the increase in pressure causes the temperature to rise. The temperature increase, in turn, results in an expansion that partially offsets the compression associated with the pressure increase. It folJows that in
the adiabatic case the volttme change is less and Ks is smaller.
It is interesting to note that the speed of sound in a gas is given by
c=
1 -
pK
=
(7.22)
'
where p is the density of the gas. The adiabatic compressibility is used since the process of compression and rarefaction takes place very fast; there is no time for heat to be exchanged with the surroundings. Using Equation (7.22), we can estimate the speed of sound in air. If air is
regarded as an ideal diatomic gas, then 'Y atmospheric pressure, and p
c=
:::=:
�;;;
1.4,
K ::=
1/P
=:::
10-5 Pa-1 for
1.2 kg m -3• Thus 1.4
•
l/2 �-
��
340 ms-1
•
This compares favorably· with the measured value of 343 ms -t at 20°C.
Chap. 7
118 :>
.
. .
::::.·
· .
.
. . .
.
· .
,
:
.
·
·.
·:
. .
.·
.
Applications of the Second Law
. ·.·
.
. .··
. _...; . ::
.· ·
7.6 ENTROPY CHANGE IN IRREVERSIBLE PROCESSES How do we calculate the entropy change for an irre\'ersible process when the entropy is only defined in terms of reversil)le heat flow? We resort again to an earlier argument. The entropy is a state variable. and the entropy difference is the same between any two equilibrium states regardless of tl1e nature of the process. Thus� we can find as for an irreversible process by choosi11g any con venient reversible path from the initial to the final state and be certain that it is equal to the change produced by the actual. irreversible path. Consider the example of Figure 7 3 A body at temperature T, is in thetmal .
.
contact with a single reservoir at temperature T2 with T2 > T1• The system is
allowed to come to thermal equilibrium with the reservoir. The process is irre versible since there is a finite temperature difference; the process cannot be reversed by an infinitesimal change. We assttme that the process is isobaric. Then
/
"IF u
Here we are using an equation reprcse11ting the first law for
a
reversible
process. That is" we are effecting a simple heat exchange by bringing up a whole series of reservoirs between ·r1 and 72 (keeping the pressure constant)
in su�h a way that the body passes through a series of equilibrium states. Then llq,
ds =
== c 1,
T
dT
.
T
so that T., -
•
-
bodv
.
.
.
:
.
. .
.
"'
.
� .
TJ
.
...--
-
.
.
•
.
.
•
-
. .
.
.
.
I
-
.
.
.
.
.
,
.
·.
:
. .
•
Q T.., -
.
Figure 7.3 Body initially at temperature T1 in thermal contact with a rese rvoir at temperature T2•
� L
. .
.
( 12
-
.
. .
: .
..
. .
.
r, )
>
.
.
. .
-
.
..
· . . .
-
..
.
.
. .
.
.
-
.
.
.
.
.. . . . . . . ' . . . . . . .
.
.
.
.
.
.
.
.
. .
.
.
.
.
. ..
..
..
.
.
.
.
.
.
.
.
.
. .
. .
.
.
.
.
.
. . .
.
\ r��ervoir
=
surroundings
.
.
.
· . .
; •
.
.
i
.
==
system
Sec .. 7.6
Entropy Change in lrre\·ersible Processes
119
while
As far as the reservoir is concerned, the process is isothermal (the tempera ture of the reservoir remains constant) as well as isobaric. The hea� flow out of the reservoir is
-cp(T2- T1).
Now
( �S ) universe
=
( �S ) body
+
( �S ) reservoif'
or •
(7.23)
It is easily seen that the entropy of the universe will always increase, whether T2
>
T1 or
T2
< T1• Let x
=
TJT1• Then the terrn in brackets can be written
Taking the derivative of f(x) and setting it equal to zero shows that the func tion has a maximum or a minimum at x =
x
=
1, where the function is zero. At
1 the second derivati.v e is positive, so the extreme value is a minimum.
Hence f(x) is always positive except when T2 heat exchange at all.
=
T1� in which case there is no
ltl the example, the increase in entropy of thf! body is larger than the decrease in entropy of the reservoir and the entropy of the universe is greater at the end of the process than at the beginning. In all real changes, the entropy will always increase; in other words, entropy is created in the process. The addi tional entropy is an onus that the actual universe must bear. For an indefi nitely expanding universe, there is no known process that can cause the entropy to decrease.* To illustrate the calculation of entropy changes in irreversible processes, we consider the following examples.
EXAMPLE
I
Suppose that 0.5 kg of water at 90°C is cooled to 20°C, the temperature of the surrounding room. The specific heat capacity at constant pressure of water is *
However9 a universe that expands without limit is different from a finite. closed system.
There is no final equilibrium state and energy conservation may not apply.
Applications of the Second Law
Chap. 7
110
1 4180 J kg-1 K- • The change in entropy of tl1e system (the water) is the change
that would occur if the "·ater were cooled reversibly from T1 T2
=
293K:
(�S)system
r aQ,
=
T
r!
=
mer
dT =
T
7,
mcp
In
=
363K to
T,
i 1
The entropy of the system decreases. The change in entropy of the surround
ings is equal to the heat transferred to the room divided by the temperature of
the room, which remains constant. The sign is positive since the heat flows into the room: T�
nzcr
•
dT =
mcp
363- 293
Therefore (4s)universe =
(�S)system +
(�.\')surroundings=
-448+499 = +51 J K-1•
The entropy of the universe has increased.
EXAMPLE2 •
4 x 105 J flows through a diathermal wall separating a high temper· ature heat reservoir at TH 500 K from a low temperature heat reservoir at Heat Q
=
==
TL
=
200 K. The p ai r of reservoirs constituting the system is thermally insu
lated from its surroundings. Then
( �s) universe= ( dS) S\'Stem •
=
=
•
Qr -
TH
+
Q, TL
=
( � s) reservoir H -
+
( � s) reser\'oir
4 X l(t5
4 X 105
500
200
-- +
= +
/.
1200 J K -I .
Note that if the two reservoirs are at tl1e same temperature .. no heat is exchanged and
(AS)
erse
un i v
=
0.
Sec. 7.7
Free
Expansion of an Jdeal Gas
Ill
EXAMPLEl A 5 kg mass falls to the ground from a height of 50 m;The temperature is con
stant at 20°C. 'The process is irreversible but we can imagine the mass being slowly and reversibly lowered by a string and pulley arrangement. Since no heat exchange is involved,
( 4S)system
=
0, but W,
mgh
We note that for the surroundings, which are ttnchanged if the mass is small, T(�S)
=
Q,
=
�U + Wr
=
0 + mgh.
7. 7 FREE EXPANSION OF AN IDEAL GAS
We have seen that in the free expansion of an ideal gas, du t h at aw
=
=
0 and llq
=
0 so
0 also. Referring to Figure 7.4, we observe that the equations
describing the equilibrium end states are
P0
RT0 =
Vo
and
A reversible isothermal process would be described by the equation Pv
=
constant or P0v0
=
P1·v1• However, the latter equation does not describe
the free exp ans i on because P1 is initially zero and the process is irreversible.
vacuum
gas
(a) Figure 7.4
(b)
Free expansion of an ideal gas: (a) initial state; and (b) final state.
Applications of the Second law
Chap. 7
Ill
Nonetheless, the entropy change can be calculated by assuming that a reversible, isothermal expansion takes place between the initial and final states of the system. Thus,
ds
=
Cv
dT T
+ R
dv V
' y I
.
=0
so
( ds)svstem .
=
v1
R ln
(7.24)
'
Vo
and
( ds) universe
=
(7.25)
( ds) system > 0..
since the system is isolated (adiabatically insulated). In an irreversible free �xpansion, the available energy levels become more closely spaced .. leading to greater randomness and increased entropy. We note that in a reversible. isothertnal expansion .. the work done would be I
u1,
Since du
=
0,
w,
=
=
RT0 In
v,
.
'Vn
q, so that ..
.is
q, =
Tu
=
R In
·vJ Vo
. •
In a free expansion no work is done but the change in entropy in the irre versible process is as if work
were
done in a reversible, isothermal process
between the same end points. If the situation involves the mixing of gases
( P1
=F
0)" the problem is
quite different and will be treated in Section 9.5.
7 .8·
ENTROPY CHANGE FOR A LIQUID OR SOLID
.
The equation of state of
a
liquid or solid is, to a
v = v0[1
+
first approximation,
P(T- T0) - K(P- P0)]
(7.26)
Problems
Ill
(see Section 2.5). The second Tds equa ti on is:
a
(7.27)
p
Setting
av iJT
=
p
'llQ . {)IJ'
we obtain
Integrating, we have
- voP(P - P0).
s-
(7.28)
The entropy increases if the temperature increases and decreases if the press ure Inc reases. •
PROBLEMS
7-1 The latent heat of fusion of ice at a pressure of 1 atm and ooc is 3.348 x 105 J kg-1• The density of ice under these conditions is 917 kg m-3 and the density of water is 999.8 kg m -J. If 1 kilomole of ice is melted, what will be
(a) the work done? (b) the change in internal energy? (c) the change in entropy?
•
7-2 Ten kg of water at 20°C is converted to ice at -l0°C by being put in contact with a reservoir at -l0°C. The process takes place at constant pressure. The heat capacities at
constant
pressure of water
and
ice
are 4180 and
2090 kg-1 K-1, respectively. The heat of fusion of water is 3.35
x
lOS J kg-1•
Calculate the change in entropy of the universe.
7-3 Calculate the change in the entropy of the universe as a result of each of the fol lowing processes:
(a) A copper block of mass 0.4 kg and heat capacity 150 JK -l at l00°C is placed in a lake at l0°C.
Applications of the Second Law
Chap. 7
124
•
(b)
The same block at l0°C is dropped from a height of 100m into the lake.
(c) Two similar blocks at 1 ()(r)C and 1 ooc are joined together. (Hint: See Problem
(d)
7-8.)
One kilomole of a gas at 0°(" is expanded reversibly and isothermally to twice its initial volume.
(e) One kilomole of
a
gas at ooc is expanded reversibly and adiabatically to
twice its initial volume.
7-4 Suppose that the specific heat c a pa city cp of the body discussed in Section 7.6 of the text is 10 J kg-1 K- 1 and T,
=
200 K. Assume that the mass of the body is 1
kg.
(a) Calculate the change in entropy of the hody and of the reservoir if T.,
..
(b)
=
400 K .
Make the same calculations for T2
==
100 K.
(c) Find the entropy change of the universe in both cases. 7-5 One kilomole of an ideal gas undergoes a free
e x pa
nsion tripling its vol um e. .
What is the entropy change of
(a) th e gas? (b) the universe? 7-6 An ideal gas has a specific heat given by
c,.
==
+
A
BT. wh e r e A and Bare con
stants. Show that the change in entropy per kilomo1e in going from state
( o1 T1) ..
to state ( ·lh T.,) is ..
-
-
;ls
=
A In
1�
.
t I
8(1�
+
-
Td + R In
v,
-
l' I
.
7-7 A 50 kg hag of sand at 2S'�C' fa ll s 10 m onto the pavement and comes to an abrupt stop. Neglect any transfer of heat het\\'ccn the sand and the surroundings and assume that the thermal capacity of the sand is so Jarge that its temperature is unchanged.
(a) What is the dissipative w o r k done on the sand? (b) What is the changt! in the internal energy of the sand'? (c) What is the entropy change associated \Vith this �U at constant T? (The sand docs no \\'Ork as it defor1ns when it hits the pavem�nt: only its shape changes.. not its volume.)
7-8 Two equal quantities of water, each of mass
m
and at temperatures T1 and T2, are adiabatica11y mixed together, the pressur� remaining constant.
(a) Show that the entropy change of the universe is . I �5 �nzcp In -
'[I
+ 1
]� \ __
-:
•
2\/T1 T�
where cp is the specific heat capacity of the water at constant pressure.
(b)
Show
th a t
�.s· > 0 fo r any
finite
temperatures T1 and T�.
(Hint:
(a- b)?> Oforallrcalaandb.)
7-9 Tw o identical blocks of copper are held at constant vo l um e with a constant hea t capacit y CF 380 J K 1• One is at an initiat tempe ra ture of 320 K the other at a temperature of 280 K. The two h1ocks arc thermally isolated and placed in con =
·
tact with each other. What is the entropy change of the system?
..
Problems
125
7-10 (a) Assuming that P and v are independent, show that in a reversible process, as
Cv =
aP
v
iJT
-
T
and
iJP
v
•
as av
P
-
cp
ar
T
ri'lJ
P
•
(b) Use these results to derive the third Tds equation (Equation (7.12)). 7-11 Use the first Tds equation to calculate the entropy for a van der Waals gas. 7-12 Calculate the difference in the spe.cific heat capacities cp - cv of copper 4.9 x 10-5 K-1, K = 7.7 x 10-12 Pa-1, and v = 7.1 x 10-3 at 300 K. Take {3 m3 kilomole-1• How does your answer differ from the result of Section 7.5? =
7-13 (a) Show that the difference between the isothermal and adiabatic compressibilities is K- K.\' =
--
Cp
.
(b) What is this difference for a monato'.:tic ideal gas? 7-14 Using the fact t h at dvjv is an exact differential, prove that iJ/3
riP
T
=
dK -
iJT
. p
7-lS (a) Show that the Joule coefficient may be written
dV
K
Cv
u
•
(b) Show that the Joule-Thomson coefficient may be written J.L
=
iJT aP
h
=
v cp
(T/3- 1).
(c) Using these results,find 11 and 1-L for a van der Waals gas and show that both are zero for an ideal gas.
7-16 Calculate the velocity of sound in liquid He4 using Equation (7.22). T'-.ke y = 1.48, p = 162kgm-3,andK = 9.43 x 10-8Pa-1. 7-17 (a) The temperature of a block of copper is increased from T0 to T without any appreciable change in its volume. Show that the change in its specific entropy is .:1s =
Cp
T
In T,
0
-
vo{32
K
(T
- T0).
(b) Calculate �sin units of J kg -l K -·I if the temperature increases from 300 K to -12 x K-1, x 10-5 Cp = K-t, f3 = 4.9 K = 7.7 3901 kg-1 10 and 310 K. Take . Pa • The density of copper is 9.85 x 1 cY kg m- 3. -
.
•
;;. :;::; :: . ; ·:'...::.:·:;. .:.; : : ; :: : · :::: · :;:. · :; ::::::::
fiX
:-; : · ·: :
•
•
IC
8.1
Introduction
129
8.2
The Legendre Transformation
130
8.3
Definition of the Thermodynamic Potentials
132 •
8.4
The Maxwell Relations
134
8.5
The Helmholtz Function
134
8.6
The Gibbs Function
136
8. 7
Application of the Gibbs Function to Phase Transitions
137
8.8
An Application of the Maxwell Relations
142
8. 9
Conditions of Stable Equilibrium
143
•
127
.
.
. .
.
:
:
.
. .
.
.
. .
:
. .
.
.
8.1 INTRODUCTION In the context of the first law, we have defined two functions of the state vari ables with the dimensions of energy: the internal energy U and the enthalpy H. Since neither of these is well suited to the analysis of certain processes, it will be convenient to introduce two additional functions the Helmholtz function F and the Gibbs functio11 G. Because of their role in detertttining the equilibrium states of systems under prescribed constraints, they are known as
thermodynamic potentials, in analogy with the potential energy in mechanics. The enthalpy, Helmholtz function, and Gibbs function, are all related to the internal energy and can. be derived from it using a procedure known as a Legendre differential transformation . To see how this is done, we consider the combined first and second laws, written as dU
=
(8.1)
TdS - PdV.
•
The two independent variables S and V are intrinsically extensive quantities. The two intensive variables, T and - P, are said to be canonically conjugate to them. That is, the canonically conjugate pairs are T, S
and
- P, V.
Note that T and S are thermal variables, whereas P and V are by nature mechanical variables. In Equation (8.1) \\'e assume U
dU
=
au \ as
=
U(S, V) so that
dS + v
au av
dV.
(8.2)
s
Comparing Equations (8.1) and (8.2), we see that
au as
v
=T '
au av
=
s
-P.
(8.3) 129
Chap. 8
130
Thermodynamic Potentials
However" the selection of the two independent variables is a matter of choice. There are four possible ways in which a thermal variable can be paired with a mechanical variable: (8.4) We consider the feasibility of defining a thermodynamic potential fclr each of the four pairs.
8.2 THE LEGENDRE TRANSFORMATION
Consider the function Z ·
·
=
Z ( .r v) and write the differential �
..
d Z ( .r y ) �
=
X tl .t + Y d y.
(8.5)
where x, X and},' Yare, by definition, canonically conjugate pairs. We wish to replace (x, y) by (X, Y) as independent variables. To do this we transform the function Z to a fun ction lv/. The equatiorl of transformation is Nf(X. Y)
=
Z- xX- y},P.
(8.6)
Then tiM
=
tiZ- X d.t -- Y llv•
.\�
dX
- v liY.
·
•
Since. by Equation (8.5) the first three ter1ns on the right-hand side sum to zero, we have {/NI
= -.r
dX-
v
•
dY .
Thus Equation (8.6) is the transformatic)n that takes us from
(8.7) a
function of one
pair of variables to the other. Equations (8.5) and (8.7) gi ve recipr(JCity rela
tions as follows: az
ax
==
az
X.
ay
aM
aM
ax � -x.
aY
If we wish to replace onl y
lJne
=
==
Y�
- Y·
(8.8)
( 8.9)
()f th e variables. say }', by its canonically
conjugate variable Y, we must consider the function
N(.r_ Y)
=
Z
-·
)'Y.
(8.1 0)
Sec. 8.2
The Legendre Transformation
Ill
Then dN
=
dZ
-
Y dy - y dY�
and, using Equation (8.5), we obtain dN
=
X dx
dY, y
-
(8.11)
with reciprocity relations iJN ax
-
=
X'
aN aY
= -
(8.12)
y.*
The equations of transfortnation, Equations (8.6) and (8.1 0), may look mysterious at first glance. Howe\'er, they are based on a very sirnple idea: a curve in a plane can be equally well represented by pairs of coordinates (point geometry), or as the envelope of a family of tangent lines (line geometry), as
depicted in Figure 8.1. Z()'), where y is the Suppose that the curve is given by the relation Z independent variable. Consider a tangent line that goes through the point (y, Z) Y. If the Z intercept is N, the equation of the line is and has a slope. d ZI dy =
y
Z-N =
y-
0,
z
Figure 8.1 A curve represented as the envelope of a family of tangent lines.
y
*It is note\\'Orthy that in classical mechanics. the transfonnation from the Lagrangian to the Hamiltonian is
Here the qJc, q"
.
a
cegendre transfortnation:
and P1c are generalized coordinates, velocities, &nd momenta, respectively.
Chap. 8
Ill
Thermodynamic Potentials
or
which is Equation (8.10). If now we differentiate this expression. we have dN = dZ
v
-
dY - Y dv. .
;
Y dj', so
But dZ =
dN
=
-
vd}"�
which is the reciprocal relation
.v = -
dN dY
Thus the Legendr e transformation is a mapping from a
(}', Z) space to a
( Y, N) space, from a point representation <>f a curve to a tangent line repre
sentation. The extension of the transformation to functions of more than one independent variable is straightforward. as shown.
8.3
DEFINITION OF THE THERMODYNAMIC POTENTIALS U(S� V) to the enthalp.y H(S� P), we r epla ce V by its conjugate varia hie - P. Thus
1. To transform the internal energy .
Since
dU
=-
TdS
+
(
-
P)dV.
•
we make the fo ll owing associations with the variables of Equations (8�5) and
(8. 1 0): Z=U,
Equations
X=T
t=S.
.
.
Y=·-P.
v==V_ •
(8.1 0), (8.11) and ( 8.12) immediately give ,
N=H.
Sec. 8.3
Definition of the Thermodynamic Potentials
Ill
H = U + PV, dfl=TdS+VdP, aH
OH aP
(8.13)
=V
·
s
These are the fundamental relations involving the enthalpy. •
2. Next we replace the thennal variable ..�in the function U(S, V) by its conjugate variable T. This leads to a new potential, the Helmholtz func tion F. The Legendre transformation is U(S, V)-+ F(T, V). With the identification z =U,
X
=
-P ,
X =V,
y = T,
y
=
S,
N =F,
we obtain the expressions •
F dF OF
=
=
-
U- ST,
= -P '
av
(8.14)
P dV- S dT,
T
OF aT
= -S. v
3. Finally, in U (S, V) we replaceS by its conjugate T and V by its conjugate -
P, yielding the Gibbs function G, the last thettnodynamic potential
required by the pairing of thermal and mechanical variables. The trans formation U(S, V) �G(T, P) is effected by using Equations (8.5) and (8.6) and setting
Z = U,
X
=
T,
x =
S,
y =V,
Y = - P,
M
=
G.
The resulting expressions are G =U- ST + PV, dG = S dT+V dP, -
OG iJT
= -S ' P
OG aP
=V. r
(8.15)
Thermodynamic Potentials
Chap. 8
134
These potentials have interesting propcrtie� which will be discussed later in this chapter.
8.4 THE MAXWELL RELATIONS Each of the four thermodynamic potentials is a state variable whose differen tial is exact. We can use the condition for exactness discussed in Appendix A .. which states that the value of a mixed second partiaJ derivative is independent
of the order in which the differentiation is applied. As an example_ we consider dU
=
Tt/S
+
(
-
au
P)dV
=
riS
au
dS +
·
,
dV.
riV
,
s
The exactness of dU immediately gives
--
av as
==
--
av
=
--
·-
-
.:=:
aP •
iJS
\' •
t.
The equality of the first derivatives,
-
av
-
,
a.s
.
. .
'
v
is known as a Maxwell refa1ion. I t s utility lies in the fact that each of the par tials is a state variable that can be integrated along any convenient reversible path to obtain differences in values of the fundamental state variables between given equilibrium states. A summary of relationships involving the thermodynamic potentials, including the MaxweJI relations, is presented in Table 8.1.
8.5 THE HELMHOLTZ FUNCTION We have seen that a key property of the internal energy is that the change in U is the heat flow in an isochoric reversible p rocess. Similarly, the change· in the enthalpy H is the heat flo\\' in an isobaric
reve rs ible
process. We ask: With
what quantities can we associate changes in F and G as a result of reversible processes? Suppose that a system is in thermal contact "'ith a reservo ir environment that is at a constant temperature T. Let tht system undergo a process from some initial state to a final stat e According t\) the second law, .
.
.
�S
+
......
�So> 0.
The Helmholtz Function
Sec. 8.5
TABLE 8. I
135
Relationships involving the thermodynamic potentials.
Thermodynamic
Independent
Reciprocity
Maxwell
Potential
Variables
Relations
Relations
T=
S, V
lntemal energy
as
v
av
aP -
=
-
·
-
as
v
a2u =
·
av
'
-
5
au
-P
dU = T dS - P d'l
u
ar
au\
s
--
av as
•
H
=
F= U- TS
=
-
=
-P
dG = -SdT + V dP
U- TS + PV
as
s =
s
--
riP as aP
as
aF
-
ar
v
av
-
ar
r
V=
--
av
T
ar
aG
p
aP
-
--
r
-T
av ar av
as
aG
liP
v
a2F
aF =
p
a2H
aP
S= -
T,P
Gibbs function =
S dT - P dV
aP
p
-
aH
S=-
T,V
dF
as
V=
dH = T dS + V dP
U + PV
Helmholtz function
G
T=
S.P
Enthalpy
av
ar
riH
ar
p
a2G aP ar
H - TS
where �S is the entropy change of the system and �So is the entropy change of the reservoir. If the reservoir transfers heat to the system, then �So Q/T and therefore, =
-
Q
<
T �S,
where the equality sign applies if the process is reversible. If we substitute this in the first law expressed in the form . W
=
-
AU +
Q,
we obtain W
or, since F
=
U
-
<
-
dU ·f. T �S,
TS, W
<
-
�F
(no change in n.
(8.16)
Thus the change in the Helmholtz function in an isotherttla l reversible process is the work done on or by the system. More generally, the decrease in F equals
Chap.8
136
Thermodynamic Potentials
the maximum energy that can be freed in an isothermal process and made available for work.* The function is therefo re often called the HelmJtoltz free energy. (Elsewhere it is frequently given the symbol A. an abbreviation for the German word for
work, Arbeit.)
If the system held in thermal cont act with the rese rvoir has a uniform pressure throughout its volume and the latter is also held constant, then the work performed will be zero and •
(8.17)
(8.18) where i and f denote the initial and final values, respectively. Hence, if the Helmholtz function is a minimum, any change in the state of the system would increase
F,
which would be cont radi c tory t«l Equati on
the condition for
equilibriun1 in
a
(8.18). It follows that
system in thermal contact with a reservoir
and kept at cons t an t volume is
(8.19) ·
with Fa minimunz. S ince T and V are the most convenient fundamental state variables for the purposes <)f statistical Jnechanics, as
we
shall see. the
Helmholtz function becomes a very significcrnt means of specifying a system's ·
properties. •
8.6 THE GIBBS FUNCTION The Gibbs function is useful in prohlems in which T and P are the fundamen tal variables. Consider a system in a surrounding environment that C< )ns t itutes
a temperature and pressure reservoir. By this we mean that the reservoir is so large that its temperature and pressure remain unch anged. Most chemic a) reactions and some phase changes (icc melting in a heakcr e x p osed to the atmosphere� for example) take p lace in this 'vay. As before, .
wi th
Q
=
�u
+
P :!1 v. •
•
Actually. it is not necessary that the temperature remain constant during the process: it is
only required that the two end points be at tcmperatur('
T.
Sec. 8.7
Application of the Gibbs Function to Phase Transitions
137
reverstb _ le. Co1nbining these expressions, we hav e
&U + P� V
-
T AS < 0.
Since G
=
U + PV- TS
'
and T and Pare constant here, we obtain
( AG)r.P
<
0,
(8.20)
or •
(8.21) Spontaneous processes occur in the direction of decreasing
G.
A system in
thet·tnal contact with a heat and pressure reservoir moves to a state of mini
mum G for which dG
=
(8.22)
0.
•
It can be shown that if nonmechanical forces (electric forces, magnetic forces, etc.) act on
a
system, doing work "''"m' then Wnm <
-
aG
(no change in Tor
P).
(8.23)
The decrease in the Gibbs function is equal t<.l the maximum energy that can be freed in an isothermal, isobaric process and made availabJe for nonmechan ical work. Again, T and Pneed not be fixed tltroughout the process; they need only have the same initial and final values. Table
8.2
lists the applicable conditions for various system states or
processes.
8. 7 APPLICATION OF THE GIBBS FUNCTION TO PHASE TRANSITIONS
Consider a system consisting of the liquid and vapor phases of some sub stance in equilibrium at temperature T and pressure P. Let n'{ be the number of kilomoles in the liquid phase and n'i' the number of kilomoles in the vapor
•
Thermodynamic Potentials
Chap. 8
138
TABLE 8.2* processes.
Conditions on thermodynamic variables for different systems or
Valid
Valid
Equilibrium
Equation
Inequality
Condition
dU :S 0 dll � 0 dF s 0 dG :50 dS � 0
L' Minimum H �finimum f' Minimum (j
State of System or Type of
Process
dS dS tiT dT 71Q dU
V constant Sand P cons ta n t T and V constant T and P constant Sand
=
==
=-=
=
Adiabatic
=
dV 0 dP 0 dV 0 dP 0 + P d\1' =
=
=
==
0
=
Minimum
MaximumS
___;. --.1 • ...... . _ •
•
.
:
.
•
• •
·
.
·.
.
.
•
•
0
•
••• .
'
:
.
.·
•
.
. . . . •
•
. • •
____ � ._... _.... .: . � . . . ·.. ..· . .
•
. . ·. . . .
n.,
•
•
.
••
.
.
• ..
• ••
.
.
.
.
·•.
.
-
.
,
...
.. . . . .. . . . . :·:._::·.-· . .. . ·.:.: ..: : . . . . . .. . . '. -� . · : .:.'. .t: . . .. ' . ' . .. . : . .. :·.· . . . : ' . . .
Figure 8.2
•,.
Liquid and vapor phases of a substance in equilibrium at temperature T
.. -
.
.
.
.
.
.
.
.
.·
.
.
.
:·
. :
.
.
;. .
.
.
.
-
.
..
.
.
.
.
.
.
.
.
and pressure P� (a) initial
·.
·'
.
.
'
.
.
:
·
.
.
.
..
.
.
.
.
..
.
�
.
.
.
. ..
.
. .
:
.
.·
.
.
.
.
.
.
.·
.
.
.
-
·
·.. •.
. .
.
.
.
. .
.
.
. . .
.
·.
·'
.
.
. .
.
.
. .
. .
.
.
.
.
.·
.
·.
.
.
.
.
.
.
.
.
.
. ·: . .
.
(a)
state: (b) final state.
(b)
phase.� 'Tne state of the system is defined in terms of the variables (T, P,
Consider a second state differing from the -first only in the
n]', n'(').
number of kilomoles of liquid and vapor and defined by
(T. P. n3, n2') (Figure
8.2). Mass is conserved so that 1z
}' +
n
j"
== n
� +
n
2' .
(8.24)
'' "' We define g and g as the specific Gibbs functions of the liquid and vapor. respectively, associated with the particular substance under investigation. Noting that the Gibbs function is an extensive variable. we have for the two states: G1 G2
== 11
�' g ,,
=: n �
g
"
,
,
+ n +
tg
"'
,,,
,,,
�
112 g .
(8.25) (8.26)
7.2 in Tht'rnlodynanJics t�nd Statistical Mechanics by P. L. Landsberg. Publ icat i ons New York, 1990.
*Adapted from Table Dover
.
is that
•
section 4.3: one. t\VO. and three prim es denote the soJid.liquid. and vapor phases. respectively. Here 1 refers to the initial state and 2 to the final state. t
The notation
used in
Sec. 8.7
Application of the Gibbs Function to Phase Transitions
139
p
liquid dP Figure 8.3 Relationship between temperature and pressure for a iiquid and vapor in equilibrium. The derivative dPIdT is the slope of the vaporization curve.
vapor
\
""
dT
•
T
I
Suppose that a reversible transition takes place from state 1 to state 2. Since
( �G)r,P
=
0 for a reversible process, it follows that
G1
=
G2. Equating
Equations (8.25) and (8.26) and using Equation (8.24), we find that g"
=
g'".
(8.27)
The specific Gibbs function is the same for the two phases. This is true for all phases in equilibrium, that is, for all points on the curve of the phase transfor mation (Figure 8.3). Since at a temperature 1' + dT and a pressure P + dP we still have equilibrium, it follows that g" + dg'' Equation (8.27), we have dg"
•
=
=
g"'
+
dg"'. Combining this with
dg'" .
Using the expression for the differential previously derived, we can write
-s
"
dT +
v
"
dP
=
-
"' s
dT +
'" v
dP,
or " " ( s ' - s ) dT
=
(v
"'
- v
"
) d P.
Thus dP -
dT
s"' =
--
s
"
---
v
"'
- v
"
•
(8.28)
Chap. 8
140
Thermodynamic Potentials
From the definition of entropy�
(8.29) where t23 is the latent heat of vaporization. Since heat is absorbed as a liquid becomes a vapor, f23 is positive and Equation
(8.28) gives
dT 23
=
s
"'
> s
' Substituting Equation (8.29) in
'
-
v"
(8.30)
(liquid-vapor).
----
T ( v"'
.
)
This is the famous Clausius-Clapeyron equation. It gives the slope of the curve denoting the boundary between the liquid and vapor phases, that is� the vapor ization curve. Similar expressions hold for the sublimation and fusion curves: •
dP dT
=
-
dT
--
T(v'''
1.1
dP
£13
=
12
-
v')
f12
--
" T ( t'
-
-
v')
(solid-vapor),
(8.31)
(solid-liquid).
(8.32)
The latent heats in these expressions are positive, and the slopes are all posi tive for substances that expand on melting. A notable exception is water, which contracts when ice melts into liquid; for this case
( dPI dT) 12
< 0.
The Clausius-Clapeyron equation� combined with the appropriate equa tions of state, can in principle yield equati(>ns for the phase transformation
curves. A simple example is the vaporization curve describing, say� the conver sion of liquid water to stea m Here .
v"' >> v''
dP
dT .,,
:=:
....
(see Chapter
f 2.1
-
T
-
"' t'
If we treat the vapor as an ideal gas.
v "'
::::::=
RT -
p '
so that
•
dP dT
-
::...:::
t23 P -
0
-
R T2.
•
2)� and so
Application of the Gibbs Function to Phase Transitions
Sec. 8.7
141
If f2 is assumed to be temperature-independent, upon integrating we obtain 3
(8.33)
where
(T0, P0)
denotes some fixed poin.t on the curve.*
The Clausius-Clapeyron equation can also help us understand why the
(273.15 K) is 0.01 K below the triple point (273.16 K). Solving for dT in Equation (8.32), we have
ice point of water
T( v" - v')
dT
=
t12
dP.
For the small temperature change involved, we can write 4T
T(v" - v') =
f12
and assume that the specific volumes
v''
and
�p
v'
are constant. This equation
gives the change in temperature as the pressure is increased from the low value at the triple point
(Prp
=
4.58 Torr) to atmospheric pressure, the pres
sure at which the normal melting point of water is defined. The density of pure liquid water is 1000 kg m -J and the density of ice is 916 kg m-3• Noting that the specific volume in m3 kg-1 is just the reciprocal of 1 1.00 X 10-3m3 kg-1 and v' 1.09 X 1 0 3 m3 kg- , the density, we have v" :=
=
-
to three significant figures. The latent heat of melting for water is f12
=
3.34
X
liP
1051 kg-1• Writing Prp in Pascals, we have =
Patm - Prp
=
1.01
X
105 Pa
-
610 Pa
::::
1.01
X
105 Pa.
Then
AT
273(1.00
X
10-3
-
=
1 . 09
3.34
X
X
10-3) (1.01
lOs
X
lOS) ..
=
-0.0074 K.
The presence of dissolved air in a mixture of ice, and water further lowers the temperature at which ice melts by 0.0023 K. The total reduction of temperature below the triple point of water is therefore 0.0097 K, or approximately 0 01 K. ..
*The more generaJ solution is P
=
P0 exp
1
R
T
Chap. 8
141
Thermodynamic Potentials
8.8 AN APPLICATION OF THE MAXWELL RELATIONS As an example of the use of a Maxwell relation, consider an ideal gas that
undergoes an isothertttal reversible change from pressure P0 to pressure P. By definition of the entropy, aQ,
=
T dS.
We express S as a function of the fundamental state variables of the prob lem
namely, T and P:
L�
S(T, P).
=
Then
dS
=
as ar
Since the process is isothermal, dT
as aP
dT + p
=
dP. r
0, and
'
dP.
ap
'�
From Table 7.1, we select the Maxwell relation <·
as
riP
iJ\' -
-
- ·-
a1
T
..
•
P
Thus
l/Q,
=
-T
av
iJT
dP. r
For an ideal gas,
riV
aT
nR -
,)
P '
and p
Pn
•
Sec. 8. 9
Conditions of Stable Equilibrium
143
If P > Po, the negative sign indicates tltat heat flows out of the system in the process.
8.9 CONDITIONS OF STABLE EQUILIBRIUM The second law states that in spontaneous processes the entropy increases. Maximum entropy corresponds to thermodynamic equilibrium. Furthermore,
maximum entropy is a point of stable equilibrium: if the system is perturbed slightly from its equilibrium state, it will return to that state spontaneously. We wish to examine the implications of this fundamental law of nature. Since U = U( S, V), it follows that S == S(U, V). The maximization of S is a problem in multivariate calculus.* A theorem states that the function
f ( x,
1.
y)
has an absolute maximum if, for all points in the domain, < 0,
fxx
2. D
•
>
0,
where
(In
fxy
this
for example,
iJ2fI ox a y.) Thus the conditions on S are
=
1.
section, the subscripts indicate differentiation;
Suu
2. D
=
< 0,
Suu Svv
-
(Suv )2
>
0.
From the combined first and second laws,
dS
=
Su dU
+
Sv dV
I =
T
dU +
p
dV. T
(8.34)
Thus •
Su
1 =
T
'
Sv
p
=
(8.35)
' T
Taking the differential of the first of these equations, we obtain
Calculus: One and Several Variables, E. Hille, Wiley. New York, 1982, pp. 792 ff. •
See, for example,
4th edition, by S.
L.
Salas and
Chap. 8
144
Thermodynamic Potentials
Then riU
Suu
=
iiT
1
--
r.,
�·
·
Since au
Cr
=
riT ,.
"
it follows that (8.36)
The temperature Tis always positive. Therefore the first condition for ity, Suu
<
0 implies that
stabil
..
(8.37) Now let''s take the tion (8.35):
d iffe r e nti als of the two equations labeled Equa
Svv dU
Sl't: dV
+
=
-
where D- Suv
=
dT � T � T
-
dP
Cramer·s rule:
-
V d
•
P
,
We can solve for dV (or tiU) using
tiT �f2
-
T
(8.38)
-------
D
Sv\-' - (Scv )2. Expanding the determinant in the numerator
and rearranging terms. we obtain
D tlV
Svu -
=
-
T -
dP
+
(S"'u
--
Suu
P)
dT "'T
·
Sec. 8.9
Conditions of Stable Equilibrium
145
Then
iJP
T '
T
or
D=
Suu
aP
T
av
-
•
T
Now av
1 I(
-
-
v
aP
'
T
and
from Equation (8.36), so that 1
(8.39)
Since Cv, T, and V are all positive, the second condition for stability, D > 0, implies that
{8.40)
I(> 0.
We conclude that for thermal stability, the addition of heat to a body causes the temperature to increase:
Cv
=
For mechanical stability, a decrease of pressure causes an increase in vol.ume:
K
-
1
v
iJV aP
> 0. r
That is, for a closed system, stable-thermodytzamic equilibrilJffi consists of ther mal and tnechanical equilibrium. There are a few exceptions to this otherwise general result. At critical points the heat capacities and the compressibility can diverge and the stability conditions are then violated. It is also interesting to note that long-range forces can give rise to negative heat capacities. Such modifications have by no means been fully investigated as yet.
Thermodynamic Potentials
Chap.8
146
PROBLEMS
8-1 A van der Waals gas and an ideal gas are originally at the same pressure, tem perature .. and volume. If each gas undergoes a reversible isothermal compres sion.. which gas will experience the greater change in entropy?
8-2 Show that for the an ideal gas
(a) f
=
C0(T
-
To)
c]
-
In
T
To .
(b) g = c,.(T
-
T0)
c,.T In
-
- RT In
T
T
�
v
Vo .
-
p + RT ln p, '
-
soT. s0T.
·
0
8-3 A cylinder contains a piston on each side of which is one kilomole of an ideal gas. The walls of the cylinder are diathermal and the system is in contact with a heat reservoir at a temperature of 0°C. The initial volumes of the gaseous sub systems on either side of the piston are 12 liters and 2 liters. respectively. The piston is now moved reversibly so that the final volumes are each 7 liters. What is the change in the Hemhohz potential? (Note that this is the work delivered to the system by the reservoir.)
8-4 Derive the following equations: (a) F
(b) Cr
=
U + T
=
aF
:
ar
\
-T
(reversible process)-� . •
(c) H
=
T
G -
(d) C,
=
-
·
rJG aT
a 2G T 2 ar
p:
(reversible process):
-
p
(The third relation is the Gibbs-Helmholtz equation. alluded to in Chapter
10.)
8-S The Helmholtz function of a certain gas is F
n2a =
-
V
-
nRT ln(V- nb) + J(T).
where J is a function of T only. Derive an expression for the pressure of the gas.
8-6 The Gibbs function of a certain gas is G where A� 8,
C.
=
nRT lnP
+
A + BP
+
CP2 2
DP3 + --3 .
and D are constants. Find the equation of state of the gas.
Problems
8-7
147
The specific Gibbs function of a gas is given by ' g
p
RT ln
=
- AP.
Po
where A is a function of T. Find expressions for:
8-8
the equation of state;
(a) (b) (c)
the specific Helmholtz function.
(a)
A van der Waals gas undergoes an isothetrnal expansion from specific vol
the sp ecific entropy ;
ume v1 to specific volume
v2•
Calculate the cha nge in the specific Helmholtz
function.
(b)
Calculate the change in the specific internal energy in terrns of
8-9 Start with the first Maxwell relation li�ted in Table
t't
and v2•
8.1 and derive the second by
using the cyclical and recipirocal relations (Appendix A) and the identity
az
ay
ax
ay I oz
I
iJx
=
f
1.
(The remaining Maxwell relations can be derived in a similar manner.)
8-10
(a)
Prove that Cp
=
T
and use the result to show that
(b)
Prove that Cp for an ideal gas is a function ofT o nly
.
8-11 ln Figure 2.2 depicting the isotherms for a van der Waals gas, the shaded areas were shown by Maxwell to be equal. The path
bed is the segment of an isotherrn
along which liquid and vapor are in equilibrium. Prove Maxwell's result by not ing that Ag
=
0 for a phase change (P and T constant) and calculating �f
8-12 The P-v diagram in Figure 8.4 shows two neighboring isotherrns in the region of a liquid-gas phase transition. By considering a Carnot cycle .between tempera tures T and T + d1. in the region shown, derive the Clausius-Clapeyron equa tion
dP/dT
=
f23/T(�v).
Here
�vis the
specific volume change between gas
and liquid.
8-13 The equations of the sublimation and the vaporization curves of a particular material are given by In
P
In
P
=
=
0.04 0.03
-
-
6fT
(sublimation),
4/T
(vaporization),
148
Chap. 8
Thermodynamic Potentials
p
+
T ·.
.
. .
.;-· �. . · . � .· . ....
T Figure 8.4
P- v diagram for
dT
·
. .
.
·. · .:... ·
· �
·
···. . . .. .
.
.
.
= const.
.. .
.
.
· . . ·. .·.
.
·. : ...-:� . . .. . .. . . . .
·
.
.
.
.
. ·
·.#
.
..
.
..
.. . . · .· · .
. .
.
I
I
= const.
'
an infinitesimal Carnot cycle.
�----· ------·
t'
where P is in atmosp heres
.
(a) Find the temperature of the tri ple point. (b) Show that the specific latent heats of vaporization and sublimation are 4R and 6R, resp ective ly (You may assume that the specific volume in the vapor .
phase is much larger than the specific volume in the liquid and solid
phases . )
(c) Find the latent heat of fusion. 8-14 (a) Calculate the slope of the fusion curve of ice in PaK _, at the normal melting point . At this temperature, the heat of fusion is 3.34 X 1 a� Jkg. i and the change in specific volume on melting is - 9.05 x 1 o- � m3kg- 1• (b) Ice at 2°C and atmospheri c pressure is compressed i so th e rmally Find the -
pressure at which the ice starts to melt (in atmospheres).
.
e en
•
9 .I
The Chemical Potential
lSI
9.2
Phase Equilibrium
ISS
9.3
The Gibbs Phase Rule
157
9.4
Chemical Reactions
160
9.5
Mixing Processes
162
149
9.1 THE CHEMICAL POTENTIAL
Until now we have confmed our discussion to closed physical systems, which cannot exchange matter with their sutToundings. We turn our attention in this chapter to open systems, in which the quantity of matter is not fixed. Suppose that
dn kilomoles of matter are introduced into a system. Each
kilomole of added matter has its own internal energy that is released to the rest of the system, possibly in a chemical reactio11. The added energy is propor tional to
dn and may be written as p,dn. The quantity #L is called the chemical
potential. The chemical potential is associated with intertttolecular forces. An elec trically polarized molecule experiences a Coulomb attraction when it is brought into the vicinity of another such molecule.* 'Ibis force is expressed as a negative potential energy, a sort of "potential well." As the new particle approaches its neighbor, it gains kinetic energy while losing potential energy. The kinetic energy is imparted to other particles through collisions, so the sys tem gains internal energy in the process. Consider a motionless molecule infinitely distant from other molecules. Its kinetic energy and potential energy are both zero. The molecule is moved into the force field of a second molecule. This can, in principle, be done slowly so that the kinetic energy is negligibly small. Left by itself, however, the mole cule picks up kinetic energy equal in magnitude to the depth of the potential well (Figure 9.1). Quantitatively, E=K+V (r), where E is the total energy, K is the kinetic energy, and V(r) is the potential energy; r is the distance between the molecule and its neighbor. At r = oo, K=
0, and V
•
=
0, so E
=
0 everywhere. At r = r0,
See Section 11.1 for a discussion of the molecular interaction potential. lSI
The Chemical Potential and Open Systems
Chap. 9
152
V(r)
r
Figure 9.1 Schematic diag ram of a potential well due to intermolecular forces. A more descriptive graph is given in Section 11.1.
I I· K I I
Energy is conserved� but a conversion from potential energy to kinetic energy takes place. The kinetic energy is added to the internal energy of the system. It is reasonable to ask what the magnitude of J.L is. In a standard labora tory experiment� sulfuric acid is added to water, producing an increase in tem perature. Imagine that 10-5 kilomoles of acid at room temperature are added to a liter of water
(5.56
x Jo·-:! kilomoles). also at room temperature. The tem-
perature is observed to rise
0.1 °C. We wish to determine the chemical poten
tial of acid in water. We shall assume that the interaction among the acid molecules is small compared with their interaction with tl1e water molecules (both acid and water molecules are eJectrically polarized). 'llle specific heat capacity of water is
7.52
x
104 J kilomole- 1
K -1• Thus the heat gained by the water through the
addition of the acid is
Q
=
lnc,�T
=
(5.56
�)(7.52
X 10
X
10-')(0.1)
=
4181.
Since the mass of the water is more than 103 times the mass of the acid .. we can ignore the heat capacity of the latter. The chemical potentiaL then� is
" ,-
=
41X
Q
-
· �n
=
-
1()-) _
=
-
7 4.1 8 x 10 J kilomole-
1• _
The sign i s negat i ve because heat is transferred jro111 the acid to the \Vater. We are interested in the chemical energy per particle, which is essentially the depth of the potential well. Since a kil()ffiole has 6.02 x 1026 n1olecules. this value is -6.94 x 10· 20 J or -0.43 e\t'. (One electron volt is equal to
1.6
x
10-19 joules.) Most chemical potentials are of this order of magnitude.
To account for the effect of adding ma ss to
a
system. we need to add a
term to our fundamental equation of thermodynamics: dU
=-
Ttl�'
-
P dV
+
J.Ldn.
(9.1)
Sec. 9.1 Here
153
The Chemical Potential
dn is
the increment of mass added (in kilomoles) and J.L is the chemical
potential in joules per kilomole. If, in an open system, U
iJS
dS
au
+
OV
V,n
dV
+
S.n
=
U (S, V,
au dn
n)
and
(9.2)
dn, S,V
then iJU
J.L=
an.
(9.3)
•
s.v
That is, the chemical potential is defined as the internal energy per kilomole added under conditions of constant entropy and volume. If there is more than one type of particle added to the system (say types), then Equation
(9.1) becomes
m
m
dU
=
TdS - PdV +
j=l
J.Li
(9.4)
d nj,
with
(9.5) The subscript nk means that all other n's except ni are held constant. Another way of seeing the relationship between U and J.Li is to integrate Equation
(9.4). This
can be done by using Euler's theorem for homogenous
functions. Euler's theorem states that if
Af(x, y, z)
=
f(Ax, Ay, Az),
(9.6)
at y-ay
(9.7)
then
at f =X ax
+
+
at . Z az
The theorem can be easily proved by differentiating Equation (9.6) with respect to
A
and then setting
A
equal to unity. Now U
=
U(S, V,
n1
•
•
•
nm)·
Suppose the amounts of all the types of substance, called constituents, in the system were doubled or halved or, more generally, changed by the factor
A
without changing any of the fundamental state variables. Then the extensive variable U would be changed by
A
and all the independent, extensive state •
The Chemical Potential and Open Systems
Chap. 9
154
variables would also be changed by the factor A. Thus U is a homogeneous function and Euler's theorem can be applied to it:
u
=
au
s
au
+ v
as
\-"
111
+
rlv
,/l 1
,·= 1
s·"
1
iJU
lli
•
rin,·
(9.8)
, v.fl J.. �)' .
From the differential of U we know that au as
=
V.nJ
iJU
T ..
av
= I
-
P
au ..
s.v.n ..
S.n,
= J.lj .
(9.9)
Substituting the relations of Equation (9.9) in Equation (9.8)� we have
U
=
ST- PV +
Jl;n;.
..
(9.10)
r= I
Recall that the Gibbs function is defined as
G
==
U
-
ST + PV It is
therefore immediatelv clear that �
111
(9.11) If only one constituent is present .. then G
==
p.11 or JL ==
Gjn: so 1-L in this case is
simply the Gibbs function per kilomole of the substance. Finally.. if we take the differential of Equation (9.1 0). we obtain .
I
J
Equating Equations
(9.4) an d (9.12) we have ..
Ill
SdT- VdP
+ (=· I
11;dJ.L,
=
•
0..
(9.13)
a relation known as the Gibbs-Duhem equation. Taking the differential of Equation
(9.11) gives
1
.
Note that
G
=
G(T P n1 ..
..
•
•
•
(9.14) . .
I
11,,). Thus if we have
a
process that takes place
at constant temperature and pressure, the first two terms of Equation
(9.13)
Sec. 9.2
Phase Equilibrium
155
vanish so the sum also vanishes. Equation
(9.14)
then yields the important
result
(9.15) •
1
We now have the mathematical tools to look at some applications.
.
.
.
·
.
.
. . .
.
.
. .
.
·. . .
·..
.
.
. .
'
.
:
.
.
.
.
9.2 PHASE EQUILIBRIUM We Ylish to find the condition of equilibrium for two subsystems under particle exchange where the subsystems are two phases of the saane substance. An example would be ice melting in liq.uid water; an exchange of H20 molecules occurs between the two subsystetns_ We assume that the total system is enclosed in a rigid adiabatic wall (Figure 9.2). Since there can be heat and particle flow across the boundary between the two phases, the boundary will move as the process evolves. The various interactions are subject to the foUowing conditions:
nA
+
n8
VA
+
V8
UA
+
U8
=n = constant
= V =constant =
U
=
constant
(conservation of mass),
(9.16)
(conservation of volume),
(9.17)
(conservation of energy).
(9.18)
The total internal energy is constant, according to the first law, since no heat flows into or out of the combined system and no work is done. At equilibrium, the entropy of the combined system will be a maximttm:
SA
+
S8
Let all the qu�ntities n, V, U, and
dS
=
S
(maximun1).
(9.19)
S change by infinitesimal amounts.. Then
=
dS.4 + dSB
=
0.
phase boundary
Figure 9.2 1\vo phases of a substance in thermal equilibrium.
(9 20) .
156
Chap. 9
The Chemical Potential and Open Systems •
But
(9.2 1 ) and
(9.22) We can eliminate three infinitesimals by invoking Equations
(9.16), ( 9.17), and
(9.18):
.
Combining these ex pressions with Equations (9.21) and (9.22), and substitut-
ing in Equation (9.20) gives the result 1 -·
1
(9.23)
--
Since Equation
(9.23) must be true for
ar b i t ra ry
increments
dUA, d\lA· dnA all ..
of the coefficients must be zero: TA
PA
=
=
J.L 11
PH
=
(thermal equilibrium) ..
T8
(9.24)
(mechanical equilibr ium)
J.L B
( diffusive tquilibrium)
(9.25)
.
(9.26)
.
We conclude that if we have two phases or s u bsys t ems in thermal and mechanical equil i bri um then they will also be in .
diffitsive equilibri u m ( equi
librium against particle exchange) if their chemical potentials are equal. Suppose that the two interacting systems are not yet in equilibriu m
.
Then the entropy of the combined system \l:ill not have reached its max i mum value bu t will still be increasing:
·
(9.27) Consider a state very close to an equilibriurr1 state and differing from it only in that there is a small excess of mass �'zA(>O) in s ubsystem A over the equ i l ib
rium value
nA,
and a correspondi n g defici� (>f n1ass
-
d11A in subsystem B. As
Sec. 9.3
The Gibbs Phase Rule
157
the combined system moves toward equilibrium, subsys tem A will give up its excess mass with an accompanying entropy change
(9.28) Simultaneously, the entropy change of subsystem B as a result of acquiring the additional mass will be
(9.29)
The temperature is the same in both expressions. Substituting Equations (9.28) and (9.29) in Equation (9.27) gives
or (9.30)
J.LA > J.LB·
Thus, in the approach to equilibrium, heat energy must flow from the hotter to the cooler body, volume will be gained by the body under lower pres sure at the expense of the other, and mass will tend to flow from the body of higher chemical potential toward the one of lower chemical potential. All of these tendencies follow from the second Jaw of thertnodynamics. •
9.3 THE GIBBS PHASE RULE Consider a system in equilibrium with k constituents in 7r phases
.
*
We note the
following: 1. Only one gaseous phase can exist because of diffusion . .
2. Several liquids can coexist in equilibrium if they are immiscible. 3. Several solids can coexist. 4. Only rarely do more than three phases of a given constituent coexist.
*The symbol here
with 3.14. .
.
.
1r,
traditionally used to denote the number of phases, is not to be confused
158
Chap. 9
The Chemical Potential and Open Systems
Let P.l be the chemical potential of the ith constituent in the yth phase. According to Equation (9. 1 1) the G i bbs functi on is
(9.31) where n? is the number of kilomoles of the ith constituent in the yth phase. For equilibrium�
(dG)r.P
=
0. so Equation (9.15) gives
1-L? dn? ..,..
=
0� 1· and P fixed.
i=] y=l
(9.32)
It also follows from the Gibbs-Duhem equation (Equation (9.13)) that for constant temperature and pressure.
(9.33)
t=J y= I
If we consider the special case of a closed �ystem in which mass
dn; of the ith
constituent is transferred from one phase t(> another with the total 1nass of the constituent unchanged.. then -
"
dll? --
:=:
0, i
=
(9.34)
1, 2. . . . k.
