Lecture notes on
Physics 18324312
STATISTICAL PHYSICS
Tjipto Prastowo and Mita Anggaryani
Department of Physics Faculty of Mathematics and Natural Sciences State University of Surabaya
Physics Department
11 February 2008
TO THE STUDENT WE LOVE Here it is, lecture notes on Fisika Statistik, a series of lectures taught once a week in tahun ke tiga, Jurusan Fisika, Fakultas Matematika dan Ilmu Pengetahuan Alam (FMIPA), Unive Universi rsitas tas Negeri Surabay Surabayaa (Unesa) (Unesa).. Please Please note that we do not intend intend to replace replace any standard book of statistical physics physics as listed in the Bibliograph Bibliography y. Rather, we would like to help students understand materials covered in this course through simple, but systematic guidance. To use math and other basic theory needed in this course, you need not just knowledge, but skill. This can be obtained obtained only through continual continual practice. You may obtain a superficial knowledge by listening to lectures in the class, but you cannot reach the skill expected by that way. It is common to come across student conversation outside the class, for example, something like “I understand it but I can’t do the problem !” This student feels uncomfortable to do some problem although it looks so easy when the lecturers do it in the class. Such Such a statemen statementt shows shows lack lack of practice practice and hence hence lack lack of skill. skill. Our dearest dearest studen students, ts, please please alway alwayss study with pencil and paper at hand. You will find that the more able you are to choose effective methods of solving problems the easier it will be for you to master new materials. materials. This costs you you nothing but practice, practice practice and again practice. Please do remember that the best way to learn to solve problems is to solve them. We welcome good comments for further improvements because penyusunan materi ajar ini merupakan merupakan salah satu upaya upaya untuk meningkatk meningkatkan an kualitas kualitas layanan layanan publik. Semoga hal ini bermanfaat bagi segenap sivitas akademik di lingkungan Jurusan Fisika FMIPA, Unesa.
Best wishes, Kampus Ketintang, 12 Mei 2008 ita nggaryani, jipto jipto rastowo rastowo
M A
T
P
ii
TO THE STUDENT WE LOVE Here it is, lecture notes on Fisika Statistik, a series of lectures taught once a week in tahun ke tiga, Jurusan Fisika, Fakultas Matematika dan Ilmu Pengetahuan Alam (FMIPA), Unive Universi rsitas tas Negeri Surabay Surabayaa (Unesa) (Unesa).. Please Please note that we do not intend intend to replace replace any standard book of statistical physics physics as listed in the Bibliograph Bibliography y. Rather, we would like to help students understand materials covered in this course through simple, but systematic guidance. To use math and other basic theory needed in this course, you need not just knowledge, but skill. This can be obtained obtained only through continual continual practice. You may obtain a superficial knowledge by listening to lectures in the class, but you cannot reach the skill expected by that way. It is common to come across student conversation outside the class, for example, something like “I understand it but I can’t do the problem !” This student feels uncomfortable to do some problem although it looks so easy when the lecturers do it in the class. Such Such a statemen statementt shows shows lack lack of practice practice and hence hence lack lack of skill. skill. Our dearest dearest studen students, ts, please please alway alwayss study with pencil and paper at hand. You will find that the more able you are to choose effective methods of solving problems the easier it will be for you to master new materials. materials. This costs you you nothing but practice, practice practice and again practice. Please do remember that the best way to learn to solve problems is to solve them. We welcome good comments for further improvements because penyusunan materi ajar ini merupakan merupakan salah satu upaya upaya untuk meningkatk meningkatkan an kualitas kualitas layanan layanan publik. Semoga hal ini bermanfaat bagi segenap sivitas akademik di lingkungan Jurusan Fisika FMIPA, Unesa.