'Y =� 1
•
This equation is simply a statement of the conse rvati on of mass: No con stituent is created or destr oy ed. Suppose that there are two constituents ( i
(y
=
cr,
=
1. 2 ) and two phases
J3). Then Equation (9.32) becomes
and Equation ( 9 34) giv es .
(9.36) .
Combining these e q ua tion s gives •
(9.37) Since the differentials dn1u and £in2o are arbitrary, the coefficients must vanish. Hence (9.38)
Sec. 9.3
The Gibbs Phase Rule
159
In other words, the equality of the chemical potentials for the two phases of each of the constituents is the condition for equiljbriutn with T and P fixed. From the conservation of mass it follows that the number of kilomoles of each constituent is fixed provided no chemical reaction occurs. Thus
·
(9.39) where
n1
and
n2
are the given nu·mber of kilomoles of constituents
and 2,
1
respectively. It. is convenient to introduce for each constituent the concept of the kilomole fraction, the ratio of the number of kilomoles of a given phase to the total number of kilomoles of all constituents in that phase: .
. .
(9.40)
-
From Equations (9.39) and (9.40), it is evident that
(9.41) Therefore, in our example of k
=
2, 1r
=
2, the state of the system is detertnined
by four independent variables: T, P, x1a, and x2a. tlowever, since there are two equilibrium conditions ( Equation (9.38)) , the ttumber of variables is reduced to two. The generalization of this specia l case leads to the so-called Gibbs phase rule. For a multiconstituent, multiphase system with T and P fixed, the condition for equilibrium is II. Q
r-i
-
·
ll.f3 - ll. y r-i
- ,.-,
-
.
•
=
.
J.L/r'
i
1' 2,
=
.
.
.
k.
( 9 42 )
7T.
(9.43)
.
The kilomole fractions are:
xY l
=
n;·y
k
i= 1
'
i
=
1' 2 ..k; .
1'
=
1' 2
...
,,Ir •
Thus, there would b� k1r k ilomole fractions were it not for the identity k
x? i
=
1
=
(9.44)
1,
which subtracts 1r kilomole fractions from the total. Counti ng the two prescribed auantities. T an d P. there are therefore 2 + k1r
-
1T
inde oenden t variables
The Chemical Potential and Open Systems
Chap. 9
160
and k ( 1T
l) equilibrium conditions (Equation (9.42)) . The number of
-
remaining .. degrees of freedom," called the variance f, is
f
==
[2
+
k7i- 1T]- [k(1T-
J)l
or
1·
=
k
-
1r +
(9.45)
(no chtmical reaction).
2
•
This is the usual form of the Gibbs phase rule. If f
f
==
=
0. the system has zero variance and is completely determined. If
1, the system is monovariant and is not determined until one additional
variable is specified. Consider the following examples: 1. A homogeneous fluid with one constituent and one phase. Here k 71'
=
1
�
f
=
=
1,
2. Thus we can choose T and P arbitrarily. This is the case of.
an ideal gas� for instance. 2. A homogeneous system consisting ot two chemically different gases in
the same phase. Here k
=
2,
r.
=
I, f
=
3. Here we can choose
T, P. and
one kilomole fraction. 3. Water in equilibrium with its saturated vapor. In this case k
f
=
1. Thus we can choose only the temperature
trarily. Equilibrium is defined by
a
=
1,
7T
==
2,
(or the pressure) arbi
p<")int on the vaporization curve for
water; specification of either Tor P gives the value of the other variable. 4. Ice, liquid water, and water vapor in equilibrium. Here k
f
=
=
1,
1T =
3,
0. The three phases can coexist in equilibrium onJy for the fixed set
of values T and P defining the triple point.
9.4 CHEMICAL REACTIONS
As we have noted, the chemical potential is a measure of the chemical energy per kilomole that a substance can generate in a reaction. Chemical thermodynam ics involves, in part, the measurement of th e chemical potentials of compounds. A chemical reaction is usually described by an equation, such as
(9.46) Such an equation can be generalized
as
m
(9.47) •
Sec. 9.4
Chemical Reactions
161
which states that a certain number of molecules of the initial reactants will com bine to yield a certain number of final product molecules. Here the Mi denote chemical symbols and the vi are so-called stoichiometric coefficients. The coeffi
ciettts are positive or negative integers; negative values represent initial reactants ·and positive values correspond to final products. Thus, in the previous example, and
vn,c) -
·=
+ 2.
For a chemica! reaction to occur, more than one substa11ce must be present either initially or fmally. The number of kilomoles ni of each of the constituents will change in a manner consistent with the conservation of atomic species. In
the example, for every two kilomoles of hydrogen and one kilomole of oxygen that disappear, two kilomoles of water will appear. The change in the number of kilomoles will therefore be proportional to the stoichiometric coefficients:
Suppose that a chemical reaction takes place under conditions of con stant temperature and pressure. This would occur when , say, the reactants are immersed in a water bath at atmospheric pressure. However, it is not neces sary that the temperature and pressure be constant throughout the process. It is sufficient that they return to their initial values after the reaction happens, which is virtually always the case. Under these Ctlnditions, the Gibbs-Duhem equation (Equation
(9.13)) states that the chemical potentials of the reactants
are fixed. Then, at equilibrium
(dG)r.P
=
0, so that
.
J
where the �j's are constants. Finally, since dnioc"i' we can write
•
(9.48)
J
for chemical equilibriztm in a reaction. Once initiated, most chemical reactions proceed spontaneously either until some reactant runs out or until the reac tion arrives at equilibrium. If the reaction is being used to determine the chem ical potentials of the constituents, the experiment will be designed to reach equilibrium. In the reaction of Equation
(9.46), we obtain the useful result (9.49)
This relation can serve as a check on the measured values of the individual p,'s.
The Chemical Potential and Open Systems
Chap. 9
162
9.5 MIXING PROCESSES
In Section 9.2 we considered the problem of particle exchange between two phases of a given substance. Here we shall address the phenomenon of the mixing or
interdiffusion
of two different gases. We assume that the mixing
takes place at constant temperature and pressure.. so that the Gibbs function is again a central concept in the description of the process. Dalton's law of partial pressures states that the pressure Pj of the jth con stituent gas of a mixture of gases is given by (9.50) where xi is the kiJomole fraction of the jth constituent gas defined in Section 9.3 .. and Pis the pressure of the mixture.* Equation (9.50) is a definition of the par tial pressure; the relation foiiO\\IS immediately from the ideal gas law for which
�
n·RT 1 v
=
p
.
:=
nRT v
�
since then
l!liaat . tl ,
p. J
=
p 11
-
.
·'
The Gibbs function for a mixture is (9.51)
J
where gj is the specific Gibbs function for the jth constituent. To calculate the
specific Gibbs function for an ideal gas we start with the first law expressed in the forrn Tlls
With
v =
cpdT
- ·
vdP.
RT/P, we obtain ds
•
=
= c
P
dT T
-
dP Rp.
We omit the superscript on the kilomole fraction because there is only one phase in this
process. Here
Ln, J
=
n.
the totaJ number of kilomoles of the system.
Sec. 9.5
Mixing Processes
16]
Integration gives s
Cp In T - R In P + s0•
=
Since g
= u
+ Pv
-
Ts
=
h
-
Ts,
and
it follows that g
=
cpT - cpT ln T
+
Kf In P
-
Ts0
+
h0,
or g
=
RT(ln P
cp),
+
(9.52)
where c/J is a function that depends on T only. We can now calculate the change in the Gibbs function as a result of the mixing of two ideal gases (Figure 9.3). The initial Gibbs function (before mixing) is
(9.53) where gli
=
RT(ln P +
1),
Ku
=
RT(In P +
cPz),
so that
(9.54) Figure 9.3
The mixing of two ideal gases. (a) The initial state; the gases are at the same temperature and pressure and are separated by a diaphragm. (b) The final state; the diaphragm has been removed, allowing the gases to mix. The final temperature is T and the final pressure is the sum of the partial pressures.
�
·
: initial state
i
T,P
T,P
diaphragm (a)
final state f
T, P
=
P1
+
P2
(diaphragm removed) (b)
The Chemical Potential and Open Systems
Chap. 9
164
Although c/>1 and c/J2 are functions of the same temperature T, the functions themselves are different because
Cp
will in general not have the same value for
the two gases. The final value of the Gibbs function is
with
Since T doesn �t change in the mixing process, c/>1 and c/>2 will be the same for the final state as for the initial state. Using P1 g11
=
RT( ln P + 1 +Inx1)�
g21
=
=
x1P and P2
=
x2P, we obtain
RT(ln P + c/J2 +Inx2),
so G1
=
n1RT(In P
+
1 +In x1) + 1z2RT(ln
P
+
4>2 +In x2).
(9.55)
Suppose that we define J.l
=
RT(ln P + 4> + lnx) g
==
+
RTin x.
Then
Comparing this with Eq u ati on (9.11) we see that J.L is the chemical potential! Since g1 and g2 are characteristics of the constituents, they are the same for the initial and final states and we can drop the subscripts on gin Equation (9.53), which becomes
(9.56) Our final result is therefore �G
=
G,. - G,
�
nt (p.1 •
-
g,) + n2(J.l2
-
g2)
•
or
(9.57)
Sec. 9.5
Mixing Processes
165
�s
==
-
a(aG) aT
•
p
Hence (9.58) Clearly
as > 0, as must be the case.
Entropy increase is to be expected when two different gases are mixed. But what if the two gases are the same? Does the removal of a diaphragm sep . aratmg the two halves of a volume V filled with one kind of gas change its entropy or not? Then
with
so that
and LlS
=
nR In 2 > 0.
(9.59)
Sttppose that the properties of gas 1 approach those of gas 2. The pre\'l ous result gives a finite entropy increase. However, if the two gases are identi cal there can be no change in entropy when the diaphragm is removed ( LlS
=
0). This is the Gibbs paradox.
TI1� resolution will be fully discussed when we treat statistical thermody namics. Suffice it to say now that there is a positive change in entropy when different, distinguishable gases are mixed. The transition from differettt to identical gases is not a continuous change, so Equation (9.59) does not apply to the case where the gases are the same.
The Chemical Potential and Open Systems
Chap. 9
166
. .
:
• '
•
.c
.
:
• .
.
•
•
.
.
.
.
•
.
'
:
.
:
:
•
.
.
.
. .
.
.
.
.
•
•
.
.
.
.
.
•
.
•
•
.
•
.
PROBLEMS
9-1 For a one-component open system tlU
==
.
TdS
PcJV
-
dn.
+ 11
(a) Use Eule r s theorem for homogeneous functions to show that ·
U
(b) Prove that (dG) r.r
==
:-::
TS
-
PV
+ JLn.
J.Ldn or� equivalently, aG
J.l
fin
=
.
-:-·-
r.P
(c) Prove the alternative definition of the ch e nt ical potent i al iiF
-:--
li=
dn
f.\·
.
9-2 (a) E xpress the chemical pote n t ia l of an ideal gas in terms of the te 1n pe rature T and the volume V.· •
J.L
==
crT
-
<·,.1· In T
(Hint: find the e n t ro py S
-
RJ' 1n V
-
snT
�·( ·r. � ): use G .
=
.
=
+ constant.
U
-
TS
+
Pl' and write
Gjn.) (b) S imilarly. find 1-L in terms of T and P. Sho\v that the chemical po1entia1 at the fixed temperature r· v aries with press ur e as J.1. =
J.L
-=
JJ..o +
( p\ RT In P,
.
u
\\'here J.Lo is the value of J.l at the rcfc_!rcncc poi n t ( P.,. T). This
ex p res si on
is
of great use in chemistry.
9-3 A con t a i ne r of volurne \t' is divided by p artit i on s into three parts containing one kiJomole of helium gas two kilomoles of neon gas. and t hree kilomoles of argon .
gas, respectively. The te m pera t ure of each gas is initiaHy 300 K and the pressure is 2 atm. The p artit ions arc removed and the gases diffuse into one a n other
.
(a) Calculate the kilomolc fraction and the partial pre ssure of ea ch gas in the mtxturc. •
(b) Calculate the cha n ge of the Gihhs function and the change of the entropy of the system in the mixing process. 9-4 Show that for
a
closed system consisting of t\\·o phases coex isti n g in equilibrium
at a temperature T and under
a
p ressure P.
;,p i�V
T dP
==
--
�
\
.
Here dP/dT is the s l ope of the ph as e equ il i brium curve.
Problems
167
9-5 A mixture of gold :1nd thallium can exist in equilibrium with four phases pre sent: solution, vapor, solid gold, and solid thallium. What is the variance? (This point is kno\vn as the
eutectic point.)
9-6 The Gibbs phase rule can be generalized to systems in which there occur
r
chemical reactions: f
=
k
1T
-
-
r
2.
+
Determine the number of degrees of freedom at equilibrium of a chemically reactive system containing solid sulfur
S and the three gases 02, S02, and S03•
The elements S and 02 appear in the reactions
S
3 �02
+
-+�
S03.
-
9-7 (a) Show that as
=
dP
v -
. T
-·-
H.n
(b) For an ideal gas, the entropy S(H, P, n) has the form
S(H, P, n) where
(H,
=
nR In
,
,
n) is a function of the enthalpy and the number of kilomoles.
Use the result of part (a) to derive the equation of state.
9-8 A container is initially separated by a diathermal wall into two compartments of equal volume. The left compartn1ent is filled with 1 kilomol� of neon gas at a pressure of 4 atmospheres and the right with argon gas at 1 atmosphere. The gases may be considered ideal. The whole system is initially at temperature T 300K, and is thermally insulated from the outside world. Suppose that the =
diathermal wall is removed.
(a) Wl)at is the nev• temperature of the system? l11e new pressure? (b) What is the change in the Gibbs function of the system? (c) What is the change in the entropy of the system?
9-9 (a) Show that for an open system with one component,
dG
=
-SdT
+
VdP
+
JLdn.
(b) Using this result. calculate G for a van der Waals gas� assuming a fixed amount of material at a given, fixed temperature. Show that
G
=
-nRT ln(V-
nb)
+
n2bR 2n2a V nb- V
+
C(T),
_
where the integration constant C(1) is, in general, different for different temperatures.
9-10 Show that, during a first-order phase transition: (a) The change of entropy of the system undergoing the transition is a linear function of the volume change.
•
The Chemical Potential and Open Systems
Chap. 9
168
(b) The change of internal energy is given by AU
d In T
L 1
=
d In p
-
.
where L is the latent heat of transfom1ation.
9-11 (a) Sh ow that the Gibbs-Duhem equation 1eads to the relation
L x;dJ..l.;
=
0
( 7. P constant)�
.
I
where
X;
6
is the kilomole fraction of the ith component.
(b) Sh ow that for a two-component system -
.
-
d )n
Xl
.
dIL2
-·
--
T.P
(c) Show that for a two-component liquid phase whose vapor can be treated as an ideal gas mixture, .·
d In P... L
-
d In
x1
r.P
.. d In
,
x2
r.t
where P1 and P2 are the partial pressures of the components of the vapor,
and
x1
and
x2
are the kilomoJe fractions of the liquid. (This is known as the
Duhem-Margules
e
q ua tio n ) .
•
I •
0
ICS
I 0.1 Statements of the Third Law
171
I 0.2 Methods of Cooling
174
I 0.3 Equivalence of the Statements
175
I 0.4 Consequences of the Third Law
178
'
169
·
· ·
.
.
.
.
·
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·
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-
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I 0.1 STATEMENTS OF THE THIRD
·
· ·
·
.
·
. · . .
· .
.
:
·
. .
. .
LAW
The third law of thermodynamics is concerned with the behavior of systems in equilibrium as their temperature approaches zero. The definition of entropy given by
Tf1Q '+So S= 0 T
(10 1 ) .
is incomplete because of the undetermined additive constant
S0, the entropy at
absolute zero. In this short chapter we shall introduce a principle that will enable us to determine
S0• The
principle, which was discovered by Nernst in
1906, is
often referred to as Nemst's postulate or the third law of ther1110dynamics. As long as we deal only with differences of the entropy, a knowledge of
S0 is unnecessary. However, the absolute entropy is by no means unimportant. Suppose, for example, that we wish to determine the change in the Gibbs func tion. Since
dG = -SdT + PdV, it is not suffjcient to know dS; we must know
S itself. In chemistry, moreover, knowing the absolute entropy
makes it possi
ble to calculate the equilibrium constants of a chernical reaction from the ther mal properties of the reactants. We have seen that a spontaneous process can occur in a system at con stant temperature and pressure if the Gibbs function decreases. The Gibbs function is related to the enthalpy by the equation reciprocity relation
S = - ( aGI aT) p,
G = H - TS.
Using the
we obtain an expression known as the
Gibbs-Helmholtz equation:
aGG=H+T aT
•
.
p
(10.2)
If this relation is applied to the initial and final states of a system undergoing an isothermal process, it takes the form
(10.3) 171
The Third Law of Thermodynamics
Chap. I 0
171
-
--·
-.----
�H
Figure 10.1 Variation of .lG and tlH in the vicinitv of absolute zero. •
L.....----�
T
which shows that the change in enthalpy and the ch ange in the Gibbs function are equal at T
=
0 for an isobaric process. Based on the results of experi
ments, Nernst postulated that as the temperature tends to zero .. not only do
llG and llH become equaL but their temperature rates of change approach
zero as well: A(�G) lim r-t, AT
=
()
ri(�H)
lim
,
iiT
T-+0
I'
=
0.
( 10.4)
p
This behavior is illustrated in FiQure 1 0.1 . We may write the first of the two equations as .
.....
iiGl
lim
ac1
-
-
il1'
T -+0
==
-
·-
aT
p
r
0.
( 1 0.5)
where the subscripts 1 and 2 refer to the initial and fi11al states, respectively. From the reciprocity re]ation� we have
J1jm, ( 5' 1
-
-+I,
5� ) ·= 0.
( 10.6)
This is the Nernst formulation of the third law : All reactio1zs in a litJUi{/ (Jr solid in tl1er1nal equilibriutn take place lvith no change o.f entrop_r in the J1e(�hh£JriJlJlJtl o_f'a/JsrJ/ute zero.
Although the Nernst theorem is quoted for solids
and liquids, it is believed to
hold for all sy ste ms in equilihrium states. (ll1ere arc certain quantum systems •
that constitute gas-like
aggregations at absolute zero.)
In 1911 Planck extended Nernst's hypothesis by assuming that it holds for G1 and G2 separately. Specifically, he prOJ.1osed that lim G(T) r�o
==
lim
T �o
lf(T),
(I 0. 7)
Sec. I 0.1
173
Statements of the Third Lav1
and
I.1m iJG
r-.o
.
Ilffi
=
aT p
ali
aT
r-.o
p
(10.8)
.
These statements lead to an important conclusion. For convenience, we tem porarily introduce a variable
G
T dG iJT
H. Equation (10.2) then becomes
-
-= 0.
(10.9)
P
By Equations (10.7) and (10.8), we have in the limit when
Adding the term
-
=
0
a
and
=
iJT
T
=
0,
0.
(10.10)
P
T ( aHI aT) P to both sides of Equation ( 10.9), we get T
d
_
, aT p
=
_
dH
T
ar p'
or
a aT p
-
-
T
--
ctJfT
aH aT
•
P
T = 0 is the same as the (iJ/ aT)p, which is zero, according to Equation (10.10). Therefore,
By L'Hopital's rule, the limit of
lim
r_.o
for
aH aT p
==
0.
( 10.11)
limit of
(10. 12 )
Finally, by Equation (10.8), we obtain the result
G lim .a r-.o aT p Since
(iJGfiJT)p
=
=
0 .
(10.13)
-S, it fqllows that lim S
T-+0
=
0.
(10.14)
This is Planck's statement of the third law. In words:
The entropy of a true equilibrium state of a system at absolute zero is zero.
Chap. 10
174
The Third Law ofThermodynamics
This statement. stronger than the Nernst theorem. needs qualification. Certain
glasses have a nonvanishing entropy as the temperat�re tend� to absolute zero. This makes sense.. since statistical thermodynamics assoctates entropy
with disorder. and glass is a disordered structure. The Planck statement can be shown to be valid for every pure crystalline solid. Quantum statistics is necessary for a complete understanding of the absolute entropy of a system. Still another statement of the third law is: It is inlplJSSible
(0
retlllc.:e the le111perature oj'a swvstet11 to absolute zero
llSi11g a finite nttnzber lJfprocesses.
In Section 10.3 this so-called unattainability principle will be shO\\'n to be equivalent to Nernst"s postulate.
.
.
.
.
..
'
'
.
.
.
.
.
.
:
.
. .
.
. .
.
I 0.2 METHODS OF COOLING A widely used method for cooling
a
substance is to isolate it from its sur
roundings (with no heat flow in or out) and reduce its temperature in an adia batic reversible process. Then work is done J,y the system solely at the expense of its internal energy. This can be accomplished by varying some parameter such as the magnetic field. The process is called adiabatic demagnetization and is discussed in Chapter 17. In the 1970s laser coolitlg was proposed� a process in which the decelera tion of a beam of atoms is accomplished t'y directing
a
laser beam so as to
oppose the atomic heam. Under the action of the laser beam, momentum is transferred to an atom by absorpti<>n of a photon. The laser light frequency is chosen to excite the lowest order at(lmic transition. If the laser frequency is lower than the atomic resonant frequency and the atom is moving against the laser beam. the laser frequency in the rest frame of the atom is Doppler shifted , toward resonance. The key point is that the laser light ·�pushes on , <>nly those atoms that are moving into the laser beam. -
Following absorption , the atom spontaneously emits the photon in a random direction. This absorption and isotropic reradiation result in an average driving force in th e direction of the incident light. Since the force is pr(JportionaJ
to the velocitv, the atomic ffi(ltion is slov.'ed and the atoms are therebv cool ed . �
In this process, the atomic vapor is cooled without becoming
�
a
liquid or a solid.
Various methods, including laser detuning, hav e been used to compen sate the changing Doppler shift as the atorns decelerate so as to keep them near resonance. Three-dimensional cooling results if ()rthogonal pairs of laser beams are used in a so-called h<.>ptical m<.>lasses" configuration.
A thorough understanding
Sec. 10.3 I O.l
Equivalence of the Statements
175
EQUIVALENCE OF THE STATEMENTS
By considering the adiabatic reversible process, we can show in a few �teps the equivalence of two statements of the third law. We assume that absolute zero is unattainable and prove that in a change from equilibrium state
1 to equilib
rium state 2,
(10.15) where the naught subscript refers to T
=
0. We can examine the process in the
entropy-temperature plane. The shape of the S versus T curve can be found from the relation
S
=
TztQ
So +
T
0
r
(10.16)
.
•
Now Cv
=
(71QfdT)v
and Cp
=
(71Q/dT)p. We assume
that either P or Vis
held constant and drop the subscript. Thus
S =So+
T dT 0
(10.17)
CT .
We note that the Debye law for the heat capacity of a solid gives Cva.T3• In this case, S increases \Vith T according to some power law, as Figure 10.2
reflects. The curve S1 applies to an initial value of the variable parameter and s2 to some final value.
s •
Figure 10.2
Entropy temperature diagram for a hypothetical cooling process in which S1 < 52• (The entropy curve for the initial value of the varied parameter lies below that of the final value.)
~
____
_
L 1 I I I I l
_ ____ ____
I I I I t I I T
Chap. I 0
l76
The Third Law ofThermodynamics
Cooling from T1 to T2 by an adiabatic (and isentropic) reversible process occurs along the horizontal line connecting the sl and s2 curves. Then s2
=
sl
gtves •
(10.18)
Transposing terms.. we have Tz
dT
(10.19)
()
Now, Cis always positive: if it were negative, a heat gain would result in a low ering of the temperature.. and the attainment of absolute zero would be triv ially easy!
501) > 0, as shown in Figure 1 0.2� and because C �> 0, there If ( S02 must be some temperature Tj for which the right-hand side of Equation (10.19) is zero. Then -
( 1 0.20)
where
T� is the final temperature corresponding to T·,.
But
T� must be zero if
Equation (10.20) is to hold. Thus absolute zero could be achieved through the · adiabatic change from state 1 to state 2 if 50� > S01• This violates the unattain ability principle. We conclude that
(10.21) Thus we need to redraw the 5- T di a gram with S02 abat s2
=
sl gives Equation
r2
0
dT C,- J1·- -
0 there is some value of 1"2 hand side of Equation ( 10.22) is zero. Then ..
5.,1 (rigure 10.3). The adi
(10.18) as before .. \Vhich can be rewritten as
•
If (SoJ - So2) >
<
( 10.22)
say T2"
such that the left
-
Sec. I 0.3
Equivalence of the Statements
177
s
•
-
Figure 10.3
Entropy temperature diagram for a hypothetical cooling process in which sl > s2.
- - ---------------
S'r.1 _____ L_________ I f I I I t I I I I I r1"
I I I I I I I
--��----�1�1---.�
�----· =
o
r1
T2" T2
T
0. This also vioiates the unattainability principle and we conclude that for this case requiring that T1"
=
So2 >Sot·
(10. 23)
The only way that Equations (10.21) and (10.23) can both be true is if
(10.24) which is the Nernst postulate. According to Equation (10.24)� the curves of �� 2 and S1 versus T must come together at T- 0. The final sketch, Figure 10.4, nicely illustrates the eq�uivalence of the two statements of the third law. The curves have been drawn to reflect the Planck statement, though the proof does not place the entropy at absolute zero. A series of isothermal and adiabatic processes (demagnetizations) are represented by the zigzag path in the figure. Each successive process reduces the temperature. It is clear, however, that because the curves intersect at
s
Figure 10.4 Entropy tetnperature diagram for an actual cooling process. The unattainability of absolute zero is illustrated by the indefinitely increasing number of steps required to achieve a given temperature reduction as absolute zero is approached.
T
•
T
=
The Third Law of Thermodynamics
Chap. I 0
178
0, an infinite num ber of steps wouJd bt required to reach absolute zero.
\\7e conclude tha t it is impossible to achieve
te m peratu re of absolute zero by
a
any practical method. This fundamental fact of nature should not be espec i al l y disquieting, however. given that cry(lgenics has enabled us to get ex t re m ely close to T
=
0.
I 0.4 CONSEQUENCES OF THE THIRD LAW
1. Expansivit.'· The expansivity (coefficient of volume expansion) is defined as
1
av
From Section 8.5 we have the Max\\'ell relation
' -
--
·-
iiT
-
-
-•
, (J p
I'
r
The Nernst postulate implies that the change in e nt ro p y is zero in any process at absolute zero: that is.
lim /'_,.()
(J�� --
=
-
·jp
(1
0.
r '
Since the volume V remai n s finite as T Jim [3
T --+(l
==
·
�
0, it is therefore true that
0.
( 10.25) •
We conclude that in the case of all kno\\'n solids, the expansivity approaches zero as the temperature approaches absol ut e zero.
2. Slope of the phase transf'or11IlllirJn curt·cs •
In con n ect i on with any phase change that takes place at low ten1perature we can invoke the Clausius-Clapeyron cquat !_<)n (Section
tl p
--
8.7):
•
==
-- ---
The Nernst theorem states that
lim �5
T--+0
=
0. ,
Since � V is not zero for so-caJJed first·order transitions.. it follows that
Problems
179 dP . I1m T --+0 dT
=
0.
(10.26)
The slope of the boundary between two phases is zero at absolute zero. This has been verified for all known sublimation curves, for the vaporiza tion curve of Helium II, and for the fusion curve of solid helium. 1 ·Experimental results show that for solid helium, dP/dT'XT , which approaches zero at a very fast rate indeed. 3. Heat Capacity From the definitions of the heat capacities at constant volume and con stant pressure, we have
Cv = T
as
Cp = T
ar v'
as ar p
.
Integrating, we obtain T
T
S- S0 =
dT Cv , T 0
S- S0 = 0
dT
Cp T.
The left-hand sides of the equations will remain finite down to absolute zero, where, by the third law, they will vanish. The right-hand sides must • •
behave in the same way. The integrals show that the heat capacities must decrease to zero at least as rapidly as T. Othenvise the integrals would diverge because of the lower limit of T = 0. We conclude that limCv T --+0
=
0,
lim Cp T -+0
=
0.
(10.27)
That is, the heat capacities approach zero as the temperature approaches absolute zero. With respect to this behavior, extensive experiments pro vide a full confirmation of the third law.
PROBLEMS
10-1 Show that the heat capacity at constant volume can be written Cv It follows that as T
•
0, InT
• -X)
as =
dInT v.
and S
•
0, so that Cv
Equation (10.18). By the same argument, Cp
•
0 as T
•
•
0.
0, in confirmation of
The Third Law of Thermodynamics
Chap. I 0
180
10·2 Show that. in view of the third law. neither the id e al gas equation nor the van 0. der Waals e quation can hold at T =
10·3 (a) Show that the third Jaw cannot pr�dict the value of the isothermal com pressibility K at absolute zero. (It remains finite.)
(b) The isothermal bulk modulus B is the reciprocal of K. 8
aP - v -av
=
7
.
Show that
riBlim T o dT .....
�-
0.
=
te-4 According to De bye's theory of heat capacities. to be discussed in Chapter 16. th e specific heat capacit y of an insulating solid at very low temperatures is given by
12
c,.
where
c,.
=
5
1r.t
T
R
-- · -
=
0.01 K and T
=
.
tJD
is in the usual units of J kilomol·..!
such a solid at T
.�
K 1• Find the specific entropy of I OK, given that 8D 300K. 1
=
10-5 (a) Show that if c\. hTU at low tempt·ratures. the third law requires that a> 0. .. (b., If ( \.' aT + hT·-"' at lo\\· temperatures, calculate the variation of the =
=
entropy with temperature.
10-6 Consider a solid whose equation of state is PV where
�
.f(\1).
=
AU.
f ( V) is a funct io n of the volume nnlv and A is a constant. Show that
•
•
Cv •OasT �o. 10-7 A low-temperature physicist \\'ishes
to
publish his experimentaJ result that the
heat capacity of a nonmagnetic dielectric material between 0.05K and 0.5K varies as AT112 publication?
+
BT3• As editor of the journaL should you accept the paper for
ases
I 1.1 Basic Assumptions
183
I 1.2 Molecular Flux
186
I 1.3 Gas Pressure and the Ideal Gas Law
188
I 1.4 Equipartition of Energy
190
11.5 Specific Heat
191
acity of an Ideal Gas
I 1.6 Distribution of Molecular Speeds
193
I I.7 Mean Free Path and Collision Frequ en cy
198
I 1.8 Effusion
201
I I.9 Transport Processes
203
Ill
11.1 BASIC ASSUMPTIONS In the nineteenth century, scientists realized that in order to go beyond the limitations of classical thermodynamics, we must take into account the structure of matter_ The recognition that bulk matter is composed of parti cles led to the kinetic theory, which assumes that atoms and molecules obey the same laws of mechanics that apply to macroscopic systems. This assump tion was ultimately challenged by quantum theory and, indeed, statistical thermodynamics is best treated from the quantum mechanical point of vtew. •
Nonetheless, kinetic theory yields far deeper insight into such concepts
as pressure, internal energy, and specific heat than classical therttlodynamics provides, and is therefore well worth investigating. In addition, it gives
an
explanation of transport processes such as viscosity, heat conduction, and dif fusion that is quite satisfactory for most purposes. Kinetic theory, as the name implies, is concerned witl1 the motion of mol ecules rather than the forces between them. It is therefore especially suitable •
for describing the behavior of gases at relatively lo\V pressures.
The. fundamental assumptions of the kinetic theory of gases are: 1. A macroscopic volume contains a large number of tnolecules. What do we mean by this statement? We know that Avogadro's number is
NA
=
6.02 x
lc¥6 molecules kilomole-1
and that at standard temperature and pressure (STP),
ums down to 10-11 atmospheres of pressure, at ordinary temperatures there are still many millions of molecules in each cubic centimeter.
2. The separation of molecules is large compared with molecular dimensions and with the range of intermolecular forces.
183
The Kinetic Theory of Gases
Chap. II
184
Take a volume of 22.4 m3, divide it into cubical cells with one molecule in each ce)J. The volume of each cell (at
3 22.4 m f6.02
is
STP)
�<
2fl 0 1
!!:::;
30 x to--27m3• Thus the distance between molecules in this case is equal
to the cube root of this volume.. or roughly 3 X Io-·"J m. This is a factor of about 10 greater than the diameter of a typical gas molecule.
We must also consider the range of the forces of interaction between molecules.. which
._
are
electrical in nature. The force between two mole-
cules is strong and repulsive at separations approximately equal to the
molecular diameter and is associated with the overlap of electron clouds.
At greater distances the force is attractive and is due to induced electri
cal dipoles. The behavior is often characterized by the total potential energy d
12 -
r
d
6
(11.1)
•
r
where fl.£ and d are empirically determined parameters. Equati<>n
is known as the Lennard-Jones. or
··6--12""
potential. and is pJotted in
Figure 11.1. The curve has a minimum at V ( r)
=
( 11.1)
fd
r
=
2 Lh
=
1.12..
where
-�E. To the left of the minimum. the curve is very steep; to the
right it is comparatively flat. Typically. �E lies in the range of 2 x 10 4 2 eV, and d v aries between 2.2 x 10 -Jn m and 4.5 x 10-10 to 5 X 10eV
m. For molecular separations of an order of magnitude greater. the inter action force is evidently negligibly small.
In a liquid or solid, the molecules have nearest neighbors at a distance
r
only slightly greater than d. Thus there is substantial interaction between them. In contrast. the molecules of a p.as under normal conditions are
most of the time traveling freely through space. An ideal gas is a good 10
I
6 \l(r)i.1E
�-
4
, -
Figure 11.1
Variation of the
intermolecular potential V ( r)
with the intermolecular distance r for a pair <>f gas
molecules.
() _.,
'
-
--
\.
-
-J.o
1.0
1.4
rltl
l.R
Sec. I 1.1
Basic Assumptions
185
approximation at these low densities; dry air at
STP deviates from the
. ideal gas law by iess than 0.1 percent. For high density gases, on the ott,er hand, it is essential to include the intermolecuiar forces, as
is done in the
van der Waals model. 3. No forces exist between molecz,les except those associated with collisions.
That is, molecules move in straight-line paths between collisions with constant velocities.
Each molecule moves forever in a zigzag path, for it
collides many million times per second with other mo�ecules of the gas or with the molecules that make up the walls of the container. 'lltese col lisions are so brief that in our analysis we entirely ignore their duration. F urtherrnore, the distance a molecule tnoves between collisions can lie between zero and thousands of times the molecular diameter. However, �he very short and very long distances are unttsual. The average distance
between collisions is called the mean free path. In oxygen at
STP, the
mean free path is about 10-7m. 4. The collisions are elastic.
Both momentum a11d mechanical energy are conserved in the collision
process.
S. The molecules are uniformly distributed within a container. •
6. The directions of the velocities of the molecules are uniformly distributed. The last two assumptions characterize the randomness of the motion. The basic idea of kinetic theory is that the n1olecules of the gas are moving rapidly and randomly, colliding \Vith each otl1er and with the walls of the con
tainer, thereby creating
a
pressure due to their countless impacts. We are con
cerned with a gas that is in thertnal equilibrium at a fixed teu1perature and in mechanical equilibrium with the walls. It follows from these assumptions that
the average molecular speed and the probability distribution of speeds are unchanging. v +
Let f( v )dv be the fractional number of molecules in the speed range v to
dv. The probability density function f ( v) has the dimensions of v- 1 since •
f(-v )dv
=
1.
The mean speed is given by X
v=
-
·vf( v )dv,
(11.2)
and the mean square speed is V"' .,
=
(11.3)
Chap. 'I I
186
The Kinetic Theory of Gases
-
The square root of
v2
is ca11ed the root mean sqt1are or rms speed: ( 11.4)
The nth moment of the distribution is defined as
We shall also be interested in the most probable speed, obtained by differentiat ing f ( t') and setting the derivative equa1 to zero. The speed distribution wiJI be derived in Section 1 1 .6. We can develop the fundamental aspects of the kinetic theory, however, without knowing the exact form of the distribution function.
11.2 MOLECULAR FLUX
In many applications of kinetic theory.. we need to know the molecular flux, the number of molecules striking a unit area per unit time. Consider a slanting cylinder of slant height vdt at angle fJ m ea su red with respect to the normal to the elementary area dA (Figure
11.2). We want to find the number of mole
cules that strike the area dA in time dt. z /
'
/
/
/
/ I
udt
/
/
'
/
/
/
/
/
/ / .·
/
./
/
/ /
8
/
I
/ '
dA
/
,_.---.,.'---- . ---+-------... '
.......
I
.....
/cb
/
--<.... •
X
Figure 11.2
Slant cylinder geometry used to calculate the
number of molecules that strike the area dA in time dt.
\'
Sec. 11.2
Molecular Flux
We define n
=
187
NJV, the number of molecules per unit volume. Now, the
number of molecules with speeds v to v + dv in the cylinder is ��( v) dv X (volume of the cylinder). Since the volume of the cylinder is (d A cos8)( vdt), the number of molecules with speeds v to v + dv in the cylinder is n v ::ra
f ( r) dv
cos8 dA dt. Only a few of these molecules are traveling toward
dA. Let the velocity of each molecule be represented by a vector, and imagine that all the velocity vectors are moved parallel to themselves to a common ori gin (Figure 11.3). Assumptio11 6 states that the head ends are distributed uni formly in a spherical shell of radius v and thickness dv. The element of area of the surface of this sphere is
du
=
v2 sin8
d8 d
with
•
sphere
Therefore, the fraction of the total number of molecules with angular distribu tions 8 to 8 + dO, to
sinO d8 d
du ---
sphere
=
----
41T
du
•
•
The number of molecules in the cylinder at time t traveling with speeds v to
v
+
dv toward dA (i.e., the 8, 4> molecules) is dN
=
sin8 d8 dl/> . nv f(v)dv cos8dAdt
(11.5)
47T
;:s::
Figure 11.3
(a) Randomly oriented molecular velocities; (b) the same vectors referred to a common ortgtn. •
•
dv (a)
(b)
I
188
Chap. II
The Kinetic Theory of Gases
The corresponding particle flux is� by definition�
d
n
=
dA dr·
411"
Tr/2
sin8 cos8d8
vf ( v) dv
==
=
(11.6)
(11.5) in Equation (11.6) and integrating� we obtain
Substituting Equation
dN
0
(11. 7)
()
or
(11.8) where vis the mean speed, defined in Equation .
.
. .
.
.
· . .
..
.
·
.
.
.
. ·
.
·.
(11.2).
.
11.3 GAS PRESSURE AND THE IDEAL GAS LAW
Next we consider a single molecule colliding with a surface. The normal com ponent of the change of momentum is (Figure
mt' cosO
-
(
-
n1v
cosfJ)
1 1 .4)
=
2mv cos8.
The change of momentum dp produced by all the molecules with speeds
v
'
+
dv traveling toward an element of the surface dA dp
v to
in time dt is
2mv cosO dN
=
2
=
mv
cosfJ
•
ll1'f(t')d·vcos6 dA
dt
sin6 dO dQ> ·
----
4�
=
nmdA =
x
---
27T 1
dt
., lf. -
v2f ( v )dv ()
0
tic/>
sin8cos28 d9 ()
From Newton's second law and the defi11ition of the pressure P. we know that dp
dAdt
=
P.
•
189
Gas Pressure and the Ideal Gas law
Sec. 1 J.3
dA
6 ------
' I l I \ I
6
\ I I
Figure 11.4 Change in momentum of a molecule striking a surface element dA.
I
m"
mv
'
•
so that
P We note that
nm -=a
=
1 -nmif.
=
(11.9)
3-
N m/V. Then
PV
1 =
1Vmv
·-
3
2
2 =
-
3
1
2
(11.10)
.
-Nmv
2
The right-hand side is two-thirds of the average translational kinetic energy of the molecules of the gas. Equation (11.10) is highly suggestive of the ideal gas law
PV
=
nRT.
(11.11)
The number of kilomoles is simply related to N:
n
=
no. 0 f k.l 1 omo 1es
no. of molecules =
N _
no. of molecules kilomole- 1
-
-N-A'
where NA is Avogadro's number. Thus
PV
=
NkT,
(11.12) •
where
Here k is Boltzmann's constant; it is the ratio of two universal constants and is therefore also a universal constant. We see that Equation
(11.12)
will agree
190
Chap. I f
The Kinetic Theory of Gases
with Equation (11.10). the result we derived from kinetic theory� if we make the association
NkT
==
I -Nmv2 .. 3
or
(11.13)
Thus, the theory has provided a statistical interpretation of the absolute tem namely the te m pe ra tu re is proportional to the aver
perature of a dilute gas
..
age kinetic energy of a molecule. The r esu lt is one of the great insi ghts given to
us by the kinetic the or y
.
From Eq u ation (11.13) it is apparent that the mean kinetic energy is
independent of the pressure .. the volume.. or the molecular species. At a given
temperature, the mean kinetic energies of different gases are all the same.
It is instructive to note that at room ten1perature (293 K), .
3
(11.14)
\Ve can compa re this energy with �E. the depth of th e potential
I>�
well"' dis
cussed at the beginning of the chapter There we observed tha t � E has a mag .
nitude ranging from about 0.0002
eV
to ().05
e V.
I 1.4 EQUIPARTITION OF ENERGY Let the mol ec u lar velocity have cornponents ( vx� v2
=-=
+
v2,
vY, vz ) .
Then
v2Y + v2,.
since taking an average is a linear op e r at i on By assumption, there arc no pre .
ferred directions, so •
-2 V
X
-2
== V
y
::_:
-2 V
l
-=
I , - V". 3
Thus the mean kinetic energy per molecule aSS(lCiatcd with any one component the x-component, for example I
is
-- nzv.;.X
2
l
'\
=
1 - n1v
6
..
l =
2
·-
kT.
Sec. I 1.4
Equipartition of Energy
191
stnce •
1
-mv2 2
=
3 , -kr 2 ,
(11.15) .
It follows that the average translational kinetic energy associated with each
component of velocity is one-third th� totai and has a magnitude Each component represents a
of kT /2.
degree of freedom f of a mechanical system. An
atom or a monatomic molecule, for example, has three translational degrees of freedom so that f
=
3.
A diatomic molecule is modeled as tv.'o masses connected by a s pri ng
.
As such, it has seven degrees of freedom: three translational, two rotational, and two vibrational. There are two rotational degrees of freedom, because t\\·o angles are needed to specify the orientation of an axially symmetric mol
ecule. The vibrational motion is represented by· a one-dimensional harmonic oscillator. For simple harmonic motion, the energy is partly kinetic and partly potential; the average values of the two are equal, and both are expressible as quadratic forms. Thus these are two vibrational degrees of freedom. The equipartition theorem (proved in Chapter 14) states that (1) there is one degree of freedom associated with every form of energy quadratic fortn, and
(2)
of the energy of
the mean value
sponding to each degree of freedom is
expressible in
a molecule corre
kT/2. Thus the total average energy of a
molecule 8 is f kT/2. It follows that the internal energy of a gas must be the total average energy of its N molecules. That is,
(11.16)-. ..
or
u
u
=
n
=
f
1_RT.
(11.17)
11.5 SPECIFIC HEAT CAPACITY OF AN IDEAL GAS For a reversible process� the specific heat capacity at constant volume is given by
v
c
From Equation
au =
(11.17), this gives Cv
=
d f__RT T 2 d
The Kinetic Theory of Gases
Chap. II
192
TABLE
11.1
Ratio of specific heat capacities of various
values the velocity of so u nd in the gas.
gases at temperatures close to room temperature. The a re obtained hy measuring Gas
Gas
y
Gas
co,-
He
1.66
H"'
1.40
Ne
l.M
0"
1.40
NH�
Ar Kr
1.67 1.69
f\., co
1.40 1.42
CH_. Air
.
-
-
Mayer's equation for an ideal gas is
Cp
=
and the ratio of specifi� heats
f-R 2
a
+
R. Hence,
c,. ·+
=
j' + 2\
R=
-
2
R
,
is
y= For
cp
1.29 1.33 1.30 1.40
Cp C I'
=
f.
2
+
.
,
•
.
•
mo n ato mi c gas,
f
•
=
3�
c,.
3
= iR.
Cp
=
5
:;R. I-
5
'Y = 1 =
1.67 .
...
These results are in good agreement with mfasured values at tem per atures in the vicinity of room temperature ( Ta ble 11.1 ). For a diatomic eas.. '-
7
9
'Y = 7 =
1.28.
Here the agreement is poor: the number of degrees of freedom is too large. We try -
.f
=
5.
c,.
)
= 2_R.
c I'
==
7 -
·
2
R
..
y =
7 -
5
==
1.40.
Good agreement is obtained wi th this adjustment. Evidently� near room tem perature the rotational degrees of freedom
(Jr
dom are excited, but not blJth. Maxwell
C(lnsidered
the vibrational degrees of free
the resolution of this
q ue st io n the most pressing challenge confronting the kinetic theory. It took statistical thermodyn ami cs to provide the final answer (see Ch ap ter 15).
Sec. 11.6
Distribution of Molecular Speeds
193
Two other observations are noteworthy. For large molecules� f will be large and y � 1. In general,
1
<
}'
<
1.67,
in pretty fair agreement with experimental resu.lts.
The kinetic theory predicts that We have seen that this is not true
c.,
and
cr
are constants for an ideal gas.
at low temperatures
when
both must
approach zero. Kinetic 1 heory, which is a classical theory, is incapable of accounting for this behavior. A quantum mechanical description is required.
DISTRIBUTION OF MOLECULAR SPEEDS
11.6
We have yet to derive_ the probability distribution of molecular speeds from which various averages can then be found. In Chapter 14, the so-called Maxwell-Boltzman distribution will be determined from a knowledge of the distribution of particles among energy levels. We incl11de here a more
physi
cally satisfying derivation that makes reference to the molecular collisions
through which the speed distribution is maintained.* In a dilute gas, it is reasonable to assume that collisions are. binary events, involving only two molecules. Simultane,Jus collisions of more than two molecules are exceedingly rare. In a l'inary collision, the vel(lcity of parti cle 1 is changed from
v1 to v� and the
velocity of particle 2 is changed from
v2
to v�. (Here we use boldface symbols to denote vectors.) For identical point masses, conservation of momentum implies coplanar collisions, with +
Vr •
v2
=
v]
+
(11.18)
v2.
•
The assumptions of kinetic theory ensure that under equilibrium conditions, for each binary collision there is an inverse collision in which the original velocities are restored (Figure 1 1.5). The velocity vectors are collinear with the original set. If we let F ( v) range
v to
v
1 and 2 per
dv
be the number of molecules with velocities in the
dv, then the number of collisions occurring between molecules unit time will be aF ( v1) F ( v2), whe.re a is a constant of propor +
tionality. For tlte inverse collisions, the number wili be a' F ( v' 1) F ( v' 2). Because the velocity distribution is unchanged by the total of all the elastic collisiorts, the two rates must be the same. A.lso, the collisions will be completely equiva lent when viewed in
a
center of mass frame of reference, so a
*See. for exampJe. Thermal Ph vsic'i, York.l991. ..
by
M. Sprackling, A1nerican
=
a'. T hus
Institute of
Physics. New
The Kinetic Theory of Gases
Chap. II
194
I
th
,
..
•
(a)
v.,
I
-
v,
(a) A binary molecular collision; (b) the
I
Figure 11.5
•
(b)
corresponding inverse collision.
or (11.19) Since kinetic energy is conserved and .,
v
·
v
.,
.
==
l v 12
..,
.
=
.,
� v,
we also have
(11.20)
vl- + v2- = vl- -1- v2-·
The functional relation of Equation ( 11.19), with the constraint of Equation {! 1.20), can be solved by inspection.* A solution is F ( v)
=
A exp (
- crv
·
v),
(11.21) .
where A and
a
are constants. This is the Maxwell-Boltzman velocity distribu-
tion. To obtain a
speed
distribution, we let N ( v )d1) be the number of mole-
cules whose speeds lie in the range Figure
v
to
v +
dv, irrespective of direction. As in
11.3, we can visualize the velocity vectors as displaced parallel to them
selves with their ends at a common point. Since the directions are random, the velocity distribution has spherical symmetry. 'Therefore N ( ·v )dv is the number
*The solution can also be found hy Chapter 13.
the method of Lagrange multipliers� discussed in
Sec. I 1.6
Distribution of Molecular Speeds
195
of velocity vectors whose tips lie \Vi thin the spherical shell between radii v and
v
+
dv. It follows that
(11.22) The constants A and a can be found from two conditions. The total number of
particles N is given by
X.
N=
N(v)dv,
(11.23)
0 .
and the total energy of the molecules in the container is
3
X
1
(11.24)
0
To determine
a we
can divide Equation
Equation (11.22). The result
7.
3
r t a v4e dv �
v2N(v)dv 0
0
-
-
-
-
-
X
X
m
by Equation
(11.23) and use
is
X
3kT
(11.24)
v2e -av·dv
2 1 ) 8a 1T
(11.25)
2a
•
2 1 J 4a
--
0
0
3
-
..
t-J(v)dv
1T
(The values of the definite integrals are given in Appendix D.) Thus, a
=
mj2kT.
The constant A can be then found from Equation (11.23): X.
N (v)dv
N=
=
(11.26)
47T A 0
0
Hence
A=
a
�12
N
=
1T
In
--
-
2rrkT
!t/2
N.
(11.27)
Finally,
N(v)dv
=
41TN
m
21rk T
The function N( v) is shown schematically in Figure 11.6.
( 1 1 . 28}
The Kinetic Theory of Gases
Chap. II
196
,
N( l') r,
Graph of the Maxwell speed distribution for three temperatures: T3 > T2 > T1.
Fig11re 11.6
-
If we divide Equation (11.28) by N� \Ye normalize N(v) and obtain a N ( v) I .V that can then be used to obtain probability density function f (v) =
some useful average values. As noted in Section 11.1, the definition of the
mean value of vn is
·v"
V11j' ( v) d·v,
=
·0
where f ( v) is the probability density f u nc ti t:>n
.
Three
speeds that characterize
the Maxwell-Boltzmann distribution arc of 5pcciaJ importance: '
1.
Mean speed:
,
I
-
·v
=
..
vN(·'-')tlv
N
o
m
.
-
•
--
2r.kT
8kT 1Tn1
( 1 I .29)
--
sqttare speed: The mean square speed is given
41T --
.
by
1'11
21TkT
3kT_ -
m
•
nz
2. Root mean
=
)
=
47T -
=
.
•
� .
1 -
. .
v4e··nl'.tlv
.
Sec. I I .6
Distribution of Molecular Speeds
197
The root mean square ( rtns) speed is, then, 3kT m
m
(11.30)
•
3. Most probable speed: This is the value
Vm
for which N (v) j N is a maximum, or for which
d -(v2e-av�) dv
a=
"
The derivative is zero for
vm
v
=
0
'
m --
2kT.
12 , which yields 1 ) a 1/ (
=
112 2kT =
=
m
112 kT 1.414 m
(11.31)
•
It is evident that (Figure 11.7) Vm
: V:
V,ms =
For a nitrogen molecule
=
1.414 : 1.596:
1 : 1.128
(m
=
4.65
:
1.732
1.225. X
(11.32)
10-26 kg) at T = 273
K,
vrms
=
493 ms -t, which is slightly greater than the speed of a sound wave in nitrogen gas. Numerical values of the mean speed and root rnean square speed for vari ous gases are given in Table 11.2.
N(v) N
0.6 0.4
- ... --6---"'- .... �---+---
�----'-_....__._ ��--+--· -6--+----�.---+---4---+-_.... .....____.___�
o� 0
Fi
� 2.0
��������.
0.4
0.8
1.0
1.2
1.6
11.7
The Maxwell-Boltzmann speed distribution function at a particular temperature showing the most probable speed vm, the mean speed v, and the root mean square speed Vrms·
The Kinetic Theory of Gases
Chap. 11
198 TABLE 11.2 Molecular Speeds at 273 K. in ms 1 Gas
v
v,ms
1690 1210 570 530 450 420 400
1840 1310 620 580 490 460 430
'
•
There are other speed distributions of less interest to us� one is the num ber of particles whose �t-component of velocity lies in th e range
vr f()
v.\
+
dv,
regardless of the values of the other components:
(11.33) ( Fi g ure 11.8). Clearly�
vx
=
vY
vz since the velocity directions are assumed to
=
be un i formly distributed over all a ng 1 e s in s p ace
.
I I. 7
MEAN FREE PATH AND COLLISION FREQUENCY
We are interested in ho\\' far the molecules of a gas travel bet\\'cen collisions and how often they undergo collisi<)ns. These concepts are called .. respectively, the mean free path ( and the coJJisi<)n frequt-ncy f .. Suppose that the radius of
a
m ole cul e of the gas is
R. Then a collision
occurs when one molecule approaches another \\'ithin a center-to-center dis tance 2R. In time t� a moving molecule sweeps out a cylindrical ��exclusion'' volume of length vt and cros-s-sectional area
a =
rr( 2R)2 (Figure 11.9). If the
nei gh bo ri n g molecules were at rest. the nun1ber <)f molec ules with centers in this vol ume would be
na ·vr
- .
and the mean free path would therefore he
-
(
l
VI -
-
-
•
-
na vr
-
nu
-
This answer is only approximately correct hecause we have used the _mean speed v for all the molecules instead of performing an inte gration over the Maxwell-Boltzman speed distribution. If that is done, and the most probable
Sec. 11.7
Mean Free Path and Collfsion Frequency
199
N(v_r)
--------�- -------� V�
0
Maxwell-Boltzman distribution function for the x component of velocity for three temperatures: T.1 > T2 > T1• (The Figure 11.8
The
distributions for they- and z-components are identical with this distribution.)
speed is used in place of v, then e
1
=
( 11 34)
--;::=--- ,
.
8
nu 7T== '
which is the correct expression for the mean free path. The corresponding c
f -
Vm
-
c
(11.35)
-
f
Note that the mean free path is independent of the molecular speed and therefore of temperature. Since the nun1ber of molecules per unit volume
!:! is --
directly proportional to the pressure, the mean free path increases as the pressure decreases. For an oxygen molecule, 2R temperature and pressure,n
=
-
f
�:;
3.6 x 10-10 m. Under standard
2.7 X 1025m-�andvm -=.-;
5 7 x lo-s .
=
370ms-1.Thus
m
and
The average time between collisions, the reciprocal of the collision frequettcy, is approximately 1.5 x 10-10 s. Note that f is the order of 300R, validating our kinetic theory assumption that the separation of molecules is large compared with molecular dimensions.
The Kinetic Theory of Gases
Chap. II
200 2R
I
I
•
•
I
I
I.
I
'
'
Figure 11.9
Exclusion volume swept out in time t by a particle of radius R traveling with average speed v.
The actual distance that a molecule \\·ill travel between collisions is sub ject to a large statistical variation, independent of time. Let the probability that a collision will occur in a distance tlx he collision will -
alLt.
not
adx.
Then the probability that a
occur \\'·hen the molecule travels a distance
dx
will be 1
If p(J�) is the probability that a molecule \\'i1J travel a finite distance�'·
before making a
co1Jision, then p(J.-) ( 1 ad�t) is the probability that a cule wi11 travel a distance x + d�t before eXJ)eriencing a collision . Thus
Expanding p(:r have
+
-
a Tay lo r se ries a nd r et a i ning the first two terms, we
dx) in
p(J:)
mole
+
d p ( .'( ) 1- --,J.r
=
lX
p(.r)- aJJ(.t)tlx.
or •
clp(x) - -
-- -
tlx
=
-
aJ> ( .r ).
Integrating. we obtain
\\'here A is a constant. Since p(O) must he
p(x) To determine the
f(x)
m ean
=
I d p( .t) I dxI
=
.
"'
A
=
1 and
���
free path. we need the probability density function
p(x). Because p(.r) f f(.r)dx. it follows since f{�t) 1nust he positive. Then
associated \Vith the probability
that f( .x )
==- e
u n it v
ae
ax,
=
Sec. 11.8
201
Effusion 1.0
p(x)
=
N 05 No .
o
�----�----�
0
0.5
1.5
1.0
xlf.
x=
-
0
xf(x)dx=
a
0
so that (11.36) The probability p(x) can be interpreted as the ratio N I N0, where N0 is the initial number of particles in a molecular beam, and N is the number of particles remaining in the beam after a distance
x. Figure
11.10 is a graph of
this so-called "survival equation." We see that the fraction of molecules with free paths longer than f is
e -t
or 37 percent, while the fraction with free paths
shorter than the mean is 63 percent. Equation (11.36) shows how
a
beam of
particles is attenuated due to collisions as it passes through a low density gas. Use of this curve is independent of the starting point: the probability of a colli sion in a length t is independent of where the last coJlision occurred.
11.8 EFFUSION
If a container of gas, surrounded by a vacuum, has a small hole in its wail, the contained gas will leak through the hole into the surrounding space. If the hole is sufficiently small, the equilibrium state of the gas in the container will not be appreciably affected. In that case, the number of molecules escaping through the hole will be the same as the number that would impinge upon the area of the hole if the latter were not present. The emergence of molecules through the hole is a process known as
e
ion.
How small must the hole be? Molecules move unimpeded through a dis tance approximately equal to the mean free path f. Thus the density of mole cules at a given point is maintained by the flow of molecules from the surface
202
Chap. II
The Kinetic Theory of Gases
2 of an imagined sphere of radius { and a rea 47Tf . If the flow from the area A of
a circular hole is to be cut off \Vithout making much of a difference to the mol'-
ecular density at the center of the sphere, then A << 47Tf2. Let the diameter of
2 the hole be D. Then rrD2/4 << 41Tf � or /.)2 << 16f2. 2 somewhat arbitrarily as D2 < O.l6f or D ·( 0.4f. At atmospheric pressure, e
We may interpret this room temperature and
10-7m. Thus� the opening must be extre me ly small
=
if the sta t e of the gas in the container is to remain virtually undisturbed. Since the mean free path is inversely proportional to the pressure, it follows that there is an upper limit to the pressure for which effusion occurs. Effusion is essentially a low pressure phenomenon. In Section 11.2 we found that the number of molecules with speeds in the range
v
to
v +
dr that strike a surface, per unit area and per unit time, is
( 11.37) Here tz,.tl-v is the nun1ber of molecules per unit volume with speeds lJetw·een and
·-
·v +
v
dv. Assu1ning that the molecules have a Maxwell--Boltzn1an speed
distribution, we know that
( 1 l.38) Substituting this in Equation ( 11.37 )_we have mv'-!2k .
T d V.
(11.39)
This is the number of molecules that escape through the hole in the container wall that have speeds from
t'
to
v +
(/I', per unit area and per unit time.
\Ve wish to determine the mean speed 1)c of escaping molecules. Following the usual procedure, we multiply
ble values of
l\
1'
by
and divide by the i n t e g ral of ¢1.dV. The last step is necessary to
ensure that the weighting function used in calculating the mean value is a true probability density function. Thus ·.::t
vp
--
-
V4e -nzv�j2kTdV
vet> v dv
0
-
)':
·0 X
cl>vdv
-
•
3eV
�
mn·-;2kT
dV
0
"'lJ
Values of the integrals arc given in Appendix D. The result is
-
1J
(J
==
3 21rkT
--
4
111
l,i2 •
( 11.40)
Sec. 11.9
Transport Process\!s
203 •
Comparing this with the mean speed v of molecules within the container (Equation
(11.29)), we note that ve is slightly higher than v: -
Ve
-
31i=
8
V
=
1.18v.
(11.41)
This is because the faster molecules have a greater probability of ''finding" the
hol e in a given tim e
.
Effusion has found an important application in experimental physics. Molecules allowed to escape from a small hole in a container into surround ings maintained at low pressure can be collimated to form a well-defined mol ecular beam. (Standard methods involving electric or magnetic fields that are
used to guide charged particles are not ap pli cable to neutral molecules. ) The
effusion technique makes it possible to study individual molecules whose interactions with other molecules are negligibly small. Molecular beams have also been used to verify the Maxwell-Boltzman velocity distribution .
Another use of effusion is in the se·paration of isotopes. The principle is
based on the fact that the rate of effusion of a molecule is dependent on its mass.
1j_V/4; (8kTj1riii)112•
The rate per unit area is the molecular flux cf> PV
=
NkT,
so
n =
N/V
=
P/kT
-
and v
=
=
for an ide al gas, Combi ning these
exp ressions we see that ,
p
=
(11.42)
--;::==-:::=::: .
21rmkT
Thus light molecules escape m ore ra pidly than heavier ones.
A gas mixture of two is otopes is permitted to effuse through tiny holes in
th e wall of a container. After a period of time� the relati ve concetttration 9f the heavier isoto pe will exceed that of the lighter i�otope in the container. In the gas that has effused , the reverse will be the case. It is interesting to note tltat the effusion method was one of the methods used to separate the isotop es 218 in the making of the aton1ic bomb. U235 and U .
I I. 9 TRANSPORT PROCESSES Transport processes involve the re action of
a
gas to small departures from
thermal equilibri um due to gradients in the medium. From a knowledge of the average speed and the mean free path of the molecules. it is r•ossible to gi"·e an elementary derivation of three common p roperties of a dilute gas: its vi scosity,
thermal conductivity, and coeffi ci en t of dif fusion
.
In a transp
Chap. t I
204
The Kinetic Theory of Gases -
�
--d
'-
. • .
•
.
.
. · .
.
.
.
.
.
.
.
. .
.
.
�
.
-
\'
•
z=O
-
-
--
-
-
--
-·
.
-
X
z =-d
. .· • • • .
• •
•
.·•_.
•
•
.
�.
-
.
. .
.
. ••
.
.
.
·.
• 0
.
0
•
.
:
.
·
.
. • •
0 .
•
•
•
..
-
•
•
.
.
-
.
•
.
.
.
.
• . .
.
.
Figure 11.11 Flow of a viscous fluid bet\\'een a stationary lower plate.
movin g
upper plate and a
collision. Hence� as it travels along the next free path, it carries a sample of the properties of the gas in the vicinity of its previous collision. If there are gradi ents present� the properties in the region of its next collision may differ from those carried by the molecule. During the next collision the molecule will give up this difference to the gas in the new region. In their motion from one colli sion to the next, the molecules will couple the properties of the gas at one point with those at another. The ultimate resttlt is an equalization of the prop erties throughout the gas.TI1is is the e s s en ce of transport processes. Consider a fluid such as a gas between two large parallel plates. The upper plate is moved horizontally while the lower plate remains at rest (Figure 11.11 ). Since real fluids are viscous� the fluid is pulled parallel with the moving plate. The magnitude of the 11uid velocity
in the l1orizontal
ll
direction changes from the speed of the upper plate to zero at the stationary
plate. In other words, a velocity gradient d1tldz exists in the z direction. per pendicular to the plates. Owing to the gradient .. the molecules that are continually moving from positive values of z are passing through the z == 0 plane.. transferring their higher momenta in the x-direction t(> the layer below.. thereby increasing the average momentum <)f the layer. Similarly� the molecules moving from nega tive values of
z
through the plane are carrying \\'ith them the lower momenta
corresponding to the region where they last underwent a collision. The net momentum transfer per unit time through unit area gives the tangential stress P-:.x (the viscous force per unit area) .. \vhich a layer below the t)t plane exerts on ..
the fluid above it due to the velocity gradient. For small gradients� we expect a relatic)n of the form
(11.43)
where the proportionality fact or TJ is the C
·
s em-�, called a
Transport Processes
Sec.ll.9
205 z
f
cos
8
l I f I I I I
I I I I
dy X
dx Figure 11.12 Molecule crossing the z 0 plane after its last collision at a distance e cos8 above the plane. =
"'poise," after Poiseuille*, who studied viscous flow. The Sl unit is Pa is equal to
10 poise.
·
�which
.
The coefficient of viscosity for a dilute gas can be calculated using kinetic theory and Newton's second law. We must first determine the average distance z above the xy plane at which a molecule makes its last cc,Jlision before crossing the plane. From Figure 11.12, z= f cos8. Then z can be found by multiplying this quantity by the incremental flux d4>8 in the 8 direction, integrating over 8, and dividing by the total flux <1>. Using Equations and
(11.7)
(11.8), we obtain
z=
--= -
-
(11.44)
---
Thus, on the average, a molecule crossing, the xy plane makes its last collision before crossing at a distance equal to two-thirds of a mean free path. If
u(O)
is the forward velocit}' of the fluid at z =
0,
then at a distance
2tj3 above the plane, the fluid velocity is
u
•
2 -e 3
2
du
(11.45)
Jean Louis Marie Poiseuille. a French anatomist.. was interested in the physics of the circu
lation of the blood. His work on the flow of liquids through capillaries was published in
1842.
Chap. II
206
The Kinetic Theory of Gases
.
Here we assume that the gradient does not vary appreciably over the distance of a mean free path. The momentum of a molecule with this vei<)City in the .x-direction is mtt(2f/3) and the total momentum transferred across the plane per unit time and per unit area by a11 of the molecules crossing from above is
1 r . - n tn v u (0 ) 4=
2 lill . -;; f , d�
+
_
-
�
Similarly, the total momentum transporte.d per unit time and per unit area across the plane by the molecules crossing from below is 1
n1n v
4= ·
u ( 0)
_
L
2 du . -f 3 (Jz
-
The net rate of momentum transport per unit area is the difference between these two quantities, or l
till -- 111n v f ·d 3 :: -
·-.
=
e 4 ua l to
From Newton ,s second law.. this is
the viscous force per unit area
given by Equation ( 11.43 ) It follo\vs that .
TJ
=
J -. -- 111111'{. 1 =
( 11.46)
-
..
R ecal l i ng that ('
b e w ri t te n
==
(
H/rrna) ·-I, --
we
see
t h at the co ef ficie nt of viscositv can �
.-
T}
1 :..::.
/1r
I
6'v
-·
2
mi' •
(11.47)
.
a ·
•
The mean kinetic energy of Thus vx ( T/ n1) 1/2 and
a
ffi()(ecule is proportional to the te 1nper atur e.
) 1 (con st T fT
2
(11.48)
•
This result may seem surprising. Intuitively, \VC might expect the viscosity to
decrea se with i ncreas ing temperature and to increase with i n creas i n g collision cross section. But higher temperatures mean higher molecular speeds and a small cross section means a large mean free path. B oth factors contribute to a more effective transfer of momentum from one fluid laver to the next and hence to a higher viscosity. This behavior differs markedly from that of the viscosity of ..
.,
Sec. 11.9
Transport Processes
207
•
liquids, which decreases rapidly with increasing temperature. That is because in a liquid the molecules are closer together and momentum transport across a
surface involves interaction forces between molecules on the two sides of the surface as well as molecular motion across the surface. An especially useful expression for calculating the viscosity is obtained by solving for n in Equation ( 11.9) and substituting the result in Equation (11.46). -
Then, assuming that v
2
-
::.::: ( v)2,we
get
•
')")'=>� .,
For oxygen under standard conditions
11
=
(1.01 X lo-'Pa)(5.7 X
Pf
-
v
(P
.
( 11.49) =
1.01
8 10- m)
(420ms-1)
105 Pa,
X
�
T
273K),
=
6
14 X
10- Pa s. ·
This compares with the measured value of close to 19 x 10-6 Pa s.TI1e discrep· ancy is due to the approxin1ate nature of kinetic theory, which treats the mole·
cules of a gas as hard spheres. Also, the factor 1/3 in Equation
(11.46) is not to be
taken too seriously, because it depends on how various quantities are averaged.
The calculation of the them1al c9nductivity proceeds in a similar way. The only difference is that it is the thermal energy that is transported rather than the momentum. The upper and lower plates in Figure 11.11 are both assumed to be
stationary but are held at different temperatures so that there is
a
temperature
gradient rather than a velocity gradient in the gas. The themtal conductivity results from the net flux of molecular kinetic energy across a surface. The aver age kinetic energy of a molecule is cvT / NA, where cv is the specific heat capacity
and NA is Avogadro's number. The thert11al conductivity A is found to be A=
1
Cv
-3�ve _
NA
=
1
-
1T
--
-
2
6
CvV
(11.50)
--
lVAa.
It is interesting to note that the ratio of the thertnal conductivity to the viscosity is A
-
=
=
cv -
•
•
so that AM= l 1JCv
'
where M is the molecular weight of the gas. Measured values are closer to 2,
but kinetic theory gives the right order of magnitude.
The Kinetic Theory of Gases
Chap. II
208
Finally.. the transport of mass across a surface as a result of random mol ecular motion gives rise to diffusion if a concentration gradient exists. The treatment of diffusion is complicated when there is more than one type of molecule present, but its main features can be understood by considering the motion of molecules in a single constituent gas.. a phenomenon known as self
diffusion. The coefficient of self-diffusion
l> can be easily calculated by mean
free path arguments like those used in the discussion of ''iscosity and heat con ductivity. The result is v
-
Since!!=
2 n(T
6
3
(11.51)
-
3Pjmv2 and both Vand (v�)l:2
1 are proportional to T
..
2,
-
D
( const) T3 2
=
.
.
.
at a fixed pressure an d
D
( CfJnst) =
P
-
-
at a fixed temperature. The ratio of the roefficie·nts of viscosity and self diffus ion is ·
D
--
.=:
11
I
-
t1n1
1
-�
p
--
where p is the mass density of the gas. Thus Dp -
-
.
17
Experimentally. DpfTI l ie s
in
=
I.
the range 1.3 to 1.5. The extent of the agreement
between theory and experiment must be COilsidered satisfactorv in view of the �
approximations made in our physical model.
-
IIIIII
PROBLEMS:
' 11.1 The distribution of particle speeds of a certain hypotheti cal gas is given by
tv' ( l' )d v where A and
v.)
are constants.
=
A ne-,.
r..
d v..
Problems
209
(a)
Determine A so that f ( v)
(b)
Find v and
i.e.
.
�;� .f ( v )dv
V,m.�
(c) Differentiate
=
N ( v) IN is a true probability density function;
=
1. S ketch f ( v) versus v.
in tem1s of v0•
f ( v)
with respect to
the most probable speed
·v
and set the result equal to zero to find
Vm.
(d) The standard deviation of the speeds from the mean is defined as
- _), 1/"2. (
-
-
u=
v-v -
·
,
where the bar denotes the mean value. Show that u
=
[ v 2 - ( v) 2] J :'2
in gener al . What is ufor this problem?
11-2 At standard temperature and pressure the mean speed of hydrogen molecules is 1. 70 X 1 OJ ms -l. What is the particle flux? 11-3 What is the number density of molecules at a temperature of 77 K in an ultra high vacuum at 10-10 torr? 11-4 Compute the
rms speed of helium atoms at 2 K; of
and of mercury atoms at 100°C.
nitrogen molecules at 27°C;
11-S Compute the mean energy in electron volts and the rms speed in ms -t of an elec tron at 1000 K. At 10,000 K, what fraction of the speed of light is the rrns speed? 11-6 At what values of the speed does the �iax wel l speed distribution have half its maximum value? Give your answers as a constant time
11-7
(a)
Show that the Maxwell speed distr ibuti on can be written as
N ( X) dX where x
(b)
(kT,Im) 112•
=
=
4N
1T
:
x-e-·',. dX, .,
vfvm.
Show that the number of molecules with speeds Jess than some specified
speed v0 is given by
�)_.X., where x0
=
=
N er/( Xo )
v0f·v,. Here erf(x0) erf(x)
2
. xoe
-
1T
-
.. 1 a
'
is the error function, defined by
2
=
·--;:= 1T
()
e
2
u du.
(c) Find the fractional number of molecules -with speeds less than 11·8 Compute
vm, v, and v,ms
vm.
for an oxygen molecule at 300 K. What are the corre
sponding values at lOJ)OO K?
11-9
(a) (b)
Show that the mean speed of a nitrogen molecule at 300 K is 476 ms- 1•
What is the ratio of the probability of finding a nitrogen molecule w hose
speed is 476 ms-
t
at 300 K to the probability of fi nding a nitrogen molecule
with the same speed at 600 K?
The Kinetic Theory of Gases
Chap. II
110 11-10 Find the
rms
free path in terms of the mean free path. What is the most probable
free path ? Use t
1e
.t'tas
the probability density function.
11-11 For carbon dioxide (molecular weight 44! molecular diameter 4.6 X 10 10m) at 1 atm and 300 K, find the mean free path f using Equation (11.34 ). What is the ratio of f to the molecular diameter? What is the collision frequency?
11-12 The number density of the gas. mainly hydrogen� that fiBs interstellar space is one molecule per cubic centimeter ( W' m l). If the moleci.Ilar diameter is 10
10
m .. what is the mean free path in interstellar space? What is the collision fre quency of a hydrogen mo1ecule in collisions per century if the tetnperature of interstellar space is
10 K?
11-13 A beam of oxygen molecules start tog e th er The pressure is 1.8 x 1 OJ to rr and .
the temperature is
3.6
x
300 K. The diameter of an oxygen molecule is approximately
10-10 m. How 1ong will half the molecules remain unscattercd? (Assume
that all the particles have a speed equal to the mean speed.)
11-14 A thin-walled vessel containing 1 liter of carbon dioxide is kept at a tempera ture of 0°C. The gas slowly leaks out through a circular hole of diameter 100 p,m. The outside pressure is low enough that leakage back into the vessel is negligi ble. (The diameter Of a CQ� molecule l" abOUt 4.6 X } 0 HIm and ifs moleCUlar weight is 44 kg kilomole- '.)
(a) Estimate the upper limit of the pressure in the container for effusion to occur through the hole.
(b) Estimate the time, starting •
at
this pressure for the pressure to drop to one..
half its initial value.
11-15 A vessel is divided into two parts of equal volume by means of a plane partition, in the middle of which is a very small hole. Initially� both parts of the vessel con tain ideal gas at a ternperature of 300 K and a lo\\· pressure P. The temperature of one-half of the vessel is then raised to 600 K while the temperature of the other half remains at 300 K. Determin� the pressure difference in terms of P between the two parts of the vessel \vhen steadv conditions are achieved . •
11-16 A vessel has porous walls containing many tiny holes. Gas molecules can pass through these holes by effusion and then be pumped off to son1e collecting chamber. The vessel is filled with a dilute gas consisting of two types of mole cules that have different masses
nz1
and
nl �
by virtue of the fact that they contain
two different isotopes of the same atom. The concentrations of these molecules are c1 and c2 respectively. (The concentration
c,
is the ratio of the number of
molecules of type ito the total number of molecules.) The concentrations can be kept constant in the vessel by providing a steady flow of fresh gas through it so as to replenish any gas that has effused. If
' c1
and
' c:
denote the concentrations
of the l\\'O types of molecules in the collecting chamber. \\'hat is the ratio
c/ Ic1 '?
11-17 Diffusion can be regarded as a random walk prohlem in \Vhich the successive displacements of a gas molecule are statistically independent. In this model. the distance IJ traveled by a molecule after N displacements is related to the mean free path ( by the expression L2 tance L is given by t
=
=
N f�. and the time
t
required to move a dis
N fj t;. Using these relations� estimate how long it \vould
Problems
211
take a molecule in a room of macroscopicnlly "still'' air with unifortn tempera ture and pressure to move
a
distance of 5 meters.
11-18 The experimental value of the viscosity of argon gas is found to be 22.0 x 10-6 Pa s at l5°C and atmospheric pressure. The atomic weight of ·
argon is 39.94. Estimate the diameter of an argon atom.
11-19 The radius of an air molecule is approximately 1.8 x 10-10m, its mass is about 4.8 x 10-26 kg, and it has 5 degrees of freedom. The molecular weight of air is 29 and its density is 1.29 kg m- 3 under standard conditions. (a) Estimate the values of the coefficient of viscosity TJ, the therrnal conductiv ity A, and the diffusion coefficient D under these conditions.
(b) Check that for air, and
Dp =
11
1.
Also, check that these quantities are indeed dimensionless.
•
12.1 Introduction
215
12.2 Coin-Tossing Experiment
216
12.3 Assembly of Distinguishable Particles
221
12.4 Thermodynamic Probability and Entrop)#
223
12.5 Quantum States and Energy Levels
225
12.6 Density of Quantum States
229
213
12.1 INTRODUCTION
The union of the energy and entropy concepts led to a science of therttlody namics that combines great generality with reliability of prediction. Classical thermodynamics, however, provides little insight into lvhy thermal phenom ena occur the way they do. The object of statistical thermodynamics is to present a particle theory resulting in an interpretation of the equilibrium properties of macroscopic sys tems. The foundation upon which the theory rests is necessarily quantum mechanics. Fortunately, a satisfactory theory can be developed using only the quantum mechanical concepts of quantum states, energy levels, and intermole cular forces. The central idea is the probability density function applied to a large col lection of identical particles. A thermodynamic system is regarded as an assembly of submicroscopic entities in an enormous nutnber of ever-changing quantum states. The basic postulate of statistical thermodynamics is that all possible microstates of an isolated assembly are equally probable. A few new concepts are needed. We shall use the term assembly to
denote a number N of identical entities, such as molecules, atoms, electrons, photons, oscillators, etc. The word system is used synonymously. The macrostate of a system, or configuration, is specified by the number of particles in each of the energy levels of the system. Thus
� is the number of
particles that occupy the jth energy level. If there are n energy levels, then n
� .... =1
j
=
N.
The macrostate of statistical thermodynamics is another word for the therttlo dynamic state of the classical theory, specified by a pair of state variables. A microstate is specified by the number of particles in each energy state.
In general, there will be more than one energy state (i.e., quantum state) for each energy level, a situation called degenerac}'· A microstate is the most spe cific description one can get. In general, there will be many, many different 215
Statistical Thermodynamics
Chap. 12
216
microstates corresponding to a given macrostate. We \\·ill be interested in the number of microstates but not in their detailed specification. The number of microstates leading to a given macrostate is called the thermodynamic probabili(v. It is the number of ways in which. a given configu
ration can be achieved. This is an Hunn<>rmalized"" probability. an integer between zero and infinity" rather than a number between zero and one. For the kth macrostate, the thermodynamic probability is taken to be
1L'k·
A true prob
ability Pk could be obtained by dividing wk by the total number of microstates
.n available to the system. For much of what we shall do, the thermodynamic probability will be adequate.
12.2 COIN-TOSSING EXPERIMENT
By way of introducing the centra) features of statistical thermodynamics"� we shall apply some elementary concepts of probability theory to a coin-tossing experiment. We assume that we have N coins that we toss on the floor and then exam ine to detertnine the number of heads N1 and the number of tails N2 We consider the case of N
=
=
N - N1•
4. Using the terrns we last introduced" we can pre
pare a table listing the possible outcomes of the experiment (Table 12.1 ) Each macrostate is defined hy the number of heads and the number of .
.
tails. There are five. For each macrostate there are one or more microstates. A microstate is specified by the state. heads
\)T
ta·ils. of each coin: it is the most
detailed description possible. We are interested in the lllllnber of microstates for each macrostate. the thermodynamic J'robability. The true probability is the number divided by the total number of microstates. Thus
( 12.1) where '
•
n
=
k=l
1vk
==·
16.
( 12.2)
We might also wish to calculate the average occupation numbers this case, the average number of heads and tails. Let j
==
in
I or 2. where N1 is the
number of heads and N2 the number <>f tails. Let Njk be the occupati<>n number for the kth macrostate. Then the average occupation number is .
N.J
k
Njkwk -
-
=
Ltvk k
k
Nj*·wk -
n
k
NikPk·
(12.3)
Coin-Tossing Experiment
Sec. 12.2
TABLE 12.1
217
Possible outcomes of a coin-tossing experiment using four coins. We introduce the
language of statistical therm
mics.
TherJnoMacrostate
Macrostate
Label
Specification
Microstate
dynamic
True
Probability
Probability
k
N.
N�
Coin 1
Coin 2
Coin 3
Coin 4
1
4
0
H
H
H
H
1
1/16
2
3
1
H
H
H
T
4
4/16
H
H
T
H
H
T
H
H
T
H
H
H
H
H
T
T
6
6/16
T H
T
H
H
T
H
T
T T H
H
T
H T H
H
T
T
T
T
H
T
T
T
T
H
T
T
T
T
H
T
T
T
T
l
2
2
3
.;.
T
H
3
1
4
0
5
4
H
•
T 4
4/16
1
1/16
The average number of heads is therefore
16 Similarly, N2
2, as we would expect. Then N1 + N2 4 Figure 12.1 is a plot of the thertnodynamic probability =
=
ber of hellds N1• The curve is symmetric about N1
=
=
N. w
versus the num
2, the most probable
configuration. Suppose now that we want to perform the coin-tossing experiment with a larger number of coins. We need a way of computing the thermodynamic I'
.
probability without tabulating the actual state of each coin. We assume that we haveN distinguishable coins and ask: h<>w many ways are there to select from
the N candidates N1 heads and N - N1 tails? The answer is given by the bino mial coefficient N
N! =
-----
Nt!(N- N1)!
.
(12.4)
Chap. 12
218
Figure 12.1
Thermodynamic probability versus the number of heads for a coin-tossing
Statistical Thermodynamics
l
experiment with four coins.
This result can easily be seen by noting that there are N ways to pick the first coin, leaving ( N
-
1) for the second, ( N
-
2) for the third, etc. Thus the total
number of ways of picking N1 heads is N!
However, this counts the N1! permutations of the N1 coins separately. The ordering of the coins is not important, s<) we divide by this factor, giving Equation (12.4). If we increase the number of coins from four to eight .. we obtain the plot
12.2. We see that the smallest value of w is again unity, b�t its maxi mum value has grown to 70� the peak has become considerably sharper. of Figure
The maximum value of the thermodynamic probability is of paramount interest to us. Its calculation is straightforward for N peak occurs at N1
=
=
8. We know that the
N /2. Thus, Equation ( 1 2.4) gives
'IVmax
8! =
4!4!
==
70.
Suppose now that we perform a more ambitious experiment, with N
=
1000. In this case, 100()! 'LOmax
==
500! 50_0_! '
(12.5)
and the computation becomes much more challenging. Fortunately, for such large numbers we can use Stirling's remarkable approximation (Appendix B) and write ln n! ::::: n In n - n.
Sec. I 2.2
Coin-Tossing Experiment
219
50 40
Figure U.2
Therm odynamic
t--
10 1--
probability versus the number of heads for a coin-tossing experiment with eight coins.
0
0
2
4
N1
8
6
Taking the natural logarithm of both sides of Equation {12.5), we then obtain In
Wmax
=
ln( l OOO ! ) - 2ln(500!)
=
693.
But log10
x =
(log10 e)ln
x =
0.4343 In
x,
so log 1 o
Wmax
=
(0.4343)(693)
=
300.
Thus Wmax
-
10300
·
(The error in using Stirling s approximation here is extremely small: even for n as '
small
as
50, the error is less than 2 percent, and decreases rapidly
For N
=
1000 we find that
Wmax
as n
gets larger ) .
is an astronomically large 011mber. We
conclude, then, from our coin-tossing experiments that the peak in our curve of thermodynamic probability versus the number of heads always occ u rs at
N/2 but grows rapidly and becomes much sharper as N increases. Also, t h e tails of the curve always terminate at
w =
1. The result is what we would expect:
for ..'honest" coins, we get ever closer to an exact 50-50 split between heads and tails as the number of coins gets larger. In other words, the most probable con fig uration
is that of total randomness.
Conceptually, this behavior for N
=
1000 is shown in Figure 12.3. The
most nearly random configuration (macrostate) N1
always occurs. .
=
N2 is the 011e that almost
Statistical Thermodynamics
Chap. 12
220
totally random region
10300
---
----
·----
��-·---
ordered region
ordered region I
Figure 12.3 Thermodynamic probability versus the number
I
1000
0
of heads for a coin-tossing experiment wit� 1000 coins.
The ''ordered regions�' almost never occur: pared with
Wmax·
w
is extremely small com
We are led to a very important conclusion
namely, that the
total number of microstates is very nearly equal to the maximum number:
!l
The figure a1so suggests
a
( 12.6)
•
=
method for finding
the curve is zero. Thus we need to find
'lVm",.
At the peak the slope of
·u�( /\'1 ), take the derivative of the func
tion with respect to N1 and set the result equal to zero. This wi11 give ��1 at �
10
=
·u'm ax
and hence N2• The o·utc
able state. For the thermodynamic probJem .. the ��outcomes'� arc the c,ccupation numbers for each of n
energy levels. Our task is to find 10( N1 N2 ..
..
•
•
•
Ni� ... N,)
..
possibly subject to constraints.. and set its derivative equal to zero. In this way we wil1 obtain the particular set { �',.} of occupation numbers corresponding to the most probable ntacrostate. The most probable macrostate is the equilib
rittm state of the assembly. This is the fundamental problem of statistical thernlo dynamics
to determine the equilibrium state of the syste m
.
Before proceeding� we need to extend our ··counting formula'" of Equation
(12.4) from two "levels"' (heads and tails) to
11
levels. We �sk: in how
many ways can N distinguishable objects be arranged if they are divided into n
groups with N1 objects in the first group� N2 in the second. etc.? The answer is
10 =
N!
N! '
NI·'N' 2···· Nn·
=
•
II
N,! i =I
(12.7)
Sec. 12.3
111
Assembly of Distinguishable Particles
represents the extended product, analogous to the sum...
Here the symbol .
mation sign Equation
(12.7)
can be easily understoocl. Suppose we want to place
N
distinguishable objects in
three boxes, with N1 in tlte first box, N2 in the sec ond, and N3 in the third. The number of ways of selecting N1 (out of N) for box 1 is
N
N! =
N1 We are left with
-----
N1!(N- N1)!.
N - N1 objects. We can choose N2 of them for box 2 in (N- N1)! =
---
---
�
==
ways. The number of ways of putting the rerr1airting Thus the number of ways
w of
._
N3 objects in box 3 is one.
selecting objects to occupy the three boxes is
the product of the three factors:
w=
N! ---
N, ! (N - N1 ) !
X
(N
-
N1 ) !
N2!'A)!
Xl=
N! N1 ! N2! N3! .
---
Equation (12.7) follows by induction. One additional important point: the coin-tossing model assumes that the coins are
distinguishable (by a date, perhaps, or a mint mark). The ther
modynamic analogy is with particles in a lattice, whose location distin guishes them. This would be the case for a crystalline solid, as an example. But for other cases, such as molecules in a gas, the particles are identical and indistinguishable and the number of microstates available to the assembly will be correspondingly fewer. We need to develop statistics for both situations.
12.3 ASSEMBLY OF DISTINGUISHABLE PARTICLES
Th.e constituents of the system under study (a gas, liquid, or solid, for exam ple) are considered to be a fixed number N of distinguishable particles, occupying a fixed volume V. Vle limit ourselves to isolated systems that do not exchange energy itt any form with the surroundings. This implies that the internal energy U is also fixed. The macrostate in which the system finds itself is defined by
(N, V, U), ira correspondence with a classical thermody namic state defined by (n, V, T), say, where n is the mass in kilomoles. Note that the temperature Tis simply a measure of the internal energy for a sys tem in thermal equilibrium.
222
Chap. 12
Statistical Therntodynamics
We assume that the particles are weakly ittteracting. By this we mean that they interact sufficiently so that the system is in thermal equilibrium, but do not experience strong coupling through \!lectromagnetic or nuclear forces. We seek the distribution among energy levels
{Ni}
for an equilibrium
state of the system. That is.. we want to determine the number of particles
Ni
with energy ei for all n energy levels of the system, subject to the restrictive conditions ll
"""-�
j =l
Nj
}l
=
(conservation of particles).
( 12.8)
(conservation of energy)�
(12.9)
ll
��1\/iei
U
=
j=}
where Nand U are ·consta n ts As with our coin-tossing experiments, we can clarify these ideas with a .
simple example. Consider three part i c l es labeled A. B. and C� distributed ,
among four energy l e ve l s 0 e.. 2e� and 3e, such that the total energy is 3e. Thus ,
..
..
.l
·'
LN,
=
3
LNje;
and
i::. (I
=
3e.
f =U
The possible microstates and macrostates are listed in Table 12.2. The occupation numbers are: N0 particles with energy 0, N1 with energy e_ N� with
TABLE 12.2
Microstates and macrostates for N
=
3. U
=�
3,.; with
t:::1
=
j£. j
=
0, I, 2, 3.
Thermo Macrostate
Macrostate
Microstate
dynamic
True
Label
Specification
Speci fie at ion
Probability
Probabilitv
B
C
0 0 3e
0 3c 0
3£
()
£
()
_F.
e
F.
()
2f
£
2£
2e
0
0
k t
?
-
2
l
0
1
()
1
1
u
2£ 3
0
..,
.)
0
0.
e
)
.
0 0
o'
0.3
2£ 0
0.6
1
0.1
E
0
Thermodynamic Probability and Entropy
Sec. 12.4
123
TABLE 11.3 Distribution of three particles among four energy levels with total energy U 3e. =
3e a e 0 k
l
2
1 1 1
3 0
1 3
2 6
3 1
energy 2e, and N3 with energy 3s. The total number of microstates is 10 and the number of possible macrostates is 3. For the most probable macrostate, the number of available microstates is 6. Another way of displaying the results is given in Table 12.3, where the occupation numbers are listed for each of the energy levels. sor "di st mo the t tha ent evid is it ll, sma are e her s ber num the ile Wh dered" macrostate is the state of highest probability. For the very large n•Jm ber of particles in a physically meaningful assembly, this state will be sltarply defined and will be the observed equilibrium state of the system. In classical thermodynamics we have seen that as a system proceeds toward a state of equilibrium the entropy increases, and at equilibrium attains its maximum value. Similarly, our statistical model suggests that systems tend to change spontaneously from states with low thertnodynamic probability to states with high thermodynamic probability (large number of microstates). Thus the second law of thermodynamics is a consequence of the theory of probability: the world changes the way it does because it seeks a state of higher probability!
12.4 THERMODYNAMIC PROBABILITY AND ENTROPY
It was Boltzmann who made the connection between the classical concept of entropy and the thermodynamic probability. His argument is as follows. Assu·me that the entropy is some function of w, that is, S
_
=
f(w).
(12.10)
HereS and ware properties of the state of the system (state variables). To be physically useful, f ( w) must be a single-valued, monotonically. increasing function. Consider two subsystems, A and B (Figure 12.4). The entropy is an extensive property; like the volume, it is doubled when the mass or number of
Chap. 12
224
system
A
Statistical Thermodynamics
system B
Figure 12.4 Two systems with entropies SA and S8.
particles is doubled. Thus the combined entropy of the two subsystems is sim ply the sum of the entropies of each subsystem:
or ( 12.11) However, one subsystem configuration can be combined with the other to give the configuration of the total system. That is. (12.12) This follows from the fact that independent probabilities are multiplicative. To return to our coin-tossing experinlent._ suppose that the two subsystems each consist of two distinguishable coins. The possible configurations of the subsys tems are listed in Table 12.4. The. probability of two heads in subsystem A is clearly 1/4, as it is in subsystem B. From Table 12.1, the probability of obtaining all heads when fottr coins are tossed is 1/16, which is equal to 1/4
><
1/4. Thus
Equation (12.12) holds. and therefore (12.13) TABLE 12.4 Possible configura tions of tossed coins for subsystems A and B, each consisting of two coins A
B
HH
HH HT TH TT
HT TH TT
.
Sec. JlS
Quantum States and Energy Levels
225
· .•.
· · :�..
.
·:·.. .
.
.
.
.
. ·.:..
.
.
�- : ,
.
.:.
-�
.
�-:· :-
.
.;
.
• .
.
,.· ··:· · •• ·• o:··:': -:;: ...
'/•-::::::. ;-:· .
.
.
..
.
-
.
. ·. . ..
.
.
.
..
. t·
.
·.. .
. •
•
�-:�: ::��f�J: � :ft '
..
.·
.
.
.::
. .
;> ••
••
..
•
U.S
Ludwig Boltzmann's tombstone, showing his famous equation. (lnstitut fur Theoretische Physik der Universitat Wien/Central Library for Physics in Vienna.)
.
. ..
�=
. - .: ..r .
.
.
·:
. ·.
:
.
Combining Equations (12.11) and (12.13), we obtain •
f( WA )
+
f( Wa )
=
f( WAWB ) ·
(12.14)
The only function for which this statement is true is the logarithm. Therefore • •
S =kIn
w,
(12.15)
1-
where k is a constant with the units of entropy. It is, in fact, Boltzmann's c o n stant :
k
=
1.38 X 10-23 J K-1•
Boltzmann's tombstone in Vienna bears this fundamental relation as its
inscription (Figure 12.5).
12.5
QUANTUM STATES AND ENERGY LEVELS
We have tacitly assumed that the energy levels of our assembly of particles fotin a discrete set rather than a continuum. This is a consequence of the particles' confinement in a container of finite volume. Only in the case of completely
_#
226
Chap. 12
l I
I
I
J
l
I
_j
=
Statistical Thermodynamics
!...._ ... ---JI
L.____.l
83 = 5
I
·�
_
/
Figure 12.6
Energy levels and quantum states. 'The degeneracy g1 is the number of quantum states whose energy level is e1.
Hfree" particles can a continuous spectrum t.>f energy exist. To understand how the discreteness arises, we must resort to quantum mechanics. In quantum theory� to each energy level there corresponds one or more quantum states described by a wave function '}1. For so-calJed stationary
states, ')1 will be a function d e pendent on the position coordinates and the time.* When there are several quantum states that have the same energy� the states are said to be degenerate. The quantum state associated with the lowest energy level is called the groLlnd state of the system: those that correspond to
high e r energies are called excited states.
The energy leveJs can be thought of as a set of shelves at different
heights� while the quantum states correspond to (Figure
12.6).
a
set of boxes on each shelf
For each energy level £;the number of quantum states is given
by the degeneracy gi.
C
As an instructive example..
mechanics, that of
a
particle of mass
n1
basic proble m in quantum
a
in a one-dimensional box with infi
nitely high walls. (The particle is confined within the region 0
c� X :5
L.)
Within the box the particle is free� s u bje ct e< i to no forces except those associ ated with the walls of the container (Figure 12.7). ll1e time-independent part of the wave functi<)n 1/J( .r) is
a
measure of
the probability of finding the particle at a positi on .r in th e box. The problem of finding
.P ( .r)
is analogous to the classical problem of transverse waves on a
string with both ends fixed. The "'·ave function sa ti s fyin g the boundary condi tions
t/1 ( 0)
=
0 and
l/1 ( LJ)
=
0 is
t/l(.r)
=
A �in k.r, 0
<
.r �
LJ.
( 12.16)
with •
k
= II
7i
l.
.
11 =
J 2 3. . •
.
.
*A It hough 'V is timc·dependent. the obscrvahle quantity '1'.
the complex conjugate of 'I'.
(12.17)
.
'I'
is constant in time.
w·
Is
Sec. 12.5
Quantum States and Energy Levels
127
00
12.7 A one
x=O
dimensional potential well.
x=L
The boundary conditions simply state that the wave function must vanish out side the box. The probability of finding the particle there is zero. Now, the de Broglie relation of wave m�!chanics is p
==
1tk,
where p is the m o ment um of the particle, k is the wavenumber, and 1i is
Planck's constant divided by 21T. The particle's kinetic energy is therefore
(12.18)
Substituting Equation (12.17) in Equation (12.18), we obtain
(12.19)
Here the integer
n
is the quantum number of the one-dimensional box. The
important result is that the energy is proportional to the square of the quan tum number. We can easily extend this result to the case of a three-dimen s i onal box with dimensions Lx, Ly, and Lz, for which the energy becomes
2 7T21i2 nx2 ny2 nl. + + . 2 2 2 2m Lx Ly Lz In this case, any particular quantum state is designated by three qua n tu m num bers nx, ny, and nr.. If Lx = Ly = L, = L, then e· J
�
(n 2+ n 2+ nZ 2) X
.V
=
nJ' .2
(12.20)
ni is the total quantum number for states whose energy level is et The energy levels depend only on the values of n/ and not on the individual values of the integers ( n "' n V' n z) . where
•
First three states of a three-dimensional infinite potential well.
TABLE 11.5
Energy State
Level
j j
j
Statistical Thermodynamics
Chap. I 2
228
=
=
=
l
Ground state
2
First excited state
3
Second excited state
The volume Vof a cubical box equals L3• so/_� , 1T*"1t- -, V n� 1 2nl
e. = 1
=
3 3
V2,'-' and hence
..,
,
·
I
3 6 9
(Ll.l) ( 1. 1.2 ): ( 1.2J ): ( 2.1.1 ) ( L2.2): (2,1.2); (2.2J)
·
..
,
-· ·
(12.21)
.
This result applies to a container of any shape whose dimensions are large compared with the de Broglie wavelength
2rrn/ p. We observe that as the vol
ume decreases, the value of the jth energy level increases. The lowest energy leveLj n?
=
=
3 and
1, is that for which
nx
=
n_,.
=
11�
This is the ground state. There is only one set of quantum numbers
(n_,
. ..
that has this energy. Thus� the lowest level is nondegenerate and g1 d�generacy for the first three states is shown in Table
the degeneracy increases fairly rapidly �·ith
12.5. As increasing j.
1. Then
=
n\
.
..
n::)
•
=
1. The
a general rule,
Now� each energy level is occupied by a number of particles
!vj.
Returning to our analogy of shelves with boxes on them, the degen eracy gi of level j is the number of boxes on the jth shelf. The number of particles in any one box is the number in a particular quantum state. Those particles on any
one shelf are in different states but have the same energy. Our interest is in the total number of particles in the boxes at the various JeveJs.. that is.. in the occu
pation numbers
numbers { Ni] then defines a macrostate of the
N1. The set of
system. A fictitious example is sho\\'n in Figure 12.8. It is instructive to estimate the quantum
n umb e r II;
for a on e li te r vol -
ume of helium gas at room temperature. The mass of a helium atom is 6.65 x I0-27kg,and1liter 2ti2
iT2
2m
1 054
2(6.65
X x
==
1 0 �m·�.Also�n
1 0- 34 27 10- )
==
1.054
x 10-34Js.Thus
2
At room temperature the mean kinetic energy of a helium gas atom is .
e1
�
kT
=
( 1.38
X
23 10- ) (293)
=
4.04
X
10-11 j
::=::=
2.5
X
10-2 eV.
Sec. 12.6 e3
l
Density of Quantum States
1
•
'�-_....._j
I
J
•
I
81 l
e
e
e
• 8
229
I
a
f
•
L
a
J
83
==
5
NJ
1
Nt
=
3
=
5
J
J
8t
=
•
Distribution of e1even particles among three energy levels. •
Then nJ
.
2.5
;;:::-! -
s
X 10
x
-
1 2 12 =
to-21
2
X 10
9•
Thus, for the vast majority of the molecules of a gas at ordinary temperatures, the quantum numbers nj are large indeed. In other words, most of the mole cules occupy highly excited states. This means that the energy levels are very closely spaced and that the discrete spectrum may be treated as an energy continuum. That is the subject of the next section . •
12.6
DENSITY OF QUANTUM STATES •
Under the conditions that the quantum numbers are large and the energy levels are very close together, we can regard then's and thee's
as
fotrning a con
tinuous function rather than a discrete set of values. We are interested in
finding the density of states g(e).
Dropping the subscripts in Equation ( 1 2. 2 1) , using 1i
ing for n2, we obtain
n2
=
n X2 +
nY2
ny, nz ( nx, ny, nz )
The possible values of nx,
+ n l2
8mV213 =
h2
e
aU positive integers
R2•
=
h/2Tr,
and solv
(12.22)
correspond to points
space. According to Equation (12.22), for a in a cubic lattice in given value of e, the valttes of nx, ny, nz that satisfy this equation lie on a sphere of radius R. A cross-section through the positive octant is shown in Figure 12.9. We let g (e)de be the number of quantum states whose energy lies in. the range
e
to
e
+
de.
This is the number of states whose quantum numbers
( nx, ny, nz ) lie within the infinitesimally thin shell of the octant of a sphere with radius proportional to the square root of the·energy (Figure 12.10). Evidently,
g(e)de
=
dn(e) n(e +de) - n(e) "'"' de. de
(12.23)
Statistical Thermodynamics
Chap. 12
130
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Figure 12.9
A cross-section
through the positive octant of
y
radius R showing lattice points corresponding to the quantum
num bers nx
and ny-
/
•
•
/
.
./ .
•
•
•
•
•
•
R/
"--
0
_,__ _ .,..
________
11.\ •
Here we have expanded the function n(e +
de) in a Taylor series and
retained only the first two terms. Now n (e) is the number of states cont ained within the octant of the sphere of r a d ius R; that is,
n(e)
1 4
=
8 3 7T R .
3
1T =
6
8m V h2
3/2 .
3/2
e .
( 1 2 24) .
so that
g(e)de
Equation
dn(e) =
de
de=
4
(12.25)
(12.25) is not quite the result w� seek. It takes into account the
translational motion only of a particle of the assembly. But quantum parti
cles may have spin as well. The result for g ( e )dr, is valid for bosons of zero spin (bosons are particles witl1 integer spin). For fermions of spin one-half (fermions are particles with half-integer spin)� there are ll·vo sp in states for
each t ranslat i o nal state. Thus we must multiply Equation factor y5:
(12.25) by a spin
(12.26) where Ys
=
1 for spin zero bosons and 'Ys
=
2 for s pin one-half fermi
For molecules .. states associated with rotation and vibration mav ex ist -
in addition to translation and sp i n . We shall consider such cases in later c h ap t e r s .
231
Problems
..
' '
.
.. .·
.
.
.
e +de . ..
.
.
.
·..
.
Figure 12.10
Shell enclosing quantum states with energies between e and e + de.
..
. .·.
·.
•
0
0 •
PROBLEMS 12-1 Suppose that you flip 50 "honest" coins.
(a)
How many microstates are there? Give your answer as a factor of the order of unity times an integral power of 10.
(b)
Ho\\' many microstates are there corresponding to the most probable macrostate?
(c) What is the true probability of achieving the most probable macrostate? Note: Use a calculator that gives you n! or a table of gattuna functions (f(n + 1) n!). Stirling's approximation will not give you sufficient accuracy. =
1.2-2 Do Problem 12-1 for the case of 100 coins. 12-3 Using whatever computing means you have at your disposal, investigate the error in Stirling's approximation
In n!
�
n In n
-
n.
(You might display the percentage error as a functionof n for n
>
10, say. Be
creative.) For what value of n is the error less than 2 percent (approximately)?
12-4 Suppose you flip 1000 coins. We found that the thermodynamic probability of getting 500 heads and 500 tails is w = lQ-'00• Show that for the case of 600 heads 9 is smaller by a factor of 400 tails w and 10 • ll-5 For N distinguishable coins the therntodynamic probability is w=
N! --
N1 ! ( N
where Nl is the number of heads and N
(a)
-
N1 ) ! '
N1 the number of taiis.
Assume that N is large enough that Stirling's approximation is valid. Show that lnw is a maximum for N1 = N /2.
(b)
-
Show that
Wmax
=::
eN1"2.
Chap. 12
232
Statistical Thermodynamics •
12-6 Consider a model thermodynamic assem hly in which the allowed (nondegener ate) states have energies 0. r.. 2r.. 3r.. 1l1e assembly has four distinguishable (localized) particles and a total energy U
=
6e.
(a) Tabulate the nine possible distributions of the four particles among the energy levels ne where ..
(b) Eval u ate wk (c)
0 1 ......
n =
..
for each of the ma crostates and calculate fl
=
Calculate the average occupation numbers
·,;v,
=
L N � 1{'�/0 J
of the four particles in the energy stat·�S.
12· 7 Six distinguishable particles are to be distributed among energy levels 0 e, 2e 3e 6e. ..... each with a degeneracy of three. The 1 otal energy is U (a) Tabulate the 1 1 po ss ible macrostates nf the assembl y. (b) Calcu late 1v� (the number of microstates) for each of the 11 macrostates as a factor times 3tl. (c) What is the total number of available microstates fl for the assentbly? (d) Find the at'erage occupa ti on numher for each energy level. (e) What is the sun1 of the average occupation numbers (it should be six)? ..
..
=
12-8 Two distinguishable part i cle s are to be distributed an1ong nondegenerate energy 2£. levels 0� e and 2e such that the tota l energy U (a) What is th e entropy of the assen1hly? (b) If a di st i ng u i s h abl e part ic l e \\·ith zerc energy is added to the system .. show that the entropy of the assembly is i nc reased by a factor of 1 .63. =
..
U-9 In reference to E quati on ( 12.1.+) prove th a t if .
j(xy) wher e
x
=
f(xl
+
j(y).
and y are indepe nd e n t variables. then .f' ( x)
=
C In
x�
(
"'
==
c ons tan t .
(Hint: Take parti al derivatives of the equation.) 12-10 (a) Tabul ate the values of the quantum numbers n, n , .. energy levels of a pa r ticl e in a contain er of volume V (b) What is the d egen eracy g of each level? �
n
=
,-
for the 12 lowest
L'.
(c) Find the energy of each level in units of 1r2it2j2n1L2• (d) Are the energy levels equally s p aced ? 12-11 Calculate the value of 11; in \vhich an o x ygen atom confined to a cubi cal box 1 em on each side will h ave the same ene rgy as the lo\\'est energy available to a helium 10 n1 on each side. atom confined to a cubical box 2 x 10 12-U Consider a gas consisting'of one ki lo m o le of helium atoms at standard tempera� ture and
e
=
the de g e neracy g( e)
for the energy level 1 ). What is the approx in1 ate ratio of g (E) to the num
pressure. Calculate
( 3/2) kT
(take y,
ber of atoms N?
=
..
.
.
.
•
•
dSSICd
an
•
•
13.1 Boltzmann Statistics
235
13.2 The Method of Lagrange Multipliers
236
13.3 The Boltzmann Distribution
238
13.4 The Fermi-Dirac Distribution
242
13.5 The Bose-Einstein Distribution
244
13.6 Dilute Gases and the Maxwell-Boltzmann Distribution
246
13.7 The Connection Between Classical and Statistical Thermodynamics
248
13.8 Comparison of the Distributions
253
.
13.9 Alternative Statistical Models
254
233
.
13.1 BOLTZMANN STATISTICS
We wish to detertnine the equilibrium configuration for an assembly of N distin
guishable noninteracting particles subject to the constraints (Equations and
(12.9))
(12.8)
n
,_,Ni
j=l
=
N,
(13.1)
n
(13.2) Here
� is the number of particles with single-particle energy sj; N and
U are
assumed to be fixed quantities. Our goal is to find the occupation number of each energy level when the thertnodynamic probability is a maximum. We need to allow for degeneracy. Consider the first energy level, j
=
1.
The number of ways of selecting N1 particles fro1n a total of N to be placed in the first level is
N N1
=
N! -----
N1! ( N
-
N1 ) ! .
We ask: in how many ways can these N1 particles be arranged in the first level? There are g1 quantum states in the first level, so for each particle there are g1 choices. That is, there are (g1)N• possibilities in all. (Suppose that g1 = 2 and N1
=
3. For each particle there are two choices, so there are
23
choices total
(Figure 13.1).) Thus the number of ways to put N1 particles into a level containing g1 distinct options is
N N!g1 , --
Nl ! ( N
-
N1 ) ! .
235 '1
Chap.
236
Stitte b
State a 123
123 3
12 23 31
Figure 13.1 The eight arrangements of three distinguishable particles in two quantum states, a and b. The particles are labeled 1, 2 and 3 .
Classical and Quantum Statistics
13
1
., '-
23 31 12
1
.,
-
.
For the second energy level .. the situation is the same, except that there are only ( N - N1) particles remaining to deal with:
Continuing the process" we obtain
- N'.
_
g /·V 'g N,�g N3
II
g ·'".
' �-· ' N. . ' N N)•'N2·'N3'···· j=J j• 1
2
3
·
_
(13.3)
Having developed the statistics 1v8 for the so-called Boltzmann distribution, our
task is to maximize w, subject to the conditions of Equations ( 13.1) and ( 13.2).