Best wishes, Kampus Ketintang, 12 Mei 2008 ita nggaryani, jipto jipto rastowo rastowo
M A
T
P
ii
iii
General Guidance
PHYSICS 18324312: STA STATISTICAL PHYSICS
Pre-requisites: Fundamental Physics (I and II) and Mathematical Physics (I and II)
Prastowo, Mita Anggaryani Anggaryani Lecturers: Tjipto Prastowo, References: An Introduction to Statistical Physics for Students (A. J. Pointon) Time and Place: Monday, 07.00 - 09.40am, D1 Marking Marking Scheme Scheme:: NA = 20%P + 20%UTS + 30%T + 30%UAS NA=Final Mark, P=Presence, UTS=Mid-Exam, T=Homework, UAS=Final Exam Notes:
1. Students are not allowed to join the class for being late (a maximum of 15 minutes from the starting time), except for reasonable arguments. 2. Each lecturer lecturer contributes contributes an equal proportion of mark to the final mark. 3. P is possibly reduced to a minimum. 4. UTS = 50%Q1 + 50%Q2 5. T = 50%T1 + 50%T2
Q = Quiz T = Assignment
6. UAS normally contains 4, but possibly has 5 problems. 7. Assignment Assignmentss will be distributed to class members and all students students are required to hand the completed assignments assignments in within a limited limited time. Various penalties will be given given for any delay, i.e, 25% discounted mark for a one-day delay and 50% for a two-day delay. There will be no mark given for more than two-day delays. 8. No additio additional nal assign assignmen ments ts or examina examination tionss after after Final Final Exam, Exam, except except for specified specified reasons with very limited permission or medical examination required. 9. Students Students are allowed allowed to have their own notes (2 pages only), written in a folio paper, in both Mid-Exam and Final Exam. 10. Other Other issues issues will be determi determined ned during the course, course, includin includingg details details of problem problemss and exam time-tables. Students are strongly encouraged to be active and well-prepared. If possible, tutorial is available for a further, detailed description of each topics.
Course Contents:
1. Chapter One: Introduction (Lectures 1 and 2)
• overview, system of many particles, distribution function, description of assembly and phase space, the scope of statistical physics
2. Chapter Two: Maxwell-Boltzmann Statistics (Lectures 3, 4 and 5)
• fundamental concepts, classical systems, applications of MB statistics, Boltzmann partition function
3. Chapter Three: Bose-Einstein Statistics (Lectures 6 and 7)
• fundamental concepts, boson systems, applications of BE statistics 4. Chapter Four: Fermi-Dirac Statistics (Lectures 8 and 9)
• fundamental concepts, fermion systems, applications of FD statistics 5. Quiz 1: Chapters Two, Three and Four (Lecture 10) 6. Chapter 5: Thermodynamics of Gases (Lectures 11 and 12)
• concept of entropy, Gibb’s paradox, semi-classical perfect gas 7. Chapter 6: Applications of Statistical Thermodynamics (Lectures 13 and 14)
• system of two-energy level, harmonic oscillator, diatomic molecule 8. Quiz 2: Chapters 5 and 6 (Lecture 15) 9. Final Exam Preparation: Discussion on Quiz 1 and 2 (Lecture 16)
iv
Contents 1 Introduction
1.1 1.2 1.3 1.4
1
Overview . . . . . . . . . . . . . . . . . . Distribution function . . . . . . . . . . . Description of assembly and phasa space The scope of statistical physics . . . . .
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2 Maxwell-Boltzmann Statistics
5
2.1 The speed distribution function . . . . . . . . . . . 2.2 The momentum and energy distribution functions . 2.3 Applications of Maxwell-Boltzmann statistics . . . 2.3.1 The mean, rms and most probable velocities 2.3.2 Equipartition principle of energy . . . . . . 2.3.3 Specific heats . . . . . . . . . . . . . . . . . 2.4 The Boltzmann partition function . . . . . . . . . . 2.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . .
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3 Bose-Einstein Statistics
3.1 3.2 3.3 3.4 3.5 3.6
Weight of configuration . Population of particles . The Bose-Einstein gas . Black-body radiation . . The specific heat of solid Exercise . . . . . . . . .
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Weight of configuration . Population of fermions . The Fermi-Dirac gas . . The electron gas . . . . . Exercise . . . . . . . . . Konsep entropi Gas klasik . . . Paradoks Gibbs Gas semi klasik Latihan soal . .
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5 Termodinamika Gas
5.1 5.2 5.3 5.4 5.5
5 9 10 10 11 13 14 15 17
4 Fermi-Dirac Statistics
4.1 4.2 4.3 4.4 4.5
1 2 3 4
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vi
CONTENTS
6 Aplikasi Distribusi Statistik
6.1 6.2 6.3 6.4
Sistem dua tingkat energi Osilator harmonik . . . . . Gas diatomik . . . . . . . Latihan soal . . . . . . . .