13.2
THE METHOD OF LAGRANGE MULTIPLIERS
The so-called method of Lagrange multipliers is tailor-made for our problem. Suppose we wish to maximize a function the constraint 4>( x
...
y)
=
f ( .x y) ..
of two variables subject to
constant. We must satisfy the condition
df
rif =
ax
dx
+
at av •
dv �
=
0.
Sec. 13.2
If
The Method of Lagrange Multipliers
237
dx and dy were independent, we would conclude that iJfjax
=
iJffiJy
= 0.
However, they are not independent but are related by the constraint equation, which can be differentiated to give
dt/>
iJc/J dx + dy ax oy
de/> =
=
0.
It follows that •
iJffiJx iJcfJfiJx
iJffiJy iJt/J/!J y
-
·
•
If we let the common ratio be the constant
iJf ax
iJ
+
= 0,
iJf ay
-
+
a, we then have
iJtjJ a ay
(13.4)
= 0.
These are exactly the expressions we would get if we attempted to maximize the function
f +
ac/J without the constraint. The method adds a new unknown
a called the Lagrange multiplier; it is the price we pay for the constraint.
The extension to a function of
n
variables is straightforward. Let
(13.5) be subject to the condition
(13.6) For a maximum, df
=
0. Since cJ> is a constant, dc/J = 0. The method gives
iJf OX·
-
iJc/> a dX·
+
'
.
1, 2, 3
= 0, l =
.
.
.
n.
(13.7)
'
(13.6), constitute a set of n + 1 equa xn tions that can in principle be solved for the independent variables x1, x2,
These equations, together with Equation
•
and the Lagrange multiplier .
If there are
(13.2)),
two
•
•
a.
constraint relations, as in our case (Equations
cP
=
.p
=
cf>( X1, X2, t/I(Xt, x2,
•
•
.
.
•
.
(13.1) and
)
(13.8)
)
(13.9)
Xn , Xn ,
Classical and Quantum Statistics
Chap. I 3
138
we must introduce two arbitrary multipliers a and {3. In this case.. we have a set of n
+ 2 equations
the two constraint equatio11s and the
at
- + a
dX;
a�
a� f3-:il.ri
+
..
dX1
=
0,
.
1
1 =
,
n
2.. 3 . .
equations of the form
(13.10)
. 11.
Thus the problem is fully specified. The multipliers are often called ��undeter mined," but they can be found.
13.3 THE BOLTZMANN DISTRIBUTION
Because In
w
is a monotonically increasing function of
equivalent to maximizing Equation
u1.
1v
..
maximizing In
w
is
The logarithm is much easier to work with. From
(13.3) we have ..
"
In
lV
=
ln.N!
+
N. In I ...,
II cJ. '"''
(13.11)
�
i= l
Since we are concerned with very large numbers, we Stirling's approximation and write
safely use
can
.
In
w =
InN!
(13.12)
+ I •
.
I
I •
We needn't approximate InN! because it is a constant that vanishes ltpon dif ferentiation. In applying the method of Lagrange multipliers.. we take partial deriva tives with respect to a particular valueNj of the set (N,,N'!. ..... N; ... Nn). Thus
(13.13) •
where
(13.14) .
I
.
I
So
aN.J
.
I
(13.15)
Sec. 13.3
The Boltzmann Distribution
239
In working out the derivatives, we note that the only terms of the sums whose j. derivatives with respect to � are other than zero are those for which i Therefore, Equation (13.15) reduces to =
(13.16) =0
Then
or
(13.17) Equation (13.17) gives, for every energy level of the system, the number of particles per quantum state for the equilibrium configuration of the assembly. The result, obtained under the assumption of distinguishable particles, is called the Boltzmann distribution function. The constants a and f3 are related to the physical properties of the assembly. We consider {3 first. If we multiply Equation (13.16) by � and sum over j, we obtain
•
•
•
I
I
I
•
I
or
•
•
J
J
The two terms on the left-hand side appear in the expression for In Equation (13.12). Making the substitution, we get ln
w
=
In N! + N -aN- {3U.
SimpJifying, we have In
w =
C- {3U,
w
of
240
Chap. 13
Classical and Quantum Statistics
where C is a constant. Using the statistical definition of entropy, S
k In
=
·1v
=
S0
-
(13.18)
k{3U,
where 50 is the constant kC. From classical theory, riS
T
T
dU +
riV
v
as av
c
dV,
g1vtng •
•
1
as au
From Equation
=
-
T.
v
(13.19)
(13.18), ( 13.20)
It follows that
1
.
{3
�·
=
-
. kT
(13.21)
The association of the classical expression with the statistical expression for the entropy suggests that the temperature is a Lagrange multiplier!* Note that the partial derivative in Equation
(13.19)
is taken with the volume held con
stant. The constancy of V is implied when \Ye write Equation (13 .2 ) , since e/x v-213• as we saw in Chapter
12. Substituting Equation (13.21) in Equati<>n ( 13.17), we obtain
of a can easily be found from the first constraint equation (Equation ( l 3 1 ) ) :
The value _
.
.
1
*Some texts give
{3
=
+
J Ik T. The
Lagrange multiplier. which is arbitrary.
.
I
positive sign comes from the choice of sign of the
Sec. 13.3
The Boltzmann Distribution
241
so that
a e =
N ----
'
•
1
and hence
�·= 1
�= gj
T k e;/ Ne
(Boltzmann distribution),
--
(13.22)
•
1
where
fi is the probabilit)' of occupation of a single state belonging to the jth
energy level. The sum in the denominator has a special significance. It is called the partition function or sum-over-states (Zustandsumme in Ge1tt1an), and is rep.. resented by the symbol Z: ,,
z j=!
T k i , e e . i
(13.23)
g
The partition function depends on the temperature and on the parameters that detertnine the energy levels and quantum states. If the energy levels are crowded together very closely, as they are for a
gi may be replaced by g(e)de, the number of states in the energy range from e to e + de. Correspondingly, � can be rep1aced by N ( e)de, the number of particles in the range e to e + de. We then gaseous assembly, the degeneracy
obtain the continuous distribution function·
f(e)
=
N(e) g(e)
=
T I< / �; Ne-
----
k ) / P. g(e e Tde
.
(13.24)
The equation is completely analogous to the expression for the discrete . energy spectrum. It is worth noting that the occupation numbers are fully determined by the temperature and the volume (whiclt established the energy levels and the degeneracies). The set of occupation numbers, in turn, that maximize w specify the equilibrium macrostate. Thus two state variables define a thermodynamic state, exactly as in the classical theory. The maximization of
w
does not absolutely require the use of Lagrange
multipliers; an alternative method is given in Appendix C.
Chap.
241
.
Classical and Quantum Statistics
13
.•••·• :: ·.:· :· ·: :·=-·: :: : ...•.·.·::-· -.·:· ·: ···:· · -: ·-:.-:.• ·- ::·.:··.·:·.-:.· -·.· . ..-:.: :·:·• ·•· :·:.·. -:: :· :·.·.·-: - • .·:·.•..· · ·.: ·.·: · :·... : ... :- .· : ·· .·: ···.·:-: --:;:: :·.· ·::.:;.•:: ... ::::: •· •· ·:·· .· :.•:· ••-:- ••••·:.:..-•.:- ·.•.• ...·. .· . · : : ·.. :-: -::-: ·.·:· ·· · ··:·::.· :··· .; · : :·..... :· ·. :·:· ·• : :..:. < -:-·� �::;v · :·.:-:.; · : · :. :··� :··:··1':..:· ·: .·: · · :.· :-.· :-·..· :· • ::-.: : .-:.•: -.:: -· . : -.: :.· .. -: :-: -·. :-· .·:·. -·.·.·: :· :. :-·• : :... : ::.• ::�::::.:.:.:::. ·.· .·:-:• -: :-:·..· : ::. .. : ::: ;:::..::: . -·.:.·· :··•··: :·:· ·: :: :.:: :·:·:.:-· .-:.·..:·:..· ·.:···. :: .· ·.·:-.· :: · :· ··..··:.·:·: ·:•:·.;; ·· • : -:-.: -:; :·.:··:.·<·.:·:-. -· .-:·.-· : -:· ...:.·.-:.· -.·:: . · :: .. .· · •.· · •. . · · .. · . · . · •.· . . · · . . · · :: · · ·: .: :· · . · • .• . • · . . • .. · .. . . . . -• .•. · . . . , , . · . •• • .·• .• · ·. · . . · · ·. · • · . . · · . · . .... : :· · . · . · :.. . . · ·.:..· ·..·:· .: • - · :: ..· · :· . . • · • · · ·.·
.
•
13.4 THE FERMI-DIRAC DISTRIBUTION As we have discussed .. Boltzmann statistics can be applied when the particles of the assembly are distinguishable a solid lattice. In this section we assume and
fl)
say, by their positions i n
that the particles are identical
indistinguishable, and (2) that they obey the Pauli exclusion principle. This
means that no quantum state can accept ffi(>re than one particle, taking spin into account. Such particles have half-integer spin and are given the generic name
fermion.
Examples of fermions are electrons, positrons, protons, neu
trons. and muons. An important application of Fermi-Dirac statistics is to the behavior of free electrons in metals and semi-conductors. Invoking once again our picture of energy level shelves with quantum state boxes resting on them., we emphasize that in the Fermi-Dirac case, one particle or no particle occupies a given state so that
N;
gi for all j (Figure 13.2).
<
In determining the thermodynamic probability for a given macrostate we note that the group of 8; states is divisible into two subgroups: are to contain one particle and
(g1 - Ni)
N; of the states
of the states must be unoccupied. The
counting problem is precisely the same as that of the coin-tossing experiment .. where we divided N particles into
N1
heads and
N - N1
tails and obtained
N!
Here
(13.25) for the jth energy level. The total number of microstates corresponding to an allowable configuration is simply the product of the individual factors of the forrtl of Equ_ation
(13.25) for all the levels:
IJ
To maximize
WFo
gI.'.
.
(13.26)
we follow exactly the same procedure as in the case of
Boltzmann statistics. We use the method of Lagrange multipliers� assuming .
} th energy lever
f
•
Figure 13.2
1 l
___
__,) {
·
•
f l
•
) L.
._I ___.lj 1L...,_-..Jj
_
__
1
e1•
-
NI
·
•
Fern1ions in quantum states with energy
- 7
gi
=
3
The Fern1i-Dirac Distribution
Sec. 13.4
143
conservation of particles and energy, to calculate the occupation numbers for the equilibrium macrostate. Taking the logarithm of Wro, we have In
wFo
(13.27)
= •
•
'
•
'
l
Using Stirling's approxi111ation, this becomes
In wro
(g; InN;+ g;g; In N; N; [g;
=
-
N;)ln(g;- i'V;) + (g;
-
N;)]
'
•
[g; In g; - N; In N;
-
-
(g; - N;) ln(g, - N;) ].
(13.28)
•
I
'
•
•
'
a(In Wfo ) ----
aNJ.
iJf/1 iJc/J +a +� aN:.J aN.J
=
0,
j
=
1, 2, .
.
.
n.
lead to -
N.·ln N· +
-
iJNI.
I
I
'
,__
(g;
-
N;)ln(g; - N;)
•
•
'
a
a
+a aN
. N; + f3 aN ' J
I
(13.29) .
'
Here we have taken into account that
-
g; In g;
-
aN.I
=
0,
•
l
since the g;'s are not functions of the N;'s. Working out the derivatives, we obtain
Ni l
•
(gi- �) \,
I
4
-I
1
or
NI·
a
<'
Chap.
244
Classical and Quantum Statistics
13
or
(13.30) As before, we set
{3
=
- 1/kT.
H owever the determination of the other .
Lagrange multiplier by summing over the N/s doesn't work for this distribu tion. Provisionally, we associate
a
with the chemical potential J.L divided by
kT,
and reserve for later the physical interpretation of this connection. Thus we set
(13.31) so that
( 13.30) becomes (13.32)
The corresponding result for the con tinuous energy spectrum is 1
(13.33)
13.5 THE BOSE-EINSTEIN DISTRIBUTION
Not all elementary particles o bey the Pauli exclusion principle; ph<>tons are the most notable exception. Thus we need st at is tics for indistingtlishable parti
cles, any number of which can occupy a given quantum state. Such particles have zero or integer spin and are called
bosons.
The sta tist i cs developed for
this case leads to the Bose-Einstein distribution, named after its inventors. Counting microstates here is slightly more complicated than it is for Ferrtli-Dirac statistics. For the jth energy level there will be g1 quantttm states containing a total of N1 identical particles with no restriction on the number of particles in each state. It is convenient to depict the arrangement of the N1 parti cles among the g1 states by
(g1
-
1) _partitions or lines and N1 dots (Figure 13.3).
jth energy level
•
•
Fig11re 13.3
•
•
•
•
•
•
•
•
•
•
•
•
N-I = 14
Bosons in quantum states with energy
e,.
Sec. 13.5
The Bose-Einstein Distribution
145
Now, we can obtain new microstates b�' shuffling the lines and dots while keep
gi and � fixed. We ask: in how many ways can the 1) symbols (lines and dots) be arranged into (gi- 1) lines and�
ing the numbers
(�+gi
dots? Again we have the binomial problem of the coin-tossing experiment. Clearly the answer is
W· =
(�+gj-1)! �!(gj- 1)!
1
for the jth energy
level. The
(13.34)
---
--
of
number
microstates for a
given
macrostate is,
then, the extended product
(�+gj-1)!
n
}. g}
j=l
Taking the logarithm of Equation
In
WsE
(13.35)
•
(13.35), we have
= '
•
•
I
•
'
Invoking Stirling's approximation, we obtain In
w8E
= •
[ ( lV;
+
g;
-
1) In ( N;
+
8; - 1 )
- ( N;+ 8;- 1) - N; In N;
l
1)+ (g;-1)]
+ N;- (g;- l)ln(g,
[(N;
=
+ g;-
l •
l)ln(N;+g;-1) - N; InN;- (g;- l)ln(g;-1)] . (13.37)
With the constraints taken into consideration, the method of Lagrange multi pliers gives a (In
WeE
a NJ.
)
+a
dc/J aN.J
+
J3
iJ«fi aN.1
=
0, j
=
1, 2, 3, .
.
.
n,
which leads to
iJN1.
l.
( N;+g;
-
1 )In ( N;+g;
-
1) I
•
(13.38)
Chap. 13
246
Classical and Quantum Statistics
or
1
1
N·
+
g· -
1
N1
N;
+
g;
1
J
-
1
I
This gives
In
N·J + g·J
l
-
Q = -a- tJej.
N·1
Neglecting unity compared with Nj + gi we obtain the result ..
-
With f3
=
-1/kT and
a =
=
----
(13.39)
J.lfkT. we have
(Bose-Einstein distribution).
For the continuous energy spectrum
(13.40)
..
1
(13.41)
13.6 DILUTE GASES AND THE MAXWELL-BOLTZMANN
DISTRIBUTION Since elementary particles have half-integral or integral spins.. the particles of a gas are either fermions or bosons: they are clearly one or the other. However, it is useful to consider another kind of statistics pertinent to a so
called dilute gas. The word --dilute·· means that for all energy levels.. the occu pation numbers are very small compared with the available number of quantum states (most quantum states N , << �,.
are
empty). We assume that
for all j.
This condition holds for real gases except at very low temperatures.
(13.42)
Dil ute Gases and the Maxweii-Bolamann Distribution
Sec. 13.6
247
If in this region very few states are occupied at all, it is extremely unlikely that more than one particle will occupy a given state. Thus it is irrelevant whether or not the particles obey the Pauli exclusion principle, and we might therefore expect Fermi-Dirac and Bose-Einstein statistics to be approx imately identical in the dilute gas limit. This is indeed the case, as can easily be seen. Recall Equation (13.26),
(13.43)
Wfo=
Now
,
gj! ----
(gj- �)!
=
gj(gj - l)(gj- 2)... (gj- �
+
l)(gj- Nj)!
._
J N
so that
(13.44)
The approximate value is slightly greater than the exact value since factors such as (g· - 1}, (gi- 2), etc. are written as Ki·
(gi •
J
+
Ni- 1)!
(13.45)
�!(gj- 1)! '
where
(gj
+
�- 1)!
=
(gj
+
�- l)(gj
+
�- 2)... (gj
+
�- �)(gj- 1)!.
We see that in the numerator of Equation (13.45) there are
(gi- 1)! so that for � << gi,
� terms ahead of
which is slightly less than the exact expression. Thus
gfj(gj- 1)! i �!(gj- 1)!
(13.46)
Classical and Quantum Statistics
Chap. 13
248
It is apparent that for a dilute gas� Ferrni-Dirac and Bose-Einstein statistics give virtually the same therrnodynamic probability This ""classical limit_" called .
Mttrwell-Boltzmann statistics, was investigated long before the development of quantum mechanics. Note the difference between Maxwell-Boltzmann statistic& Equation (13.46), and Boltzmann statistics_ Equation (13.3 ):
(13.47)
The much larger Boltzmann proba bility includes the pernltltation N! tlf the N identifiable particles, giving rise to additional microstates. The distribution of particles among e11ergy levels can be found by the usu al method of Lagrange multipliers. However, the result can be written down immediately by obser ving that
'lVMB
and
1.08
differ only by a c<>nstant
the factor N!. Since maximizing the thermo tlynamic probability involves tak ing derivatives and t he derivative <>f
a
constant is zero_ we get pre cisely the
Boltzmann distribution Nj !'""""'
=
N e-
.
-
)
This should not come as
cJ/kT -
z
a
• -
(Max\vcii-B<)Itzmann distribution).
( 13.48)
surprise: two functions differin g only by a constant
will have maximum values at the same point. Because the two distributions are identical, Boltzmann statistics and MaxweJJ-BoJtzmann statistic s are frequently C<>nfused with each other. Boltzmann statistics assumes distinguishable (localizable) particles a11d there fore has limited application_ largely to solids and some liquids. For gases. either Fertni-Dirac or Bose-Einstein statistics applies_ depending on the spin of the particles. Maxwell-Boltzmann statistics is a very useful approximation for the special case of a dilute gas_ which is
a
good model for a real gas under most
conditions. It so happens that the cor responding distribution is the same as the Boltzmann distribution.
13.7 THE CONNECTION BETWEEN CLASSICAL AND STATISTICAL THERMODYNAMICS The statistical expression for internal energy is
.
I
Sec. 13.7
The Connection Between Ciassical and Statisticai Thermodynamics
149
Taking the differential, we have
(13.49) •
•
J
1
where si is some function of an extensive property X such as the volume:
Since dei -
(dBi/dX)dX, it follows that
•
N·de· } J
.., =
dei
.J �dX
J
dX.
Let
Then
dU
= •
e·dN· 1 I
Y dX.
}
For two states with X the same, dX
=
0 and
(13.50) •
I
The classical analog to Equation (13.49) is
dU
=
TdS- YdX,
so that
(dU)x
=
TdS.
Comparing Equations (13.50) and (13.51), we see that
(13.51)
Classical and Quantum Statistics
Chap. 13
250 Accordingly,
Furthermore, since aQ,
=
e ·dN· � J J
T dS and aW,
=
aQ,
=
}' dX� it follows that
and
(13.52)
The first equation states that heat transfer is energy resulting in a net redistri bution of particles among the available energy levels. involving no work. In Figure 13.4 heat added to a system shifts particles from lower to higher energy
levels. The second part of Equation (13.52) can be interpreted in terms of an adiabatic reversible process (no heat flow) in which work is done
ott
the sys
tem (note the negative sign). An increase in the system"s internal energy could therefore be brought about by a decrease in volume with an associated increase in the e1 's. The energy levels are shifted to higher values with no redis tribution of the particles among the levels (Figure 13.5 ). The relationships of the two parts of Equation ( 13.52) provide a deeper understanding of the fundamental concepts <>f thermodynamics.
j�
�
'
'
'
' '
N,
--
'
'
-.... ...- .....
\
'
"�
, ..... ....
'
-
'�
--
•
.
'
'
'
'
�
Figure 13.4 Heat added to a system moves particles from lower to higher energy levels.
\
.... .
'
'
'
\ • �
'
NJ
'
' '
' '
'
'
---
'
'
' '
Figure 13.5
Work done on a
system moves energy levels to hig her values.
-
'
'
'
£· J
£· 1
Sec. 13.7
The Connection Between Classical and Statistical Thermodynamics
251
We turn our attention next to the chemical potential, the quantity appear ing thus far in two of the distribution functions. In Chapter 9 we saw that when
matter is added to or taken from an open system, the change in internal energy is dU
=
TdS- PdV +.p,dN,
(13.53)
where 1-L is the chemical potential defined on a per particle basis.* Now the
Helmholtz function is defined as F dF
=
U- T S so that
TdS- PdV + p,dN- TdS- SdT,
=
or dF
=
-SdT- PdV + JLdN.
Therefore, JL
iJF =
aN
(13.54)
.
T.V
Let us calculate firstS and then F fo r Maxwell-Boltztnann statistics:
(13.55)
with
Then
S
=
k ln
k
w =
N,.In g. � J 1 •
J
•
I
--- k ,_..
•
•
J
J
•
(13.56)
•
1
Substituting the distribution function of Equation (13.55) in Equation (13.56), we obtain
*Equation (9.1) is dU
=
T dS- P dV
is Avogadro's number. Thus, if we p,dN;p, is the chemical potential energy per particle.
defined as energy per kilomole. But n de fi ne J.' as JJ.*/N A• then p,*dn
=
+ JJ,*dn, where J.L* is the chemical potential
=
NINA, where N,.
Classical and Quantum Statistics
Chap. 13
252 or
S
==
u T
+
Nk ( ln
Z
-·
In
N
+
1)
(13.57)
.
Then
•
F Using Equation
=
U- TS
=
-NkT(ln
Z
-
In N +
1).
(13.58)
(13.54)� we find that J.L
=
- k T ( ln Z- In N + 1) +
NkT N
.
or
11 = r-
kT
In
N -·-
z
( 13.59)
.
For gases, the partition function Zis proportional to the volume� as we shall see in Chapter 14. Thus /J- is an increasing function of both the number of particles per unit volume and the temperature. This makes sense since particles flow from higher to lower chemical potential and fr{>rn hi gher to lower concentrations. The difference in chemical potential is a true potential energy: for two systems it is the potential barrier that will bring the systetns into diffusive equilibrium. An example is the coexistence of two r>hases of the same substance
a liquid and its vapor at the same temperature. In Chapter
say..
9� we observed that
the chemical potentials for two phases of a given constituent must be equal for the system to be in diffusive equilibrium. This means that there is no net transfer .
of particles from one phase to the other: the total number of particles/Vis a constant. This is precisely the condition that allowed
us
to determine the Lagrange
multiplier a. Thus the association of a with Jl/k Tis completely plausible. Finally we note that Equ ation ,
N
_
z
(] 3.59) can be written ==
e�-LikT
(13.60)
Substituting this in the Maxwe11-B())t7.mann distribution gives
g
--
---
-
'
e(t,- J.t}rkT.
(13.61)
We are now in a position to summarize a11d compare the distribution functions we have derived in the previous sections of this chapter.
153
Comparison of the Distributions
Sec. 13.8
I I 18
I IIIII-I II 111M
IU
S N IO T U IB R T IS D E H T F O N O IS 13.8 COMPAR
The distribution functions for identical indistinguishable particles can be rep resented by the single equation
Ni
-
gi
-
1
(13.62)
-----
e
a
'
where + 1 for FD statistics a =
-
1 for BE statistics
.
0 for MB statistics
For
� <<
gi, the denominator in Equation (13.62) is very large compared with
unity, and the MB distribution is an approximation to the FD and BE distribu tions, as we have seen. Plots of y N;/g1 versus x ( ei �-t) /kT are shown in -
Figure 13.6.
A word about the sign of JL is appropriate. The value of the chemical
potential in some sense is arbitrary. When particles are being transferred from one system to another, it is important to establish a common zero-energy ref•
erence point from which all potential energies are measured. The problem is exactly analogous to defining a reference level for the gravitational potential in classical mechanics or for a potential well in quantum mechanics. Thus ei
-
p,, the thei"mal energy of a particle, is the energy measured relative to the
lowest point of the potential well. If a particle which itself has zero energy is inserted into a system, the ensuing interactions with its neighbors are most often attractive, in which case the chemical potential is negative. When
MB
BE
-
--
FD Figure 13.6 Comparison of the FD, BE a·nd MB distributions.
-3
-2
-1
•
X=
0 ei
1 -
kT
J.'
2
3
I
I
Chap. I 3
254
Classical and Quantum Statistics
systems are not exchanging particles. we can choose any convenien1 reference level, which is usuallv taken as zero. .
J
Examining the curves of Figure 13.6
1. BE curve: y
=
1/(ex- 1 ) :
.t
for
The distribution is undefined for wher e
centrate) in regions
curve: y For .x � -
2. FD
=
\VC
..
note the fo)Jowing:
)0 y ) x and for lar ge ..
..
x
x,
y
i=s
e-x.
< 0. Particles tend to condense (con
e is small. that is, in the lo\ver energy states. ; -x + l ) ; for x = 0, y = 1/2, and for x large, y :=�e .
1/(e�
y- ) 1. A t the lower levels with eJ J.L negative, the quantum states are nearly un i form ly populated with one particle per state. x,
-
3. MB curve: The curve y
e-x lies between the BE and FD curves and is
=
only valid for y << 1. This is the dilute gas r egi on : many.. many states are unoccupied. Note that
�
g,. =
g ,
--
N; At high temper at ure s
mately unity so that
.
k1r
�
( cJ
�
t
e- AI
( E.
c. ,
,
C'
{.;;)
.
<
c. r �
£1) << I and the cxponentia1 is approxi gJ. Since �- tends to increase with increasing -
energy.. so does the occupa t ion number. At /(J»' temperatures.. on the other hand, the population of the )<)Wcr states is fa\'ored: for g, = g1 N; > �- where i < j. ..
StatisticaJ equilibrium is a balance between the rando mizin g forces of thermal agitation, tending to produc e a uniform populati on of the energy lev
els.. and the t endency of mechanical systetns to sink to the states of lowest energy.. resulting in highl y ordered populatir>ns. For a negativ e chemical poten-
tial, Equation
(13.60) gives
z
=
N
When many states are available
e
to
P-
kr
=
N
e
4
JJ.
kT.
the system, Z is large ()nd 1-L is highly nega
t i ve. A particle seeks a position <>f lo\vest potential energy� choosing to be where the chemical p ot e nt ia l has its largest negative value.
Having developed the statistics and equilibrium state distribution func
tions for all situations of physi c� l interest.
\\'e are
no\\' in a positi on to apply
the results to some specific prc)hlems in subsequent chapters.
13.9 ALTERNATIVE STATISTICAL MODELS
The statistical approach d i sc ussed in the previous t\\'O chapters describes an
assembly of Nparticles with total energy U. whe re both Nand U are fixed. The assembly is
con si der e d
to he isolated in the sense that it does not exchange
Sec. 13.9
Alternatjve Statistical Models
255 •
energy in any form with an external system. Such an assembly is known as
a
microcanonical ensemble. In many experiments the system under investigation is not isolated. In chemical reactions, for example, the reaction vessel is (lften held at constant temperature by being kept in thertnal contact with a heat reservoir such as a water bath that absorbs energy generated in the reaction. In this case it is con venient to consider an ensemble of
a
large number
say
NA
of identical
assemblies in contact with a reservoir. The ensemble can be regarded as a sin gle assembly in a heat reservoir provided by the remaining NA
-
1 assemblies.
The reservoir fixes the temperature of the system rather than the energy. The individual members of this so-called canonical ensemble can exchange heat and work with the reservoir and hence the energy of the system can fluctuate. This is in contrast with the microcanonical ensemble in which there is no inter action with a reservoir at all. In spite of this difference, the two et1sembles give essentially the same results. The reason for this coincidence can be seen as follows. The assemblies of the canonical ensemble are separate macroscopic objects, like distinguish able particles, and can therefore be assumed to obey Boltzmann statistics, with the distribution
N.I
=
N . -e;fkT gIe ' Z z
(1 3 63)
=:
.
I •
The corresponding distribution for the ensemble involves the energy states of a single sample assembly, whicl1 alone has a very large number of states with a given energy. The allowed values of the energy E are extremely close together, so that the degeneracy
g(E)dE,
gi
can be replaced by a density of states
the number of one-assembly states (microstates) with energy
E and E becomes N(E)j1VA energy E. Thus
between
+ =
dE. Sums can be replaced by integrals and NilN P(E), the probability that the sample assembly has an
E
-E/kT
ZA
IX
(13.64)
.
0
The density of states g (E) is a rapidly increasing fttnction of
E,
whereas
the exponential factor is a rapidly decreasing function. For definiteness, sup pose that g(£)
a
En; then (13.65)
P(E) is found by setting its derivative equal to zero; the result (3/2)NkT for an ideal U. But U �ives n = Ejk T at the peak E (3/2)N, a very large number. Thus one factor in P(E) monatomic gas, so n
The maximum of
=
=
=
Chap.
256
13
Classical and Quantum Statistics
is large and increasing at an astronomically high rate and the other is extremely small and decreasing. As a result
P( E) has an exceptionally sharp peak. determining the probable fluctuation of E around ..
We are interested in the mean value
U. To do this� it is convenient to expand the logarithm of P(E) in a Taylor series about U: ln
P(£)
=
In P(U)
dlnP(£)
+
dE
E .:.u
. E=U
(£
-
(£- U)
2 U) + .
.
.
(13.66)
.
Here In P(E)
nln
=
n
dlnP(£) ---
=
.
n
lnP(£)
------
d E2
The first derivative is zero at E,
=
1 kT.
---
£
dE 2 d
£ -- EfkT,
:=.
.....
£
. 2 -
U and the second derivative is
-
2j3N k2T2
•
Using these values and neglecting terms higJ1er than second order�we find that
(13.67) where 3 u £2 :::;: - Nk2T2. 2
(13.68)
This is the usual form of the Gaussian distribution in terms of the standard deviation
u £,which
our purpose is the
is a measure of the peak width of
P(E).
More useful for
fractional width 1 -
u
-
--
Thus for a kilomole of an ideal gas. the deviation from the n1ean value
( 13.69)
U is of
the order of one part in 1013• It follows that predictions of the physical proper ties of a system using the canonical ensemble are virtually identicalwith those derived from the microcanonical ensemble in which the energy fluctuation is zero. When fluctuation effects are important, the canonical ensemble yields
Problems
157
information not obtainable from the microcanonical ensemble. Otherwise the two are closely related and either one can be derived from the other.
In the canonical ensemble we remove the restriction that the energy U is fixed. ., but the assemblies are still assumed to have a fixed number
N of parti
cles. As a further generalization, \Ve can consider assembl ies with an indefinit e number of particles. The resulting grand canonical ensemble is a system i n co n tact with a reservoir with ·which it can exchange not only energy but particles
as well. In this model only the mean number of particles is fixed. The grand canonical ensemble is useful when the constraint of a fixed number of parti cles is too restrictive, as in the problem of chemical equil i b rium among a num
ber of different species. Sumn1arizing the three approache� the microcantlnical ensemble treats a single material sample of volume V consisti ng of an assembly of
N particles
with fixed total energy U. The independent variables are V, N, and U The cen tral concept is the distribution Nil N, the fractional number of p art icles occu pying the jth energy level. The method is the least abstract of the thtee.· -It is -�·
mathematically the simplest but is completely adequate for solving the prob ,
lems usually discussed i11 undergraduate texts. The canonical ensemble considers a collection of NA i dentical assem blies, each of volume V. A single assembly is assumed to be in contact through
1 assemblies. a diathermal wall with a heat reservoir of the remaining NA The independent variables are V, N, and T, wi1ere Tis the temperature of the -
reservoir. The av eragi ng process involves the probability distribution function
P(E)
=
N(E)/NA. The
approach is more mathematically abstract than the
microcanonical ensemble, but one can be derived from the otl1er. 'The most general and most abstract of the three models is the grand canonical ensemble, which consists of open assemblies that can exchange both energies and particles with a reservoir. The independent variables describing the ensemble are V, T, and 1-L� where 1-L is the chemical potential. The grand canonical ensemble permits the treatment of more complicated problems than does either of the other methods and is correspondingly more mathemati cally dem�nding.
PROBLEMS
13-1 Use the method of Lagrange multipliers to solve the following problems: (a) A rectangle has a base x and a height y, where x + y 8. Find the values of x andy that give the maximum area. =
(b)
Find the area of the largest rectangle that can be inscribed in the ellipse X
2
az
y2
+
•
fj
=
1
.
What percentage of the area of the ellipse does the rectangle occupy?
Chap. 13
258
Classical and Quantum Statistics
13-2 Five identical noninteracting particles occupy the kth energy level, which is ten fold degenerate. How many possible microstates are there if
(a) the particles are bosons? (b) the particles are fermions? 13-3 A model therrnodynamic system in which the allowed nondegenerate states 6e. have energies 0. e., 2e .. 3e, . , consists of .four parti c les with total energy U .
=
.
Identify the possible distributions of particles, evaluate n
U'k,
=
and work
k
out the average occupation numbers for the various energy Jevels
(a) (b)
when the particles are gaseous bosons: when the particles are gaseous ferrnions.
13-4 Show that for a system of lV particles obeying Maxwell-Boltzmann statistics, the occupation number for the jth energy l eve l is give n hy •
�e.I
.
.1.
13-5 Show that it is poss i bl e to write the the rmodynamic probability in the general form
where -+--
a =
-
for FD statistics
1 for BE statistics
0 for MB statistics 13-6 Substitute the Maxwell-Boltzmann distribution function into the equation Le_;dNi
=
T dS
to obtain
Hint: note that
LdN1
=
0 and dN1 In
N-
-' �,
=
-
13-7 For the Fermi-Dirac distribution .. sketch
d Ni In
lv;/g1 versus e, ·
for T
=
0 and for T
sJightly greater than zero.
13-8 Show that for a system of a large number N of bosons at very lo"'· temperature (such that they are all in the nondegenerate lowt!st energy state chemical potential varies with temperature according to
kT J..L
� -··
r4o
N
--
•
£ =
0), the
Problems 13-9
(a)
259 At room temperature (300 K), what is the equilibrium number of bosons per state (N ilgi) for an energy ei (1) 0.001 e\' above the chemical potential J.L? (2) 0.1 eV above the chemica) potential p,?
(b)
At room temperature
(300 K), what is the equilibrium number of fermions
per state ( �/ gi) for an energy ei
(I) 0.1 eV below the chemical potential p,? (2) 0.1 eV above the chemical potentia) JL?
13-10
(a)
Using results from Chapter 9, sho\\' that
iJN Jv.v (b)
It follows from the statistical definition of the entropy that
IJ. Alnw=-k AN. T Consider a system with a chemical potential p,
=
-
0.3 eV. By what factor is
the number of possible microstates of the system increased when a single particle is added to it at room temperature?
13-11 A system has energy levels at e
=
0, e
=
300k, and e
600k, where k is
=
Boltzmann,s constant. The degeneracies of the levels are 1, 3, and 5 respectively.
(a)
Calculate the partition function, the relative populations of the energy lev
els, and the average energy, all at a temperature of 300 K.
(b)
k At what temperature is the population of the energy level at 600 equal to the population of the energy level at 300k?
13-12 A system with two nondegenerate energy levels e0 and e1 (et
>
eo
>
0) is popu
lated by N distinguishable particles at temperature T.
(a)
Show that the average energy per particle is given by u =
(b)
U N
Show that when T
and when T
=
·
�
E
o
+
1 +
e-!Jle A t P f; e-
e
de
•
=
€t
-
£o,
�
=
ljkT.
0,
)lo ·)0 ,
(c) Show that the specific heat at constant volume is cv
(d)
Compute
cv
=
k
in the limits T
versus �e/kT.
.
�e
2
-��:fkT e
- ---
kT �
(1
+
·--
e-�,;/kT)2
•
0 and T --+ ::,: and make a careful sketch of
cv
rea
14.1 Thermodynamic Properties from the Partition Function 14.2 Partition Function for a Gas
263 265
14.3 Properties of a Monatomic Ideal Gas
266
14.4 Applicability of the Maxwell-Boltzmann Distribution
268
14.5 Distribution of Molecular Speeds
269
14.6 Equipartition of Energy
270
14.7 Entropy Change of Mixing Revisited
271
14.8 Maxwell's Demon
273
261
14.1 THERMODYNAMIC PROPERTIES FROM THE PARTITION FUNCTION The power of statistical thertnodynamics is beautifully demonstrated in the application of the theory to an ideal gas. We shall begin by showing that all the thermodynamic properties can be expressed in terms of the logarithm of the partition function and its derivatives. Thereafter we need only to evaluate the partition function for a gas to obtain its thermodynamic properties. In Section
13.7
we obtained the following relations for the Max,vell
Boltzmann distribution, valid for a dilute gas:
S
u =
F
T
=
-
+
Nk (ln Z -
NkT(ln
In
N
+
Z- InN+
p, = - kT( ln Z - In
N)
1), 1),
.
(14.1) (14.2) (14.3)
To these we can add the other thermodynamic potentials and the pressure. • •
•
1. Internal energy Using the Maxwell-Boltzmann distribution function, we can write
(14.4) with
Z=
(14.5) •
1
Differentiating Equation (J 4.5) with respect to T, we have iJZ -
aT
v
1
�j
kT
(14.6) 263
The Classical Statistical Treatment of an Ideal Gas
Chap. 14
164 Since
e·
=
e·(V),
V constant
keeping
means treating e1 as a constant in
tuting it in Equation ( 14.4)�we obtain N
U
=
z
az
.,
(kT�)
aT �:
or
U
2.
a In z
1VkT2
=
(14.7)
•
aT
v
Gibbs function The calculation of G is trivially simple; in Chapter 9 we saw that
G
=
J.LN
for a single component system. Using Equation (14.3), we have G 3.
-NkT(In Z- InN).
=
Enthalpy The enthalpy is defined
G
U
TS
-
+ PV, so that G
Substituting Equations H
=
H
as
H
=
U + PV and H TS_ or
-
=
.
the Gibbs function is
-·
G + TS.
( 14.1 ) ( 14. 7), ,
U + NkT, or H
4.
( 14.8)
and ( 14.8) in this equation yields
�,
=
Nk T 1 + T
a In z
(14.9)
•
ar
.
Pressure A reciprocity relation from Section
p
8.5 gives
aF ==
-
Taking the derivative of Equation
av
.
-r •
( 14.2)
with respect to
V with
T held
constant gtves •
P
iJ ln Z NkT a\'
·-
=
•
T
(14.10)
Sec. 14.2
Partition Function for a Gas
265
14.2 PARTITION FUNCTION FOR A GAS
The definition of the partition function is
•
1
Suppose that gi
=
1 for all j. For convenience, we assume temporarily that the
sum runs from 0 to n
-
1 and we take e0
=
0. Then
Suppose that the energy levels are far apart. What does Z look like? For low ' temperatures, with the ei s differing significantly from one another, the expo nentials in the series get successively smaller quite rapidly, whereas at higher temperatures the falloff is more gradual (Figure 14.1 ) The partition function is a measure of the states available to the system and the distribution gives the occupation numbers, which are proportional to the factors e-e/kT. We see that .
for low T, essentially all the particles are in the ground state; for high T, the populations of the excited states are sigitificant. For a sample of gas in a container of macroscopic size, the energy levels are very closely spaced and many states are available to the system (Figure 14.2). It follows that the energy levels can be regarded as a continuum and we can use the result for the density of states derived in Chapter 12. In Equation (12.26) we use a spin factor l's
=
1 since the gas is composed of molecules rather than
-t=·lkT e J
1
\ \ \
'
1
low T
'
'
'
'
1\ \
high T '
'
'
,,
, ,_ .... .... .
. ... ...
, ' -.. ., ,
._,_._,_
...... ...... .... .... __
.._.__ ___,_____... �=__
so
el
(a) Figure 14.1
e2
ej
: ___._____...._,.. � _ ...._____ _
60
e1
£2
(b)
Successive terms of the partition function sum for (a) low temperature; and (b) hi gh temperature. The energy level spacings are comparatively large for a small volume.
83
ei
166
Chap. 14
The Classical Statistical Treatment of an Ideal Gas - &·ikT
e
J
�
l
low T
high T
1 '
I •
'
I • ' I '
r
'
. .
I
I I •
j
I
I ' '
-
I
I I •
I • '
E· J
I
(a)
•
•
I
I I
I I I I
•
. I '
I ) I .
' ' •
•
I
I
I
I
.
I
.
I
I
I
l
I '
I
'
'
I I
I
•
•
I
I
I
I
I I I I
I
•
. • •
I
I •
j
e· 1
(b)
Figure 14.2
Successive terms of the partition function sum for (a) IO\\' temperature; and (b) high temperature. Here the energy levels are very closely spaced when the volume is large.
spin 1/2 particles; that is� the quantization associated with the spin does not apply. Thus
g(e)de
4
=
-
27TV h3
---
1'2 l/2d e. m·l e ·
(14.11)
Then � · .
Z=
4
0
(I .
The integral can be found in tables and is
This gives the partition function for a gas under the assumption that the energy levels are so closely spaced that they form a continuum:
•
(14.12)
14.3 PROPERTIES OF A MONATOMIC IDEAL GAS The partition function· depends on both the volume V and the temperature T. We need In Z and· its partial derivatives:
Sec. 14.3
Properties of a Monoatomic Ideal Gas
In Z
3
= 2" In T
In V
+
a In
Z
2 In
=
v
2�mk h2
-
1
' V
3 1 2 r·
-·-
Substituting these expressions in the equations of Section
PV
=
NkT
and
,
-
T
a In Z
iJT
3
+
=
av
267
U
=
14.1 gives
3 zNkT.
(14.13)
However familiar these equations are, their appearance here is an exhilarat ing development. It represents the capability of statistical thermodynamics to produce relations that occur in classical thermodynamics as empirical generalizations. The calculation of the entropy is even more dramatic. It shows that the statistical theory can give results not at all obtainable in classical thermodynam ics. Using Equation (14 1 2) and U .
=
(3/2)NkT in
Equation
V (21TmkT)312
5
•
(14.1), we obtain (14.14)
This formula is known as the Sackur-Tetrode equation for the entropy of a monatomic gas. A rearrangement of terms yields
S
=
3 Nk - In T 2
+ In
v
+
N
-
S0,
(14.15)
where
S0 Recalling that s
=
Nk
= S/n and Nk/n s
=
Cv
5
2 =
In
+
ln
•
R, we obtain
T
+
R In
v
+
so,
(3/2)R. This is the equation for the specific entropy we obtained earlier. We note that s0 is an undeterrnined constant in the classical theory. T he Sackur-Tetrode equation shows us how to calculate the number S0, where
cv
=
which could only have been previously estimated on empirical grounds. We
The Classical Statistical Treatment of an Ideal Gas
Chap. 14
268
therefore now have two different ways of evaluating the entropy <>f an ideal monatomic gas. From measurements of the heat capacity at various tempera tures we can make an empirical detertnination and we can also make a theo retical calculation by using Equation (14.15). Excellent agreement has been found of the results obtained bv the two methods. We note.. however� that Equation (14.15) is obviously not valid down to absolute zero since S does not -'
approach zero as T
�
0.
14.4 APPLICABILITY OFTHE MAXWELL-BOLTZMANN DISTRIBUTION
Maxwell-Boltzmann statistics is valid u11der the dilute gas assumption
�·
<< gi. How good is this approximation f<>r gases under average conditions?
We can write Equation (14.12) in the form
where
1z0
2 1rm kT =
312
h2
has the dimension of the reciprocal of the volume (Z is a pure number) .. and is called the quantzlm ctJncentratiolz of the gas. Thus we can express the Maxwell Boltzmann distribution as
N-}
-
gj
-
-
N --·e
'
- r
z
Consider helium gas under standard
and T
=
273 K, so that
=
N V
1 -- e n0
·-·£ J
;kT •
( STP) conditions. Then 111
=
6.65
><
10--27 kg
The quantity N/V, the number of particles per unit volume, can be f<>und from Avogadro "s Ia\\·:
Sec. 14.5
Since
Distribution of Molecular Speeds
269
e/kT is of the order of unity and so therefore is
e-e/kT,
•
•
The result indicates that on the average only
.
a
few states in a million are occu-
pied. Helium gas is thus highly dilute under normal conditions. When we say that the gas is in the
Ni
<< gj,
classical regime. An ideal gas is defined as a gas of
noninteracting molecules in th� classical regime.
14.5 DISTRIBUTION OF MOLECULAR SPEEDS
The Maxwell-Boltzmann distribution of particles among energy levels leads directly to the speed distributions of molecules of an ideal gas. For a contin uum of energy levels,
N(e)de
=
f(e)g(e)de,
N(e) de is the number of particles having de. For the Maxwell-Boltzmann distribution
where
e
+
(14.16) energies between
e
and
(14.17) where, for an ideal gas,
Z=V
2TTmkT h2
---
312 •
(14.18)
The density of states is given by
(14.19) for molecules. Combining these equations we obtain
(14.20)
Chap. 14
270
Here
The Classical Statistical Treatment of an Ideal Gas
e is the single particle kinetic energy.lbus 1
�
=
Ci
- mv2
2
.
de
=:
mvdv.
and
N ( e)de� N ( v )dv, where N ( v )dv is the number of particles whose speeds are in the range v to v + dv.
Converting from kinetic energy to speed. we have
The computation gives the Maxwell speed distribution
N(v)d·v
4 N =
r.
n1
(14.21)
21rkT
This is precisely the result derived in Section 11.6 by using a more heuristic argument. The various averages discussed there need not be repeated here.
��
i!1888�1 8(. 1118�
14.6 EQUIPARTITION OF ENERGY In the kinetic theory t1f gases" an important result is that
(14.22) where e is the average energy of a molecule and f is the number of degrees of freedom of its motion. Thus. for a monatomic gas there are three degrees of freedom� one for each direction of the molecu]e ,s translational motion._ and
U
=
Nf:
=
( 3/2) NkT. It should be possi b l e to obtain Equation ( 14.22) from
statistical thermodynamics� and, indeed. that is the case. Let
z
be a parameter such that e = e( z). Each
a degree of freedom. For example, if velocity.. then z
=
l'x
z
is to be associated with
is the _r-component of the molecular
v and e = e ( v,{). From Equation ( 11.33) it is evident that x
we can \Vrtte •
(14.23) where N(z)dz is the number of molecules with parameter z +
z
in the range
z to
dz and A is independent of z. From the definition of an average, we have
_
Sec. 1..4.7
Entrop)' Change of Mixing Revisited
271
e(z)N(z)dz e(z)
=
----
,
(14.24)
N(z)dz where the denominator insures that N (z) is a probability density function. We note in passing that £(z) = UIN. We ass••ttte that e(z) is a quadratic function of z, that is, e(z) = az2, where a is a constant. This is certainly the case when eo: vx2• Therefore Equation (14.24) gives oc
a
-
2
7r
112
a
where a= ajk T. Thus (14.25) The principle of the equipartition of energy states that for every degree of freedom for which the energy is a quadratic function, the mean energy per par ticle of a system in equilibrium at temperature T is (1/2)kT.
14.7 ENTROPY CHANGE OF MIXING REVISITED
In Section 9.5 we used the results of classical thermodynamics to determine the change in entropy due to the mixing of two different gases. We considered a volume V divided into two equal parts by a partition. On one side of the par tition there are n1 = x1n kilomoles of an ideal gas and on the other side there are n2 = x2n kilomoles of a different gas, both gases being at the same tem perature and pressure. The partition is removed, each gas diffuses into the other, and a new equilibrium state is obtained in which both gases occupy the total volume. The change in entropy as a result of the mixing process was found to be (14.26) where x1 and x2 are the kilomole fractions of the constituents.
•
Chap. 14
271
The Classical Statistical Treatment of an Ideal Gas
Let's examine the problem from the point of view of statistical thermo
dynamics. First. we observe that n
That is,
x1
= Nk/R, so that
is the fractional number of molecules of gas 1 and x2 is the corre
sponding quantity of gas 2, with Xi +
x2
= 1.
The effect of the mixing is to itzcrease the number of configurations
available to the system by
a
facto r of
au'=
N! N) .'Nt �.
N!
=
(14.27)
the number of ways of arranging the N molecules with N1 in one container and N2 in the other. The correspt)nding entropy change due to mixing is �S
=k ln
=kIn N! - k ln(x1N)!
A1v
-
k l n(x� N ) ! .
Using Stirling,s approximation� we obtain
1
1
or
(14.28) Noting that Nk
=nR, we see that this is precisely the classical result.
We ask: For what proportion of the two constituent molecules is the.
entropy of mixing a maximum? Let x1 u
=
-
.lS k N
x
=
In
x
=x and x2 = 1
+
-
x, so that
(I - x)ln(l - x).
To maximize u. we take du
-d-x
=0 = In
Thus xj ( 1 - x)
=
x
+ I
-
1 and x
1 =
1 _
x
-
ln ( I
- x) .
-
+
I
x _
x
=ln
1
X
-
·- .\"
•
1/2. The entropy change due to mixing is there
fore a maximum when the two gases are present in equal proportions. This is
Sec. 14.8
Maxwell's Demon
273
in no way a surprise since the mixing problem is exactly equivalent to the
tossing experiment.
coin
The statistical analysis of mixing sheds light on the Gibbs paradox described in Section 9.5. Clearly, if the gases are identical, there is no increase in the number of microstates when the partition is removed and the occupa tion numbers remain the same. As we have seen, the sharp distinction between distinguishable and nondistinguishable particles is fundamental in quantum mechanics and in statistical thermodynamics.The Gibbs paradox was viewed as a serious problem in the purely classical context in which it first arose.
ELL'S DEMON
14.8
Jatnes Clerk Maxwell viewed the secottd law of thermodynan1ics as statistical even though it was left to Boltzmann to develop its theory. Maxwell, however, illustrated the statistical nature of the law with his famous ''demon." In 1867 he suggested a conceivable way in which if two things are in contact, the hotter "could take heat from the colder without external agency."* Consider a gas in a container divided by an adiabatic diaphragm into two sections, A and B (Figure 14.3). The gas in A is assumed to be hotter than the
0
0 0
0
•
0
A
0
B
0
0
0
0
Fi
14.3 Maxwell's demon allows a molecule to pass from A to B if its kinetic energy is less than the average molecular kinetic energy in B. Passage from B to A is allowed only for molecules whose kinetic energies exceed the average kinetic energy per molecule in A. The demon is thumbing his nose at the second law of thern1odynamics.
1867. Reprinted in C.G. Knott, Life and Scientific Work of Peter Guthrie Tait, Cambridge University Press, 1911. *
J.C. Maxwell to P.G. Tait, 11 December
Chap. 14
274
The Classical Statistical Treatment of an Ideal Gas
gas in B. From the molecular point of view, the higher temperature means a higher average value of the kinetic energy <)f the molecules in A compared with those in B. Maxwell wrote� �4Nf a slide � without mass.'' Maxwell s beitlg is assigned t
rm.s
velocity of the molecules in B. At the same time, a mole
cule from B is allowed to pass through the hole into A only when its velocity exceeds the
rms
velocity of the molecules in A. These two procedures are car
ried out in such a way that the number of molecules in A and B do not change. As a result, Hthe energy in A is increased and that in B is diminished; that is, the hot system has got hotter and the cold colder and yet no work has been done; only the intelligence of a very observant and neat-fingered being has been employed." In this way the being vi
X
X
o o
X
X
0 X
0 X
X o
0
X
0
.I
0 Maxv.etrs
X X
X
Demon
.
X X
0
0
X
0
0
X
c
•
0
0
X Maxwtfi
s
Demon later
Figure 14.4 Maxw e ll s demon in action. In this version the demon ope rates a valve. allowing one '
species of a two-component gas (hot or cold) through a partit ion separating the gas from an initially evacuated chamber. Onlv fast m ol e cul es are allowed through .. resulting in a cold gas in one chamber and a hot gas in the other. (Reproduced .
�
from Gasser. R.P.H. and Richards. W.G., l:.�ntropy and Energy Levels, Clarendon Press.. Oxford 1974 .. by ..
pertnission of Oxford University Press.)
we
Problems
275
can't," added .Maxwell. "not being clever enough." Kelvin immediately gave the being the name ''demon" by \Vhich it has been kno�Nn ever since.
Maxwell's demon has attracted an enormous amount of attention since it appeared well over a century ago. It has been said that the ''demon is catlike;
several times it has jumped up and taken on a ne\v character after being left for dead."* Because of its connection with information theory and computer science, the subject of the demon continues to generate irtterest. We shall return briefly to it in the last chapter.
PROBLEMS
•
14-1 (a) Calculate the entropy S and the Helmholtz function F for an assembly of distinguishable particles. (b) Show that the total energy U and the pressure Pare the same for distin guishable particles as for molecules of an ideal gas while S is different. Explain why this makes sense.
14-2 Show that for an assembly of N particles that obeys Maxwell-Boltzmann statis tics, the occupation numbers for the most probable distribution are given by:
14-3 (a) Show that for an ideal gas of 1V molecules,
where
z
(kT)Si2 -
-
-
N
(b) Forei
=
(3/2)kT, T
=
p
300K,P
=
103 Pa,andm
=
10-26kg,calculategi/hJ.
14-4 Calculate the entropy of a kilomole of neon gas at standard temperature and pressure. The mass of a neon atom is 3.35 x 10-26 kg. 14-5 Calculate the chemical potential in electron volts for a kilomole of helium gas at 27 standard temperature and pressure. The mass of a helium atom is 6.65 x 10- kg.
•
See
Maxwell's Demon: Entropy, Information. Computing,
by H.S. Leff and A. F. Rex,
Princeton University Press, 1990. This bock contains a chronological bibliography and reprints of
25 papers on the subject.
Chap. 14
276
The Classical Statistical Treatment of an Ideal Gas
14-6 A tank contains one kilomole of argon gas at 1 atm and 300 K. The mass of an argon atom is 6.63 x to-:?tt kg. (a) What is the internal energy of the gas in joules? What is the average energy of a molecule in e V?
(b) What is the partition function Z? (c) What is the chemical potential J.! in e v: (d) What is Nijg1? 14-7 The partition function of a system that obeys Maxwell-Boltzmann statistics is aVT\ where a is a constant. (..alculate U. P, and S. given by Z =
14-8 An ideal monatomic gas consists of N atoms in a volume V. The gas is allowed to expand isothern1ally to fill a volume 2V. Show that the entropy change is Nk In 2.
ea •
I I
•
I
15.1 Introduction
279
15.2 The Quantized Linear Oscillator
279
15.3 Vibrational Modes of Diatomic Molecules
282
15.4 Rotational Modes of Diatomic Molecules
284
15.5 Electronic Excitation
287
15.6 The Total Heat Capacity
288
177
.
..
v�
15.1
��
� 0'> ���001!1
)')1> .. · �·· ---�-· � ·· � ��
INTRODUCTION
n u co en e w m le b ro p y rn o t h a e lv so re n ca el d o m l ca ti is at st e th therefore that s di s s. ga c o� at di a to s jc arr yn od errn th l ca si as cl of n io at lic p ap e t in d re te e th t ve g to tls fa gy er en of on iti rt a i p qu e cussed in Chapter 11, the pr inciple of p re sc di is th of n io at an pl ex e Th y. cit pa ca at he c ifi ec sp observed value of the e th ing fac ge en all ch nt rta po im st e mo th be ll to we ancy was considered by Max statistical theory. In thi s chapter we shall see how the problem is solved.
�
h
15.2
THE QUANTIZED LINEAR OSCILLATOR
Until now we have confined our attention to systems of particles that have translational degrees of freedom only. To deal with particles such as diatomic molecules, we need to investigate so-called
internal degrees of freedom such as
vibrations, rotations, and electronic excitations. We shall address vibrations frrst. We consider an assembly of N one-dimensional harmonic oscillators. We assume that the oscillators are
loosely coupled in that the energy exchange
among them is small. This means that each particle can oscillate nearly inde pendently of the others. We further assume that eaclt oscillator is free to
vibrate in one dimension on ly with some natural frequency
v.
From quantum mechanics, the single particle energy levels are
given by
•
(15.1) Note that the energies are equally spaced and that the ground state has "zero point" energy equal to hv/2. The states are nondegenerate in that
gi
For the internal degrees of freedom, Boltzmann statistics
=
1 for allj.
applies
. The
assumption may seem questionable since Boltzmann statistics characterizes
distinguishable particles (localized in a crystal lattice, for example) and would 279
280
Chap. IS
The Heat Capacity of a Diatomic Gas
therefore appear to be inappropriate for the treatment of an assembly of diatomic molecules. However.. in the dilute gas approximation .. the number of
translational quantum states is so much larger than the number of particles
(gi
>>
N;)
that the great majority of states are unoccupied, a few are occupied
by a single particle, and virtually none have a population greater than one. Thus, in our treatment of the internal degrees of freedom of diatomic molecules, \Ve can regard the particles tum states that they occupy.
•
differentiated by the translational quan-
as
We begin by evaluating the partition function :r.
z
=
gJe £jkT
=
j
j'=O
The temperature at which kT
e
�J-+ t/2lh� /KT.
(15.2)
-0
ht' is caJ1ed the characteristic temperature 8:
=
JJv
=
H
k .
Using this in Equation (15.2), we have 2.
X
j
=
=
..
I
)
e-o 2r
(1
e-eflT(1
+
e;r
+ e
e/T +
+ e
e
2e1r
+ .
(e-etr)2
. .
)
+ ...
)
.
The sum in this expression is just an infinite geometric series of the form I
•
which equals ( 1 times
(1
+ y
+
y
y i ..... 2
I
y) -'.This can easily be seen by taking the product ( 1
2 + y -
.
+
+
.
.
.
). Here
.v
z =
=
e
at�T
1-e
-B;i·
( 15.3)
..
the fraction of the total number of particles \\·ith energy =
1 is
.Y)
exp( -0/T)� so
We shall also be concerned with the occupation numbers tribution for g,
-
e,.
or
with
Ni/ N �
The Boltzmann dis-
(15.4) •
Sec. I 5.2
The Quantized Linear Oscillator
281
The exponent of the term outside the parentheses can be written -
e
()
.
kT
-
2T
_
+ (15.4)
Thus Equation ,
=
. 1 hv hv + ]+2 kT 2kT
hv
=
-
j kT
8
=
-
j T.
(15.5)
becomes (15.6)
A sketch of Equation (15.6) for two temperatures shows that the lower the tem
. perature, the more rapidly the occupation numbers decrease withj (Figure 15.1). At higher temperatures, more particles populate the higher energy levels.
Next we compute the internal energy of the assembly of oscillators. The
expression we obtained for U in terms of the derivative of In w (Equation (14.7)) applies here because the Boltzmann distribution is the same as the Maxwell-Boltzmann distribution. We have
(15.7) v
Substituting Equation (15.3) in this equation and carrying out the differentia---tion, we obtain
u
1.0
=
1 Nk6 - + 2
� T
0.8
=
e
8/T
- 1
(15.8)
•
1.0
9 2
T
-
0.8
=
28
· 0.6
· 0.6 NJ N
1
NJ N
0.4
0.4 •
0.2
0.2
v
,
I
0
0
2
1 (a)
(a) T
15.1 =
1
0
3
0
•
0
1
2
3
1
(b)
Fractional occupation numbers for quantized linear oscillators with 8/2, and (b) T = 28.
Sec. I 5.3
Vibrational Modes of Diatomic Molecules
283
or
(15.9) At very high temperatures T/8 >> 1 or 9/T << 1 and 2
8
1 T
That is, the heat capacity approaches the constant Nk. In the low temperature limit, forT/8 << 1 or 8/T >> 1, we have e81T >> 1 and
•
so that 2
e
-8/T
.
The rate at which the exponential factor approaches zero as T--+ 0 is greater than the rate of growth of (8/T)2 so that Cv
�
0
as
T
�
0,
consistent with the third law. The variation of Cv wit� Tis shown schematically
in Figure 15.3.
k .... .. N
-----------------
Cv
Figure 15.3 Variation with temperature of the he.at capacity of an assembly of linear oscillators.
...,_ �----_._.
0
___.__ .. �
____
28
T
The Heat Capacity of a Diatomic Gas
Chap. 15
284
At an extremely low temperature
T c(<
0 or
kT
<<
hv;
the oscillators
(particles) are Hfrozen "" into the ground state., in that virtually none are ther mally excited at any one time. We have U
Nhv/2
==
and C\.�
=
0. This is the
extreme "quantum limit.'" The discrete quantum nature of the states totally dominates the phy sical properties. At the opposite extreme of temperature for which
T
we reach a "classical limit."" Here the expressions for U and
hv.
>> 8 or
Cv
kT
>>
hv,
do not involve
Planck"s consta11t .. the scale of the energy level separation, is irrelevant.
Oscillators of any frequency have the same average energy dence of
U on kT is associated
k T.
The depen
with the classical law of energy equipartition:
each degree of freedom contributes
NkT/2
to the internal energy. But the
question left unanswered by the classical law is: when is a degree of freedom excited and when is it frozen out? The total internal energy of a diatomic molecule is made up of four- con.
tributions that can be treated separately: ( 1) lhe kinetic energy associated.with the translational motion of the center of the mass of the molecule (this is (3/2)kT .. the same as that for a monatomic molecule); (2) the rotational energy due to the rotation of the two atoms about the center of mass of the molecule; (3) the vibrational motion of the two atoms along the axis joining them; and (4) the energy of excitation of the atomic electrons. The last three are internal modes of possessing energy. Because the four contributions can vary independently� it follows that the partition function is a product of the corresponding factors. This can be
se e n
by n<)ting that the internal energy is a
function of the logarithm of Z. Having considered the vibrational motion in Section 15.3. we turn our attention to the rotational contribution.
15.4 ROTATIONAL MODES OF DIAT OMIC M OLECULES The rotation of a diatomic molecule is modeled as the motion of a quantum mechanical rigid rotator. The rotation takes place about an axis thr
..
and r0 is the equilibrium value of the distance between the nuclei. Quantum mechanics states that the all<>wed values of the square of the angular momentum are 27T and
I
=
0 1 2 3 . . ..
..
energy is ( 1/2) 1
..
2 w "
.
I (I + 1) .
1f2 where tz is Planck�s constant divided by ..
Recall frotn classical mechanics that the r
where w is the angular velocity. The angular momentum L
is I w so the energy is L 2/21. The quantized energy levels are therefore
e,
=
I (I + 1 )
n
·,
2i .
(15.10)
Sec. I 5.4
Rotational Modes of Diatomic Molecules
285
We define a characteristic temperature for rotation:
(15.11) so
that (15.12)
Experimental values of 8rot are given in Table 15.1 for several gases; they are found from infrared spectroscopy, in which the energies required to excite the molecules to higher rotational states are measured. The energy levels of Equation (15.12) are deg\!nerate; quantum mechantcs gtves •
•
g,
=
(2/ + 1)
states for Ievell corresponding to different possible directions of the angular momentum vector. Given these results, we can write down the partition func tion for rotation:
(15.13)
The important quantity in this expression is the argument of the exponential. For T << 8r01, virtually all the molecules are in the fe\v lowest rotational states, and the series can be truncated with negligible error after the first two or three tertns. It can be st1own that in this. case both the internal energy and the heat capacity at constant vblume vanish as the temperature approaches zero (see Problem 15-4)
.
For all diatomic gases except hydrogen, the rotational characteristic tem perature is of the order of 10 K or less. Since these gases have liquefied
TABLE IS. I Characteristic temper ature of rotation of diatomic molecules. Substance
8rot (K)
H2 0., N2 HCI co NO Cl2
85.4 2.1 2.9 15.2 2.8 2.4 0.36
•
I Chap. 15
286
The Heat Capacity of a Diatomic Gas
(indeed, solidified) at such low ten1peratures� it is always true that T >> Brot at ordinary temperatures. Hence a great many closely spaced energy states are excited and the sum in Equation write dx
=
x
(15.13) may be replaced by an integral. We x as a '�ontinuous variable. Note that
l (l + 1) and treat (21 + 1 )dl. In this approximation,
T
X
Z=
[exp( -9roJT)x]
dx
�
0
()rot
(15.14)
.
This result is too large by a factor of 2 for homonuclear molecules such as
H2,
02, and N2• For two identical nuclei, two opposite positions of the molecular axis correspond to the same state of the molecule. This slight modification has no effect on the thermodynamic properties
energy and the heat capacity. Using Equation (15.14) in a ln Z ar
v
(15.15)
•
we find that for T >> Hr01, •
Urot
NkT.
=
(15.16)
and Cv.rot
=
T >>(}rot·
N k.
(15.17)
�us, at significantly high temperatures, the rigid rotator exhibits the eq uipar
tltion of energy between two degrees of free{lom. At very low temperatures such that Orot >> T, the partition function. Equation (15.13), can be expanded as Z
=
1
+
3e- 2Bro)T
+
5e
--
h#J,.,,/T + .
.
.
.
Retaining only the first two terms, we can write
Since the exponential term is small compared with unitv. we can use. the approximate relation --
,
In ( 1
+
c)
�
e,
£
<< 1
Sec. I S.S
Electronic Excitation
187
I
Nk ..,..._ __ _
.. ____ _ ... ,..-.-...
_
_ _ --- __ ._.... _ _ __
Cv. rot
Variation with temperature of the heat capacity of an assembly of rigid rotators. Fig11re 15.4
0
28rot
T
and obtain the result In Z
��
JT. 29ro 3e-
Hence a In z iJT
=
28roJT e-
61Vk8rot
'
v
(}rot>>
T
(15.18)
and CV,rot
-
-
au aT
=
3Nk
v
1'£J ._
rot
T
2
28 JT e- n .
.... .... ... ... T 8 rot//
(15.19)
Again, we see that Cv approaches zero as T approaches zero, as it must. Fig ure 15.4 shows schematically tl1e contribution of rotation to the heat capacity
as a function of temperature.
I 5.5 ELECTRONIC EXCITATION
The electronic partition function can �e written
(15.20) where g0 and g1 a re respectively. the degeneracies of the gro un d state and the ,
first excited state, and Be is the energy separation of the two lowest states
divided by Boltzmann's constant k. For most gases, the higher electronic states are not excited (6e
--
120,000
K for hydrogen) and Z
;�
g0• Thus the partition
Chap. IS
288
The Heat Capacity of a Diatomic Gas
function at practical temperatures makes no contribution to the external energy or heat capacity.
'
I 5.6 THE TOTAL HEAT CAPACITY We are finally in a position to account fully for the behavior of Cl/ as a func tion of T for a diatomic gas at temperatures in the usual range of interest. Assuming that electronic excitations are negligible.. we have
Cv
==
C'v.tr + Cv.rol + CV.vih·
Thus, at ordinary temperatures,
or
C,
5
=
Nk -
2
+
(15.21)
T
Table 15.2 gives values for 8,ih for several diatomic gases. At the lowest temperatures at which the system exists in the gaseous the
phase,
(3/2)Nk
=
molecular
(3/2)1zR
motion
is
solely
translationaL
contributing
to the heat capacity. A.s the temperature increases and
approaches Or01.. rotational quantum states are excited. EventuaJiy, for T >> 8r(lt\ the rotational contribution be�omes equal to nR. corresponding to two
rotational degrees of freedom. Therefore, the heat capacity steps up to
( 5/2) nR
in the region 8n,1 << T << 8,.;�· Since room temperature lies within
this range.. this is the value usually given for the heat capacity of a diatomic gas. ------- -
TABLE
15.1
Characteristic tempera
tures of vibration of diatomic molecules. Substanc�
H1 02 N2 HCl co NO Cl2
8-Jtb ( K; 6140
2239 3352 4150
3080
2690 810
Problems
289
4
I
'
.
'
:
'
'
., -
I '
3
.
I I '
. '
Cv nR
2
-4 1
( I
-
I
•
I
I
.
•
I
10
50
100 500 1000 8rot Temperature (K)
5000
•
•
Figure 15.5
Values of CvfnR for hydrogen (H2) as a function of temperature. The temperature scale is logarithmic. (Adapted from Thermodynamics, Kinetic Theory, and Statistical Thermodynamics, 3rd edition. by F.W. Sears and G.L. Salinger. Addison-Wesley. 1975.)
At elevated temperatures, the higher vibrational states are excited and the heat capacity exhibits two additional degrees of freedom, rising asymptot ically to a value of (7/2) nR for T >> 8vib· The characteristic temperature of vibration depends on the bond strength between the two atoms of the mole cule and on their masses. In some cases, the diatomic molecules disassociate as the vibrational energy overcomes the bonding energy. Experimental values of the heat capacity at constant volume are close to predicted values over a wide range of temperature. Values of Cv/nR are shown for hydrogen on a logarithmic scale in Figure 15.5. An understanding of the temperature variation of the heat capacity of diatomic gases is surely an •
outstanding triumph of quantum statistical theory.
PROBLEMS 15-1
(a)
quantum states
IS-2
�/ N
Calculate the fractional number
Nil N
(j
=
0,
1, 2)
of oscillators
in
the three lowest
forT= 8/4 and forT= 8.
(b)
Sketch
(a)
For a system of localized distinguishable oscillators, Boltzmann statistics
versus j for the two temperatures.
applies. Show that the entropy S is given by
S
=
-k�� ln •
I •
N.J N .
The Heat Capacity of a Diatomic Gas
Chap. 15
190
(b)
Substitute the Boltzmann distribution in the previous result to show that
v
S
(c)
+
T
=
ln Z.
Nk
Using the expressions derived in the text for U and T, prove that
S where 8
=
hv/ k.
8/T
==
N k - 8IT -
e
- 1
- In ( 1
H1' e- 1
-
Examine the behavior of S as
)
4
T approaches zero.
15·3 Consider 1000 diatomic molecules at a temperature Ovib/2.
(a) (b)
Find the number in each of the three lowest vibrational states. Find the vibrational energy of the system.
15-4 (a) In the low temperature approximation of Section 15·44 show that the Helmholtz function for rotation is
Ff(lt
=
(b) Use the reciprocal relation 5
..
-
=
same approximation. Note that
e - �8"-L/T
3 N kT
•
(aF/aT)v to find the entropy Srot in the S � 0 as T--+ 0., in agreement with the third -
law.
15·5 As an alternative evaluation of Zrot to that given in the text, assume that for T >> 6rot the numbers I in the sum of Equation ( 15.13) are large compared with unity and replace the summation by integration with respect to I. Show that Zrot
=
T/ (Jro1
•
15·6 Use the data of Table 15.1 to determine
r0
..
the equilibrium distance between the
nuclei.. for
(a) (b)
an H2 molecule� a CO molecule.
15-7 Consider a diatomic gas near room temperature. Sho\v that the entropy is S
=
312 512 7 V -T 2 rrn1k Nk -+In N 2 h2 28rot
if the atoms of the diatomic molecule are identical.
lS-8 For a kilomole of nitrogen (N2) at standard temperature and pressure. compute (a) the internal energy U; (b) the Helmholtz function F; and (c) the entropy S. 15-9 Using the relation p
==
NkT
a
In
itV
� .
T
.
show that the equation of state of a diatomic gas is the same as that of a monatomtc gas. •
15-10 Calculate the specific heat capacity at constant· volume for hydrog�n (H2) at T 2GOO K. =
f�j;fY!i?t
·= \),: :\?·::::,':/"'
-
;
•
..
16.1 Introduction
293
16.2 Einstein's Theory of the Heat Capacity of a Solid
293
16.3 Debye's Theory of the Heat Capacity of a Solid
296
•
291
l!!lliill
'
16.1 INTRODUCTION
'
' '
'
I '
f ' •
�
e t f o le p m a x � � a is ic p to is th , 15 r te p a h C f o t c je b su e densed matter. Like th cs m a ch e m l a tc s tt a st r o d e e n e th d n a ry o e th c ti e in k l a ic ss a cl f o shortcomings
!
�
nt ta ns co at ty Ci pa ca at he c ifi ec sp e th at th ed rv se In 1819, Dulong and Petit ob -t K -I e ol m lo ki J 10" X 9 2.4 y el at im ox pr ap is s lid so volume of all elementary the d by ine pla ex be n ca ult res is Th e. ur rat pe tem om ro ar at temperatures ne a lin as id sol the of m ato ry eve g atin tre by rgy ene of ion rtit ipa equ of ple princi ear oscillator with six degrees of freedom and then assOCiating an energy of 3R, in agreement (l/2)kT with each degree of freedom. Then cv = 6 (R/2 ) =
with the observation of Dulong and Petit. However, additional measurements showed that the specific heat of solids varies with temperature, decreasing to zero as the temperature approaches zero. This behavior cannot be explained by the "freezing" of degrees of freedom when the temperature is decreased since the specific heat varies gradually with temperature and does not exhibit abrupt jumps by any multiple of R/2 (in contrast to the specific heat of a diatomic gas). Even at room temperature the specific heat capacities of certain substances such as beryllium, boron, carbon, and silicon were found to be much stnaller than 3R. . Quantum statistics is needed to explain these discoveries.
I �.2 EINSTEIN'S THEORY OF·rHE HEAT CAPACITY OFA SOLID ·,
It was Einstein who developed the first reasonably satisfactory theory of the specific heat capacity of a solid. He assumed that the crystal lattice structure of a solid comprising N atoms can be treated as an assembly of 3N distinguish able one-dimensional osci�lators (three oscillators for each atom since the atoms of a solid are free to move in three dimensions). From Equation (15.8), the internal energy of a solid made up of N atoms is . 293
The Heat Capacity of a Solid
Chap. 16
294 U
=
1 3N kfJE -2
1 / 8 e 1 T
+
-
where 8E is the Einstein temperature given by fJE constant volume is
v-
ar v
=
3Nk
T
1
(16.1)
hv/k. The heat capacity at
=
8 e ' /T
2
(J £
,
(
eBEIT
_
( 16.2)
1 )2 •
For temperatures very large compared with the Einstein tem�erature. Cv � 3Nk
=
(16.3)
3nR.
Thus the high temperature limit of Einstein's equation gives the value of Dulong and Petit. The failure of their law becomes evident when we examine the low temperature limit. For 8F./T >> 1,
( 16.4) As T approaches zer(l, Cv
powers the growth of
a lso ,
( OE/T)-.
goes to zero since the exponent ial decay over.
Einstein �s theory also explains the lov.' heat capacities of some elements
at mo d erately high temper at ures. Jf an elen1 ent has a large Einstein tempera will be large even for temperatures well above absolute l1vj k must be very large zero and Cv wil l be small. For such an element (JE
ture, the ratio
8E/T
..
-
and accordingl y
,
,
v
must be larg e. For an oscillator with force constant
K
and
reduced mass p., the oscillator frequency is ..
v =
A large frequency value suggests
a
]
K
2rr
( 16.5)
•
s mall reduced mass or a large force con
stant. corresponding to lighter elements an(l elements that produce very hard crystals. The theory correctly predicts the failure of the law of Dulong and Petit
for those elements. As an example, the h ea t capacity of diamond approache s
3Nk only at e xtremely high te mp e rat ures
(8�:
=
1450 K for diamo nd )
.
The essential behavior of the specifi c heat capacity of solids is incorpo
rated in the ratio
OE/T. When
this ratio is large_ the p artit i on function reduces
to the zero-point term\ i m ply i n g that all the atoms are in the ground state and th e vibrational degrees of freedom are not excited ( Equation
(15.6)). The spe
cific heat remains close to zero since sm all temperature increases are not suffi
cient to excite a si gn ificant number of at()ffiS to the first vibrational state.
When 8E/T is smalL the difference b e t we en e ne rgi es corr espo ndi ng to various
Sec. 16.2
295
Einstein's Theory of the Heat Capacity of a Sotid
vibrational states is small contpared with thettnal energies. Thus the vibra tional states can be approximated by
an
energy continuutn and treated by
classical theory. For values of 8JT between the two extremes, there is a transi tion region of partial excitations. A consequence of tlte fact that c,,; N depends
only
on the ratio 8EfT is
that a single measurement of the heat capacity at one temperature detertnines its value at all other temperatures. ln additit:n, different elements at different temperatures will possess the same specific heat capacity if the ratio 8EfT is the same in each case. The elements are said to be in ''"corresponding states." The graph in Figure 16.1 clearly shows this behavior. Careful measurements of heat capacities show that Einstein's model gives results that fall slightly below experimental values in the transition range of 8EfT between the two limiting values. The discrepancy can be seen in Figure 16.2 in which the heat capacity of lead is shown for temperatures in the range 0-50 K. 2.5'
2.0
t--
1.5 ....... 1£r' I kilomole-1 K--l cv
X
X
Pb
•
Ag
o
Zn
6. Cu
1.0 .....
+AI 0.5
•c
�
'
0
0.5
1.0
1.5
2.0
2.5
Figure 16.1
The specific heat capacity of various solids as a function of T/9E. Note (NAJN)Cv.(Adapted from Elements of Statistical Th ermo dynamic s' that Cv zud edition, by L.K. Nash, Addison-Wesley, 1972.) =
2.5
r-
2.0
t-
•
•
Figure 16.2
The specific heat capacity of lead showing the disagreement with Einstein's theory at low temperatures. (Adapted from Elements of
104 J kilomole-1 K-1 c, X
0.5
..,_
Statistical Thennodynamics,
Einstein
2nd edition, by L.K. Nash, Addison-Wesley,
1972.)
0
25
50
T(K)
Tl8e
The Heat Capacity of a Solid
Chap. 16
196
16.3 DEBYE'STHEORY OFTHE HEAT CAPACITY OFA SOLID The disagreement between Einstein's result and the experimental data is due to the fact that Einstein's assumptions about the atoms in a crystal do not strictly apply to real crystals. The main problem lies in the assumption that a single frequency of vibration characterizes all 3N oscilJators. Debye improved •
on Einstein �s theory by considering the vibrations of a body as a whole.. regarding it as a continuous elastic solid. He associated the internal energy of the solid with stationary elastic sound �·aves. Each independent mode of vibration is treated as a degree of freedom. In Debye's theory a solid is viewed as a phonon gas. Vibrati<>nal waves are matter waves� each with its own de Broglie wavel ength and associated par ticle. The particle is called a phonon .. with characteristics similar to those of a
photon. We are interested in determining the number of possible wavelengths or frequencies within a given range. For quantum waves in a one-dimensional box. we saw in Chapter 12 that the wave function is l/J A sin kx, where =
•
I" 2 .. 3.. .
n =
.
.
.
( 16.6)
Here A is the de Broglie wavelength and /_j is the dimension of the box. Using the fundamental equation of \vave motion,
•
where cis the wave velocity and
v
the frequency, \ve obtain
tl
2L =
-
v.
('
If we consider an elastic solid as a cuhe of volume V
n = •
c
v.
=
L3, we get
( 16. 7)
nx2 + 1zy2 + n j/ . The quantum numbers tlx, n.r, and IZ: where, in this case n2 are positive integers. Thus the possible values that they can assume occupy the =
first octant of a sphere of radius
(Figure 16.3).
Sec. 16.3
Debye's Theory of the Heat Capacity of a Solid
297
dn
•
I
/
n.{
16.3 A shell of thickness dn of an octant of a sphere of radius n.
Let
g(v)dv
n
l
be the number of possible frequencies in the range
v
to
Since n is proportional to v, g(v)dv is the number of positive sets of integers in the interval n to n + dn that is, within a shell of thickness dn of
v
+
dv.
an octant of a sphere with radius n:
Substituting Equation
(16.7)
for n, we obtain
(16.8) In a vibrating solid, there are three types of waves: one longitudinal with velocity c1 and two transverse with velocity c1• All are propagated in the same direction. When all three waves are taken into account, Equation (16.8) becomes
g( v)dv
=
4·n.Y
1
3 c,
+
2
3 c,
v2dv.
(16.9)
Since each oscillator of the assembly vibrates with its own frequency, and we are considering an assembly of 3N linear oscillators, there must be an upper limit to the frequency spectrum. The maximum frequency
vm
is deter
mined from the fact that there are only 3N phonons:
3N
=
(16.10)
The Heat Capacity of a Solid
Chap. 16
298
Combining this result with Equation (16.9). we get
(16.11)
Equation (16.10) provides us with some insight into the cutoff frequency and wavelength. Since
vm (N/V)113 and A min r:x.lfvm, it follows that �.
\:
/\mtn .
X
1/3
v
•
N
The minimum possible waveJength is determined by the average interatomic spacing. Thus the structure of the crystal sets a lower limit to the wavelength; shorter wavelengths do not lead to new modes of atomic vibration. The principal difference between Einstein's description and Debye's model is in the assumption about the frequency spectrum of the lattice vibra tions. This is shown graphically in Figure 16.4. More rigorous calculations have led to more complex spectra. Now, there is no restriction on the number of phonons per energy level jhv, where j is an integer. Thus phonons .are bosons. *This means that the occu pation numbers must be given by the Bose-Einstein distribution. For the con tinuum , Equation (13.41) applies:
N(e)
l =
g(e)
-----
e(t-p.)/kT-
1.
I I I I
..
.
.
.
g(v)
... .·.. ..
,•,
3N�
g(v)
. .
.
·.
.
. . . .. . . . . . .· :. '·. . . .·. . ·· · , : : ·· : : . . . .. . . . . . · · . · .., : · · : · · · , · · . . .··· ... · . . . · . . ·. . ··. 2: :::. ::. � ·.· · L . · . . · :· . · ;: · · ; · · · . . . . -· . ·: . . .; : . : : . .:· :--:·
.- . ... ... . - . ==:::�l·. ;:- .�· -� •
•
t
0
••
.
..
. .
.
.
.
. :'.
.
. .
..
·
.
.. .
.
.
.
.
•.
•
0
•
. .
••
,,
._
.
(a)
• • •
• •
•
•
.
;
'
.
•
•
0
•
:> :
. :.
.
•
.
0
+
•
..
�
•
.
•
•
.
•
•
•
•
·. . :
.
•
•
•
.
•
•
•
•
••
•
.. •
..
.
.
•
· : . . •. ·... .
: ,
•0
.
.
>
'
.
.
•
•
•
•
•
•
•
•
•
. .. :· ·
; ; .-:··
·:·
•
:·
••
·
.
•
.
•
•
.'
•
.
.
•
. .
.
::
(b)
Frequency s pectra of crystal vibrations: (a) Einstein model: · (b) Debye model.
Figure 16.4
*Some physicists like to call them "·phony bosons."
Jl
Sec. 16.3
Debye's Theory of the Heat Capacity of a Solid
299
In this expression the chemical potential J.L must be set equal to zero. This is because the total number N of phonons is not an independent variable but
rather is determined by the volume and temperature of the particular crystal being considered. Specifically,
N is
the number of phonons that causes the
Helmholtz function to be a minimum at equilibrium. Since p. follows that J.L
=
0. With e
=
=
( aF1aN)r,v, it
hv, we have
N(v)dv
g(v)dv /kT - , h e 1
=
(16.12)
"
N( v) dv is the number of phonons with frequencies in the range v to dv. When we substitute Equation (16.11) in this result, we obtain
where
v
+
N(v)dv=
v
-· --- --
(16.13)
•
0
,
V
>Jim
The total energy of the phonons in the frequency range
v
to
v
hv N(v)dv. Hence the internal energy of the assembly is U
+
dv
is
hvN(v)dv
=
0
9Nh =-JIm 3
v3dv
"m
O
/kT hv e
_
(16.14)
1.
(We leave out the constant zero-point energy since this terrn has no effect on the heat capacity. Its value is calculated in Problem
16-2.)
The Debye temperature 80 is defined as
(16.15) That is, the Debye temperature is proportional to the cutoff frequency Some values are given in Table 16.1 To obtain the heat capacity, we need to differentiate Equation with respect to the temperature. Now
d dT
1
h fkT v e
_
1
hv -
h, fkT e
kT2 ( eh"/kT .
_
l )2
·
"m·
(16.14)
300
Chap. 16
The Heat Capacity of a Solid
TABLE 16.1 Debye tem peratures of some materials. Substance
80 ( K)
Lead Mercury Sodium Silver Copper Iron Diamond
88 97 172 215 315 453 1860
Thus
Cv We let
x =
=
dU dT
hvjkT and Xm
9Nh2
( 16.16)
-v m- 3"-·-Tk -:!
=
=
1
hvmfkT
=
80/T. With the change of variable we
have
(16.17) .
high temperatures. 00/T << 1 and .t << 1. So e"" 1 = 1 in the numerator� so the integral becomes tor and ex For
-
.r
in the denomina-
=e-
..,
8o J T
0
•
Hence
Cv=9Nk
T 3 ·1 Oo J
80
3
T
=
3Nk.
This is the law of Dulong and Petit. For low temperatures� 00/T is large and we can let the upper limit of the integral be infinity. Then
( 16.18)
Problems
301
and T
5
3
6o
(16.19)
•
T3 law. It is valid when the temperature is lower than about 0.180, which means for most substances about 10-20 K. The
This equation is known as Debye's
relation gives a better fit to experimental data at very low temperatures than the Einstein model, and is valid for
all monatomic solids. When the tempera
ture is above the De bye temperature, the heat capacity is very nearly equal to the classical value 3Nk. For temperatures below the Debye tetnperature, quantum effects become important ·and Cv decreases to zero. Note that dia•
mond, with a Debye temperature of
1860 K, is a Hquantum solid" at room
temperature. Recent work has centered on the behavior of solids at low temperatures.
3 Experiments suggest that amorphous materials do not follow the Debye T law even at temperatures below 0.0180. There is more yet to be learned.
PROBLEMS
16-1 The partition function of an Einstein solid is Z
e-erJ2T =
1
-
e
rr' E
-8
where 8E is the Einstein temperature Treat the crystalline lattice as an assembly ..
of 3N distinguishable oscillators.
(a) (b) (c)
Calculate the Helmholtz function F.
Calculate the entropy S..
Show that the entropy approaches zero as the tem per ature goes to absolute zero. Show that at high temperatures, S � 3Nk[1 + ln(T/9E}]. Sketclt Sf3Nk as a function ofT/BE.
16-l Show that the inclusion o� the zero-point energy in Equation (16.14) gives a term (9/8)Nk60 that is added to the integral. Since the terrn is a constant, it does not contribute to the heat capacity.
16-3
(a)
Referring to Equation (16.14) and Problem 16-2, show that the internal energy of the assembly of pbonons can be written
9
•
U where
D(80/T)
=
Nk00 g
+
3NkTD
Oo T
,
is one of several Debye functions. Its definition is
The Heat Capacity of a Solid
Chap. 16
302
9n D T (b)
-
-
---
---
For very low temperatures. the upper limit of the integral can be taken to be infinity. Show that
1=
--
e�
0
1
-
=
-
15
. rm te by rm te g in at gr te in d an -.r e of s er w by expanding the integrand in po n. ow kn l el w is m se su ho w s rie se ite fin in an The result can be expressed as Note
(c)
that D( x)
=
0.
Show that at low temperatures
11"
r3
r 4 •
16-4 (a) The heat capacity can be expressed in terms of the Debye function
D(90/T) by noting that
x �ex
fll, .'T
(ex
J )2
---
0
-
tlx
=
0
and integrating by parts. Sho\\·' that
(d) Calculate and plot' CvI ( 3N k) for 1·.1 H0
=
0.2. 0.4. 0.6 0.8, and J .0. The ..
cor
responding values of the De bye function are:
0.2
0.4
0.6
0.8
1.0
0.117
0.354
0.496
0.608
0.674
16-S The partition function of a Dehye solid is
ln Z
==
-9
T
,8o
J
Hp/T a
x2
Jn( 1
-
e-x)dx. -·
Show 1hat the Helmholtz function can be written
F
=
9Nk7� In( I
-
8 o1r e)
3NkTD(00/T).
16-6 The experimental value of c,. for diamond is 2.68 x 1{)3 J kilomole -l K- 1 at a temperature of 207 K. For diamond the Einstein temperature is 1450 K and the Debye temperature is 1860 K. Calculate c,. at 207 K using the Einstein and Debye models and compare the results with the experimental value.
Problems 16-7
(a)
303 The Debye temperature
80 can
be determined from measurements of the
speed of sound in the solid. Use Equation (16.10) to show that .a
un -
1/3
3N
he
_
k
'
where
·
(b)
Show that -
00(K) -
X
2.51
10
p
-2
M
1/3
c,
where p is the density in kg m-3, M is the molecular weight, and c is the sound speed in ms-1• For copper, p = 8.95 x 103m-3, M 63.6 and =
c
= 2.24
X
l
Calculate
,
90.
16-8 The equation of state for a monatomic solid is
Pv where
v
+
f(v)
u
is the specific internal energy of the lattice
is the specific volume and
= fu,
vibrations of the crystal. Tne so-called Gruneisen constant r is important in solid state theory. (a) Show that
f
(b)
Bv . CJ<
=
Show that
y = 1
+
fBT,
where i' is the ratio of specific heats.
(c)
The Grtineisen constant is defined as
f Assun1e that
80 in the
=
-
d ln
00 dIn v
·
Debye model is a function of
v only and
�how that
this definition leads to the result of (a). Hint: Use Maxweil's relation
as av
-
r
aP aT
t'
and assume that the specific entropy sis a function of T/80.
.::>· ·:· (\l :;;';; •· · · ..-..
:
�::�:�:_:
.;:::::::� ::::::::::::::::
•
•
ICS •
17.1 Introduction
307 •
17.2 Paramagnetism
308
.
17.3 Properties of a Spin-1/2 Paramagnet
315
17.4 Adiabatic Demagnetization
318
17.5 Negative Temperature
321
17.6 Ferromagnetism
325
. -.
lOS
. :-:·:·:·:·:·:,.·.::·:<.;·>.·>:·:·:·:.:·:···.·.· :: .
.
:-
.·.: .: .··:
.;.;.:-:·:-::·:·:·:-:-:-:-:·:·:·:· ;.;·. ;.;.· :-:-:·:·. : .. ...::·: :··· : -
:-·
::
: - ·:- :
; .
: .. ·:·.
.
. /·:·:··· :·:·:· :: · · : - :-: : ·. · > •· . llliiil iiiiBBii 818i!B�
17.1 INTRODUCTION Magnetism comprises phy sical phenomena involving magnetic fields and their effects on rrtaterials. All magnetic fields are due to electric charges in motion. On an atomic scale, individual atoms can give rise to magnetic fields when their electrons have a net magnetic moment as a result of their orbital and/or spin angular momentum. Each electron of an atom contributes a m.agnetic dipole that experiences a torque when placed in an external magnetic field. The phenomenon of para magnetism results from the tendency of unpaired dipole moments to align themselves parallel to the applied field. Paramagnetism in solids occurs in com pounds containing transition metal ions that have either incomplete d shells (the iron, palladium and platinum groups) or incomplete f shells (the lan thanide and actinide groups). Both the net spin and orbital magnetic moments may contribute to paramagnetism, but for an ion with an incomplete d shell, the effective orbital moment may be "quenched" by electrostatic interactions with its neighboring ions, leading to a predominant spin moment. Paramagnetism is therefore due mainly to the spin angula� momenta of the electrons. In an external magnetic field, all the orbiting electrons of an atom speed up or slow down, depending on the direction of the field. In either case, a change in each e]ectron's magnetic moment is produced antiparallel to tl1e field. The effect is known as diamagnetism. It is present in all atoms, but is usu ally much weaker than paramagnetism. In ferromagnetic materials, each atom has a comparatively large dipole moment caused primarily by uncompensated electron spins. Interatomic forces produce parallel alignments of the spins over regions containing large numbers of atoms. These regions, or domains, have a variety of shapes and sizes (with dimensiotts ranging from a micron to several centimeters), depend ing on the material sample and its magnetic history. The domain moments are generally randomly oriented, so the material as a whole has no magnetic moment. Upon application of an external field, however, those domains with moments in the direction of the applied field increase their size at the expense of their neighbors, and the internal field becomes much larger than the external
307
The Thermodynamics of Magnetism
Chap. 17
308
field alone. When the external field is removed .. a random domain alignment does not usually occur, and a residual dipole field remains in the sample. This effect is called hysteresis. The only elements that are ferromagnetic at room temperature are iron.. nickel, and cobalt. At sufficiently high temperatures.. a ferromagnet becomes paramagnetic. The modern quantum mechanical theory of magnetism was to a large extent developed by J. H. Van V leck. One C)f the trium phs of this theory is the remarkable agreement between the observed and theoretically calculated magnetic moments of paramagnetic salts. "These are the magnetic systems of primary interest in statistical thermodynamics.
17.2 PARAMAGNETISM
Most experiments on magnetic materials are performed at constant pressure and involve insignificant volume changes. Thus.. pressure and volume varia
tions can be ignored, and equilibrium states can be represented by a relation involving three thermodynamic variables: the magnetic field B, the magneti zation M, and the temperature T. The methods of statistical thermodynamics are appropriate since the alignment of large numbers of atomic magnets
is a
statistical process. •
Paramagnetic effects are produced in a crystal when some of its atoms have a net magnetic dipole moment associated �·ith the electron orbital angu lar momentum, the electron spin .. or both. To understand the relationship between the magnetic moment and the angular momentum. it is instructive to consider first the hydrogen atom and then give the result for more complex atoms or ions. The
basic ideas are the same for other systems as for hydrogen.
In the quantum-mechanical description, the magnitude of the orbital
angular momentum vector of an isolated hydrogen atom is V /(/ + 1 )1i, where the allowed values of the quantum number I are 0. 1 2 . . The z com.. ponent of the orbital angular momentum is n111i where -I < m1 < I. It fol ..
lows that there are 2/ + I values of
.
.
m1 for a given value of /. The magnetic
dipole moment p, is proportional to the angular momentum: the constant of proportionality is ef2m�, where e is the charge of the electron and m(� is its
mass. Thus
ett
and
-
1
<
m1 <
l.
Sec. 17.2
Paramagnetism
309
e1if2me is the so-called Bohr magn�ton and g is a number of the Here J.Ls order of unity referred to as the Lande g factor; g is difficult to deterrttine =
exactly but is known to be
1 for orbital ang ular momentum.
The analogous expressi ons for the electron's intrinsic angular momen
tum or spin are
IPI
s(s + 1),
=KILn
-s < ms <
s'
2 for spin. For an electron the spin quantum number s is 1/2, so that ms can assume only two values, -1/2 and 1/2. (Both m1 and ms where g is very close to
vary i n integral steps.) The magnitude of the total angular momentum , the vector sum of the orbital and spin parts, is
ll
Vj(j + 1)1i,
where the possible values of j are s J. The z component of angul ar momentum has
f fi - sf + 1, ... ll + the value mili, where the possible values of mi are -
s ,
-
j, j
+
-
1, ... j
-
1, j.
These relations are restricted to one orbital electron. For a many-electron atom, similar expressions apply, with capital letters instead of small letters to indicate that they pertain to the tot al number of particles. The generalization is
jp.
=
(17.1)
KILB v'i( J + 1), -J < m < J,
(17.2)
Sl, IL St + 1, . . IL + Sl and the . . J- 1, J. Here LandS are the total
where the possible values of J are fL
-
-
.
allowed values of mare -1,-J + 1, orbital and total spin quantun1 numbers.* For simplicity, the so-called mag netic quantum number m is written without a subscript. Since the total angular .
momentum includes spin,
J can
be integral or half-integral, depending on
whether the atom has an even or an odd number of electrons, respectively. To a good approximation, the Lande g factor is given by
J(J
+
1)
+
S(S + 1)
-
L(L
*For paramagnetic ions in solids it is often convenient to
+ 1)
use
(17.3)
an effective spin
S' to
describe the magnetic behavior of the lowest group of energy levels. In this case� we may write
fl'l
=
' g
p.s
S'(S'
+
1),
w here g', the effective value of g, is close to 2 for many ions with incomplete d shells. However, we shall continue to use S and g rather than S'
and g'.
The Thermodynamics of Magnetism
Chap. 17
310 l
p.
H H
Figure 17.1 Magnetic moment p, in a magnetic field B.
A dipole with magnetic moment p. placed in an external magnetic field by en giv N e qu tor a ce i en per e x l wil 1) 17. re gu (Fi ion ect dir z the in ing int po B the vector prod uct
N
p, X
=
( 17.4)
B.
The magnetic p otential energy when the dipole is in the angular posi t ion 8 (the angle b e tween B and p,) is the work that must be done to rotate the dipole from its zero energy p os i t ion fJ
TT/2 t o the given posi tion 0:
a
e
NdO
€ =
=
=
sin Bd8
J.LB
rr/2
• TT
=
-
�B cos e
2
( 17.5) By substituting Equation ( 17.2) in this equation, we get
( 17.6) the magnetic energy of the atom in the state designated by the quantum num be r
nz. *
As an example, for an electron with a single net electron spin� there
would be only t wo poss i bl e energies.. wit h n1
=
-1/2 (spin antipara1lel to
n1 =
1/2 (spin parallel to
B). Note that the
B) and
energy is lowest when the
dipole moment is parallel to the field. In a paramagnetic material. because the atoms occupy definite sites in the crystal lattice� they can be regarded as distinguishable particles and there fore obey Boltzmann statistics. Accordingly, the probability
in an energy state
e,,
P,11
that an atom is
is equal to the fractional occupation number correspond
ing to that energy level (Equation (13.22)). Since the degeneracy is removed *
\Ve used the
classical argument
to derive
as the result obtained from quantum theory.
E quat ion ( 17.) )
.
hut Equation
( 17.6) is the
same
Sec. I 7.2
Paramagnetism
311
by the applied magnetic field, the degeneracy factors are unity for all possible states. Thus
m
N
'
Z
(17.7)
where
Z-
J
(17.8)
m=-1
It follows that the mean z component of the magnetic moment of the atom is
(17.9) with I
T Jl"'11Bm/k e Z= . m=-J
(17.10)
We can write the expression for J.Lz more compactly as follows. Differentiating Equation (17.10) with respect to B, we obtain ,
az aB T
=
1
J
kT m=--J
gJLsmeg�£8Bm/kT.
Then
kT dZ P-
z
=
Z
dB T
kT
=
d In Z iJB
r
.
(17.11)
The next step is to evaluate the partition function Z. We introduce the abbreviation KJLsB
kT ' and write J
Z=
eflm.
m=- J
(17.12)
312
.
The Thermodynamics of Magnetism
Chap. 17
Sett ing
x
-
e", we have J
Z
=
Xm
X J
=
--
+
+
X J+ 1
·
·
·
+
X1 -I
+
:c1
m=--1
The terms in parentheses constitute a finite geometric series whose sum is 1 -
x2J+'
1-
•
X
•
Hence •
Z
=
e-JTJ
1 - e(:!J + 1 >TJ 1
=
efl
----
-
1
e11
-----
-
•
If we multiply both the numerator and the denominator by e-.,;2 we obtain ..
the following result:
Z=
<1 + 1/2)71 e
'' � 1;2)'1 e I"
-
12
e -TJ.
e"
-
=
sinh ( J +
! )11
sinh YJ/2
4.
(17.13)
•
Fmally,. using Equation (17.11).. we have p,l.
•
=
kT
a In Z aB
r
=
kT
(J + ! )cosh(J sinh(J +
iJ
11 aB +
a In Z d1) 1
�)11 -
!)17
a In Z
-·
2
T
cosh(71/2) ----
sinh(71/2)
·
Therefore,
(17.14) where •
1 1 J +- coth J -#· 2 2
-
11
-
1
1
2
2
- coth -1J
is called the Brillouin function. Its variation with 71 is sketched in
(17.15) Figure 17.2.
Sec. 17.2
Paramagnetism
313 J
1.0 1--
=
712 J j
o
Figure 17.2 The Brillouin function for several values of J.
=
=
112
1
----'..._ __ _ _ ..._� _ ---. .... . ..
o
2
l
3
4
5
Let us examine the behavior of 81 ( 11) for large and small values of its argument. For 1J >> 1, the hyperbolic cotangent approaches unity and
J+
1 2
-
1 -
2
=
1,
, >> 1,
(17.16)
corresponding to magnetic saturation. For small values of y,
l11y + y 3; hence BJ ( , )
Sd
1 J+2
1 J
1 (J+ !)11
+
-1 1 1 1 1 +- J+- TJTJ 2 3 I 1)
1 -
3
=
-
31
1
1
4
4
1
-
2
2
71
+ 2 TJ 6
-
11
2
11
1 1+-,
-
12
]2 + J + - - -
J+l 3
,,
, << 1.
(17.17)
From Equations (17.12), (17.14), (17.16), and (17.17), we obtain KJLBB >>
(17.18)
kT,
.
and
'
KILsB <<
k T.
(17.19)
The Thermodynamics of Magnetism
Chap. 17
314
Of particular interest is the second region, co�res�onding to high te�
peratures and a weak magnetic field. The magnettzatlon M of a magnetic material is defined as the total mean dipole moment per unit volume. Thus, for
��
a material sample of N atoms and volume N g2J.LB2J (J + --
=
1) B
( 17.20)
3VkT
V
This is the experimentally observed Curie law� discovered by Pierre Curie., which states that the magnetization of a paramagnetic solid is proportional to the ratio of the magnetic field to the temperature. For B
=
1 T, * T
=
300 K and g
=
2�
-
8JLsB kT
=
5 (86 . 2 X 10- eV/K) (3
-----
----
----
X
102 K)
so the Curie law certainly applies under these conditions. Even at g
=
2 and J
=
3
K. with
1/2, a field of 1 T (a comparatively strong laboratory magnetic
field) would produce Jess than 25 percent magnetic saturation. The paramagnetic salts most widely used contain paramagnetic ions sur r o un de d
by a large number of nonmagnetic particles. A typical example is
Cr2(S04)3
•
K2S04
•
24 H20 (chromium p()tassium alum). Its magnetic prop-·
erties are due solely to the chromium atoms that exist in the crystal as ions. The chromium ion cr+
t +
has three unpaired electron spins and therefore a mag
netic moment of 3 J.tB· Besides the two chrotnium ions there are 4 sulfur atoms, 2 potassium atoms, 40 oxygen atoms, and 48 hydrogen atoms. Hence there are a total of 94 particles that are nonmagnetic. TI1e magnetic ions are so widely sep arated in the molecule that the interaction between them is negligibly small. At· the same time. the effect of the orbital motions of the valence electrons is quenched by the fields of neighboring ions. What remains is a net electron spin. The configuration of the chromium atom �jCr is
5 ( Ar) ( 4s) 1 ( 3d) • Argon
has 18 electrons in closed shells, so chromium has an additional6 electrons•
one in the 4s subshell and 5 in the 4s electron and two of the
3d
3d subshell. In the chromium ion Cr-t-+ 3d electrons that
produce p�r��agnetism.
Since the total orbital angular momentum is essentially zero� J (1/2 for each of the unpaired electrons). With g z
the
electrons are misS'iag. Thus the structure of
Cr+++ is (Ar) (3d)3. It is the three
in the
T
==
=
3/2
2� the mean dipole moment
direction is
.. :··
( 17.21)
1 T
*The SI unit
=
1
kg/ A·
of
the magnetic
s2. Also. 1 T
=
field 8
104 gauss.
is the tesJa
(T):
in terrns
of
fundamental unit�
Sec. 17.3
Properties of a Spin- I /2 Paramagnet
Figure 17.3
The mean
magnettc moment •
•
tn
l15
Uillts of •
the Bohr magneton for t he
chromium ion cr+++ in chromium potassium alum. The s olid curve is the
theoretical result. The experimental values are those
1
2
3
4
reported by W. E. Henry, Phys. Rev. 88,561 (1952). In Figure 17.3
JLz/#Ls is plotted against BjT for chromium potassium alum.
Note the excellent agreement between the theoretical result and the experi mental data.
17.3
PROPERTIES OFA SPIN-1/2 PARAMAGNET
The simpl�st paramagnetic system is an atom or ion with a spin-1/2 ground state and zero orbital angular momentum. Then J = 1/2 g = 2, and there are just two energy levels. The thermodynamic properties of a two-level system exhibit the ,
same features as those of more complex systems and are easier to calculate. For this case, the partition function of Equation (17.13) takes the fortn Z=
sinh 11
11 =
--
TJ .h SID
.
'
,
2p,BB kT
(17.22)
.
2
This can be simplified by using the identity .
sinh 11 = 2 sinh
1J
2
'T1
cosh
2
•
in the numerator and canceling one of the factors with the denominator, giving Z = 2 cosh
where
e
11
2
==
2 cosh
e
kT ,
( 1 7 . 23 )
J.L8B. The tota l magnetic energy U8 can then be found from a In Z aT
•
8
(17.2 4)
The Thermodynamics of Magnetism
Chap. 17
316
This is Equation
calculated for the Maxwell-Boltzmann distribution
(14.7),
and also valid here. (Boltzmann statistics and Maxwell-Boltzmann statistics differ by a constant factor, but the distributions are the same.) Then iJ In Z
e
-
iJT
kT
8
sinh(ker)
2
·
---
cosh (£)" 0·
so that
U8
Since tanh(O)
=
=
Ne tanh
-
0 and tanh( x )
e
( 17.25)
k.r .
1. the energy is evidently
=
-
Ne at T
=
0
and approaches zero asymptotically at high temperatures. This behavior, shown in Figure 17.4, is markedly different from that of the corresponding energy of an assembly of harmonic oscillators treated in Section
15.2. The lat
ter increases without limit as the temperature increases. Here higher tempera
tures produce increased randomization <>f the dipole moments and the magnetic potential energy goes to zero. The magnetic contribution to the heat capacity also has distinctive char acteristics, given by
au8 iJT
.
H.N
=
t:
Nk
-·-
kT
2
e
.,
sech-
·
(17.26)
-
kT .
This expression can be put in an alternative form from which the limiting behavior is more readily seen (Problem
17 -X): •
8
e2t·ikT +
-
kT
(
-
)
(17.27)
2 · ·
()r--
•
u
---·
-0.5
t--
- 1.0 Fig•••e 17.4 of a spin-1/2
1
-
Internal energy paramagnet.
()
1
'
I
2
3
kT £
4
5
Sec. 17.3
Properties of a Spin- I /2 Paramagnet
317
Then it is clear that c
8
2e =:;
kT/B
kT
<< 1,
•
and the exponential factor reduces C8 rapidly to zero as the temperature decreases, in accordance v;ith the third law. At the other limit,
kT/e For
efkT in
known as a
>> 1.
the v icinity of unity, the heat capacity has a fairly sharp peak
Schottky anomaly
(Figure 17.5). The anomaly is useful for deter
mining energy level splittings of ions in rare-earth and transition-group met als. Like the energy, the heat capaci ty is qualitatively different from other systems in its variation with temperature. The most important thertttodynamic property of two-level systems is the entropy. For Boltzmann statistics, .
( 17.28) Substituting Equations (17.23) and (17.25) in this expression, we get
S
=
Nk
In 2 cosh
Figure 17.6 is a plot of S/Nk versus
8 -
kT
e kT
·
tanh
e kT
•
( 17.29)
kTje.
At low temperatures,
Cs Nk
H eat capacity of _ a spin-1/2 paramagnet.
Figt•re 17.5
•
kT e
•
The Thermodynamics of Magnetism
Chap. 17
318
s .'Vk
0.6
t--
0.4
t-
0.2
I--
0 Figure 17.6
Entropy of a spin -1/2 paramagnet.
f "'--__.___...._�__.._____._�
0
4
2
6
kT E
and e - tanh kT kT so s
�
e I
•
-----·
kT
..
().
high temperatures. the se con d term in curly brackets approaches zero, cosh ( ej k T) � 1 � and S � Nk In 2. Th is is exactly what we would expect. At the upper temperature limit 1v = 2/v the number of equally probable .N k In 2. This corresponds to a pattern of ran microstates, and S = k In 2i,. dom dipole orientations\ involving equal numbers of parallel and antiparallel magnets in any chosen direct ion In this disordered state the entropy is a max imum. As T � 0 all the di po les are in the lower energy state pointing in a direction parallel to the applied magnetic field. There is only one possible n1icrostate so the thermodynamic probability 10 is 1 and S k In 1 0. The foregoin·g discussion is significant because many systems can be treated as two-level svstems. At
..