Bibliography
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Chapter 1 Introduction 1.1
Overview
This course is devoted to discuss systems containing very many particles (of order 1023 ), which include gases, liquids and solids. The inherent problem of these systems appears to be complex owing to the complexity of interactions of such a large number of particles. The behaviour of these particles as a whole often reveals different qualitative features from that of each of the individual particles. Since it is not possible to study the detailed behaviour of each particle, a statistical approach then becomes plausible. The question is that what kind of statistics is appropriate for describing the dynamics of the system? Can common statistics handle this problem? Or, do we need to seek for an alternative method? If so, what should we do then? At this stage, it is useful to distinguish microscopic systems (i.e., small-scale systems of order 10−9 m or less), whose description is of fundamental interest, from macroscopic systems (i.e., large-scale systems of order 10−6 m or greater). In the latter systems, macroscopic parameters, such as temperature, volume and pressure, are usually used to characterise the systems. For an isolated, macroscopic system, these parameters may not vary with time and the associated condition is said to be in equilibrium . Whereas for an open, macroscopic system where interactions between particles constituting the system and the surroundings are allowed, macroscopic parameters depend upon time. The time variations of these parameters lead to mean values over a period of large times corresponding to a final equilibrium state. The discipline in which fundamental laws of physics describing the relationship between macroscopic parameters are examined is the subject of thermodynamics. A fundamental approach, statistical physics provides insights into macroscopic systems from a microscopic point of view. In this approach, a wide range of physical principles from basic laws to advanced methods are used to examine systems under consideration. 1
2
1.2
1. Introduction
Distribution function
In this section, we discuss discrete and continuous distribution functions that represent a physical system. For any statistical data associated with this system, three important quantities involved are: the average or mean value, the root-mean-square (rms) value and the most probable value. If the data are given in a discrete form of xi , then these values are written as xave, xrms and xmax , respectively, and are in general different from each other. In practice, such values are obtained by introducing a discrete distribution function f i defined as f i = ni /N , where ni is the occurrence of the individual data and N is the total number of the data. The followings are useful definitions associated with this discrete system.
ni 1 = N N
f i =
i
i
xave =
xi f i =
i
(x2)ave =
x2i f i =
i
1 N
1 N
ni = 1,
(1.1)
xi ni ,
(1.2)
i
i
x2i ni
(x2 )ave,
xrms =
(1.3)
i
xmax is obtained when f i is maximum.
(1.4)
Let us examine further the above definitions using an explicit example: 10 students taking Final Exam on Statistical Physics. The Final Marks are 20, 20, 35, 60, 60, 60, 60, 80, 80 and 90. From these marks we have discrete data in the form of xi = 20, 20, 35, 60, 60, 60, 60, 80, 80, 90, and N = 10 for which
× × f i =
i
xave =
xrms =
1 N
1 N
1 N
ni =
i
xi ni =
i
x2i ni =
i
1 1 n1 + n2 + n3 + n4 + n5 = 2 + 1 + 4 + 2 + 1 = 1, 10 10 1 20 10
1 202 10
2 + 35
2 + 352
× 1 + 60 × 4 + 80 × 2 + 90 × 1 2
2
2
= 56.5,
× 1 + 60 × 4 + 80 × 2 + 90 × 1
= 61.1,
xmax = 60 for f 1 = n1 /N = 0.2, f 2 = n2 /N = 0.1, f 3 = n3/N = 0.4, f 4 = n4 /N = 0.2 and f 5 = n5 /N = 0.1. As statistical physics deals with systems of very many particles, the above formulations no longer hold. In these systems, a continuous distribution function f (x) is used to describe the dynamics of the systems. Hence, (1.1) through (1.4) are replaced by
f (x) dx = 1,
(1.5)
1.3. Description of assembly and phasa space
xave = (x2 )ave =
3
x f (x) dx,
x2 f (x) dx
xrms =
(1.6)
(x2 )ave,
(1.7)
xmax occurs when the first derivative of f (x) is equal to zero.
(1.8)
In addition, a useful quantity, the standard deviation σ, is introduced as a measure of the spread of data about the average value (i.e., the scatter of the measurements). The standard deviation of the distribution is defined as σ=
(x2 )ave
−x
2 , ave
(1.9)
≥
where σ is, by definition, positive since xrms xave. For a normal or Gaussian distribution, two-thirds of the values constituting the data are expected to fall within the range xave σ. There will always be random and systematic errors in the measurements. Mean values thus include some error, which could also be indicated by the standard error of the mean σm , defined as σ σm = (1.10) N
±
√
It is therefore possible to write the results of the measurements as xave
1.3
±σ
m.