..
=
�
=
=
•
"' ... ....;.( .. •. •. .. *» . .•
•
y•
17.4 ADIABATIC DEMAGNETIZATION
•
Paramagne�ic salts have been used to a tt a in very low temperatures with a process known as adiabatic lielnagJzetizatilJil, alluded to in Chapter 10. The method rests on the fact that the entropy is a monotonically increasing func tion of kT/c that is. of Tl B. When the magnetic field is increased� the degree of alignment of the magnetic moments increases an·d the disorder of the spin system decreases, thereby lowering the entropy. The entropy is also lowered if the temperature decreases� again because the moments t.end to line up. But if the field is
Sec. 17.4
Fi
Adiabatic Demagnet!zation
319
s
17.7
Entropy versus temperature for two values of the magnetic field B1 and 82, wjth � > B1• The spin systeJn
a
goes from state a to state b at
I I I I I I
a constant temperature 1j as the magnetic field is increased. When the magnetic field is reduced to its original value along the adiabatic path from
b to c, the�emperature is
lb I I t I I T·
reduced to its final value T1.
T
I
reduced adiabatically, without changing the entropy, the temperature must drop if the same degree of disorder is maintained.
This method is illustrated in Figure 17.7. A paramagnetic salt is first cooled to a temperature of 1 K or less by contact with liquid helium. A mag
netic field acting on the salt is then increased from B1 to B2 under isothermal conditions at temperature T;. The spins are in a lower energy state so heat is evolved and the entropy is lowered. At this point the salt is thermally isolated so that no heat is exchanged with the surroundings and the field is gradually reduced from B2 to B1. Since the entropy is constant,
At very low temperatures the heat capacity of the spin system is much larger than that of the crystal lattice, so the final temperature of the paramagnetic
salt is only slightly greater than T1. Temperatures of less than 1 mK have been
achieved with this technique. The reason a final temperature of absolute zero cannot be reached is discussed in Chapter 10.
Demagnetization cooling can be understood by combining results from classical thermodynamics with the statistically derived Curie law. The work done by the paramagnetic salt when an applied magnetic field changes by an amount dB is 7/W
dV
=
=
MdB where M is the total magnetization (
0, the Helmholtz function takes the form
dF
=
-SdT- MdB. •
If F is considered a function of T and B,
dF
iJF =
ar
8
dT +
OF
aB
dB. r
=
MV). With
310
Chap. I 7
Since F is a state
The Thermodynamics of Magnetism
variable� dF is a perfect differentiaL so that
From this we immediately obtain the Maxwell relation
as
-
()8
-
T
riT
-
(17.30)
•
8
•
The magnetization Misgiven approximately by the Curie law
( 17.31) where. a is a pos itiv e constant. Taking the derivative with respect to T while keeping
B constant, we have riS
---
aB --
( 17.32)
-··-
'•
Hence the entropy always decreases with an isothermal increase in the mag
netic fie l d
.
Figure 17.7, we
For the adiabatic Jeg in
can use the fundamental relation •
dU Here it is convenient to letS
T dS or, for
a
=
=
=
TtiS- MdB.
S(T, B). Then
T
(15� iJ7'
dT
+
T
H
CIS
aB
dB.
r
\
reversible process,
T ds
=
cRdT +
where C8 is the hea t
Setting dS
=
•
T
riS
. iJB
dB, T
capacity of the system in a constant magnetic 0 and using Equation ( 17.30). \Ve find that dT T
l
field.
(17.33)
Sec. 17.5
Negative Temperature
321
Because thermal motions tend to disalign the tnagnetic dipoles of the para magnetic material, the magnetization decreases with increasi11g temperature. Therefore
(oM/oT) 8 is negative and consequently the temperature drops
as
the magnetic field is lowered, according to Equation (17.33). This is known as the
magnetocaloric effect.
17.5 NEGATIVETEMPERATURE The two-level case can be used to discuss an extension of the notion of tem perature. As noted, the energy of the lower l�vel (p, parall�l to B) is B p. ==:' e1 ts to B) le ral B tpa ant el (p, lev -p.8 whereas that of the upper � . 80 n tto tta exc the , 2e ts B els lev of n tio ara sep the t tha so e, for be as p.8 e We set gap. The number of particles in the lower level e is N0 and the number in the
�
-
=
-
upper level e is N1, with N0 + N1
(17.34)
N,
=
and
IA.YTO
N -
-
z
T /k e e '
N1
N =
z
/kT -�: . e
(17.35)
The total magnetic energy of the system is (17.36) The ratio N1/ N0 is N , No
-
-2e/kT e ,
which leads to the following expression for the temperature: T
l =
2e
k In N0- In N1
•
(17.37)
In a state of stable equilibrium the occupation number N0 of the lower energy level is greater than the occupation number N1 of the higher energy level, so the absolute temperature N1 is positive in this equation. Imagine now that the direction of the applied magnetic field is suddenly reversed. The elementary magnets that were parallel to the original field and
Chap. 17
322
The Thermodynamics of Magnetism
in the lower energy state are now antiparallel to the field and in the higher energy state. Conversely, the magnetic moments originally in the higher
are now in the lower energy state. A popttlation inversion has taken place. In Equation (17.37) the sign changes and the temperature T energy state
becomes negative!
To be sure, after some relaxation time has passed, the moments in the higher energy state will flop over into the new low-energy state and a new state of equilibrium will be established. But immediately after the field is reversed, the temperature of the spin systen1 will be negative.
The situation can be made clearer by examin i ng how the entropy
a�d
temperature vary with the total energy. The entropy of the two-level system is S k ln w, where =
'U.�
Using Equat i ons
N! =
,"'t,! Nl!
·-
(17.38)
.
(17.34) and (17.36), we can write �)
N 2
:=;
1
U
-
NE
N, •
Substituting t hes e relations in Equation 1�)
N
=
-
1 -
2
·-·
f
1Ve
.
=
2
U
l + Ne .
(17.39)
u
(17.40)
( 17.3R), we have N!
u
N
•
!\'
-·-
""
L
1+
1Vc
•
' •
The use of St irli n g s approximation leads to an e x pr e ssion for the entropy in terms of the dimensionless q ua n tity ,t '
=
U/.'Vc, (Problem
17-10):
( 1 7 .4 1 ) E videntl y x varies from
-1 to
+
1: in this intl!rval
0 (Figure
S/Nk is an even function of x
17.8). Referring to Equation ( 17.39). we see that when U / N E = - 1, �l N, a nd N1 0. all of the particles occupy the lower energy level. Similarly. when 1, �> N, the upper level is fully occupied. In both U IN e 0, and N1 with a maximum at x
==
==
=
cases T
=
=
0.
==
.
=
When the particles are equally distrihuted between the two levels, the
entropy reaches its maximum v a lue of Nk In 2. Here the temperature sud
denly j umps from +x to -�as the direction of the field is reversed.
·
Sec. 17.5
Negative Temperature
In 2
s Nk
-
----------
323
-
I J I I J I I I I I I r I I I I I I
0.5
T>O
0
T
--------�--�-
-1
0
-0.5
x =
Figure 17.8
Entropy
as a
1
0.5
u iVe
function of energy for a two-level system.
To obtain a more detailed picture of the relationship between the tem perature and
the energy, we can calculate procity fortttula (see Chapter 8): T=
T as a function of x by
usin
g a reci
au •
as
N
Then as
1
=
-
T
au
N
1
as
=
-
Ne
ax
N
k
2e
In
1-
X
1 +X
•
(17.42)
UjNe. Clearly shown is the abrupt In Figure 17.9, kT/2s is plott ed versus x shift from +oo to oo when slightly more particles occupy the upper energy level. =
-
Three conditions must be satisfied if a system perature:
(1)
is to have a negative tem
the system must be in thermal equilibrium, (2) the energy spec
trum of the system must have a finite upper bound, and (3) the system
must be
energetically isolated from states that are at a positive temperature. The sec rule
s out systems such as upper limit to its possible energy.
ond condition
a harmonic oscillator, which has no
The Thermodynamics of Magnetism
Chap. 17
324 3
--------���--· --------�
I I '
'
2,__
I
I I l •
I '
kT
0
-----
-
----
-1
-
--
t 1
-1 ....
---- --·
- -------
1
{)
I • 0
I J
-2
.
-
3
. I I I I
.,._
���----�--� x =
Figure 17.9
u Ne
Temperature as a function of energy for a two Jcvel ..
system. Note the jump in temperature from particles are in the higher energy state.
+x
to
-x
when more
In paramagnetic materials the interaction between the magnetic ions and the crystal lattice, though weak, is sufficiently large that the substance can not exist in a state of population inversion for any appreciable length of time. However, in a series of elegant experiments. Purcell and Pound in 1951 used nuclear magnetic resonance techniques to show that a n�gative temperature could exist in a nuclear spin subsystem. In
a
lithium fluoride crystal, nuclear
spins were aligned by a strong magnetic field. At a positive temperature�
lithium nuclei absorbed energy supplied by a radio-frequency oscillator with
angular frequency equal to the excitation frequency Zejn. When the magnetic field was reversed, the investigators detected resonant emission of electro magnetic energy instead of resonant absorption. indicating a population inver sion and a resultant negative temperature. The relaxation time necessary to reestablish thermal equilibrium between the spins and the crystal lattice was 2-5 minutes, long eno�gh to demonstrate the existence of the inverted popula
tion. In the experiments only the spin subsystem was affected by the field . reversal; the whole crystal remain�d at laboratory temperature� In radio astronomy.. negative temperature systems have been used to amplify very weak radio-frequency signals. Also. the action of a laser depends on a population inversion, which is sustained by some external influence. In the case of a ruby laser, the method is optical pumping using light radiation . •
Sec. 17.6
Ferromagnetism
315
::: :·::::.:: :: ::::;:;:;:;:;:::::::::::::::::: :;:::: ·::=: .·::.::::::::: :::;:::: ::::::::.:::=:: :::: ::·=·:.::::-: =:::: :: :;::� ::::::::::::::: ;:;:::::.; ;:;::::::::: ::.:: : :;:·::·:.::::;:::::;:::;::::�:: =:;:: ·::::::::::::::::::: :::::;:;:::::�>:::;:::: .: :;.; · : :::::::.:::::.::: :::::::::::::;·:: ;::.:::
17.6 FERROMAGNETISM
Pierre Weiss was the first to understand that underlying the domain structure of ferromagnetism is its fundamental atomic nature. The atoms of a ferromag net have a net magnetic moment and these couple together to forttl the domains. Much of the previous discussion of this chapter is applicable to ferro magnetism, with one important modification.
A ferromagnet has a magnetic moment even when it is not in an external
field, a property called spontaneous magnetization. This observation led Weiss to postulate the existence of an internal field proportional to the magnetiza tion, and to replace B with the sum of the external and internal fields:
B-+B
+AM,
(17.43)
where A is an undetermined constant. From Equation
(17.14)
we know that
the magnetization is
NJLz
For the case J
=
1/2 and g
=
·N
(17.44)
2, this becomes •
2J.L8B kT '
TJ =
(17.45)
where the Brillouin function is
(17.46) This can be simplified at once by using the identities coth 71
cosh,.,
11
cosh ·11 +
11
1
cosh 11- 1
sinh 11'
Then
and
M=
Np,B V
tanh
�tsB kT
.
(17.47)
The Thermodynamics of Magnetism
Chap. 17
326
With the replacement indicated in Equation ( 17.43 ), we find that
(17.48)
From this equation we can Jearn a great deal about ferromagnetic behavior. To investigate spontaneous magnetization we set B equal to zero in Equation·(17.48) and obtain a transcendental equation jor M: ..
•
M
N J.LB =
V
tanh
AJL8M
(17.49)
-·--
kT
•
If we set .
VkT --·
(17.50)
-
Equation (17.49) becomes
g = tanh
g" �(II
( 17.51)
.
•
Fi gu re 17.10 is a graphical' solution of this equation. Nonzero solutions evid en tly exist for
�0 in the range 0
<:
�0 < 1. The c orrespondi n g temperature
range is 0 < T < �·-where
( 1.7 52) ..
•
This critical temperature is cal1ed the Curie tt:mperature or Curie point.
1.0
t.
r--
' /
•
\
/ '
Figure 17.10
Graphical solution of Equation ( 17.51 ) The dashed curve is the Jeft
hand side, a straight curve \\'ith unit slope. The solid
curves represent the right hand side for the indicated values of �0 Non-zero •
.
intersections occur onlv for •
go <· I.
O.H
-
().6
t--
I
tanh ? �� 0 � () �) fo (
.
0.� 0.2 .....,._. o
s;...__ .,_ ....__ ..�....-_...___ ... ...___
0
0.5
J.O
1.5
=
=
=
0.5 1.0 2.0
Sec. 17.6
Ferromagnetism
At T
=
317
0 the magnetization takes on its saturation value
Np,8fV. As T
increases, Mdecreases, until at the Curie point an abrupt change in the mater ial properties takes place. This can be seen b}? solving for the temperature in Equation
(17.49): (17.53)
Assuming that MV/NILB is small, we can approxim ate the inverse hyperbolic function using tanh-1x !!!¥ x +
x3
X<<
3'
-
1,
and write
(17.54) With a little algebra this becomes -NP,a
M=
We note that as T
� Tc,
3
v
•
T
(17.55)
the spontaneous magnetization vanishes abruptly:
M � 0 with infinite slope. For T
-+
T,, the ferromagnetic material becomes
paramagnetic. This is an example of a second-order phase transition. • A plot of the reduced magnetization as
a
function of T/Tc is shown in ·Figure
17.11,
along with experimental values for nickel. We have seen ho\v the spontaneous magnetization varies with tempera
ture in the absence of an applied field. In the paramagnetic region, T >
1;;, we
would like to know how M varies with T in an external field B. At tempera
tures above the Curie point, the magnetization is small compared with its sat uration value. Therefore the hyperbolic tangent is small in Equation
(17.48)
and can be set equal to its argument. The resultant approximate expression for the magnetization is M
=
N•'· ,-s2B
Vk(T
-
•
(17.56)
Tt)'
*A second-order phase transitio!l occurs when
a
system passes through a critical region
corresponding to incipient instability. Here the Curie temperature is the ferromagnetic critical point and the phase transition consists of an order-disorder transition of the electron pairs.
The Thermodynamics of Magnetism
Chap. 17
318
1.0
M\l �vJ.J.H
Figure 17.11
Magnetization of nickel as a function of temperature. The
-+-�r----r-
0.8
1----
0.6
....,___ -+--�-+----t-\o-•-1
0.4
.,__- ·-'--+---.---t-��
0.2
1-----
experimental values were r eported by P. Weiss and R. Forrer The solid curve is the 1/2. th eoretical result for J
0
0 '-0
-+---+---t---t1 -'--
__._
0.2
_ _ _ . _ __.
_
_
-
0.4
.
0.6
0.8
1.0
1/T(·
=
This relation is the Weiss-Curie law\ which is a modification of Curie ·s law dis cussed in Section Equation
17 .2. (17 56 ) suggests a method for .
determining the Curie temp�ra-
ture. For a given material� measurements are made of the magnetic susceptibil ity defined as X
( B/JLoM
-
1)
·
1•
For temperatures well above the Curie
point, a plot of experimental values of
1
+ .t -I versus T gives a straight line .
whose intercept with the temperature axis is proportional to J: Except for mi nor departures at very high temperatures� linear behavior is observed and .
.
extrapolation is straightforward. For many .paramagnetic substances 7;_. is less than a kelvin so that Curie ·s law gives an adequate representation at all but the lowest temperatures. A more extensive treatment of ferromagnetism leads to refinements of these results and attempts to incorporate a detailed description of domains and hysteresis in the physical model. I
PROBLEMS 17-1 Show that the B·ohr magncton is given by JJ.H = in a hydrogen atom moves
en/211lt'' A ssume that the electron in a circular orbit of radius a about the proton and
that its angu l a r momentum is 1i. The magnetic moment J.L is the product of the .
electron current and the area swept out by the orbiting electron.
17·2
(a)
For the hydrogen atom of Problem 17-1 show that the orbital magnetic
: dipole moment is ea w/2, \\'here w is the electron·s angular velocity.
(b) Show that the torque produced by a n1agne tic field para11el to the plane of
;, the orbit is ea wB 12. '
(c) By equating th e Coulomb force and the mass times the cent ripet al acceleration, show that •
w=
•
Problems
(d)
329
•
Find values for the angular velocity, torque, and orbital magnetic moment 11 m; let B 10X 5.29 = a 1 T. for a hydrogen atom, where =
17·2
Suppose that the hydrogen atom of Problems 17-1 and 17-2 is subjected to a magnetic field B perpendicular to the plane of the orbit . It can be shown that
the forces caused by by ap,
=
B result in a decrease in the orbital magnetic moment given
e2a2Bf4mt. For the relatively large magnetic field of 1 T, what is
llJ.L/JL
in parts per million? (Titis is the mechanism responsible for diamagnetism .)
17·3
The paramagnetic salt iron ammonium alum has the magnetic ion Fe+++. The spin system has S
(L
=
O).Thus J
=
=
5/2
and the orbital angular momentum is quenched
5/2andg
=
2.Findthe mean dipole moment �zof the para
magnetic salt in a magnetic field of 1 tesla at a temperature of 2 K. What is the saturation value of liz?
17-S
In a paramagnetic solid containing N particles, the total magnetic moment
JLr.TOTAL is 'NP-z· The Curie law is often written in the form TOTAL_
ILl
-
CcB T '
where C, is the so-called Curie constant, given by
Consider a kilomole of a paramagnetic material made up of molecules whose electron cloud has zero orbital angular momentum, spin 1/2, and total angular momentUIIl J = 3/2. Compute Cc.
17-6
A sa11tple of paramagnetic material containing 1()25 atoms is placed in an exter
nal magnetic field of 1 tesla at room temperature. Assume that the electrons of
each atom have zero orbital angular momentum and spin angular momentum of 1/2. Thus J = 1/2 and g = 2.
(a) (b) 17·7
Find the total magnetic moment of the sample.
What would be the total magnetic moment at a temperature of 0.1 K?
Show that the magnetic moment
able by proving that dM is
17-8
an
M of
a paramagnetic material is· a state vari
exact differential. (Use the Curie law
M«B/T.)
Consider a two-level system with an energy 2e separating the upper and lower
states. Assutne that the energy splitting is the result of an external magnetic field
B. Given that the total magnetic energy is Us= -Ne. tanh
kT •
show that the associated heat capacity is
2s
thereby verifying Equation
(17.27).
2
eafkT
Chap. 17
330
The Thermodynamics of Magnetism
17-9 The energy levels of a localized particle are 0, e and 2e. The middle level is dou ..
bly degenerate and the other Jevels are nondegenerate.
(a) Write and simplify the partition functi<,n. (b) Find the total ene rgy the heat capacity� and the entropy of a system of these ,
particles. Sketch these properties as a function of the temperature and com pare them with the corresponding properties of the spin-1/2 system. 17-10 Prove Equation (17.41) starting with Equation (17.40) and using Stirling's approximation. Show that the entropy is
a 1n ax i mum
when the total energy U is
zero, corresponding to equal populations of the two energy levels· of the system. 17-11 Verify Equation (17.42).. referring to Equation (17.41 ). 17-12
(a) For iron find the saturation value of the magnetization. (The atomic weight of iron is 55.9 and its density is 7880 kg m- ·'.) 1042 K). (b) Estimate the magnitude of the internal field AM for iron (T, =
17-13 What is the magnetization of iron in an external magnetic field of 1 tesla at a temperature twice the Curie temperature?
•
•
18.1 Blackbody Radiation
333
18.2 Properties of a Photon Gas
338
18�3 Bose-Einstein Condensation
340
I 8.4 Properties of a Boson Gas
345
18.5 Application to Liquid Helium
347
331
Statistical thermodynamics is applicable to radiant energy as well as material particles. It is a familiar observation that a hot body loses heat by radiation. The energy loss is attributable to the emission of electromagnetic waves from the body. r"Ibe distribution of the energy flux over the wavelength spectrum does not depend on the nature of the body but does depend on its temperature. Here we are concerned with the therxttodynamic properties of electro magnetic radiation in therntal equilibrium. The radiation can be regarded as a photon gas. We consider an enclosure or cavity of volume Vat a constant tem perature T. The walls of the cavity are thermally ittsulated and perfectly reflect ing. Since the system is isolated, it has a fixed energy U. However, the photons emitted by one energy level may be absorbed at another, so the number of photons is not constant. This means that the restriction
� Ni =
N does not
apply. Correspondingly, the Lagrange multiplier a that was determined by this condition is zero and e-a = 1. Photons are bosons of spin 1 and hence obey Bose-Einstein statistics. The number of photons per quantum state is therefore given by Equation (13.40) with p. set equal to zero (recall that N.I
a =
p.fkT):
1
For a continuous spectrum of energies, •
(18.1)
The energy of a photon is hv, so this equatio� can be written N(v)
1
(18.2) 333
Chap. 18
334
Bose-Einstein Gases
•
g(v)dv is the number of quantum states with frequencies in the range v to v + dv. This number was obtained in Chapter 16 for a phonon gas; however, Here
it must be doubled in this application because in a photon gas there are two states of polarization corresponding to the two independent directions of polarization of an electromagnetic wave. Each photon may be in either polar
( 16.8) becomes
ization state. With this modification� Equation
g(v)dv
81rV 1
=
c·
,
(18.3)
+
dv is the number of photons in
v--dv.
where c in this case is the speed of light.
( v)dv in the range v to v this range times the energy hv of each: The energy
u
u(v)dv
=
N(v)dl.'
X
hv.
(18.4)
But
N(v)
=
g(v)j(11).
Hence
( 18.5) Thi� is the Pla�ck radiation formula. It gives the sp ectral distribution of the . radtant energy mstde the encJosure. i.e., the energy pe r unit frequencv. The blackbody spectrum is often expressed in terms of the wa�elength. Then
li(v)dv ll(A)dA ·x
and
dv
=
d
c -
A
c
=
-
dA , ,.\2 ·
or
The wavelength spectrum is therefore •
>._5
(
ehc/AkT
_
]
)
'
(18.6)
1
I
Blackbody Radiation
Sec. 18.1
335
u(A)
Fagure 18.1 The wavelength spectrum of blackbody radiation energy for three temperatures: T1 > T2 > T3•
(Figure 18.1). Here u(A) is the energy per unit \Vavelength. Prior to Planck's analysis, various empirical formulas existed. All of them can be found from Equations (18.5) or (18.6). The Stefan-BoltZinann law states that the total radiation energy is proportional to T4 (the area under the curves of Figure 18.1}. The total energy density (energy per unit volume) is (18.7)
Setting x
=
hcjltkT, we have
The integral has the value 1T4/15 (see Appendix D).Thus _U v
=
aT4'
(18.8)
where
Since radiation inside the cavity is continually absorbed and emitted by the inner surface, we can relate the ener. gy per unit volume, which moves at the speed of light
c, to the energy emitted per unit area of the surface per unit
time. The latter is the power per unit area or energy flux. In Chapter 11 we
Chap. 18
336
found the particle flux to be
(vj4)n� �=
Bose-Einstein Gases
where vis the mean speed and
n
is the
number of particles per unit volume. In a similar way, the energy flux
e
is
( c/4) ( UjV). Therefore,
(18.9) where
(18.10)
Equation
(18.9)
is known as the Stefan-Boltzmann Jaw, and
u
is called the
Stefan-Boltzmann constant. The theoretical and experimental values of
u
agree to within at least three significant figures. The wavelength
Amax at which u(A) is a maximum satisfies a relation
known as Wien 's displacement law. It can be found by setting the derivative of
u(A) equal to zero or, equivalent1y, by minimizing the denominator in Equation (18.6):
With .t
=
hc/AkT. this
reduces to
This is a transcendental equation whose numerical solution is 4.96. Hence
( 18.11)
(18.11) is Wien ·s displacement law. The experimental values of the constant on the right-hand side and of the Stefan-Boltzmann constant can be Equation
used to determine the values of h and k, assu1ning that
c
is known.
For long wavelengths.l1c/AkT << 1 in Equation (18.6), and the exponen tial can be approximated by the first two tertns of its Taylor series expansion:
ehc/AkT
===
1 +
he AkT.
Sec. i 8.1
Blackbody Radiation
337
Then
u(A)dA
;o.::
81rkT V dA.. 4 A
(18.12)
This is the so-called Rayleigh-Jeans formula, which exhibits an ''ultraviolet catastrophe" as the wavelength approaches zero (u(A) becomes infinite). For short wavelengths, 1 >> hc/AkT e and
u(A)dA
z::
V
87ThC 5
A
. -hc/AhT e dA
(18.13)
This is Wien's law, valid in the short wavelength regiort (Figure 18.2).
To summarize, the total blackbody radiant energy per unit volume increases with the fourth power of the temperature, and the wavelength of the peak of the radiation curve u(A) is inversely proportional to T.
The temperature of the Sun's surface is approximately 6000 K, and Amax is 483 nm, a wavelength in the visible range of the electromagnetic spectrum. At the surface temperature of the Earth, roughly 300 K, Amax is about lOJLm, which is in the infrared region. The cosmic background microwave radiation that perrneates all of space was discovered by accident in 1964 by Amo Penzias and Robert W. Wilson while making measurements of radio signals at a wavelength of 7.35 centime
ters (4 GHz). They observed a nonzero value for the radiation in every direc tion in which they pointed their radio telescope. In 1990 the Cosmic Background Explorer (COBE) satellite measured the background radiation
u(A) I
\
\4 \
Rayleigh-Jeans
\
t
VJien
Figure 18.2
..,..._Planck
Sketch of
Planck's law, Wien's law and
the Rayleigh-Jeans law.
A
Chap. 18
338
Bose-Einstein Gases
energy
density 500
0
;... ___., ..._ __._ _ __.__ ____ _ ___ _ ....� ... __, _
400
200
0
600
frequency (GHz) Figure 18.3 Cosmic background measurements by the COBE satellite. The smooth curve is the theoretical blackbody spectrum at 2. 735 K. The energy density is in eV m-3 per gigahe rtz (Adapted from The Inflationary Universe by Alan H. Guth, Addison-Wesley, Reading. .
Massachusetts, 1997.)
at a large number of wavelengths. The data fit the theoretical Planck curve
18.3). The cosmic background radi ation is evidently a blackbody with a temperature of 2.735 :±: 0.06 K.
with almost unbelievable precision (Figure
18.2 PROPERTIES OFA PHOTON GAS
It is instructive to ask: what is the average value of the ratio hv Ik Tin ·a volume Vat temperature T? The number of photons having frequencies between and
v
+
v
dv is found by combining Equations (18.2) and (18.3): N(v)dv
81rV ==
1
c·
eh,./kT
(18.14)
-- ---
_
1
·
.
The total number of photons in the cavity is determined by integrating this expression over the infinite range of frequencies. The result is
N where the substitution x
=
=
R17.Y
kT he
3
,.
0
2
x dx ex
-
1-
(18.15)
hvjkT has been made. The integral has the numer
ical value 2.404. Hence
N
=
2.02
X
107 T3
V,
(18.16)
Sec. 18.2
Properties of a Photon Gas
339
where Tis in kelvins and Vis in m3 • To find the mean energy ofthe photons in the cavity, we divide the total energy U 7.55 x 10-16 T4 V J (Equation (18.8)) by =
N. The result is
3.74 X
.
10-23 T
=:;
2.7 kT.
Thus, the average value of h vjk T is ofthe order of unity.
The heat capacity is easily calculated using Equation (18.8):
c
_
v-
au
3
ar
v
v.
(18.17)
Since this relation holds down to a temperature of absolute zero, it can be used to determine the absolute entropy:
S=
kT 0
0
3
3
he
45
V.
(18.18)
Thus both the heat capacity and the entropy increase with the third power of the temperature. '
For an open system, we know that
dU
=
TdS - PdV
+
JLdN.
Substituting the differential of the Helmholtz function F
=
U
-
TS in this
equation gives
dF
=
-SdT- PdV + p,dN.
(18.19)
Therefore, •
J.L=
aF
aN
T.V
•
(18.20)
For the photon gas,
(18.21) Thus, as we discussed in the previous section, the chemical potential JL is zero, because F does not depend explicitly on N.
Bose-Einstein Gases
Chap. 18
340
Finally� Equation (18.19) also yields = P
aF aV
(18.22)
•
r.N
The pressure of the photon gas on the walls of the cavity is therefore
(18.23)
Note that this differs from the relation P
=:=
2U/3 V that obtains
for an ideal
gas of weakly interacting particles (see Chal>ter 11). Equation (18.23) can also
be obtained from electromagnetic theory if it is assumed that the radiant energy in the enclosure is isotropic.
18.3 BOSE-EINSTEIN CONDENSATION
In this section we shall be concerned with a gas of noninteracting particles (atoms or molecules) of comparatively large mass such that quantum effects only become important at very low temperatures. The particles are assumed to comprise an ideal Bose-Einstein gas. The discussion is relevant to 4He_ which undergoes a remarkable phase transition known as Bose-Einstein conde11sa tion. This phenomenon is intimately related to the superfluidity of liquid
helium at low temperatures. As we noted previously� bosons are particles of integral spin that obey Bose-Einstein statistics. There is no limit t(> the number of bosons that can occupy any single particle state. We consider an ideal boson gas consisting of N bosons in a container of volume V held at absolute temperature T. The Bose Einstein continuum distribution is
N(e)
·
l
(18.24)
---
g e
e·
..
-
Our initial concern is determining how the chemical potential J..L varies with the temperature. We shall adopt the convention of choosing the ground state energy to be zero. At T
=
0 all N bosons will be in the ground state since
0 there is no restriction on the number of bosons in a given state. Setting e in Equation (18.24) .. we see that if .f(e) is to make sense J.L must be intrinsi =
..
cally negative. Furthermore, J.L must be zero at a temperature of absolute zero and only slightly less than zero at nonzero low temperatures.. assuming N to be a large number.
Sec. 18.3
Bose-Einstein Condensation
341
At high temperatures, in the classical limit of
Boltzmann distribution applies:
a dilute
gas,
the Maxwell-
·
(18.25) From Chapter
14, 1-L =
-
kT
z ln -N ,
(18.26)
where
z =
2TrmkT 312 v. 2 h '
•
(18.27)
Thus
2TrmkT 312 V
_JL_ =-In
kT
(18.28) •
As an example, for one kilomole of a boson gas comprising 4He atoms at stan dard temperature J.L
kT
=-In
and pressure, we have
27T(6.65
X
2 10- 7) ( 1 38 .
X
10-23) (273)
(6.63 X 10-34)2
312
22.4 6.02 X 1()26
-----
= -12.43. The chemical potential per se is --0.29 eV. In comparison, the average energy of
an
(3/2)kT = 0.035 eV at 273 K. Also, 12.4 = 13.9; substituting this value in Equation (18.25)
ideal monatomic gas atom is
e
JL) /kT 1.5 + gives f ( e ) = 9.2 x 10-7, confirming
(e
-
=
=
!he validity of the dilute gas assumption.
In this classical limit,
27TinkT
h2
312 V
N
(18.29)
is a positive number that increases with temperature and decreases with the
particle density NjV.
The obvious way to determine the temperature dependence of JL is to
use the conservation of particles condition that led to the statistical definition
342
Chap. 18
Bose-Einstein Gases
of the chemical potential in the first place. For the continuum appr<>Ximation, this is X
N(e)tle
�f(e)g(e)de
=
0
=
N.
(] 8.30)
0
For g(e)� we have the result of Equation (12.26) with the spin factory_, equal to unitv: .,
g(e )de
4 =
·
2 7T V
h3
m
3/2
I /l
e de.
(1 8 3 1 ) .
Thus � �
e112de
� 0
e(e-p._t/kT
_
}
•
(18.32)
There is a significant flaw in this fortnulation. In using the integral approximation
of Equation (18.30) rather than the sum, the ground state e = 0 is left out. This terrn, in the sum of particle states at low temperature, is the largest tetm of all. It
is omitted here because the density of states function g (e) depends one' :! �which
is zero for the ground state. Under ordinary circumstances this doesn�t matter..
since, fore small compared �·ith kT. the omission of the one term makes a negli gible difference in the result. But at low temperatures, bosons condense into this
lowest state and its occupation becomes much greater than for any other state. We can surmount the difficulty in the foJJowing way. The tota1 number of bosons consists of Nc. in the ground state and Nex in the excited states. Hence (18.33) Since the ground state is excluded from Equation ( 1 8.32)� the integration only gives the number of bosons in excited states. Therefore
(18.34) The integral can be solved numerically to obtain a relationship between J.L the temperature,. and the particle density. It is n1orc instructive. however.. to note ..
that at temperatures very close to zero, N.,
-==
N. We can therefore pute
Equation (18.24) and write N=::
1 e
JJ./kT
-
1
.
=
0 in
Sec. 18.3
Bose-Einstein Condensation
343
Then
N
·
N
for N large. (Even for a small macrosc opic system N will be a huge number.) For low temperatures, then, we can safely put exp(- JL/ kT) equal to unity in Equation (18.34). With the change of variable x
-
e/kT, we get
(18.35)
---
0 ex-
The value of the integral is 2.612
1
•
TT/2. Thus 3/2
(18.36)
•
The so-called Bose temperature T8 is the temperature above which all the bosons should be in excited states. Thus we set Nex N and T = T8 in -
Equation (18.36) and obtaitt
N = 2.612V
(18.37)
Solving for T8, we get
To=
Jil
2/3
N
(18.38)
•
21rmk 2.612V
•
T8, all the bosons are in excited states. As T falls below T8, an increas 0, all the bosons ing number of bosons occupy the groltnd state until at T are in this state and N0 = N. The fractional number of bosons in the ground
For T >
=
state 1s •
(18.39)
Dividing Equation (18.36) by Equation (18.37), we have
No -=1N
T T8
3/2 •
(18.40)
Chap. 18
344
Bose-Einstein Gases
What is a typical value of the Bose temperature? Consider a boson gas made up of 6.02 x 1023 4He atoms confined to a volume of 22.4 x 10-3m3. The mass of a 4He atom is 6.65
x
Io--27 kg. Using Equation
(18.38)�
we find
that
X l
In the case of helium, the gas liquifies at before the temperature is reduced to
4.21
2/3 =
0.03 6
K.
Kat atmospheric pressure, long
0.036 K.
In fact, all real gases liquefy
before their Bose temperature is reached.
N0/ N and NeJ N v·ersus T jT8 are shown in Figure 18.4. A corre sponding graph of p,/ kT8, numerically calculated, is shown in Figure 18.5. *The Plots of
sudden "collapse" into the ground state at very low temperature is called the Bose-Einstein condensation. It is not a condensation in geometrical space but rather in what might be called momentum spac e. For many years physicists believed that a Bose-Einstein condensate did not exist in nature. In 1995 researchers created a nearly p�re condensate by
1.0 Nn 0.5 Figure 18.4
Variation with temperature of N0/ N and
NcJ N
for a boson gas.
0
Nl'X N
.N t--
0
I
() -------r.. �------� ) 1 -
T Ts
JJ.
kT8
-0.5 ._
-1.0 � Figure 18.5 boson gas.
Variation with temperature of I-ll kTn for
•
a
*See Problem 18-10 for the method of calculating p./ kT8 versus T/TH.
Sec. 18.4
Properties of a Boson Gas
345
cooling a vapor of rubidium atoms to a temperature of 1.3 x 10-7
K. The
achievement was hailed as "the most interesting development in atomic phy sics in a decade,'' opening up an entirely new field of investigation.* •
18.4
PROPERTIES OFA BOSON GAS
The bosons in the ground state do not contribute to the internal energy nor to the heat capacit y. The first term in the sum I �si is Noe0• For T < Ta, No may be large but e0 = 0, and for T > T8, N0
0 in any case. For temperatures above the Bose temperature, all the bosons are in excited states and we may expect the internal energy to approach (3/2) NkT as =
the temperature is increased. Below the Bose temperature the number of bosons in the excited states is Nex = N(T/T8)3f2. As a first approximation, we assume that each of these bosons will have a thermal energy of the order of kT.Thus 3/2 .
(T < T8),
indicating that U varies as T512 below the Bose temperature. A more nearly exact result is obtained by noting that 00
U =
eN(e)de.
(18.41)
Using Equations (18.24) and (18.31), we find that ·
U
=
21TV
2m J/2 oo
h2
0
e312de ( )/kT
e e-p,
-
1
•
(18.42)
As we have seen, if we choose the energy of the ground state to be zero, the chemical potential JL of the boson gas is very close to zero for temperatures below the Bose temperature. Setting p, = 0 and making the substitution x =
e/kT, we
obtain '
U
=
2 1T
kT
2'1TmkT 3/2 h2
v
x312dx 0
eX -
1
•
(18.43) '
*See the Richtmyer Memorial Lecture, "Bose-Einstein Condensation in an Ultracold
Gas,n by Carl E. Wieman, American Journal of Physics 64, pp. 847-855
(1996).
Chap. 18
346
Bose-Einstein Gases
the
The definite integral is equal to the product of
gamma function
f(S/2) 3TT1i2J4 and the Riemann zeta function {,(5/2) ·(· expression for T11, Equation (18.38), we get =
u
0.770 N kT
=
3/2
T
(T
T 'B
<
=
1.34. Using the
(18.44)
Tn),
not greatly different from our approximation. It is a simple matter now to obtain the heat capacity:
(18.45) Note that Cv is proportional to T
312•
A graph of the heat capacity as a function of temperature is shown in Figure
18.6. The
curve has a change in slope at T
T1h and the heat
=
capacity
1.92 Nk. At higher temperatures, Cv approaches the classical value of (3/2) Nk. has its maximum there, equal to
The absolute entropy at temperatures below the Bose tempt:rature can be calculated from the heat capacity, since Equation
(18.45) holds at absolute
zero:
(T
The entropy goes to zero at T
=
=
(18.46).
T8).
() as it should, according to the third law of
thermodynamics. The Helmholtz function F
F
<
U - T S is
3/2
T
-0.51 NkT
=
T B.
(T
<
T8),
(18.47)
2.0 � J
•
5
--
- .. ·- -
-
-
-
-
-
-
-
......
-
Cv 1.0 Nk 0.5 Figure 18.6 Variatio� with temperature of the heat capacity of a boson gas.
0
t--
�----�---
0
I
2
T
•
Ts
Sec. 18.5
Application to Uquid Helium
347
from which we can deterrnine the pressure, using the reci procity relation P
=
-
(aFjaV)r.N· Expressing Fin terms of the volume V using Equation
(18.38) for T 8, we have
3(2 V,
h
(18.48)
from which
(T
<
T8).
(18.49)
The pressure of a boson gas at low temperatures is proportional to T512 and is
independent of the volume. This is because the bosons in the ground state e =
0 have no momentum and therefore make no contribution to· the pres
sure. It is easy to show that P
=
(2/3)UfV, the same relation obtained for a
classical ide al gas of noninteracting particles.
I 5 APPLICATION TO LIQUID HELIUM The phase diagram of ordinary heliu1n (4He) is shown in Figure 18.7. The sub stance has a critical point at a temperature of 5.25 K and exhibits unique
behavior in the vicinity of 2 K. Above the critical temperature 4He can't exist
as a liquid. When helium gas is compressed isother1J1ally at a temperature below the critical temperature but above 2.18 K, it condenses to a liquid phase
30 p
(atm)
....._
20 10 .,__
solid He A-line liquid He II
liquid He I
0 L...-.-..J..-.----�..+-0 1 2 3 2.18 K
CP 4
5
He gas 6
T(K)
5.25 K
Figure 18.7 Phase diagram of 4He. The two liquid phases are labeled I and II.
Chap. 18
348
Bose-Einstein Gases
called helium I. When the vapor is compressed at temperatures beJow 2.18 K, a liquid phase called helium II re sul ts. Helittm II is a superfluid. He I and He II can coexist in equilibrium over a range of pressures and temperatures defined by the so-called lambda line. He II remains a liquid down to absolute zero. Solid helium can�t exist at pressures below 25 atm and it can't exist in ,
equil i b rium with its vapor at any pressure or temperature. Helium has two triple points. At one of them liquid He I and liquid He II are in equilibrium with solid helium. At the other, called th e A-point� the two forms of liquid helium are in equilibrium with the vapor. The transition between ordinary liquid helium He I and superfluid He II can take place at any point along the A-line. A graph of the heat capacity ver sus temperature for the two phases has the general sha pe of the Greek letter ,\ (Figure 18.8). As the figure shows. the heat capacity does not change continu ously. Its variation with temperature is vastl y different in the two phases. The properties of l iquid He II are unique. Its viscosit y is virtuall y zero� hence it is called a superfluid. In the two-tluid theory of He II it is assumed that, at temperatures below the A-po�nt.. liquid He 11 is a mixture of a no1n1al fl uid having viscosity and a superfluid that has no viscosity. The proportion of the superfluid is zero at the A-point a n d increases to
u nity
as the temperature
approaches zero. Since the forces of interaction between atoms are weak in liq ltid helium� we might assume that liquid He ll i s an ideal boson gas, to a first appro xima
tion.* If we identify the A-point with the Bose temperature, the two-fluid model s uggests
that the supc rfl uid component co n si sts of the �) atoms in 1 he ground
I
• •
f I 4
I
liquid 3
--
liquid f-Ie I
I .
j
He II
I •
I.
.. .
•
c�, Nk
I
2 I Jt--
o
----��i
.... .
0
_j__ ..
__ __
I
-.,
,.,_!
____
_
•
l
"" �'
.. r( K)
--..
-
2.1H K Figure 18.8
The heat capacity anomaly of �He.
*It might seem strange to
apply a
theory for an ideal gas to a liquid. However. in some
ways, liquid helium behaves �ore like a gas than a liquid.
Problems
349
state while the normal component is identified with the Nex atoms in the excited
states. As the temperature falls below T8, an increasing number of atoms popu late the ground state, and the superfluid component becomes predominant. Are T8 and the A-point the same? The volume of a kilomole of liquid 4He is 27 X 10-3m3 so that the concentration is
Using the value of 6.65 x 10-27 kg for the mass of the 4He atom, we can com pute T8 from Equation (18.38). The result is T8
=
3.1 K,
as compared with 2�18 K for the A-point. We conclude that the superfluid prop erties of liquid He II could be attributed, in part at least, to a Bose-Einstein like condensation at the A-point. The ideal boson gas model, however, is only an approximation to act1.1al liquid helium that neglects interatomic interac tions. The complete picture is considerably more compl�cated.
.
\ '·
PROBLEMS
18-l (a) Calculate the total electromagnetic energy inside an oven of volume 1m3 heated to a temperature of 400°F. (b) Show that the thermal energy of the air in the oven is a factor of approxi mately 1010 larger than the electromagnetic energy.
18-l (a) Calculate Amax for the Earth, assnnting the Earth to be a blackbody.
(b)
Calculate the temperature at which the hntnan eye is most sensitive to a
wavelength of 500 nm.
18-3 Assume that the radiation from the Sun can be regarded as blackbody radia tion. The radiant energy per wavelength interval has a maximum at 480 nm.
(a) (b)
Estimate the temperature of the Sun . Calculate the total radiant power emitted by the Sun. (The radius of the Sun
is approximately 7 X
lei m.)
18-4 (a) Find the frequency at which the radiant energy per unit frequency interval of a blackbody is a maximum. How does this compare with the frequency at
(b)
which the radiant energy per unit wavelength interval is a maximuJn? Find the frequency at which the cosmic background radiation is a ma?Cimum and compare your result with that shown in Figure 18.3. Assume that
T
=
2.7 K. I
Chap. 18
350
Bose-Einstein Gases
18-5 (a) Calculate the number of photons in equilibrium in a cavity of volume 1 273 K. held at a temperature T
� m·
=
(b)
Compare this number with the number of molecules the same volume of an ideal gas contains at STP.
18-6 Assume that the universe is a spherical cavity with radius 101n_ m and tempera ture 2.7 K. How many thermally excited photons are there in the universe? 18·7 For photons. where there is no restricti<)n on the total number of )Jarticles, the partition function is independent of N, as are the physical properties of a photon gas. The partition function
z
for a single oscillator is given by (see Section 15.2)
In
-ln(1-
z =
e h ...
lkT),
ignoring the zero-point energy. The number of single particle (photon) states in a volume V in the frequency range
v to �·
+
dv is
Therefore, the partition function Z of the photon gas is the sum over states given
(a)
by
Show that integration by parts Jeads to the equation In Z where x
=
=
B1rV he
J
hvfkT.
(b) Referring to Appendix D. show that -
In z
•
.
87T> k T
=
45
3
he
V
·
18-8 Use the result of Problem 1H.7 to show that
(a)
U
=
kT2
consistent with Eq u ati o n
iJ In Z
aT
( 18.8):
v.v
·
.
(b)
Jl
=
-
kT
iJ
tn-z =
riN
V.T
0�
confirming the statement in Section 18.1:
(c)
S
u =
T
+kIn Z
=
321T5k
k
45
he
3
T3V.
Problems
351
the same result as Equation (18.18); and
p
(d)
=
kT
which differs from the result P
1
�In �
=
u
3
-
v ,
-
T.N
obtained in Chapter 11 for an
( 2/3) ( U /V)
ideal gas of molecules.
18-9 In classical thermodynamics, it is found that
=
T
T iJP iJT
-
v
For blackbody radiation, the energy density does the pressure P ential equation for
u ::s
( 1/3)
=
u. 1:11;;&
P.
u = ==
UfV depends on T only, as
Use the above classical result to obtain a differ-
whose solution is the Stefan-Boltzmann law.
18.5 is a plot of�/ kT8 versusT /T8. In the region 0 < T /Tn < 1, p,/ kT8 is essentially zero. ForT/T8 > 1, Equation (18.34) applies, with Nex set equal toN
18-10 Figure
(all N of the bosons are in excited states). Using the definitions x
�
=
p,fkT8, 11
= T /T8,
=
ejkT8,
show that Equation (18.34) reduces to 2.315
..
x
=
xl.'2dx
oex � · "-1 (
-&.) /
-,
-
which can be evaluated numerically to give � as a function of 1J (Figure
18.5).
2T8 is approxi 18-11 The chemical potential of a boson gas at a temperature T mately -0.8 kT8• Deterinine the mean number of bosons f (e) in single particle =
states having energies of (a) 0, (b) 0.5 kT13,
18-12
(a)
(c)
2.0 kTu, (d) 2.0 kTn.· (e) 3.0 kT8•
Find the chemical potential of a kilomole of 4He gas at STP. Express your answer in joules and in electron-volts.
(b)
Use Equation (18.24) to show that the mean occu11ancy of a single particle state having energy
18-13
(a)
(3/2) leT is 8.8
x
10-7 at STP
An ideal boson gas consists of 4He atoms whose Bose temperature is 0.087 K. Find the boson concentration, the number of bosons per cubic meter.
(b)
What percentage of the bosons are in the ground state at a temperature of to-·2 K?
18-14 In a Bose-Einstein condensation experiment, 107 rubidiun1-81 atoms were cooled down to a temperature of 200 nK. The atoms were confined to 1!\ of approximately lo-- m3•
(a) (b) (c)
a
volume
Calculate the Bose temperature 7!3• Determine how many atoms were in the ground state at 200 nK. Calculate the ratio kTje0, where T energy
e0
is given by
=
200 nK and \vhere the ground state
Bose-Einstein Gases
Chap. 18
352
18-15 Hydrogen freezes at 14 K and boils at 20 Kat atmospheric pressure. The density of liquid hydrogen
is 70 kg m-).
Hydrogen molecules are bosons. No evidence
has been found for Bose-Einstein condensation in hydrogen. How do you account for this?
18-16 Show that for temperatures below the Bose temperature of a boson gas, (a) Cv ( d) P
=
=
(5/2) UjT, (b) S (2,/3) UjV .
=
(5/3) UfT.
(c ) F
=
-
(2/3) U,
18-17 A system of N bosons of mass tn and zero spin is in a container of volume V at a temperature T
>
0. The number of particles is
N
=
21TV
2m
�/, _, ..
2
h
�
0
1 '2
e' de ( - )lkT
e t P.
-
·
I
'
(Equation (18.32 ) ). In the dilute gas approximation, exp (
-
J.L! kT) >> 1, and the
Bose-Einstein distribution becomes the Maxwell-Boltzmann distribution. Evaluate the integral in this approximation� referring to Appendix D, and show that exp(- p.,/kT)
d =
A
-
) �
2 where A= h/(21TtnkT)1-' is the de Broglie wavelength of the particles' thermal motion and d (V I N) 11\ It follows that in the classical limit the average dis ==
tance
d between the particles is very large. compared to A.
•
I
•
19.1 The Fermi Energy
355
19.2 The Calculation of JL T
357
19.3 Free Electrons in a Metal
361
19.4 Properties of a Fermion Gas
364
19.5 Application to White Dwarf Stars
367
•
353
19.1 THE FERMI ENERGY
Fermi-Dirac statistics governs the behavior of indistinguishable particles o f half-integer spin called fermions. Fermions obey the Pauli exclusion principle, which prohibits the occupancy of an available quantum state by more than one particle. We consider an ideal gas comprising each of mass
m, in a container of volume
N noninteracting fermions,
V held at temperature T.
The Fer111i-Dirac distribution is
·
(19.1) •
or
N(e)
1
(19.2)
in the continuum approximation. The right-hand side of Equation
(19.2)
is
referred to as the "Fern1i function"; it gives the probability that a single parti cle s tate
e will
For
e
=
be occupied by a fermion. Clearly, 0
JL, f ( e )
<
f (e )
<
1.
has the value 1/2 at any temperature. Here J.L is the
chemical potential, which is a function of temperature. Its value at T
=
0, that
is JL ( 0), is called the Fermi energy. The Fermi energy is also written as eF. Consider the Fermi function at a temperature of absolute zero. ,
Evidently, at T
=
0,
(e
-
JL(O) )/kT
- �
if e <
oo
if e >
=
p,(O) � (0)
·
355
Fermi-Dirac Gases
Chap. 19
356
Correspondingly�
0 tf E
>
JL(O)
This tells us that at T = 0 all states with energy e states with
e
>
( 19.3)
<
JL(O) are occupied and all
�(0) are unoccupied. At absolute zero fermions will occupy
the lowest energy states available. The exclusion principle says that only one fettnion is allowed per state. So all
N particles will be crowded into theNlow
est energy levels. It follows that only one configuration (microstate) is possible for the who]e assembly at T S
=
=
0. Thus the thermodynamic probability w is 1 and
k ln w = 0. The vanishing of the entropy at absolute zero is consistent
_ ction at T with the third law of thermodynamics. The Fermi fun
19.1. We ask: how does the Fermi energy JL(O)
=
0 is shown in
m� N, and V? 0. We need g(e), It obviously doesn't depend on the temperature since T the density of states of Chapter 12 once again. For particles of spin 1/2, such as Figure
depend on =
electrons, the spin factor is 2� and
(19.4) •
For conservation of particles,
� N;. I
N
=
�
or
:X
N(e)de 0
Substituting Equations p.,(O)
f(e)g(e)de
N.
=
(19.5)
()
( 19.3) and ( 19.4) in Equation ( 19.5), we get J/2
2n1
N=
=
p.,(O)
·1
e'i··de
87T V = --
(19.6)
3
f)
()
2m
/(e) '
(T
=
0)
I ,._.____-.
Figure 19.1 The Fermi function at T 0. =
•
�t(O)
Sec. I 9.2
357
The Calculation of p,(T)
lt u s e r e h t in a t b o e w , Solving for #J-(0) h
#J.(O)
=
2
2/3
3N
•
2m
temperature
· t rm c F a e c u d o tr in For convenience, we s a n e tt ri w e b n ca is Th F. T k p & p,(O) =
(19.7) TF
such
that
=
� F
-
h
2
2/3
N
(19.8)
--
27Tmk
'
1.504V
analogous to the Bose temperature given in Equation
19.2 THE CALCULATION OF
(18.38).
p,(T)
•
Figures 19.2 and
19.3 show how the Fermi function changes with temperature.
To obtain these curves, we must determine J.L(T). The calculation is consider 0. We have ably more complicated than it was for T =
2 3/ 2m
:o
'
o
elf2de
�
+1
0et JL
(19.9)
Let the integral be I in this expression and equate Equation (19.6) with Equation
(19.9). The result is
O
e(s-p.)jkT + 1.
/(e) 1.0 ......
.-- (T
=
--- - ---
�
. ---------5 0
Fig11re 19.2 The Fermi function at T 0.2TF. =
0
(19.10)
0.5
1 I I I
1.0
0) T
p,
=
=
0.2 TF 0.96#L(O)
e/�(0)
358
Chap. 19
f(e) 1.0
0"
Fermi-Dirac Gases
�(T =O)
-
-
-
-
·-
-
- - - -�
T 1 J.L I I
=
=
1.2 TF -0.45�(0)
I
I -----------
·-
--1 . . .
I
Figure 19.3 The Fermi function at T 1.2 TF.
0.5
0
=
The plus sign in the denominator of
1.0
e/�(0)
integrand makes all the difference in our ability to evaluate the integral. It can be evaluated numerically. But we can the
find a very good infinite series approximation for p,(T) in the following way. Let
I=
(19.11)
where
2
•')
and
f (e )
·
I nt egra ting E q uati on
1 ==
) ·k r
(
eF.-p._.-
+ 1
·
( 19.1]) by parts, we ot,tain df(e)
X
X
de
·
0
Now F ( oc)
f(s)f (e) ..
F(O)
=
=
·
is i nfin ite
but f ( x)
11
==
0 and its de ca y is exponential,
0 at the upper litnit. At the
Jo\ver limit�
0, so the product is zero. Hence •:/_
I
=
F (e)
-
0
d_((t-;) ·
de
ll£.
so
/(0) is finite but
The Calculation of J.t(T)
Sec. 19.2
359
19.1 shows that at T 0, the functio11 f(e) has a zero JL(O), where the slope is infinite. Thus the slope everywhere except at e derivative d.f (e)/de is a Dirac deita function. At temperatures less than TF but A quick look at Figure
=
=
greater than zero, we might expect tl1e derivative to be function peaked at
e=
p,. This suggests that we expand
a
kind of �'fuzzy" delta
F(e)
in a Taylor series
about p, since the only significant contributions to the integral! will be in the vicinity of e
=
JL. * The expansion is
(19.12) The contribution to the integral of the linear term is zero since
df (e) /de
is
symmetric about JL and the areas under the curve to the left and right of p, will
cancel. Thus
(19.13) Evaluating the derivative
df(e)/de in the integrand, we obtain
= I
We set
y
=
(e
-
1 kT
�
0
kT F( e )e
(19.14)
-----
•
J.L) I kT in the integrand, so that I=
F(y)eYdy . 2 (eY + 1) -p.jkT :G
0 where y
The factor
eY is already negligible
for
at least at low temperatures, which is the region of concern. Hence
e<
Bf,
at
e
we can safely replace the lower limit by
=
-
co
.
=
-
J.L/kT, and it's small
Then, using Equation
(19.12), we
get
*This method of evaluating the chemical potential and the internal energy for low temper atures is known
as
the Sommerfeld expansion.
Fermi-Dirac Gases
Chap. 19
360
7T'2/3.
The first integral has the value 1 and the second
Thus, reminding our
selves of Equation ( 19.10), we have
A little algebra gives
JL
=
JL(O)
7T2 kT
2
kT
2
-2/3
1 + 8
-li_ :z:o ,.,., ( Q )
•
j
Finally, we replace J.L in the ''correction term'' by J.L (0)
� =
1T2
1-L(O)
T
==
kTF, to obtain
:
(19.15)
A comparison of this approximate expression with the Hexact'' numerical eval
0 at T fTF 1.1 p, (0). instead of
uation is shown in Figure 19.4. The approximation gives J.L
instead of 0.999, and JL
=
-0.85 J.L (0)
at T/TF
=
1.5
=
=
1.10
-
We especially note that J.L is positive for temperatures below the Fertni
temperature and negative for higher temperatures. As the temperature increases above
TF, more
and more of the fermions are in the excited states
and the mean occupancy of the ground state falls below
1.0
0.5 J.L
�(0)
0
-0.5 Figure 19.4
Exact and
approximate calculations of JL/p,(O) versus T/TF.
-1.0
1/2. In this region,
....
approximate
t--
exact 0.5
' '
1.5 \ \ \ \ \ \ \ \ \ \
T!TF
Sec. 19.3
Free Electrons in
a
Metal
361 1
f(O)
_ -e-p./kT+l
1
<
2'
which implies that
JL kT or J-L
<
<
0'
0. We also note that the situation is different for
a
boson gas, where J.L
is negative at all temperatures and is zero at absolute zero. At high temperatures the fermion gas approximates the classica1 ideal
gas. In the classical limit,
JL
=
z
-kT In
N
,
(19.16)
with Z N
=
2
27TmkT h2
312
V
. N
(The spin degeneracy factor is 2 for fermions.) For large negative value and exp(- J.LfkT) >>
(19.17) T
>>
TF, JL/kT takes
on a
1. As an example, consider a kilo
mole of 3He gas atoms (which are fermions) at standard temperature and pressure (Problem
T/TF
=
19-2). The Fermi temperature is 0.()69 K, so that -12.7 3900. Using Equations (19.16) and (19.17), we find that J.L/kT
and exp(- JL/kT)
=
=
3.3 x lfr'\. The average occupancy of single partit�le states
is very small, as in the case of an ideal dilute gas obeying the Maxwell Boltzmann distribution.
19.3 FREE ELECTRONS IN A METAL Statistical tl1ermodynamics provides profound insights into the behavior of conduction electrons in metals at moderate tentperatures. Electrons are spin
1/2 fermions. Each atom
in the crystal lattice of the metal is assumed to part
with some number of its outer valence electrons, \Vhich can then move freely
about in the metal. There is an electric field due to the positive ions that varies
widely from point to point. Ho\vever, the effect of the field is canceled out except at the surface of the metal where there is a strong poteittial barrier,
called the work function, th�t draws back into the metal any electron that hap pens to make a small excursion outside. The free electrons are therefore con
fined to the interior of the metal as gas molecules are confined to the interior
of a container. We speak of the electrons as an electron gas.
Chap. 19
362
Fermi-Dirac Gases
j
1--------.-----
� ......
._--
-Fern1i level
J.L(T) •
Figure 19.5 Potential well for free electrons in a metal
�lr
.
In this model, the free electrons move in a potential box or well whose walls coincide with the boundaries of the specimen. They occupy energy states up to the so-called Fermi level, which is the chemical potential p.(T). The work
cf> is the energy required to remtlve an electron at the Fermi level from the metal surface. The depth of the potential well is equal to J.L(T) + 4J function
(Figure 19.5). The Fermi level of the free electrons in most metals at room temperature J.L(O). It is often assumed is only fractionally less than the Fermi energy eF that the two are equaL and this leads to confusion. The Fermi level, strictly speaking, is p,(T), which is an approximation to the Fermi energy valid for T << TF. A more realistic picture of the potential well is given in Figure 19.6,
which shows how the potential varies in the vicinity of the positive ions in the crystal lattice. The periodicity leads to a band structure in the density of quan tum states, which is the foundation of semiconductor physics. To get an idea of the magnitude of the quantities in the electron gas -·
modeL we consider the free electrons in silver\ which is monovalent (one free
I I I
I I I '
-
Figure 19.6
-'-----'-----1.--.L-
J
Sketch of th� potential well showing
periodicities associated with the positive ions of the crystal.
Sec. 19.3
Free Electrons in
a
Metal
363
electron per atom). The density of silver is 10.5 x Hf kg m-3 an d its atomic weight is 107. The concentration is therefore
Since silver is monovalent, this is also the electron concentration. The Ferttti energy is given by Equation (19.7):
ep
=
J.L(O)
h2
2/ 3
3N
=- 2m 81rV
•
•
Herem is the electron mass, equal to 9.11 x 10-31 kg. So
e F
-
(6.63
X
3 2 2/ 3 X 5.90 X 10 8
10-34)2 X
------
2 X 9.11
- 8.85 X 10 _
10-31
_ 19
J X
87T
1 eV 1.6 X 10-
19
_
J
-
5.6 eV.
The Fermi temperature is T.
F
=
eF k
5.6 eV
= 8.62
x
10-5 eV K-1
= 65 '000 K.
The ratio T/TF at room temperature is
T TF
=
300
4
6.5 X 10
= 0.00462.
At room temperature, therefore, the electron gas is in the so-called degenerate region T << TF. The chemical potential (the Fermi level) can be found from Equation (19.15). The computation gives
J.t(T)
=
0 999eF .
This shows why J.L(T) is often identified with
.
eF.
The work function tP depends on the metal and the condition of its sur face and is typically of the order of 3-4 eV. At very high temperatures, some of
Fermi-Dirac Gases
Chap. 19
364
the free electrons may have sufficient energies to leave the metal. 1his results in thermionic emission. The condition for their escape is ,
where Px is the component of the electron's momentum normal to the surface of the
metal
.
19.4 PROPERTIES OF A FERMION GAS The function
N ( e)d£
is the number of fer1nions in the single particle ene.rgy
range e to e+de. We know that
N(e)de
=
f(e)g(e)ds,
where the Fermi function f ( s ) and the degeneracy function
g( s)
the curves of Figure 19.7 for a temperature in the range 0
·r << TF.
<
behave like
The product of the two curves gives N (e) versus E� as shown in Figure 19.8. The electrons crowd around the Fermi energy because the degeneracy increases with energy; there are th�refore n1ore available quantum states to be occupied with 0 or 1 electron( s) per state.
(It
is the electrons in the tail of this
distribution that have the best chance of escaping from the metal in the free electron model.)
/(e)
(T ....,_.
=
0)
I
___ ----
_
I I • •
.
'--
..__ _ __.;:.... _... .. ..
____ _
(a) Figure 19.7
F.
L...... --- - --- ---..
(h)
The variation with single particle e�ergy of (a) the Fermi function._ and (b) the degeneracy function. The curves are sketched for 0 < T << TF.
t;
N(e)
•
,
, ,
,'
'
__
...
I I I
--
-
-
(T I
0)
=
I Figure 19.8 The energy distribution of fem1ions for 0 < T << TF.
·
The internal energy of a gas of N fermions is oo
U
=
0
�
eN(e)de
=
0
ef(e)g(e)de
·=
47TV
3 2 / 2m h
2
e312de
oo
e(e-J.L)/kT + 1
o
.
An approximate evaluation of the integral can be carried out in the same manner in which we evaluated the similar integral in Section of order T4 is included, the result is
I
3 At T
=
0, U
=
2 2 57r T
(3/5)Nc;F;
7T4
T
19.2. If the term
4
(19.18)
this energy is large because all the electrons must
occupy the lowest energy states up to the Fermi level. The average energy of a free electron in silver at T
-
e(O)
=
=
0 is
u (0) N
3
=
5eF
3 .
=
5 (5.6 eV)
=
3 4 eV. .
Note that the mean kinetic energy of an electron, even at absolute zero, is two orders of magnitude greater than the mean kinetic energy of an ordinary gas molecule at room temperature. The electronic heat capacity Ce can be found by taking the derivative of Equation
(19.18): e
-
dT
2
-
+ .
.
.
.
(19.19)
For temperatures that are small compared with the Fern1i temperature, we can neglect the second term in the expansion compared with the first and obtain
Chap. 19
366 1r1 =
2
--·-
Fermi-Dirac Gases
kT
Nk
( 19.20)
•
BF
For silver at room temperature.
2
t:
5.6eV
=
2.2 X 10
Thus the electronic specific heat capacity is 2.2 explain s
x
.
.
2
Nk.
1 o- 2 R. This small value
a puzzle. Metals have a specific heat capacity of about 3R,
the same
as
for other solids. It was originally believed that their free electrons should contribute an additional
( 3/2) R associated with their three translational degrees
of freedom. Our last calculation shows that the contribution is negligible. Wh)' is it so small? While the kinetic energy of the electrons is much greater than the thermal energy of electrons in a gas� the energy of the elec trons changes only slightly with temperature
(dU/dT
is smalJ). Only those
electrons near the Fermi level can increase their energies as the temperature is raised, and there are precious few of them. At very low temperatures the picture is different. From the Debye the ory, Cvcx.T3 and so the heat capacity of a metal takes the form
where the first term is the electronic contribution and the second is associated with the crystal lattice. At sufficiently low temperatures, the former can domi nate, as the sketch of Figure 19.9·indicates. Whereas we have emphasized free electrons in metals� most of our results apply to any ideal gas of fermions. Using the fact that the reversible •
3/v'k
1--
-
-
-·-
-
--
- -- - - - - - -- -- ·- �-.,1�7 -2 -
-
-
heat
capa_cuy •
Celect• onic
Figure 19.9 ..
Sketch of the heat capacity of a melal as a
function of temperature showing the electronic and lattice contributions.
Sec. 19.5
Application to White Dvnrf Stars
367
heat flow into a gas at constant volume is given by TdS late the entropy from Equation
S
T
c dT'
=
Therefore
S
=
0 at T
=
Vk
.
(19.21)
0, as it must be. The Helmholtz function F
U - TS is
=
T'
0
{19.19):
1T2
e
Cl'dT, we can calcu
=
2
+ .
--
i
2
7T4
T
4
+ .
.
.
.
.
.
(19.22)
The ferttlion gas pressure is found from
p
It is left as Problem
=
iJF -
av
r.N
(19.23)
.
19-9 to prove that p
=
57r2 T 2
2NkTF 5 v
-
-
+ . .
.
(19.24)
.
Comparison with Eq uation (19.18) sho\vs that P (2/3)UfV. For silver we found that N/V 5. 9 X 1028 m-3 and TF 65,000 K. Thus =
=
=
5 Given this tremendous pressure, we can appreciate the role of the surface potential barrier in keeping the electrons from evaporating from the metal. -
19.5 APPLICATION TO WHITE DWARF STARS
Very high pressures, of the order of 1017 atm, exist in the degenerate electron gas of a white dwarf star. It is this pressure that prevents its gravitational collapse. A typical star has a hot core with a temperature of the order of
107
K.
The heat energy is supplied by thertnonuclear reactions. The atoms are com pletely ionized (kT is 900 e V at tttis terr1perature), creating a huge electron gas. In young stars the electron pressure is sufficient to withstand the weight of the material pressing on the center, thereby preventing collapse. In old stars, the hydrogen at the core has run out, fusion has stopped, and the core cools. However, the loss of gravitational energy results in an increase in the kinetic
Fermi-Dirac Gases
Chap. 19
368
In t. fse of a lly ti pa r is s es oc pr g in ol co e th d an energy of th e electrons and ions n tro ec el e h t by d te e n v p r e is t in po n i a r t a e c nd e yo b se white dwarfs the colla p of us di ra e th d an n Su e th of s as m e th ve ha ly l gas pressure. White dwarfs typ i ca the Earth.
The pressure of the electron gas in Si rius B. mated us ing the formula
white dwarf. can be esti-
a
� = P
(19.25)
5
We need the fol low ing characteristics of Si riu s B:
Mass M
=
Radius R
=
V o) umc V
==
2.09
::< IO·'f' kg
5.57
::<
106m
7. 2 3
::<
1 0 �t J 'l
We assume that nuclear fu sion has ceased
m
" ·'.
that all the core hydroge n has
been converted to helium. We further assume that the helium is completely
io nized
that each He atom has Jost two electrons. In addition, we suppose
that the number of nucleons in th e star is equal to its mass di vi ded by the mass of a nucleon. Thus
No. of nucl e ons
2.09 ·
==
x
1 o-'u
I . 66 X l ()
'l-
..
' ·
=
1 . 26
x
1 a�,. �-
--
Since there are four
n uc l eons
and t\VO electrons in a helium a torn, the number
of electrons N in the fermion gas of Sirius B is 0.63 x 1057• Then� wi t h 1n e
=
9. 11 X 1 0
-
31
kg,
3N 81rV Then TF
=
eF! k
=
3.9
X
l �
-
·
5.33
=
10<) K. If 7
..
=
X
Io-·141
==
0.33MeV.
107 K. then the condition T << TF is
satisfied. To a fair approximation. the ele ct ron gas p ressure is given by Equ a t i on
( 19.25). Thus
5- 7.23
·
X
10�0
( 5. 3 3 x l 0 ·
14
)
=
1 .8
x
1 02 � Pa
=
1. 8
x
1 0 17 atm.
We "�d lik e to show that this pressure is sufficient to prevent further collapse.
Sec. 19.S
Application to White Dwarf Stars
369
A white dwarf is stable when its total energy is a n1inimum. The energy is
(19.26) where Ue is the energy of the electron gas and Ugrav is the gravitational energy.
We need to express each in ternts of the star's radius tive of U with respect to ForT<< TF,
R and set it equal to zero. 3
But
V
=
R, then take the deriva
3 It
3 81rV
(4/3)1rR3, so (19.27)
where
(19.28) The gravitational energy of a solid sphere of mass M and radius known from classical mechanics to be
U1rav
b =
-
R
'
R
is
(19.29)
where
(19.30) Thus
(19.31) Minimizing this function, we find that
Rmin
2a =
. b
(19.32)
.
Fermi-Dirac Gases
Chap. 19
370
Substituting the values of a and b from Equations (19.28) and (19. 3()), respec tively, we find that (see Problem 19-5) ( 19.33) Since the observed value of the radius of Sirius B is roughly equal to this, we conclude that the star has contracted down to its stable minimum size and is truly a white dwarf.
. White dwarfs are at present viewed as comprising a core of degenerate
electron gas surrounded by a nondegenerate (T > TF) outer layer Nuclear .
reactions have stopped. The star is cooling slowly. The time before it becomes invisible is estimated to be 1010 years. Since the age of the universe is believed to be 1.5 X toH> years it follows that very few white dwarfs have had time to ,
become invisible. The st abi li ty of white dwarfs is but one example of the very extensive use of quantum statistics in the st udy of st ellar evolution.*
PROBLEMS
19-1 Assume that for T
=
3TE- t he value of the chemical potential is -5.6 eF.
Calculate the value of the Fermi function
of e/ et of (a)
0, (b) 0.5, (c) I .0. and ( d ) 2.0.
f (e) at the temperature T for values
19-2 Consider a ki lom ol e of -�He gas atoms under STP tonditions. (a) What is the Fenni temperature of the gas?
(b)
Calculate J.t/kT and exp(- 11/ kT).
(c) Find the average occupancy energy of (3/2) kT
19-3
.f ( f; ) of
a sin gle particle state that has an
�or a system of n oni nteract i ng electrons. show that the probability f (e) of find
A above the chemical potentialJ.L is the same as the probability of finding an electron absent from a state with energy .6. belo�' J.L at any given temperature 7� mg an electron in a state with
e n e rg y
19-4 (a) Verify that the average energy per r�rmion is ( 3/5 )r.r at absolute zero bv .. making a direct cal cu la t i on of U (0)/�·V.
(b)
Similarly, prove that the average speed of a fermion gas particle at T (3/4)t·F· where the Fermi v d oc i ty v1• is defined by r.r .
=
(
=
0 is
l/2)ml:l
this section was motivated by Chapter 12 of Statistical Ph\•sics bv W.C. V Rosser, Ellis Horwood. 1982. Rosser gives an exten si ve treatment of white dw�rfs and *The discussion of
neutron stars.
Problems
371
SectJo� �9.2. Wnte F(e) (2/5)e512 and expand F(e) in a Taylor series a b o u t JL, retamm� the first thre� terms only. You will get an expression for u in terms =
approximately Ce
where A can
be A
=
be determined by
=
AT,.
experiment. For gold, the constant A is found to
0.73 J kilomole-1 K-2• Compare this result to the value obtained from
Equation (19.20). (The atomic weight of gold is 197 and its density is
18.9 x lQl kg m-3.) 19·7
(a) (b) (c)
19-8
Calculate eF for aluminum assuming three electrons per aluminum atom.
1000 K, JL differs from eF by less than 0.01 %. (The density of alumincm is 2.69 x lQl kg m-3 and its atomic weight is 27.)
Show that for alutninum at T
=
Calculate the electronic contribution to the specific heat capacity of alu minum at room temperature and compare it to 3R.
In sodium there are approximately 2.6 x 1028 conduction electrons per cubic meter, which behave as a free electron gas. Give an approximate value for the electronic specific heat of sodium at room temperature. (The atomic weight of sodium is
23.)
In the series expansion force, show that at this temperature the
cubic terrn is negligible compared with the linear terrn.
19-9 Derive the expression for the fermion gas pressure, Equation (19.24), using Equations (19.22) and (19.23).
19·10 Calculate the isothermal compressibility of the ferrnion gas consisting of the
free electrons in silver. Compare your answer with the experimental value for
silver of 0.99 x 10-11 Pa-l.
19·11 Show that the gravitational energy of a uniformly solid sphere of mass M and (3/5) GM2/R. (Consider the l?'avitational interaction radius R is Ugrav =
-
. between a solid sphere (core) of radius rand a surroundmg shell of thtckness dr.)
19-U For the white dwarf star Sirius B, do the computation leading to Rmin
:==
7 X lW m.
Use Equations (19.28) and (19.30) with the appropriate parameter values.
19-13 Consider the collapse of the Sun into a white dwarf. For the Sun, M R 7 x lOS m, V 1.4 x 1027 m3• =
(a) (b) (c)
=
2 x 10 30 kg,
=
Calculate the Fermi energy of the Sun's electrons.
What is the Fermi temperature? What is the average speed of the electrons in the fermion gas (see Problem
19-4). Compare your answer with the speed of light. (d) What is the density of the electron gas? Compare it with the density of water. (Note: For a star with mass greater than a critical value of 1.44 times the mass of
the Sun, the collapse continues beyond the white dwarf stage as gravitational attraction overcomes the electron pressure. The result is a very dense neutron star that eventually ends in a supernova explosion.)
•
20.1 Introduction
375
20.2 Uncertainty and Information
375
20.3 Unit of Information
379
20.4 Maximum Entropy
381
20.5 The Connection to Statistical Thermodynamics
384
20.6 Information Theory and the Laws ofThermodynamics
20.7 Maxwell's Demon Exorcised
386 387
373
20.1 INTRODUCTION No reasonably up-to-date introduction to statistical therrnodynamics would be complete without some mention of inforn1ation theory. We have seen that the entropy is a measure of the degree of randomness or disorder of a system.
A system that is in its single lowest quantum state is one in perfect order. On the other hand, a system that is likely to be in any of a ntltnber of different
states is a disorderly system. The greater the number of states it might occupy, the greater the disorder. Disorder, in this sense, implies a lack of infoiiitation regarding the exact state of the system. A disordered system is one about which we lack complete information. The reason for briefly discussing infor mation theory is that it casts further light on the meaning of entropy.
20.2
UNCERTAINTY AND INFORMATION
In his classic 1948 paper, A
Mathematical Theory of Communication, Oaude
Shannon laid down the foundations of information theory.* It was further devel oped by Leon Brillouint and applied to statistical thermodynamics by E.T. Jaynes.* Shannon's single greatest accomplishment was to provide a mathemati cal measure of inforrnation. The cornerstone of the theory is the observation that information is a combination of the certain and the uncertain, of the expected and the unexpected. The first part carries no real inforrnation; the mea surable infortttation is contained in the second part, the unexpected. The degree of surprise generated by a certain event·
say, one that has already occurred
is
zero. If a less probable event is reported, the information conveyed is greater. The infortnation should therefore increase as the probability decreases. Uncertainty is reduced by relevant inforn1ation. If I throw a die rnany times, I can expect that the average score will be 3 1/2. If I throw it once there *C.E. Shannon, Bell System
Tech. J. 27,379 and 623 (1948). L. Brillouin, Science and Information Theory, Academic Press, New York, 1956. tE.T. Jaynes, "Information Theory and statistical mechanics" Phys. Rev. 106,620 (1957) . t
.,
375
Information Theory
Chap. 20
376
is an equal probability of getting any number between
1 and 6. But if someone
tells me before I throw it that the same number 3 appears on all six sides, my uncertainty about the outcome of the experiment is totally removed. I can ask
the question, before I throw the die, how much uncertainty is there about the outcome? Or. correspondingly. how much information is transmitted by the actual result? For a given experiment, consider a set of possible outcomes whose probabilities are p1, p2..
tity H (p1
•
•
•
p11)
.
. . Pn· Shannon discovered that it is possible to find a quan
that measures in a unique way the amount of uncertainty
represented by the given set of probabilities. Only three conditions are needed
H (p1
to specify the function
p,) to within a constant factor. They are: •
•
•
•
1. H is a continuous function of the p;.
2. If all the p;'s are equaL
increasing function of
p;
n.
= l jn; then
H(ljn, ... , 1/n)
is a mo-notonic
(With equally likely outcomes.. there is more
uncertainty when there are more possible events.)
3. If the possible outcomes of a particular experiment depend on the possi
n
ble outcomes of
subsidiary experiments, then
H
is the sum of the
uncertainties of the subsidiary experiments. The last condition is known as the composition law and needs some elab
oration. Consider three possible outcomes of an experiment with probabilities
1/6. We note that the sum of the probabilities is 1/3, and p3 Pt 1/2, P2 1 as must be the case. Now consider a second experiment broken down into =
=
=
�
two successive experiments. Let the first subsidiary experiment have two
equally probable outcomes, with p1
1/2 and P2 = 1/2. Let the second sub
=
sidiary experiment have two possible outc<>mes with probabilities and
Pi
=
p{
=
2/3
1/3 (Figure 20.1 ). We note that the probability of two independent
outcomes occurring together is the product of the probabilities of the separate
outcomes. We also observe that in the two cases the final outcomes have the.. same probabilities. Thus.. condition 3 requires that 1
1
1
H 2'3'6 l/2 single
expenment
l/3
=
1 1
H
2·2 l/2
two •
expenments SUCCCSSI\'C
•
l/6
(20.1)
•
l/2
213
l/ 3
l/3
116
Figure ZO.l Decomposition of an experime nt For the first expe rime nt.. the pro babi liti es of the outcomes are p1 1/2 p� 1 /3 .. and P3 1/6. In the second experiment .. the first subsidiary experiment has probabilities p1 1/2 1/3. and P2 = 1/2 while the second has probabilities J't' = 2/3 and Pi .
=
..
=
=
=
..
=
··
Uncertainty and Information
Sec. 20.2
377
The coefficient of the second term on the right-hand side is 1/2 because the second outcome occurs only half the time (see Problem 20-1
).
Shannon showed that the simplest choice for H consistent w ith the three conditions is
H (Pt
. ·
·
P�,)
=
i=l
(20.2)
1· (p;),
where the function f is unknown. Since f (Pi) is a continuous function, it suf fices to detetntine /for the special case of equal probabilities. For P; 1 / n for =
all i, Equation (20.2) gives
1
1
�
n
n
n
(20.3)
•
•
We note that condition 2 requires that d
f n dn
1
>
n
(20.4)
0.
Invoking the composition law, condition 3, we consider the case where the total number of equally possible outcomes of the first subsidiary experi ment is r and that of the second subsidiary experiment is s. Thus n rs and =
H
1
1
r
--
,
... , -
r
+
Using Equation
1 1 1 1 =H --, . H - , ... , - - H , . .., 11 n s s rs 1
.
,
rs
-
.
(20.5)
(20.3) we obtain 1
rf -
+
r
Let R
.
1
1/r and S
sf
1 -
S
=
rsf
1 TS
(20.6)
.
1/s. Equation (20.6) then becomes •
1
f(R) R Setting g(R)
1
+
5f(S)
(1/ R)f(R), g(S) g(R)
+
=
1 f(RS) . RS
(20.7)
(1/ S)/(S), etc., we have g(S)
=
g(RS).
If we differentiate this equation first with respect to R and then with respect to
S, we obtain
Chap. 20
378
g'(R)
=
Sg'(RS)
g'(S)
=
Rg'(RS)
Information Theory
and
From these relations, it follows that Rg'(R)
(20.8)
Sg'(S).
=
Since R and S are independent variables, the only way in which Equation (20.8) can be satisfied is for both sides to be equal t
Rg'(R)
A
=
(.A
=
constant)..
(C
=
constant).
or
g( R)
=
A In R
+
C
Recalling the definitions of R and g(R)� we see that this gives A
l
f
=
-
-In
-
r
r
r +
C r
.
.
If the probability is C
=
1, the uncertainty must be zero. That is f ( 1)
=
0, so
0 .. and for equal probabilities..
nf
1
-A ln
=
-
ll
(20.9)
n.
The remaining issue is the sign of (20.4), we find that
A. Applying the condition of Equation (),so A must be negative. Setting A K with K
A/n > 1/p we obtain positive, and writing n -
=
=
-
..
f ( p)
=
-
K p h1 p.
(20.10)
Using this result in Equation (20.2), we obtain the expression for the uncer tainty we have been seeking: "
H
=
-
K
p; i =I
where K is a positive constant.
ltl p;.
(20.11)
Sec. 20.3
Unit of Information
379
Let's check our example of the composite experiment. We have 1
1
1
1 2 1 + -ll 2 3'3
=
--
-
K(-0.346 - 0.346)
K -
·
2
( -0.270- 0 366 ) .
l.OlK.
=
Also, for equal probabilities, p;
=
1/n for all i, andH
=
K In
n. It follows that
Equation (20.11) satisfies all the conditions imposed on the measure of uncertainty. It is instructive to consider the
binar)' case of two possible outcomes an experiment with probabilities p1 and p2 such that p1 + P2 = 1. Then H/K
= =
-
-
P1
In
P1 - P2 In P2
Pt
In
Pt
-
(1 -
of
(20.12)
Pt ) ln( l - Pt)·
What isH when p1 is either 0 or 1? We need L'Hopital's rule in the form .
llDl
X�0
( ) V (X ) u x
'
.
=
ltm
(x ) . ' v (x )
u
X-+0
.
liere
lim(x In
X-+0
x
)
=
In x . ltm 1
X-+0
-
x
Thus the uncertainty is zero when either
=
1.1m
1 i
X-+0
1
=
0.
� -
2 x
p1 or p2 is zero and the other is con
sequently unity. Also, if we differentiate Equation (20.12) with respect to p1 and set the derivative equal to zero, we find that H/K has a maximum at p1
=
1/2 ( = p2). These results are illustrated in Figure 20.2.
These inferences can easily be extended to the case of n possible outcomes. In particular, we can use the method of Lagrange multipliers to maximize H sub-
ject to the cortdition
p;
=
1 (Problem 20.2). The result isHmax
=
K In
n.
•
I
20.3 UNIT OF INFORf1ATION
We need to choose a value for K so that H has some convenient unit. Consider the simplest possible experiment with equally likely outcomes or the binary digits 0 or 1. Then
heads or tails,
Information Theory
Chap. 20
380
HIK
Figure 20.2
Uncertainty
1'----
measure for the case of two
I
____.
___ ___
0.5
1
possibilities with probabilities
1
0.5
0
()
Pl and P2· •
H -
K
=
-
1 1 1 1 - In - - - In 2 2 2 2
-
=
In 2.
If we use 2 as the base of the logarithm and take K
==
1 �we obtain
We call the unit of information a bit for Bin a ry dig .IT.* Consider the follo\\'ing examples:
1. Decimal digit Here
Thus a decimal digit contains about 3 1/3 bits of information. 2. Foregone ·conclztsion The mess·age ''the sun will rise between midnight and noon tomorrow'� has a probability of one so that H
=
0 hits.
3. Coin flipping Flip a coin three times. The information will be three times that associ ated with one flip of the coin. Thus H
=
3 1og2 2
==
3 bits. This can be
*The ubit'' was coined by the statistician John Tukey during a lunchtime discussion at AT&T Bell Laboratories.
Sec. 20.4
Maximum Entropy
381
TABLE 20.1 Known probabilities of occurrence for symbols in the English alphabet. Symbol
P;
blank E T 0 A
0.200 0.105 0.072 0.065 0.063
•
•
•
•
•
•
J,Q,Z
0.001
written H
=
log2 23• By extension, flipping a coin N times yields N bits
of information. 4. English alphabet
Suppose that a message comprises words using the 26 letters of the alphabet and equal, then H
a =
blank, totaling 27 characters. If the probabilities were log2 27
=
4. 76 bits per character. If, on the other hand,
we use kttown probabilities of occurrence, H is reduced to 4.03 bits per character (1able 20.1). Because of the redundancy built into the struc
ture of the language, the actual information content is in the range of 1
to 2 bits per character. The calculation involves the use of conditional probabilities.
t0.4 MAXIMUM ENTROPY
Ne have seen that the amount of uncertainty represented by a discrete proba >ility ·distribution is n
H (Pt
·
·
·
Pn)
=
-
K i= I
p; In p;,
(20.13)
vhere K > 0. In Section 20.5 it will be shown that, apart from the constant, his is just the expression for the entropy in statistical thermodynamics. .
llerefore, we can consider the tet111s "entropy" and "uncertainty'' to be syn �nymous. Indeed, entropy is a tetnl used in information theory.
The connection between entropy and uncertainty. is illustrated by the
Jllowing example. Imagine that an insulated container is divided into two
Chap. 20
382
Information Theory
chambers by a partition. Initially there is gas on one side of the partition and none on the other. Suppose the partition is suddenly removed so that the gas expands to fill the whole container. We know from our study of the free expan sion of a gas that the internal energy is unchanged but the entropy increases. When the molecules of the gas are all in one of the two chambers, the entropy is less than when they are distributed throughout the container. Our knowledge of the possible positions of the molecules is greater when they are all on one side of the partition than when they are somewhere in the entire container. The more detailed our knowledge is concerning a physical system, the less uncertainty we have about it .. and the less the entropy is. Conversely, more uncertainty means greater entropy. To make inferences based on only partial information. it is necessary to determine the probability distribution that has maximum entropy subject to what is already known. Suppose that we know the mean value of son1e partic ular variable x that can assume the discrete values .xi, i
=
1" 2, .
.
.
..
n. The
mean
value is
-·
..
.t
=
i
-
·
I
(2().14)
PiX,.
where the unknown probabi1ities satisfy the condition •
II
i
:-
(20.15)
1
In general, there will be a large number of probability distributions
{p,, P2· ... Pn} consistent with the infortnation given�- Our problem is to deter mine that distribution which maximizes the uncertainty (i.e. the e-ntropy). We ..
introduce the Lagrange multipliers a and {3 and follow the procedure we used in Chapter 13. We write
(20.16) where
(20.17) I
and
.
I
(20.18)
Sec. 20.4
Maximum Entropy
383
Then
_,;
P; In P;
a +a-· apj
•
•
l
'
or
- K In Pi - K
Pi
+a +
Pi
f3xi = 0.
Solving for In Pi and simplifying the constants, we obtain
or
We can determine the new Lagrange multipliers A and p, from the constraints. Equation (20.17) gives e-A
1 =
---
,
I
•
so that
Pi=
·
(20.19)
1
•
We define the partition function z
1
(20.20)
•
Then (20.21) •
and A- In Z.
(20.22)
.
Chap. 20
384
Information Theory
Invoking the remaining constraint� Equation (20.18)� we can derive an expres sion from which we can determine J.L. Noting that az
-
=
•
J
we have 1 az
_
x
�
-
z dJ.L
=
-
a -- In Z.
i}J.L
(20.24)
Equation (20.21) gives the probabilities that maximize the uncertainty. Thus
.
J
=
K(J.Lx +In Z).
(20.25)
This result can easily be generalized to include additional information associ ated with the knowledge of the mean value of other variables.
20.5 THE CONNECTION TO STATISTICAL THERMODYNAMICS The randomness or
disorder of a system implies an uncertainty regarding
its state. Boltzmann described the entropy as a measure of the Hmissing'" information. The similarity between the development of the last section and the pre vious chapters is striking. Consider a system of N distinguishable particles obeying Boltzmann statistics. Assume that the quantum states are nondegen erate. Then the thermodynamic probability is •
U;=
N! --
n
..
(20.26)
Sec. 20.5
The Connection to Statistical Thermodynamics
385
and the entropy is
•
•
1 •
=:;
N InN- N-
N1· •
•
I
•
•
•
J
J
J\j [lnN-
=
J
In
N;]
•
J
N1·
-
-
N
•
I
.
Then
(20.27) Imagine that the system is in perfect order, meaning that all the particles are in the lowest quantum state. Then N1
1 and
=
N
N3
N2 =
N
=
N
.
.
.
=
Nn
N
=
0,
and the sum in Equation (20.27) reduces to one term equal to zero. Hence the entropy is zero for the perfectly ordered state. A disordered system would be Jikely to be in any number of different
quantum states; the larger the number of states available to it, the greater the disorder. If
N;
=
1 for N different states and Ni
=
0 for all other available
·
states, then N
S
=
kN
-
=
kN In N
.
(20.28)
This function is positive and increases with increasing N.
If we associate
N;/ N with the probabilit;· Pi that Ni particles are in the jth
quantum state, it follows from Equation (20.27) that .
S
=
-
kN •
1
Pi I n Pi
kN =
K
H.
(20.29)
Chap. 20 ... �
386
Information Theory
e th d an s ic am yn od m er th al ic ist at st of Apart from a constant, the entropy lly fu e in rm te de to pt m te at e w If e. m sa e th e entropy of information theory ar e th en th n, io at rm fo in al rti pa ly on n ve gi e ar t bu the macrostate of a system k lac r ou of re su ea m a be d ul wo in ga d ul wo expected amount of information we is it , ng ssi mi is on ati m m inf the e us ca Be . tem sys e th of te e sta th of knowledge of the , ver we Ho . ure rat lite the in y rop ent neg or py tro en ve ati neg led cal s sometime constant of proportionality between Sand H in Equation (20.29) is positive. The Boltzmann distribution for nondegenerate energy states is e-e,/kT
N j
=
N
--·�
Z
'
where •
.
1
If we substitute the distribution in Equation (20.27), we obtain N 1 Smax =T Z
.
J
The internal energy is
U
= .
J
Thus
Sm
a
x
=
u
T
,. + N k ln Z.
This expression has exactly the same form as Equation (20.25). Hence the internal energy plays the role of a variable whose mean value is known. "Partial information'' in this case is knowledge of the energy of the system.
20.6 INFORMATION THEORY AND THE LAWS OF THERMODYNAMICS
.
We liave seen that the thermo-dynamic concept of entropy as defined by Clausius is physically related to the information theory concept of entropy as defined by Shannon. This characterization is clearly more than a mere analogy.
Sec. 20.7
Maxwell's Demon Exorcised
387
Accordingly, some in vesti g ators have attempted to show that the fundamental equations of equilibri un1 thermodynamics can be de ri ved from infottnation theory. In classical ther modynamics heat, temperature, and work are taken as primary concepts, while energy and entropy are derivative ideas. The informa tion theory treatment inverts the procedure. Heat is introduced without the need to define it in terms of reservoirs, adiabatic walls, etc. Information theory
is free from all artificial constructs such as heat baths or Camot engines.The zeroth, first, second, and third laws become consequences rather than
premises. The third law, for example, is true by definition since, for a perfectly ordered state at absolute zero, there i s no missing information (S
=
0).
Infotittation theory is thus an alternative way of presenting the postu lates of thermodynamics. Its elucidation as such is beyond the scope of this
text. Suffice it to say that its fa r rea ching consequences unify our understand -
ing of thermal phenomena.
20.7 MAXWELL'S DEMON EXORCISED How seriously Maxwell took his demon is hard to say. In a11y case, he neither
conducted nor pr omot ed experiments to test his hypothesis What is certain is .
that the paradox posed by his demon botltered generations of physicists. In the early 1900s it was noted that Bro\\ nian agitation of the trap door '
(recall t hat it is massless and frictionless) would result in a random opening and closing of the door, thereby rendering ineffective its long-term operation . After the d evelop ment of quantum mechanics in the 1920s, it
\vas
suggested
that the uncertainty principle might play a role in the prob lem: later it \vas shown that this would not be the case for heavy molecules at reasonably low pressures.
In his seminal1929 paper on Maxwell's demon, Szilard observed that an amount of energy at least as great as whatever amount of energy can be gained, must necessarily have been ex1:.ended by the demon in his detection, inspec
tion, and routing activities. Demers and Brillouin (1944-51) called attention to
the fact that in an isolated enclosure it would be impossible for the demon to see
the individual molecules.To make the molecule.s visible against the back
ground blackbody radiation, the demon 'Nould have to use a flashlight. The entropy produced in the irreversible
ope ration
of the flashl ight would always
exceed the entropy destroyed by the demon's sorting procedure.Thus a real demon could not produce a violation of the second law of thermodynamics. Richard Feynman, looking at the demon as a mechanical system, wrote "If \Ve assume that the specific heat of the den1on is not infinite, it must heat up . .. Soon it is shaking from Brownian motion so much that it can't tell .
whether it is coming or going, much less whether the molecules are coming or
going so it does not work." ,
Information Theory
Chap. 20
388
Information theory provides a new framework in which to view the
�
paradox. A knowledge of the momentum and position of the molecules co� st tutes the information needed to achieve a given entropy decrease. But thts
IS
counterbalanced by the entropy increase 11ecessary to acquire this informa tion. Lehringer observes that about 1
reduce the entropy of 1 kilomole of gas by 1 calorie/K. Recently.. Landauer* and Bennettt have interpreted Maxwelrs demon as a computing automaton, subject to limitations imposed by the rules of data processing and computational logic. Their view is that the gathering of infor mation by the demon does not nece ss arily increase the entropy. But after the demon has used the information to allow the passage of fast molecules.. he must forget it if he is to continue doing his j(Jb. It can be shown that erasing the inforr11ation raises the entropy more than any previous lowering that arises from its use. The association of computation with thermodynamics suggests that computational procedures might fruitfully be expressed in terms of the behavior of some thermodynamic system. In The Natltre ofThermodynan1ics� B rid gma n states that If the Maxwell ..
demon had been invented vesterdav instead of in t he last centurv I b elieve he .
"'
...
would not have caused as much consternation.�-� Perhaps not. but the language of statistical thermodynamics is beginning to permeate fields as diverse science, complexity theory� biology, and animal behavior.
as computer
Maxwell's demon may be exorcised, but the ideas that conjured him up live on .
.•
;.·.:-:.. �··:--. :-:.:- - ;-.: ·: · !.:: :· ·.· .·:· :·.· · : : ·:
.•
· :,..:..· -.·. :? :· '.·.-.·.;,·. . .:: ...... ·. · .� .. · ·.:· · =·" ·· ·· ··· :···:-: ·· · ..... ...·.., ..... . ..· ..:v;,· .· .. .......... .. .. . � -.... · '·. ··· • · ··• - "<"··�· ·x:· · ·-· .--.
• ••• •••
_,
...
.. . ... . ..
..
.. · ·. ..
. . . .. . . . ·..·. .. ·. . ·.·. . . ...·. .·.· ..· .
·
.
.
. .... ... .
.
. .
. . ... .... . . . . .. . .
.
.
. .. . .
. . -
.
.
PROBLEMS
20-1 Refe rri ng to the·compound experiment of Section 20.2� show explicitly that
=H
Il --
:?
-
..
1 .,
4-
1 2 I +-H 2 3' 3 . -
20-2 Use the method of Lagrange multipliers to maximize H su bj ec t to the condition n
� p1
==
1. Show that Hm;r'( = K In
n.
i= 1
20-3 Make a table of the logarithn1s to thl! base 2 of the numbers from 1 to 10. In adjoining columns list the logarithms to the base e and to the base 10 .
. .; * R. Land aue r ··computation: A Fundamental Physical \'ic\\·.H A.fa.lu·e/15 De1non: Entropy. Information. Computingt edited hy H.S. Leff and A.F. Rex. Princeton lJnivc:rsity Press. Princeton. N Jersey, 1990, pp. 260 ff ,
ew
·
t
C.H. Bennett ··The Thermodynamics
of Computation
A Revie\\·��·
Ibid. pp. 213 ff .
�P. W. Bridgman. The Nature ofThermodynamic.ft, Peter Smith. Gloucester� Mass.
1969.
Problems
389
Note that
20-4 Consider a loaded die. Let the probabilities of throwing the numbers 1, 2, 3, 4 , 5,' and 6 be 0.1, 0.1, 0.1, 0.1, 0.1, and 0.5, respectively. Calculate the un certait1ty H in bits for a throw of the die. What would H be if the die were "honest"? 20·5 A perfect shuffle of a deck of 52 playing cards is one in which all of the possible orderings of the cards are equally probab le. (a) How much information h1 bi ts is there in a perfect shuffle?
(b) HO\\' many times do you have to cut and interleave a deck in order to achieve something approximating a perfect shuffle?
246.3 K. Assume the ''alphabef' of possible values as being multidimensional: �e hundreds of degrees being one dimensio n, the tens value a second d i� nsion etc. Associated
20·6 Suppose that a measurement of ten1perature gives T
=
,
with each dimension is a probability. Reasonable values might be
PH�(2)
=
l;
p10(4)
0.2;
=
p1 ( 6)
=
0.1; Po.t( 3)
=
0.1.
With these values, calculate the amount of information in the measurement
.
20-7 Consider a substitution cryptogram in which, for each letter of the alphabet, some other letter is substituted. The number of possible keys is 26!. The cryp togram can be viewed as a compound experiment consisting of two parts: x, the communication of the clear text, andy, the choice of a key from one of the 26! possibiliti es. Thus the total information associated with the compound e"peri ment is
H ( x)
+
log 26! bits. A substitution cryptogram of 40 letters can usually
2
be solved. What does this imply about the maximum information in bits per let ter of an English messag e ?
20-8 Imagine that system 1 has probability p}1> of being forced into a state j and sys tem 2 has probability p�2> of being forced h1to state k. Then ·�
H0>
=
-
K �pf0 In
pfl)� H<2>
=
-
j
K �pk<2> In P k
2> < �c .
Each state of the composite system consisting of systems 1 and 2 can then be labeled by the pair of numbers j, k. Let the probability that the composite sys tem is found in this state be P;k· Then
H
=
-
K � � Pjlc In Ppc j k
If the two systems are only weakly interacting, so that they are statistically inde pendent, then Pik
=
p}1> Pk(2). In this case show that H
=
flO>
+
H < 2>
.
20·9 P.T. Landsberg (Thermodynamics and Statistical Mechanics, Dover, New York, 1990) draws a distinction between entropy and disorder. For a system of tinguishable states, he defines the disorder D(n) as
n dis
·
Chap. 20
390
D(n) Here
5'(n) is the
entropy and
c
=
-
S(n)
.:·
ln
·
Information Theory
.
n
is a constant. The rate of ch ange of disorder can
be found by diff ere ntiating this equation with respect to the time. Show that •
. D(n) .
=
S( n)
S (n)
•
-
n
n-- l -·JD(n).
ln
n
Here the dots denote time derivatives. Since the second term in the brackets can conce iv ably be greater than the first tem1� the result suggests that it is possible
to have decreasing disorder even though the entropy increases. The expression has a possible application to the growth of biological systems.
•
•
I
•
A. I PARTIAL DERIVATIVES
Consider a function of three variables
f (x, y, z)
=
0.
(A.l)
Since only two variables are independent, we can write
y
=
y(x, z). Then dx
=
ax dy ay z
+
ax az
dz,
x
=
x(y, z)
and
(A.2)
Y
and
dy
ay =
ax ..
dx
....
+
ay dZ
dz.
(A.3)
X
The subscripts denote the variables that are held constant in the differentia tion and are often omitted. Substituting Equation (A.3) in Equation (A.2), we obt a n
i
391
Review of Partial Differentiation
Appendix A .
392 ax
dx
If we choose
t
..
=
ay
z
av
.. ax \
ay
�
dx + ax :
+
·
ay
z
az
x
ax az
dz.
(A.4)
r •
and z as the independent variables, then Equation (A.4) holds
for all values of dx and dz. Thus, if dz
=
0 and dx * 0. we have 1
iJ.x
-
av ..
(A.5)
•
-
•
dV
..
"'
...
·-
a.x
-
...
This expression is known as the reciprocal relation. If dx
=
0 and dz :f:. 0, we get ax
dX -
--
av
..
..
...
-
-
az
•
az
x
"
(A.6)
The reciproca-l relation gives 1 a.. .
· -
ri.r
\'
Substituting this in Equation (A.6) yields ax =
rlV .. ..
Equation
-1.
(A.7)
...
(A.7) is the cyclical rule, or c,vclical relation.
A function
lt
of three variables x� y, and z can be written as a function of
only two variables when they are related through Equation (A.l), that is.. when only two variables are independent. Thus li = ll ( X,
y)
(A.8)
.
Alternatively, we can write .r =
(A.9)
.X(ll, v) . •
Then
dx
=
rlX -au
,
du \'
�ax +
av ..
dv. .,
u
(A.lO) •
Sec. A.l
Exact and Inexact Differentials
193
If we divide Equation (A.lO) by dz while holding u constant, we obtain
ax az
ay
ax ay
-
u
dZ
u
•
(A.ll)
•
u
This is the chain rule of differentiation.
A.l EXACT AND INEXACT DIFFERENTIALS Consider the function z b
=
z(x). The fundamental theorem of integral C<'llculus is b
d
(A.12)
X
a
·
a
If the integration is taken around the closed path C (a to b and b to b
dz
dz
=
b
a
dz
+ b
a
b
dz -
=
dz
=
=
M(x, y)dx
+
0.
(A.l3)
a
a
This can be generalized to two independent variables. Let
dz
a), we have
z
=
z(x, y). Then
N(x, y)dy.
(A.l4)
We wish to determine the condition under which the integral of dz around a closed path is zero in this case. Consider the integral of aNfiJx over a rectangle A in the x-y plane. From Figure A.l, the integral is
aN -dxdy ax A
d
b
= c
a
iJ N -dxdy ax
d
[N(b, y) - N(a, y)]dy.
= c
y d
---- ....
--
----
-
---
c
A •
c
A.l Contour C enclosing the rectangular area A.
----
--- ----
I I I I a
l I I \ b
X
(A.15)
Review of Partial Differentiation
Appendix A
394
N(x, y)d)'
We now evaluate the·integral
around the path C in a counter•
clockwise direction:
N(x, y)dy
N(b, y)tiy
=
N(a, y)dy
+ d
[N(b,y)- N(a,y)]dy.
=
(A.l6}
The contribution along the horizontal segments of the .Path are zero since dy = 0 there. Combining Equations (A.15} and (A.l6) gives
aN -dxdy A
dX
Ndy.
=
(A.17)
Similarly, it can be shown that
aM -dxdy ay
-
Mdx.
=
(A.l8)
A
Adding these two equations� we obtain the result
(Mdx
+
N dy)
aN aM - dxdy. ax ay
d., -
=
.(.
-
·
c
(A.l9)
It follows that the condition under which the integral of dz around a closed path is zero is
aM
-
=
aN ax
iJv "'
(A.20)
•
When this condition. is satisfied, the differential dz is said to be Note also that
d� -
a;. . d.,'(
(A.21)
and .
av a.r
--
.,
-
-
-
ayiJx
exact.
a-z ·-
axay
·
az
.
-
ax ay
•
Sec. A2
Exact and Inexact Differentials
395
Identifying the partial derivatives in Equation obtain Equation (A.20).
(A.21) with M and N, we again
dz is exact, we wish to be able to integrate it to obtain z. If M = M(x) and N N(>'), the problem is trivial: If the differential
=
M(;:)dx
z =
N(y)dy.
+
(A.22)
However, this will not be the case in general. We define
Mdx,
u y
the partial integral of �.M \Vith y fixed. That the function N
- iJufay depends on
y only can be seen as follows: au a Nay ax
-
-
aN ax
-
aN
a au ax ay
-
d.X
a a Mdx ay ax y
aN aM = 0. ay ax
The last step is a consequence of our assumption that dz is exact. As a trial solution, we take
z=u+
LV-
au dy ay au dv Nay
-
.�
.Y
=
Mdx + g(y).
(A.23)
y
This gives us a recipe for finding z: carry out the integration indicated in Equation (A.23), take the differential dz including the tertn
g'(>·)dy, and com pare the result with the original differential to find g'. A final integration gives g. As an example, consider the exact differential dz
=
(x + y)dx + (x + 2y)dy.
(A.24)
Review of Partial Differentiation
Appendix A
396 Then
(x + y)dJ.· + g(.Y)
z =
So
dz
=
=
:cdx + vt/:( + xtly
+.
..
g'(y )d)'
(x + .v)dx + (x + g'(}')jd�v.
Comparing this with the original differential, we see that
g(y)
=
2 y
+
g' (y)
=
2y, so that
C. Therefore "'
z
x-
==
·
,
+ .r v + .y- +
.;
-
2
(A.25)
C.
Alternatively.. we could have written •
•
-
. "
Ntl_v
-
+
(A.26)
h(x)�
which leads to the satne result. What if dz is inexact? Consider the example
dz which doesn't satisfy Equation that
=:
vdx
-
(A.20)
..
xdv_ �
,
the criterion for exactness.lo indicate
dz is not exact, we write
dz
=
vtlx -
-
·
.rdv. .
This cannot be integrated. Ho\\'ever, we 11otice that if we multiply the equation by
2 1/y
..
we obtain an exact differential:
vdx
dz -
., -
.,
yThe term
1/ v2 ..
-
is known as an
XlfV .,
-
-
-
..,
v�
d
X -
v .
...
•
•
i11tegrating .factor. It can be proved that it is
always possible to find an integrating factor for an inexact differential that is a function of two independent variables. and hence to obtain a differential that is integrable. This fact is of great importance in thermodynamics.
Sec. A.2
Exact and Inexact Differentials
397
An integrating factor is not unique; for a given inexact differential, many integrating factors can often be found. The following are all integrating factors for the differential ydx - xdy: •
1
1
1
1
1
1
1
A formal method exists for fi nding an integrating factor J.L. Let dw
ILliZ
�tMdx + p,Ndy.
=
(A.27)
If dw �s exact, then
or iJ�t oM -M + p, ay ay
iJ�L =
ax
N + p,
iJN ax
.
In general, this cannot be solved for J.L· However, suppose that J.L is of x only. Then
=
a
function
0, and aM
=
J.L ay
dp. dx
N
J.L
+
iJN ax' •
or d �t -
aM
1 =
aN -
N
ay
a.:r
dx.
Integrating, and setting the constant of integration equal to zero, we obtain
p,
=
exp
1
iJkl
!.J
ay
_
iJN ax
dx .
(A.28)
Note that if the differential is exact to begin with, the integral is zero and
1. In pr actice the formal integration of Equation (A.28) is unnecessary, as we can see from the following example. Let
IL
=
,
dz
=
sinydx
+
cosydy.
Review of Partial Differentiation
Appendix A
398 Then
For exactness� a -(�J,siny) av
iJ =
•
ax
(�tcosy).
or rl� ay If JL
=
JL(X), iJ�L/iJy
stny + J.L cosy
d�
.
= 0, and
diL/JL
=
==
ax
d:c, or p,
cosy.
=ex.
Thus
which is exact. If we choose to select J-L = Jl.(y), then J.L = 1/sin..v and
dw
=
d.t + C<)t Vd}'· ..
Thus we have obtained two integrating factors; usuaiJy one is easier to calcu late than the other. From this analysis it is evident that the differential
dz
=
Md:c + Ndy is •
exact if the equivalent statements hold:
1.
aM -
aN =
-
ox
dV "'
2.
dz
=
0
(A.29)
h
dz is independent of the path.
3. a
This means that if a problem can be formulated in terms of ·exact differentials, or differentials made exact through the use of an integrating factor. then the path of integration between the end points is arbitrary.. and we are free to select one that leads to a simple evaluation. In thern1odynamics all state variables are, by definition .. exact differen tials whose integrals are path-independent. However, there exist certain ther-
Sec. A2
Exact and lnexaa Differentials
399
.modynamic quantities such as the differentials of work and heat wh ose inte grals depend on the path of integration. Thus the search for appropria te inte grating factors that allow us to express fundamental relationships in tetrns of variables of state is an importa1.1t goal of the mathematical theory. •
PROBLEMS
A·l Test the following differentials for exactness. For the cases in which the differen tial is exact find the function z(x, y).