Description of assembly and phasa space
Here an assembly is defined as a group of individual components forming a physical body. For example, in the case of a gas in a closed container such components are gas atoms or gas molecules while the assembly is the gas itself. The state of an assembly is determined by position and momentum of each of individual components in a six-dimensional, Γ space (known as phasa space). In a Cartesian coordinate system, the phasa space consists of Euclidean space (x, y, z ) and momentum space ( px , py , pz ). Thus, any quantity can be written in terms of position and momentum coordinates. For example, element volume dΓ of a single-particle system in phasa space is written as dΓ = dx dy dz
× d p
x
d py d pz ,
(1.11)
where dxdydz is element volume in Euclidean space and d px d py d pz is element volume in momentum space. The (non-relativistic) kinetic energy of this system is written as
1 = p2x + p2y + p2z , 2m
(1.12)
4
1. Introduction
where m is the mass of the system, and px , py and pz are the components of linear momentum of the system in x, y and z directions, respectively. The simple method above could be extended for a system consisting of many particles. The corresponding element volume dΓN and total kinetic energy E for N particles are respectively written as dΓN = dx1 dy1 dz 1 d px1 d py1 d pz1 N
=
× ·· ·· · · dx
N dyN dz N d pxN d pyN d pzN
(1.13)
dxi dyi dz i d pxi d pyi d pzi
i=1
and E = 1 + 2 + 3 +
N
=
i=1
······
N
1 p2xi + p2yi + p2zi , 2m
(1.14)
where i in both (1.13) and (1.14) refers to individual particles.
1.4
The scope of statistical physics
While modern statistical physics in general covers a broad spectrum of materials, the main theme considered in this course involves classical and quantum statistics, with their simple applications being introduced. There are fundamental differences in the basic assumptions made regarding the distinct behaviour of classical and quantum assemblies. For a classical assembly, each of individual components is considered to be completely distinguishable. For a quantum assembly, however, the particles constituting the assembly is considered to be indistinguishable. Another difference between the classical and quantum statistics is that energy and momentum of classical systems spread over a continuum spectrum while in the quantum systems particles are distributed over discrete energy levels. The quantum systems studied here are divided into two groups of particles: bosons and fermions. The differences between these systems will be detailed in the next chapters. The main reference to which we refer is Pointon (1978) as this book is much simpler for undergraduate students without losing basic ideas behind the scene. The students are also encouraged to have a read the topics considered in other sources from a fundamental level (Tipler, 1999) to an advance level (Reif , 1985; Huang, 1987), as well as from related materials in a standard book of modern or quantum physics (Beiser, 1988; Gasiorowicz, 1996).
Chapter 2 Maxwell-Boltzmann Statistics It is desired to find a continuous distribution function that characterises classical systems. These systems are allowed to have arbitrary values of energy and momentum, for which they obey Maxwell-Boltzmann (MB) statistics. We will derive this statistical distribution using a simple way, independent of the details of molecular interactions. We first construct the classical velocity distribution function for each direction, with which we derive the MB speed distribution function in 2.1. In order to obtain a full description of a classical assembly, we then examine the momentum and energy distribution functions in 2.2. This is followed by the applications of the MB statistical distribution in 2.3. The fundamental concept of partition function is also introduced in 2.4.
§
§
§
2.1
§
The speed distribution function
Here we are particularly interested in finding the speed distribution function of a classical system containing very many particles. Such a system can be an ideal gas in a closed volume. Once this function is obtained, other distribution functions (i.e., energy and momentum distribution functions) can be directly derived. We begin with describing the dynamics of molecules of such a gas using a Cartesian coordinate system for simplicity. The density ρN of the gas molecules per unit volume in velocity space is determined by the product of the total number N of the molecules and velocity distribution function for all motional directions, ρN = N f (vx ) f (vy ) f (vz ),
(2.1)
where f (vx ), f (vy ) and f (vz ) are velocity distribution functions, and vx , vy and vz are the speeds of the molecules in x, y and z directions, respectively. 5
6
2. Maxwell-Boltzmann Statistics
The total derivative dρN of (2.1) with respect to vx , vy and vz is dρN =
∂ρ N ∂ρ N ∂ρ N dvx + dvy + dvz , ∂v x ∂v y ∂v z
(2.2)
which can be, due to (2.1), rewritten as
dρN = N f (vx ) f (vy ) f (vz ) dvx + f (vx ) f (vy ) f (vz ) dvy + f (vx ) f (vy ) f (vz ) dvz .
(2.3)
It is expected to have separated parameters for each of the terms on the RHS of (2.3). Thus we divide (2.3) by ρN , resulting in dρN f (vx ) f (vy ) f (vz ) = dvx + dvy + dvz . ρN f (vx ) f (vy ) f (vz ) As the density is constant, i.e., dρN = 0, the above equation becomes f (vx ) f (vy ) f (vz ) 0= dvx + dvy + dvz . f (vx ) f (vy ) f (vz )
(2.4)
We assume that the gas molecules are separated by distances that are sufficiently large compared with the molecule diameters and that there is no external forces acting on the molecules except when they collide. In the absence of such forces, there is no preferred direction for the velocity of each molecule and the speed is, of course, constant. We can then write v.v = v 2 = vx2 + vy2 + vz2 = constant, which can be derived with respect to vx , vy and vz to obtain vx dvx + vy dvy + vz dvz = 0,
(2.5)
as a restrictive condition for (2.4). Thus, we have so far a set of two-differential equations, (2.4) and (2.5). The solution of these equations can be obtained by introducing a constant parameter λ, to be determined later, by which we multiply (2.5) to obtain λ vx dvx + λ vy dvy + λ vz dvz = 0.
(2.6)
We then add (2.6) to (2.4) to obtain
f (vx ) + λ vx dvx + f (vx )
f (vy ) + λ vy dvy + f (vy )
f (vz ) + λ vz dvz = 0, f (vz )
where dvx , dvy and dvz are all not zero for non-trivial solutions. Instead, a system of
2.1. The speed distribution function
7
homogeneous differential equations below are required, f (vx ) + λ vx = 0, f (vx )
f (vy ) + λ vy = 0, f (vy )
f (vz ) + λ vz = 0, f (vz )
for which f (vx ), f (vy ) and f (vz ) will be determined. For the x-component, the above expression means df (vx ) = dvx
−λ v
x f (vx )
df (vx ) = f (vx )
or
−λ v
x
dvx ,
which can be easily solved. Integrating both sides and applying simple calculus yield nf (vx ) =
− 12 λ v
2 x
+ n C = nCe−
1 2
λvx2
.
We thus have f (vx ) = Ce−
1 2
λvx2
,
(2.7)
where C is an integration constant, and will be determined below. The next step is thus to determine the exact value of C by normalising (2.7) such that
∞
∞
f (vx ) dvx = 1
C
or
−∞
As e−
1 2
λvx2
e−
1 2
λvx2
dvx = 1.
(2.8)
−∞
is an even function, (2.8) can then be written as
∞
2
× C
e−
1 2
λvx2
dvx = 1
0
(see Appendix 6 of Pointon, 1978 for the details of the properties of Γ function integrals). We here refer to ∞ 1 n+1 xn e−ax dx = (n+1)/2 Γ (2.9) 2a 2 0
and use Γ(1/2) =
2
√ π to calculate C as
2 C λ
1/2
√ π = 1
Hence, (2.7) becomes f (vx ) =
C =
or
λ − e 2π
1 2
λvx2
.
λ . 2π
(2.10)
8
2. Maxwell-Boltzmann Statistics
Using (2.10), we can calculate the mean square velocity as
∞
(vx2 )ave
=
−∞
∞
vx2
f (vx ) dvx =
−∞
vx2
λ − e 2π
1 2
λvx2
dvx =
1 , λ
(2.11)
where we have again used (2.9) and a simple relation, Γ(n + 1) = n Γ(n). According to the kinetic theory of a perfectly ideal gas, for every degree of freedom in each direction translational kinetic energy is equal to the thermal energy of kT/2 such that 1 1 m (vx2 )ave = kT , 2 2
1 1 m (vy2 )ave = kT , 2 2
1 1 m (vz2 )ave = kT , 2 2
where m is the total mass of molecules, k is the Boltzmann’s constant (1.38 and T is the equilibrium temperature of the molecules. We thus have (vx2 )ave =
(2.12)
× 10
kT m
−23
J K−1 )
(2.13)
for the x-component of the velocity. Combining both (2.11) and (2.13) yields λ = m/kT and similar results are obtained for the y and z directions. We can therefore rewrite the x-component of the velocity distribution function in (2.10) as f (vx ) =
m − e 2πkT
1 2
mvx2 /kT
.
(2.14)
In the same manner, the velocity distribution function for the other degrees of freedom can be obtained, f (vy ) =
m − e 2πkT
1 2
mvy2 /kT
and
f (vz ) =
m − e 2πkT
1 2
mvz2 /kT
.
(2.15)
Having derived the velocity distribution function for each direction, we are now ready to derive the corresponding speed distribution function by defining the molecule density ρN , previously given in (2.1), as dN vx vy vz ρN = , (2.16) dvx dvy dvz where dN vx vy vz is the number of gas molecules having velocity components in the range vx to vx + dvx in the x-direction, vy to vy + dvy in the y-direction, vz to vz + dvz in the z -direction, and dvxdvy dvz is element volume in velocity space in a Cartesian coordinate system. Substituting (2.16) into (2.1) gives dN vx vy vz = N f (vx ) f (vy ) f (vz ) dvx dvy dvz
(2.17)
2.2. The momentum and energy distribution functions
or dN vx vy vz
3/2
m = N 2πkT
e−
1 2
m(vx2 +vy2 +vz2 )/kT
9
dvx dvy dvz .
(2.18)
Equation (2.18) can be written in a spherical coordinate system as follows dN vr vθ vφ
m = N 2πkT
3/2
e−
1 2
mv 2 /kT
v 2 sin θ dθ dφ dv.
(2.19)
Integrating (2.19) over the whole space and using (2.17) written in terms of v yield
m dN v = N f (v) dv = N 2πkT or
3/2 −
e
1
2
mv /kT 2
π
2
v dv
2π
sin θ dθ
0
m dN v = N f (v) dv = 4π N 2πkT
3/2
e−
1 2
mv2 /kT
dφ,
0
v 2 dv,
(2.20)
where dN v is the number of gas molecules having speed in the range v to v + dv. It is clear from (2.20) that 3/2 m f (v) = 4π e− mv /kT v 2 , (2.21) 2πkT
1
2
2
which is known as the MB speed distribution function. This function is more convenient to use than the velocity component distribution functions, defined in (2.14) and (2.15), in that it is taken independent of direction.
2.2
The momentum and energy distribution functions
As previously mentioned, the speed distribution function defined in (2.21) can be used to derive other forms of distribution function. For convenience, we will first derive momentum distribution function, then energy distribution function. But first, we rewrite (2.21) as f (v) dv = 4π
m 2πkT
3/2
e−
1 2
mv2 /kT
v 2 dv,
(2.22)
where f (v) dv is the MB probability speed distribution function. Integral of this function over the whole space is, by definition, unity since such integral describes the total probability to find a molecule with speed in the range v to v + dv in the velocity space. The Euclidean volume V of each molecule is absent in the present discussion. It follows that the properties of the molecules of ideal gases are independent of the geometry of the molecules. Using a simple relation from classical physics p = mv and its associated differential form d p = m dv, we can write an expression for the MB probability momentum distribution
10
2. Maxwell-Boltzmann Statistics
function f ( p) d p as f ( p) d p = 4π
1 2πmkT
3/2 2
e− p
/2mkT 2
p d p,
(2.23)
which characterises classical assemblies having momentum in the range p to p + d p. As is the case of the probability speed distribution function given in (2.22), the integral of (2.23) over the whole space equals to one. It is understood that f ( p) = 4π
1 2πmkT
3/2 2
e− p
/2mkT 2
p ,
(2.24)
where f ( p) is known as the MB momentum distribution function. Again, using a simple formula from classical physics but this time relating momentum to kinetic energy p = 2m and its associated differential form d p = m/2 d, we can write an expression for the MB probability energy distribution function f () d as
√
f () d = 2π
1 πkT
3/2
e−/kT 1/2 d,
(2.25)
where f () d describes classical assemblies with energy in the range to + d. Again, as is the case of the probability speed and momentum distribution functions given in (2.22) and (2.23), respectively, the integral of (2.25) over the whole space equals to one. It then follows that 3/2 1 f () = 2π e−/kT 1/2 , (2.26) πkT
where f () is known as the MB energy distribution function.
2.3 2.3.1
Applications of