(a) dz (b) dz (c) dz A-2
=
= =
2x lnydx
+
(x2jy)dy.
(y - l)d x + (x - 3)dy. (2y3- 3x)dx - 4xydy.
(a) Show that dz ydx + (x + 2y)dy is exact and int egrat e it to find z(x, y). (b) Show that az xdx + (x + 2 y)dy is inexact. (c) Integrate the two previous differentials counterclockwise around the trian=
=
gle whose vertices are the three points (0,0), (1,1), and (0,1) .
A-3 Consider the differential dz (2xy- 3)dx (a) Show that dz is an exact differential. =
+
x�dy. .,
z (x, y).
(b)
Find z
(c)
Referring to Figure A.2, evaluate the line integral
=
dz. from (1,0) to (0,1) along
Path A: lines parallel to the axes; Path B: circular arc; Path C: straight line diagonal.
y
(0. l)
A
( 1 1) �
8 c Paths of integration for Problem A-3.
Flgure A.l
(0,0)
(1. 0)
X
Review of Partial Differentiation
Appendix A
400
A·4 In the following cases show that liz is not an exact differential. Find an integrat ing factor such that dw = JL71Z is an exact differential and check that t his is so. (a) i1z 2y dx + 3x dy. =
(b) az (c) llz
=
=
y dx + (2x (x4 + y4)dx
-
y2)dy.
-
xy3dy.
A-S Find an integrating factor p.. and integrate du; llz
=
(ycos3x
-
1 )dx
+
=
,.uJz, where
sinx cos2x dy.
i. .
'
·
.
. . :·.· .. ·..·. -:: • . . . . . . . . ' . : · · · . . · · : :. - . , , . ... . :- . . - ; . .. ... .. ;- ' .
·' . ··
.
.
.. :�� : '
.
:: ::::
;
'
'
·
.
-: · · .· >, · : . • ". -. .:
'.
·
·
· ·' .· : : . :::
•. .
:
:· . .
-.·.
.
.
:
· . ' -·
-
.·
- · ·-·
· .
· . . · . . . - - . . ·. ' '. · : ·; ·- ·:
..
'
.
•
'
.
_.
'
.. · .
•
The definition ofn!, the factorial of a positive integern, is n! = n(n- l)(n-
2)· · ·1.
(B.l)
From this and the fact that (n + 1)! = (n + l)n!,
(B.2)
it follows that 1 ! = 0! = 1. We are interested in the natural logarithm of n! when n is a very large number. From Equation
(B.l) .we see that n
In tz! = In 1 + In 2 + In 3 + '
y
0
Equation
(B.3)
I
·
·
·
+ Inn =
Ink.
(8.3)
k=t
is clearly the area under the step curve shown by the dashed
lines between n = 1 and n = n in Figure B.l. The rectangles all have unit width; the height of the first is In 2, that of the second is In 3, etc. This area is approximately equal to the area under the smooth curve y = Inn between the limits 1 andn ifn is large. For small values ofn the area under the step curve is appreciably different from and exceeds that of the smooth curve. As
• -·
n
401
402
Appendix 8
Stirling's Approximation
=Inn "
v "'
•
ln6 ...... In 5 In 4
I
r--I 1
ln3 ..... In 2 .....
I I I I I I I I I
r--
1 I -
I I I I I I I
--
•
I
I I
I
I I I
I
I
•
.
.
I
I I
t
I I
0
..___ ___ ...__ _ """"'_ _.__ _ ........_
__._
_
1
0
Figure B.l
2
3
4
Graph of In n versus
5
_
___
6
n
n.
increases, the smooth curve flattens out and the difference becomes negligibly small. Thus we can approximate Equation {8.3) by •
·n
Inn!
n
= n Inn - n + 1.
Inn dn = (n Inn - n)
�=
l
J
Sincen is assumed large, we can neglect unity and get In n!
�=' n
Inn - n.
(B.4)
This is Stirling's approximation, valid for large n. For n = 10, Equation (B.4) gives a value low by 1 3.8 percent and. for
n
= 30, a value low by 3.5 percent.
As an example, we can use Stirling's formula to estimate 52!, the number of possible rearrangements of cards in a standard deck of playing cards. Here In 52! :::; 52 ln 52 - 52 = 153. Then log1052! = (0.4343)ln 52!= (0.4343)(1 53 ) = 66, so
:
·
:.
/:
.
'
-
, ...
.
.
·:':· . '·
' . . ' -. . ·. .. ' - :. . _ '. . . .· . . · : : :: :
<
.
'
. ..• -
..: . >-··::•.
�
•• •. :• ·. _ •
.
'
•
.. .·
.•..•: .
'
.
·
:
.·
•
.
: . .
· : · -· _ . ': -' . ;) : ' \: . : :_ . · _ . '
.
•
. .. ·
•
I ann
The expression defining Boltzmann statistics is
(C.l)
We wish to show that
N· J
-
Ki
ex
e!Je·'
(C.2)
•
To simplify the argument we assume no degeneracy for any energy level, so that gi
=
1 for allj. Then Equations w=
(C.l) and (C.2) become N!
. . A..I ' 1Yn ·
--
-- --
Nl ' . . . N-'· •
1•
(C.3)
and
(C.4) 403
404
Appendix C
Alternative Approach To Finding the Maxweii-Boltzman Distribution
We also assume, initially at least, that the energy levels are equally spaced. We wish to make a minimal change in the given configuration while holding Nand
U constant.* Thus we transfer one particle from level j to level k (gaining energy) and one particle from leveJ I to level k (losing the same amount of energy) (Figure C.l). Then the new (primed) configuration is:
Note that we still have
N
The new value of
w
N1
=
and U
N_,·e1.
�
is then
as given by Equation (C.l), is the predominant (equilibrium) configura tion for which ln w reaches a maximum value, then the relatively infinitesimal If
w,
change in configuration described previously produces essentially no change tv'. Equating Equations (C.3) and (C.5) .. we obtain in w. Thus we write w =
N! �;! Nk! N,!
-------
Nl !
·
·
·
·
·
·
Nn!
=
-------
N, !
·
·
·
( N,
-
N! I )_! ( N�c + 2) ! ( N,
------
-
I) !
·
·
·
Nn! '
so
or
J�·!
N,! (N1-l)! (fVt-1)!
-------
-------
:=
e, �----�
'
1 Ek �--1 F.i t-----
Figure C.l
levels.
*Here N system.
=
Adjacent energy
� N1 is the total number of particles and U
=
:L ,vie; is the total energy of the
405
or
But N1c >> 2 or 1, so
to a very good approximation. This can be written
(C.6)
If we bad chosen i,j, k instead of j, k, l, we would have obtained N·I
so we evidently have a continuing geometric series
(C.7) Now, if the levels are not equally spaced, but are given by
(s1
sk)
(ek - ei)
p =
-
q'
then it can easily be shown that Equation (C.6) is replaced by
so that
.•
or
406
Appendix C
Alternative Approach To Finding the Maxweii-Boltzman Distribution
Again, we would have found equalities like this if we had used i,j, k instead of j, k, I, etc. Thus the function must be a constant
call it - {3. Finally, we have
N-1
or
or
(C.8) which is the result we have been seeking. The Boltzmann distribution and the Maxwell-Boltzmann distribution are identical.
•
·:
· .·. · ··
· - - . :. . .: : .. . .· · ···
. .-
_
. _•, ,
.
+co _2 r e - a dx
•
11'
1/2
_
=
0
-oo
.
a
(D.l)
(0.2) -oo
+'lO x2n+le-ax2dx
=
0.
(D.3)
-oo
00
xne·-axdx 0
=
n! . . l + an
(0.4)
•
•
(For Equations (D.2), (0.3), and (D.4), n oo
s-ld X X
e 0
s >
0.
=
0, 1, 2, 3 .... )
(0. 5)
407
408
Appendix D
t J;xn- e-xdx.
Here f(n) is the gamma function: r(n)
f(n
1)- nf(n)
+
Various Integrals
=
11!,
wherenisapositiveinteger.
r(t) = r(2) = 1. 1T.
-
2
:X
l(a) is the zeta function: ' (a)
n
-a
.
n=l
.
l(3/2) = 2.612.
,(1) = oo. (( 2 )
=
((3)
=
7T2/6
l(5/2). = 1.341.
1.645.
=
1.202.
((7/2) = 1.127.
((4) = 1T4/90 =-1.082.
((9/2) = 1.055.
•x
X
3dX
--
=
---
0
2 ) 1 (ex
=
(0.6)
(D.7)
-
15
·
=-
(0.8)
1.
(D.9)
---
eXdx -x(ex+l) .. X
.,
---
=
.
---
=
-
(0 .10)
409
Note that
smce
n
an integer,
•
-
---
------
0
and therefore the integrand is seen to be an even function of .t. In evaluating integrals
·ated with the Maxwell-Boltztnann speed dis-
tribution, consider the integral
r e-a dx,
I
a
> 0.
The equation could equally well be written
so that
- (il+f> e a dxdy.
=
0
Changing the integration variables to polar coordinates, we have
dxdy
=
rdrd6.
The region of integration is the first quadrant. Thus 11/2 /2
r e-a rdrd8
MID
=
0 =
=
7T -
2
0
il e-a rdr
1T
2 0
2a
(D.ll)
410
Appendix 0
Various Integrals
X
-- 'Tr e -ar� _
4a
0
1T
=
-
Taking the square root, we obtain
I=
(0.12)
Integrals of the form
•
X-ne- ux·dx, '
.,
n
0
-
·t 2 3 '
'
,
.
.
.
'
can be found by differentiating Equation (D.12) with respect to a:
dl -
-
da
X •
x2e-�'x·dx
-
··-
•
t)
v; a-.;.:.
J'"'
4
�
and
-
,
-
da--
=
•
()
Successive differentiation leads to the general formula Equation y
=
For odd powers of x
(0.2).
in the algebraic term of the integrand, we can set
x2• For example,
xe
J=
-
]
ax
dx
0
=
1
X
-
2
n
e-aydv ..
411
1 =
2a
e-
.
��y
00
0
1 -
-
2a. Then
dJ -
-
-
da With successive differentiations, we can generate the general for1nula
(0.13)
•
•
•
I
•
I
•
CLASSICAL THERMODYNAMICS C.J. Adkins, Equilibrium Thermodynamics, 3rd edition, Cambridge University Press, Cambridge, 1983. W.P. Allis and M.A. Herlin, Thermodynamics and Statistical Mechanics, McGraw-Hill, •
New York, 1952. M. Bailyn, A Survey ofThermodynamics, American Institute of Physics, New York, 1994. R.P. Bauman, Modem Thermodynamics with Statistical Mechanics, Macntillan, New York, 1992. P.W. Bridgman, The Nature o[TI1ermodynamics, Harper, New York, 1961. H.B. Callen, Thermodynamics, Wiley, New York, 1960. G. Carrington, Basic Thermodynamics, Oxford University Press, Oxford.1994.
E. Fermi, Notes on Thermodynamics and Statistics, University of Chic.ago Press, Chicago, 1966.
E. Ferrni, Thermodynatnics, Dover, N�w )·ork, 1956. C.B.P. Finn, Thermal Physics, 2nd edition, Chapman and Hall, London, 1993. W. Greiner, L. Neise, H. Stocker, Thermodynamics and Statistical klechanics, Springer Verlag, New York, 1995.
W. Kauzmann, Thermodynamics and Statistics: lVith Applications to Gases, Benjamin, New York, 1967. 413
414
Bibliography
C. Kittel and H. Kroemer. Thermal Physics, 2nd edition, Freeman, San Francisco. 1980. R. Kubo, Tl1ermodvnamics, John Wilev and Sons, New York, 1960. �
�
P.T. Landsberg. T hermodynamics and Statistical Meclzanics, Dover, New York, 1990. D.F. Lawden, Principles ofThermodynamics, Wiley, New York, 1987. M.C. Martin, Elements of Thermodynamics, Prentice-Hall. Englewood Cliffs, New Jersey. 1986. P.M. Morse, Thermal Physics, 2nd edition, Benjamin. New York. 1969 .. A.P. Pippard, The Elements ofClassical Thermodynamics, Cambridge University Press, Cambridge, 1987.
M. Planck, Treatise on Thermodynamics, 3rd edition, Dover, New York, 1 945. H. Reiss, Methods ofThermodynatnics, Dover .. New York, 1996. .
.
B.N. Roy, Principles of Moden1 Thermodynamics, Institute of Physics, London, 1995. F.W. Sears and G.L. Salinger, Thermodynanzics, Kinetic Theory, and Statistical Thermodynamics, 3rd edition, Addison-Wesley, Reading, Massachusetts. 1975.
A. Sommerfeld.. Thermodynatnics and Statistical Mechanics, Academic Pres� New York� 1964. M. Sprackling, Heat and Thermodynamics, MacMillan, London, 1993. M. Sprackling, Thermal Physics, American Institute of Phy sics, New York .. l991. M.V. Sussman
Elementary General Thermodynamics, Addison-Wesley, Reading.
..
Massachusetts, 1972.
H. C. Van Ness, Understanding T h ermodynam ics, Dover, New York .. 19 69 . M.W. Zemansky and R.H. Dittman. Heat and Thermod vnamics, 7th edition .. McGraw __
Hill, New York� 1997.
.
: ·.
.
.
:: ··
·.:
.
.
. ·
.
:
•'
·.
.
.
:
.
.
.
.
'·: .
.
. :·
:
.
.
.
KINETIC THEORY OF GASES L. Boltzmann, Lectures <>n Gas Theory, Dover, Ne\\' York. 1995. C. H. Collie, Kinetic Theory and Entropy, Longman� London, 1982.
W. Kauzmann, Kinetic Theory
of Gases, W.A. Ben.jamin, New York, 1966.
STATISTICAL MECHANICS R. Bowley and M. Sanchez. Introductory Statistical Mechanics, Clarendon Press, Oxford, 1996.
R.H. Fowler and E.A. G uggenhei m Statistical Tht,rmodynamics, Cambridge University .
Press, Cambridge, 1952. C. Garrod, Statistical Mechanics and Thermodynamics. Oxford University Pres� Oxford, 1995. -
T. Grenault� Statistical Physics 2nd edition, Chapman and Hall� Londo� 1995. .
Bibr
415
T.L. Hill, A n Introduction to Statistical Thermodynamics, D over, New York, 1986 . K . Huang, Statistical Mechanics, 2nd edition, Wiley, New Yor k, 1987.
E.L. Knuth, Introduction to Statistical Thennodynamics, M cGraw-HilL New York, 1966. R. Kubo Statistica/ Mechanics, North-Holland,Amster dam, 1965. ,
L.D. Landau and E.M. Lifschitz, Statistical Physics, A ddison- Wesley, Reading, Massachusetts, 1958.
R .B. Lindsay, Introduction to Physical Statistics, Wile y, New York, 1941. L.K. Nash, Elements of Statistical Thermodynamics, 2nd edition, Addison-Wesley Reading, Massachusetts, 1972.
'
R .K. Pathria, Statistical Mechanics, Pergamon Press, Oxford, 1972. M. Plischke and B. Bergerson, Equilibrium Statistical Physics, 2nd edition, World Scientific, Singapore, 1994.
F. Reif, Fundamentals of Statistical and Thermal Physics, McGraw-Hill, New York, 1965. W .G.V. Rosser, An Introduction to Statistical Physics, 2nd edition, Horwood, Chichester, UK,l985. K. Stowe, Introduction to Statistical Mechanics and Thermodynamics, Wiley, New York, 1984. R.C. Tolman, Principles of Statistical Mechanics, Oxford University Press, Oxford, l938. G.H. Wannier, Statistical Physics, Dover, New York, 1966.
::;:;:;:;: ;.;.;:;:;:: ·:::;: ;:;:;:;·:,:;.; :·:;:;:,:; :;:;:.:: ···::::;:; :·:;:;:;:;:;: ; :: ;: ::;: ;.:;:;:.:;.;:;:·:::;:;:·: ::;:;:;:;.;:;:;:;:;::: :;: : ::·: ·: ;:;:;:;:.·: :::; :;: ;.. ;:::;:·:;::;: : ;:;:;:;:;:;:;:::;;: :.:;:;:;:;:;.;:;;
SPECIAL TOPICS P. Coveney and R. Highfield, The Arrow of Time, W.H. Allen, London, 1990. J.S. Dugdale, Entropy and Its Physical Meaning, Taylor and Francis, London, 1996. J.D. Fast, Entropy, McGraw-Hill, New York, 1962. M . Goldstein and I.F. Goldstein, The Refrigerator Press, Cambridge, Massachusetts, 1993.
d the Universe, Harvard University
D.L. Goodstein, States of Matter, Dover, New York, 1985. H.S. Leff and A.F. Rex, editors, Maxwell's Demon: Entropy, Information, Computing, Princeton University Press, Prinr.eton, New Jersey, 1990. M. Mott-Smith, The Concept of Heat and Its Workings Simply Explained, Dover, New York, 1962 . J.R. Pierce, An Introduction to Information Theory: Symbols, Signals and Noise, 2nd edition, Dover, New York, 1980. G. Raisbeck, Information Theory, M.I.T. Press, Cambridge, M3ssachusetts, 1964. C. E. Shannon and W. Weaver, The Mathenzatical Theory of Communication, University of Illinois Press, Urbana, Illinois, 1963. M.W. Zemansky, Temperatures Very Low and Very High, Dover, New York, 19�t
Chapter 1
1·1
(a)
closed;
(b)
1-2
(a) (b) (c) (d) (e) (f)
quasistatic, reversible, isobaric.
isolated;
(c)
open. ·
quasistatic, irreversible, isotherntal. irreversible, adiabatic (if fast enough so no heat is exchanged). irreversible, isochoric. reversible, isobaric, isothertttal, isochoric. irreversible, adiabatic.
l·S
260°.
1-6
348.65 K; 75.50°C.
1·7
4.00 K.
1-8
a=
1-10
574.
1·11
-195.80°C;- 320.44°F; 139.23 R.
0.5 mV/°C;b
=
-lo-J mV/(,C)2; T =
430.3°C.
Chapter 2 10-4kg.
1-1
1.5
X
Z.Z
(a)
0.308 kilomole;
(b)
9.86 kg;
(c)
3.96
x
l
(d)
0.277 kilomole.
I
417
Answers to Selected Problems
418 2-3
(c) 800K. 2.67.
2-7 Chapter
3-1
3.69. 3 -5.74
7 10 J.
X
-
3-Z
1.91
3-3
(b) T1/4; -
X
lfr' J.
(c)
-
(3/8) nRT1•
92 1. .
3-5
(c) 4.3 0 x tonJ.
(b) 4.26 x 106J; (b) 3.12
X
10� J.
3-7
(a) 3. 45
X
106 J;
3-8
(a ) 601;
(b) -70J;
3-9
(a) -3.12
X
3-to
9.71 x
3-11
(a) A= -1.6 x (d) 3.6 X 1 0� J.
Chapter
(b) 3.45 X 1 06 J. (c) 50J,lOJ.
(b) -4.32 X 1(}� J:
105 J;
(c) 150K;
(d) 1.25
X
105 Pa.
to·-2 J. to-·)m1Pa 1.8
2.6mJ;
=
(b) 270K;
(c) 6 x l04J:
4
4-1
(a) 59.2 J kilomole -I K (c) 37301 kilomo le 1•
4-S
l0-1� approximately.
4-6
(a) 8.3 1 x 1 06 J;
4-7
Q = n[aT
--
t.
7400 J kilomole
1 K -1;
(b) 1.85 x lfr.:; J;
..
+
4-11
(a) 1 m�, flU 4.48 X 1 Q-� J.
4-13
3.66
X
106 J.
4-14
4.28
X
10·' J.
4-lS
696 m.
(b) 11.6 x 1otl J.
(3/2)bT2 - cj2T]. =
0,
-
5 62 .
x lOi J�
.
1
(b) 1.22 m·, 357K. -4.48 X 10=' J,
Chapter 5
5-3
(a) 11 = aj c��v2; (b) h u0 + c,.T - 2ajv (d) K = [RTv/(v- b)2- 2afv2]-1
S-6
(a)
5-7
340 watts.
-
-900kcal�
=
(b) -1600kcal�
+
RTvj (v
-
(c) 300kcaL-400kcal;
b); (d) 25°/o,3.
It is more efficient to lower the temperature of the cold reservoir.
Answers to Selected Problems
�19
5-9
71 K, 227 K.
5-10
(a) 2.49 X HfJ, 10.70 (b) 2.68 x 10" J; (c)
5-11
200J, -1200J.
5-13
(a) T,j(T2
5-14
-
X
HfJ, -2.49
X
20.3%.
106J, -8.02
X
HfJ;
1)); (b) 14.6.
(b) 55.3%.
Chapter6 6-1
294 JK-•.
6-2
(a) 1074JK-1; (b) 1220JK-1; (c) -6060JK-1•
6-3
127JK-1• 0, 0.893JK-1•
6-5
(a) 19 K; (b)
6-9
(a) 5.76
6-10
(b) aT
X
0.27 JK-1, 0.27 JK-1•
10"1JK-1,0;
(b) 5.76
X
1()-�JK-1,5.76
X
lfr'JK-1•
+ bT3/3.
Chapter? 7·1
(a) -165J; (b) 6.03 X 106J; (c) 2.21
7-l
660JK-1•
7-3
(a) 6.3JK-l; (b) 1.38JK-1; (c)
7-4
(a) 6.93JK-1, -5.0JK-1; (b) -6.93 JK-1; 10.0 JK-1 (c) 1.93JK-1,3.07JK-1•
7·5
(a) 9.13
7-7
(a) -4.9
7-9
1.7 JK-1•
7·11
Cv
7-12
664Jkilomole-1K-1•
7-13
(b) 0.4/P.
7-16
331 ms-1•
7-17
(b) 12.5Jkg-1K-1•
X X
1Q-1JK-1; lfr� JK-1;
In T + R ln( v
-
b)
�o;
X
lfrlJK-1•
(b) 4.9
X
1�1 JK-1;
+
s0•
8-1
The change is greaterfor an ideal gas.
8-3
-1.62 nKT +
10'»J.
BP
+
CP2
+ DJ>-'.
let'JK-1•
(d) 5.73
(b) 9.13
ChapterS
X
X
x
1()-1 JK-1;
(c) 16.4JK-1•
(e) 0.
410
Answers to Selected Problems
8-8
(a) - RT
8-13
(a) 200 K;
b
v.,-
ln
-
+ a
b
v,-
1
l -
-
v1
(c) 2R.
;
---·
v2
(b)
a
1
-
v1
-
1 -
v�
.
Chapter 9
9·3
(b) - 1 50
9-5
0.
9-6
2.
9-8
(a)
.
107 1�5
X
X
300 K, 2.5 atm ;
10�
JK-1•
(b) -1.56
x
(c)
lOfl J:
5.19
x
lOJ JK- 1•
Chapter 10
10-4
2.38
10-8 J kilomole -I K- 1 23.8 J kilomole- 1 K _,_
10·5
(b) a(T- T0)
x
�
Chapter 11
11·1 11-2 11·3 11-4
-
+
(b/3)(T3-
T03).
2.45
(c)
(a) Nj
2 v0 ;
1.14
x
1Q28 m - 2s - 1 •
1.25
X
101J m- 3.
-1 ms ,
112
(b)
517
2
ms
v0,
-1
,
v0;
v0;
(d) 1.41'V.
215 ms -1• •
11-S
0.129 eV, 2.1 3
11-6
0.682
11-7
(c)
11·8
394 ms
11·9
(b) 1.50.
11-10
v2e, zero.
x
105 ms-1 �about 0.2°/o.
(kT/m) 'I�� 2.31 (kT/nt) '12•
0.427. -
1,
445 ms-
1,
483 ms- 1: 2278 ms
2571
11·11 3.86
X
10-8m_ 83.9,
11-12 2.00
x
1013 m_ one collision everv 20 centuries.
11-13 4.0
x
6.16
1•
X
ms- 1 2790 ms- 1• _
109 s -1• "'
to-ss.
11-14 (a) 14 Pa;
(b) 975 s.
11·15 0.485 P.
112• 2 1 ) m m 11-16 ( c2/c1 ) ( / 11-17 About 10 days. 11·18 2.80
x
10 -tom.
11·19 (a) 1.10
x
10-5 Pa
�
7.39
x
10-3 Jm
1s-'K-1• 8.53
x
10-6
2 m s -1•
Answers
to
Selected Problems
421
Chapter 12
12-1
(a)
1.13
x
1015;
(b)
1.26
x
101 4 ;
(c)
0.112.
12-2
(a)
1.27
x
10:�;
(b)
1.01
x
1029;
(c)
0.0796.
U-3
49.
1?-6
(b) 84; (c)
12-7
(c)
336,798;
1'-8
(a)
k
12·11 8.66 12-U 5.6
2.73, 1.64, 0.908, 0.454, 0.195,0.065, 0.013.
(d)
In 3. 1 07
X
X
1.33, 1.00, 0.71,0.48,0.29, 0.14, 0.05.
•
letc;2, lcY6•
Chapter 13
13-1
(a)
4,4;
13-l
(a)
2002;
13-3
(a)
1.33,1.00, 0.78, 0.44, 0.22, 0.11, 0.11;
13-9
(a)
25.3,0.0211;
13-10
(b) lQ-c;.
13-11
(a)
(b) 2ab,64o/o. (b)
252.
(b)
(b)
1, 1, 1, 1, 0, 0, 0.
0.980, 0.0202.
2.78, 1:1.10:0.675,265
k; (b)
587 K.
Chape t r 14
14-1
(a) U/T + Nk In Z, -NkT In Z.
14-3
(b)
14-4
1.44
14-5
- 0.29eV.
14-6
(a) (d)
14-7
2.66
X
108•
l(}c; JK-1•
X
3.74
x
106J,0.0388e\l;
2.22
x
10-8•
(b) 6.06
x
10"1�;
(c)
-0.417eV;
4NkT,NkTfV,Nk[5 + ln(aVT4,f.V)].
Chapter 15
ts-1
(a)
0.982, 0.0180, 0.00033, 0.632, 0.232, 0.085.
15-3
(a)
865,117,16;
15-4
(b) 6Nk{8roJT)exp( -28roJT) ( 1 + T/28rot)·
15-6
(a)
7.53
x
to·-ll
15-8
(a)
5.67
X
106 J;
15-10 2.48
x
(b)
m;
(b)
656 k 6.
(b)
11.2
-4.34
104 J kilomole--t K -t.
x X
10-11 107 J;
m.
(c)
1.79
X
105 JK-1•
Answers to Selected Problems
422
Chapter 16 16-6
1.11 x 1()l J kilomole
16-7
(b)
-
K-t
t
(Einstein)� 2.68
x
1oJ J
kilomole -t K -I (Debye).
293 K.
Chapter 17 -1 1016 s • 9. 24
4.13 X
X
10-24 Nm, 9.24 X 10-24 Am2•
17·2
(d)
17·3
2.12 x to-6•
17-4
2.27 X 10 4 evT-1, 2.90 X 10-4 eVT-1•
17·5
1.87
17-6
(a)
0.212 JT-1:
17·12
(a)
7.87 ?<
17-13
507 Am-1•
-
X
104 JKT-2•
(b)
92.7 JT--1•
l
1550T.
Chapter 18 X
10-5 J.
18-1
(8) 3.89
18- 2
(a)
18-3
(a) 6040 K; (b) 4.65 x tc¥0 megawatts.
18-4
(a)
10.6 mic rons ;
(b)
5800 K.
2.82 kT/h, smaller by a factor of 1.76;
(b)
1.58 X 1011Hz.
1 1 x 1 0 1• (b) photons/molecules= .7 -
1014:
(a) 4.1 x
18-S
·' '
Approximately 10�7•
18-6 18-11
(a)
2.032:
18-12
(a)
-4.67 x 10-20 J. -0.292 eV.
(b)
1.092:
(c) 0.685;
(d)
0.327;
(e)
0.176. .
.
18-13
(a)
1.0 1 X 1026 m-3;
18·14
(a)
8.59 X 1 0-6 K;
(b) 96°/o. (c)
(b) 9.96 X 1 0(,;
242 .
•
Chapter 19 19·1
·(a)
0.134�
(b)
19-2
(a)
0.070
K; (b)
0.116;
(c)
0.0997;
(d)
-12.7, 3.28 X 105;
0.0735.
(c) 6.80 X 10-7
•
.
19-6
·
0.648 J kilomole-J K -1• Equation ( 19.20) gives a value 12 °/o lower than the
- experimental value. 1.88 eV;
(c) 1.89 x 1 oJ 1
19·7
(a)
19-8
331 J kilomole -t K -I.
kiJomoJe -l K
-
t
.
ce/3R
=
0.046.
19-10 2.9 X 10-11 Pa 1• - -
19-12 19-13
7.2 x 1 06 m
(a)
0.21
.
MeV; (b) 2.4 x
109 K;
(c)
2 x
-1 lOS ms ;
(d)
0.39 kg m-3.
Answers to Selected Problems
413
Chapter20
2.161 bits, 2.583 bits. 20-5
(a) 226 bits; (b) 5.
20·6
1.33 bits.
20-7
2.5 bits/letter.
Appendix: A
A·l
(a)
z
(b)
Z =
=
x2
lny +
y (X
-
C.
3)
-
X +
C.
(c) Inexact. A-l
(a)
z =
xy + y2 +
C.
(c) 0, 1/2. A-3
A -4
A·S
(b)
2 z = x y - 3x + C.
(c)
3 fo r all three paths.
(a)
• 2 lf y 3 1 1 o r x
(b)
y.
(c)
x-�.
(a)
sec2x, w = ysin.t - tanx + C.
Absolute temperature, 14,98-103
Berthelot equation of state, 32
Absolute zero, 14
Binomial coe fficient 217 ,
entropy at, 172-173
Blackbody radiation. 333-338
heat capacity at, 179
Blackbody spectrum, 335-338
unattainability of. 17 4
Bohr magneton. 308-309
Adiabat, 22
Boiling, 13,140, 147
Adiabatic boundary, 7
Boltzmann distribution, 238-241
Adiabatic compressibility, 117
Boltzmann factor. 403-406
Adiabatic de magnet iza tion 174, 177, . 318-321
Boltzmann's constant. 189
Adiabatic expansion of an ideal gas,
Bose-Einstein condensation, 340-345
,
Boltzmann statistics. 235-236 Bose-Einstein distribution. 244-246.
61-62 Adiabatic process, 8, 62
253-254
Adiabatic work, 40-43, 61-63� 104
Bose-Einstein statistics. 245.333
Alternative statistical models,
Bose temperature. 343
254-257
Boson. 230.244-246
Arrow of time, 97
Boson gas
Assembly, 215,221-223
chemical potentiaL 341. 3-44
Atmosphere (unit)� 9
energy, 345-346
Atmospheric pressure� 9
·
entropy, 346
Available microstates. 216
heat capacity, 346
Average speed,l85, 196,202-203
Helmholtz function. 346
Avo gadro s law, 6, 22
partition function. 341
Avogadro's number, 6
pressure.347
'
414
.
Index
425
Boundary, 7-8
of thermal conductivity,207
Boyle's law,22
of viscosity,204-208, 211
Brillouin functicn,312-315
Coin-tossing experiment, 216-221
Bulk modulus,180
Collision cross section,198-200 Collision frequency, 199
Calorie (unit), 45
Collisions with wall,188-189
Canonical ensemble,255
Combined first and second laws, 103-104
Canonically conjugate pairs, 21,130
Comparison of distributions, 253-254
Camot cycle,74-79
()o�pressibuuty
and Qausius-Clapeyron equation, 147-148
adiabatic,117 of ideal gas, 28
efficiency,76,91-94
isothertnal,28
with ideal gas,76-79
Compression ratio, 83
temperature-entropy diagram,110-111
Concentration,252
and thertnodynamic temperature, 98-103 Carnot engine,76
Condensation, 23,340-345 Condition of integrability,31
Carnot heat pump, 82
Configuration,215
Camot refrigerator,79-80
Configuration work,37-40
Carnot's theorem,91-94,98-100
Conservation of energy, 40
Celsius temperature scale,13,16
Constant volume gas thertnometer, 13-15
Chain rule of differentiation,393 . . Change of phase,7, 23,2 6, 137-141 Characteristic temperature
Constituents, 153,157-160 Constraints, 237,238 Cooling
De bye,180,299
laser, 174
Einstein,294
magnetic,174,318-321
for linear oscillator,280
Cosmic background radiation, 337-338
for rotation, 285
Corresponding states, 295
for vibration, 280
Critical constants, van der Waals gas, 23-25
Charles' law, 22
Critical point,23-25,27
Chemical equilibrium, 160-161
Critical pressure,24-25
Chemical potential,151-155,244,251-252
Critical temperature,23-25
of boson gas, 341,344
Critical volume, 24-25
of dilute gas, 251,263
Cross section,198-200
of ideal gas, 166
Curie constant,329
of feranion gas, 355- 361
Curie law,314
multicomponent system, 153
Curie point, 32.6-328
of photon gas, 333
Curie temperature,326
Chemical reactions, 160-161,167
Curie-Weiss Ia"', 327-328
Classical distribution, 248
Cutoff frequency,297
Classical limit,247
•
Cycle
Classical statistics, 248
Carnot, 74-79,98-100
Clausius-Clapeyron equation't 140-141,147
Joule,84
Clausius inequality. 94-96
Otto,83
Clausius statement of the second law,90
Cyclical process, 8, 44,58
Closed system, 5
Cyclical rule,392
Coefficient binomial. 217
Dalton's law of partial pressures, 162
of diffusion, 208
de Broglie wavelength, 227, 296,352
of expansion
(see also Expansivity), 28
of performance,79
Debye functions, 301, 302 Debye T3 law, 64, 180
Index
416 Debye temperature, 180, 299
Einstein temperature, 294
Deby e theory of specific heat capacity,
Einstein theory of specific heat capacity� 293-295
296-301
Electrolytic cell,38
Degeneracy, 215 Degeneracy function, 229-230
Electromagnetic radiation, 333
Degenerate region, 363
Electromotive force, 38
Degrees of freedom
Electron gas, 361-364,367-368
electronic excitation, 279,287-288
Electronic excitation, 279.. 287-288
Gibbs phase rule, 160
Electronic heat capacity, 365-366
internal, 279
Electrons in a metal, 361-364
rotational, 191,279,284-287
Empirical temperature, 12, 103
translational. 19 1 , 279
Energy
vibrational, 191 279,282-284 �
Demagnetization� adiabatic, 174, 177,
conservation of.. 40 equipartition of. 190-191, 270-271 Ferrni, 364
318-321
Density, 5, 10
frc e. 1 36
Density of states, 229-231
gravitationaL 369
Diamagnetism, 307. 329
heat. 53
Dia thertnal boundary, 8
of ideal gas, 56
Diatomic gas, 56.192,279-289
internal, 42 263
Diatomic molecules
kinetic, 151-152, 190
,
electronic excitation, 287
levels, 215,225-229
rotational modes, 284-287
magnetic. 310
translational modes, 28H
pot entiaL 151-152.310
vibrational modes, 282-284
states,225-229
Dielectric polarization. 38 Dieterici equation of state. 32 Differentials
ze ro
-
poi nL 279
Enthalpy definition, 57, 132-133
exact, 42,393-400
of dilute gas. 264
inexact� 37, 40,393-400
heats of transformation, 57-59
Diffusion, 208
of ideal gas. 60
Diffusive equilibrium. 156
Joule-Thomson experiment. 60
Dilute gas, 246 248 -
Entropy
Disorder, 220, 384. 388 - 389
at abs olute zero, 172-173
Dis si pati ve work,40-41
and available energy, 97-98
Distinguishable particles, 221, 235-236
Boltzmann definition. 223-225
Distribution function
of dilute gas, 252 .. 263
Boltzmann, 238-241,403-406
disorder, 220 .. 384,388-389
Bose-Einstein� 244-246,253-254
of ideal g as, 112 113
Fern1i-Dirac, 242-244,253-254
information theory, 381,384-386
Maxwell-Boltzmann, 246-248.253-254
irreversibility, 95-96,118-121. 121-122
speed. 1 9J- 1 98 velocity, 194
Ncrnst heat theorem, 172 pr i ncipl e of increase of. 96-97
velocity components. 190
reversibilitv, 88 95,109-110
-
..
.,
Duhem-Margules equation, 168
second law� 97
DuLong-Petit law� 54, 293, 300
third law, 172-173
of van dcr Waals gas, 125 Efficiency of heat engine, 76, 91-94 Effusion, 201-203
Equations of state, 21- 31 Berthelot. 32 .
Index Dieterici, 32 ideal gas. 22 liquid or solid, 29
van der WaaJs gas, 23-25 Equilibritun chemical, 160-161
mechanical,145,156 stable, 143-145 state,7 thermal, 10-12,145,156 thermodynamic, 145 Equipartition of energy, 190-191,270-271
Error function,209 Euler's theorem, 153 Eutectic point, 157 Exact differential, 42,393- 400 Exclusion principle, 242, 356
Expansion coefficient,28 (see also E xpansivi ty) Expansion, free, 70, 104, 121-122 Expansivity, 28
427 Free expansion, 70,104, 121-122 Freezing, 26 Fusion, 26,140,148 Gamma function, 231,408 Gas constant, 22
Gas thermometer, 13-15 Gaussian integral,409 Gay Lussac Joule experiment, 69-72 -
-
Gay-Lussac law, 22 Gibbs-Duhem equation, 154 Gibbs function, 136-137 chemical potential, 154-155 of dilute gas, 264 of ideal gas, 146 open systems, 151-165 phase transitions, 137-139 thi rd
law, 171-173
Gibbs-Helmholz equation, 146,171 Gibbs paradox, 165,273 Gibbs phase rule, 157-160,167
of ideal gas, 28
Gradient, 9,204
of liquid or solid, 28-29
Grand canonical ensemble, 257
Extensive variable, 5
G rav itatio na l energy, 369 GrUneisen constant, 303
Fahrenheit temperature scale, 16 Fernti-Dirac distribution, 242-244,253-254
Harrnonic oscillator, 279-282
Ferrni-Dirac statistics, 242, 355
Heat
Fern1i energy, 355-357
definition, 43-45
Fernti level, 362
inexact di fferential, 44
Fermjon, 230,242-244 Fernlion gas
latent, 57,140 mechanical equivalent of, 46
chemical potential, 355-361
reservoir, 74, 75, 79, 81,82,89, 118-120
electronic heat capacity, 365-366
sign convention for, 44
energy,364
statistical interpretation, 250
entropy, 367 free electrons in a metal, 361-364
units of,45 Heat capacity, 53,191-193
Helmholtz function, 367
at absolute zero, 179
pressure,367
anomaly of liquid heliwn, 348
Fernli temperature, 357
at constant pressure, 53-54,116,191-193
Fermi velocity, 370
at constant volume, 53-54, 116,191-193
first law of thermodynamics, 44,46
Debye theory, 296-301
combined vlith second law, 103-104
of a diatomic gas, 191-193,279-289
FIXed-point temperature, 14
Einstein theory, 293-295
Fluctuations, 255-256 �ux,186-188,202-203,205,336
electronic, 365- 366 of an ideal gas, 54-56,191-193
Free energy
Mayer's equation, 54-56
Gibbs, 136-137
ratio, 56,65, 192-193
Helmholtz, 134-136
of a solid, 293- 301
Index
428 of a spin-1/2 paramagnet, 317
definition 42 ..
Heat e n gin e.. 76
dilute gas. 263
Heat flow, direction of, 44
ideal gas, 56, 191
Heat of transfortnation
relation to partition function, 263-264
fusion, 26, 140
Inverse collision, 193
sublimat ion 26, 140
Inversion point, 74
vaporization, 26, 140-141
Irreversible process.. 8 89-90, 118-121
.
..
Heat pump. 82
di ssipat i ve w ork , 40 41
Helium, liquid and solid, 347-349
entropy change, 95-96, 118- 121 , 121-122
Hel mholtz function
Ciibbs function, 136-137
-
and chemical poten tial. 251
Helmholtz function, 135-136
definition, 133 .. 1 34 -1 36
Isen tropic compressib ilit y 117
of dilute gas, 252, 263
Isobar, 22
of ideal gas, 146
Isobaric process, 8
H y steresis.. 308
,
Isochore. 22 Isochoric process, 8
Ice phase diagram, 26
Isotope se par ation by effusion , 203
Ice point. 13
Isotherm, 22
,
Ideal gas
critical, 23-24
adiabatic process. 62
Isothern1al com pressib ility. 28
and Carnot cycle, 74 -79
Isot hermal process, 8, 47
com pr essibility, 28 diatomic, 56
Joule (unit), 45
energy. 56
Joule co effici ent .. 70-71
e n thalpy. 60
Joule cvcle. 84
entropy, 112-113
Joule-Thomson coefficient, 73
e quation of state. 22
Joule-Thomson e xperi me n t 60 72-74 (see
..,.
,
expansivity, 28
,
also Throttling process)
Gibbs func tion . 146 heat capacity, 56
Kelvin-Joule experiment, 72-74
Helmholtz fun ction, 146
Kelvin-Planck statement of second law, 75.
internal energy. 56
90
Joule coefficient, 71
Kelvin temperature scale, 14-15
Joule-Thomson coefficient, 73-74
Kilocalorie (unit). 45
•
Increase of entropy, 96 -9 7
Kilom ole (un it ) , 6
Indistinguishable particles, 242. 244
Kilomole fraction, 1 59
Inequality of Clausius, 94 -96
Kinetic energy, 151-152, 190
Inexact differential� 37 40, 393-400
Kinetic theory of gases. 183-208
,
Inflecti o n point, 23, 24
assumptions, 183-185
Informati on
ideal gas, 188-189
bit, 380
entropy, 381
Lagrange multipliers� 236-238,257
maximum entropy. 381-384
Lambda li ne 348
uncertainty. 375-379
Lambda point . 348-349
lnfortnation theory. 375-388
connection to thermodvnamics. 384-387 .,.
In tegrati n g factor, 88, 396- 398
.
Lase r cooli ng .. 17 4 Latent heat 57. 140-141 (see also Heat of
transformation)'
Intensive variable, 5
Law of D uLong and Petit, 54� 293, 300
Intermolecular forces, 151-152.183-185
Law s of thermodynamics
Internal energy
combined first and second, 103-104.
Index first, 44, 46 second,96-97 third,171-179 zeroth,10-12
Legendre transformation, 130-132 Lennard-Jones potential, 184 (see also 612 potential) Linear oscillator,279 -282
Liq uefaction of gas, 24 Liquid, 25 entropy change, 122-123 equation of state, 29
419 Microstate, 215,217 Microstates, sum over, 216,241,263-264 �uing,l62-165,271-273 Molecular diameter, 184
Molecular flux, 186-188 Molecular weight, 6 Moments of a probability distribution,186 Most probable speed,197
Negative temperature, 321-324 experimental demonstration, 324 Negentropy, 386
Loschmidt's number, 6
Nernst statement of third law,172
Macrostate, 215,217
Occupation number. 216
Magnetic cooling,174,177,318 -321
Open systems, 151-165
Magnetic dipole moment,308-315
Otto cycle,83
energy, 310
paramagnetic salt, 307,314 saturation,313, 314
Paramagnetism, 308 -314 Brillouin function,312- 315
Magnetic domains, 307
Curie constant,329
Magne tic field, 38, 174, 307-310,318-321,
Curie law, 314
325-328
energy, 316
Magnetic saturation, 313,314
entropy,317-318
Magnetic susceptibility, 328
heat capacity, 318
Magnetism, theory ot 308-328
mean magnetic dipole moment, 311-315
Magnetization, 314,325- 328
spin-1/2 system,315-318
spontaneous, 325
.
Partial differentiation, 391-393
total, 38, 319
chain rule,393
work,38
cyclical rule,392
Manometer, 9-10 Maxwell-Boltzmann distribution, 248, 253-254,403-406 Maxwell-Boltzmann speed distribution,
193-198,269-270
reciprocal relation,392 Partial pressure, 162 Partition function,241,263-264 De bye solid,302 dilute gas, 265-266
Maxwell-Boltzmann statistics,248
linear oscillator,280
Maxwell-Boltzmann velocity distribution,
rotational,285-286
194
vibrational,280
Maxwell relations, 134,135,142-143
Pascal (unit),9
Maxwell's demon,273-275,387-388
Path,7
Mayer's equation,54-56
dependence,30
Mean free path,198-200
independence, 30
Mean kinetic energy,190
Pauli exclusion principle, 242, 256
Mean magnetic dipole moment,311-315
Performance,coefficient of,79
�ean speed,l85,196,202-203
Phase,change of, 7, 23,26,137-141
Mean square speed, 185, 196
Mechanical equilibrium, 145, 146
(see also
Phase transforntation)
Phase diagram
Mechanical equivalent of heat, 46
helium,347
Melting,26
substance that contracts on freezing, 26
Microcanonical ensemble,255
substance that expands on freezing, 26
Index
430 water, 26
intensive, 5
Phase equilibrium,25-27,140-141,155-157
P-1' diagram, 22 .. 26
Phase rule, 157-160,167
P-v·diagrarn,22.. 24
Phase transformation (or transition), 7. 23. 26 .. 137-141
P-v· T surface, 25
P honon 298
Quantum concentration, 268
Photon. 333- 334
Quantum density of states,229-231
Photon gas, 333-340
Quantum number, 227-229
.
chemical potential. 333,339
Quantum of energy, 333
energy, 335
Quantum states, 225-229
entropy, 339
Quasi-static process. 8
•
heat capacity,339 Helmholtz function,339
Radiation, blackbody,333-338
pressure,340
Rankine temperature scale, 16
Planck curve,335,337,338
Ratio of specific heat capacities� 56,65
Planck radiation fortnula .. 334-335
Rayleigh-Jeans formula� 337
Planck statement of third law, 172-173
Reactions! chemical, 160-161, 167
Poise (unit), 205
Reciprocal relation of differentiation. 392
Polarization, dielectric. 38
Reciprocity relations, 130, 133, 135
Population inversion, 322,324
Reduced mass, 284
Porous plug experiment,72-74
Reduced variables, 32
Potential energy, 121, 151-152. -184
Refrigerator, Carnot, 79-80
gravitationaL 369. 371 magnetic, 310
Reservoir, heat. 74,75, 79. 81� 82. 89. 118-120
Potential well, 151-152, 184,226-227
Resistance thermometer. 17
Pressure critical,24-26
Reversiblc processes.. 8 entropy change, 109-11 0, 111-112
dilute gas, 264
Riemann zeta function, 408
ideal gas, 22
Rigid rotator,224-285
kinetic theory evaluation. 188 -189
Root mean square speed, 186� 196-197
.
partial, 162 radiation, 340
Sackur-Tetrode equation� 267
units ot 9
Saturated liquid. 26
van der Waals gas,23
Saturated vapor, 26. 160
Probability density function, 185
Saturation,magnetic,313,314
Probability� thermodynamic, 216-217
Schottky anomaly, 317
Process, 8
Second law of thermodynamics, 96-97
adiabatic, 8
Clausius statement, 90
cyc li ca l ,
and first law, 1 03 -104
8
free expansion. 70.104. 121-122
increase of entropy, 97
irreversible, 8. 89-90 .. 118-121
Kelvin-Planck statement 90
isentropic,112-113
Self-diffusion. 208
isobaric 8
Sign convention for work and heat 37 .. 41
isochoric, 8
6-12 potential, 184 (see also Lennard-Janes
..
isothermaL 8 quasi-static, 8
potential)
Solid. 25
reversible, 8, 109-110
entropy change .. 122-123
throttling, 72-74
equation of state, 29
Property, 5 extensive, 5
specific heat capacity, 54.. 293- 301 Solid helium, 347
431 Index 7 2 5 , e 2 S a h p d li o S 359 sion, expan ld mrner fe So 117 gas, in d spee Sound . also (see 93 191-1 53, ty, capac i beat c Specifi Heat capacity) , 6 1 1 , 4 5 3 5 , e r u s s e r a t constant p 191-19 3 3 9 1 1 9 1 , 6 1 1 , 4 5 3 5 , e at constant volum 9 3 -1 1 19 , 6 5 4 5 s, a g i a e of an id Mayer's equation, 54-56 ratio, 56, 65, 192-193 6 e, b l a v ari e v si n te x e n a f o e lu a Specific v Spectral density,334 Speed average,185,196 distribution, 193-198,269-270 mean,185, 196 mean square,185,196 most probable, 1 97 root mean square,186, 196-197 of sound in gas, 117
Superfluid, 348 Surface film, 38
97 , s, 5 g undin Su1ro Survival equation, 201 8 2 3 , c i t e n g a m , y t li i Susceptib System closed, 5
isolated, 5 open, 5, 151 165 and surroundings, 5, 9 7 -
thermodynamic, 5
Tds equations, 1 1 3-117 Temperature absolute, 14 98 103 absolute zero,171-178 ,
-
Bose,343 Celsius, 13, 16 characteristic, 280, 285 critical, 23 Debye, 180,299
Spin angular momentum, 307
Einstein, 294
Spin degeneracy factor, 230
empirical, 12, 103
Spin-1/2 paramagnet, 315-31.8
Fahrenheit, 16
Stability of white dwarf stars, 369-370 Stable equilibriuu1, 143-145 entropy, 96, 138 Gibbs function, 136-138 Helmholtz function, 135-136, 138 mechanical, 145
therntal, 145 thermodynamic, 145 Standard deviation, 209 State variable,7 Statistical entropy, 223-225 Stltistics Boltzanann, 235-236 Bose-Einstein, 244-246 classical, 248 Fermi-Dirac, 242 Maxweli-Boltztnann, 248 Steam point, 13
Fernti, 357 fixed point, 14 ice point,13 Kelvin, 14 negative, 321-324 Rankine, 16 steam point, 13 thermodynamic, 10-12,103 triple point, 14 Temperature-entropy diagram, 110-111 Temperature scale, 12-16,103 absolute, 14, 103 linear, 13-14 logari tlunic, 18 Thern1al conductivity, 207 Theranal equilibrium, 10-12,145,156 Thertual stability, 145 Thermionic emission, 364
Stefan-Boltzmann constant, 336
Thermocouple,17
Stefan-Boltzntann law, 236
Thermodynamic coordinates, 5
Stirling's approximation, 218,401-402 Stirring work, 40- 41
Therntodynamic equilibrium, 21,145
Stoichiometric coefficients, 160-16 .l
Therrnodynamic laws, 10-12,44,46,96-97, 171-179
Stress, 204
Thermodynamic potentials, 132-138
Sublimation, 26, 140, 1 47
Therntodynamic probability, 216-217
Sum over states, 241
Boltzmann, 235-236
Index
432 Bose-Einstein. 244-246
specific, 6
and entropy, 223-225
state. 5
Fermi-Dirac. 242
Variance, 160
Maxwell-Boltzmann, 248
Velocitv
•
.,
Thertnodynamic system. 5
distribution" 194
Thertnodynamic temperature. 10-12. 103
gradient. 204
Thermometer
l='ernti, 370
constant volume gas, 13-15
fluid, 204
resistance, 17
Viscosity� 204-208
thermocouple, 17
coefficient of. 204-208,211
Thermometric property, 12, 13, 17
Volume
Third law of thermodynamic� 171-179 Nernst statement, 172
critical, 24 expansivity. 2R
·
Planck statement 172-173
specific� 27
unattainability staten1ent, 174, 177
v-1, diagram, 22
Throttling process, 72-74 Time ·s arrow, 97
Water
Torr (unit), 9
critical point 27
Total magnetization, 38� 319
heat of fusion. 59
.
Transforrnation._ latent heat of, 57. 1 40-141
heat of vaporization, 59
Transfortnation, Legendre, 130-132
ice point� 13
Transport processes. 203-208 diffusion, 208
pressure-temperature diagram, 26 specific heat capacity, 65
thermal conductivity._ 207
steam point 13
viscosity.. 204-208
triple point, 14. 15 vapor pressure. 26
Triple line. 25 Triple point, 14.58
Wave function. 226
of carbon dioxide. 26
White dwarf stars, 367-370
of water, 14, 15
Wicn 's displacement law, 336
T-s diagram, 110-111
Wicn 's law, 337
Two-fluid theory. 348
Work
Two-level system. 315-318
adiabatic, 41-43. 61-63 c:arnot cycle� 76-79
Ultraviolet catastrophe. 337
configuration, 37-40
Unattainabilitv statement of third law. 174 .
dissipative.. 40-41
.,
177
Uncertainty.. information theory� 375-379
electrical.. 41
Units, International System of (Sl ) .. 9
free expansion, 70. 104 inexact differentiaL 37
Universal gas constant, 22
isobaric. 48
Universe (system and surroundings). 5. 97
isothermaL 47 path-dependent. 40
Van der Waals gas. 23 characteristic constants. 23
path-independent (adiabatic). 42 sign convention for. 37
critical constants. 24. 25
statistical interpretation. 250
equation of state. 23
stirring. 40-41
Vapor, 24-26, 160 Vapor pressure. 26
Zero-point energy. 279
Vaporization. 26, 140, 147� 160 Variable
Zeroth law of thermodvnamics. 10-12 Zeta function. 408
extensive, 5 intensive. 5
"'
Zustandsunune, 241
CONVERSION FACTORS
QUANTITY Length
Mass
CONVERSION 1m= 100
em
=
3.281 ft
=
39.37 in
1 kg= 10� g
= 2.205lbm Volume =
103 liter
35.31 ft-' ')
=
Force
1 N = 1 kg· m · s- 2
10� dyne -
=
= 0.2248lbr
Pressure
1 Pa = 1 N · m-2 =
10 dyne· cm-2 •
=
=
9.872 x 10-6atm 10
..
-
'
bar
= 7.502 x to-·; torr -
=
Energy
14.50 x lo--� psia
lJ=lN·m =
=:
=:
=
107 erg
2.390 x 10-4 kcaJ 6.242 X 101R eV 9.478 X 10-� Btu
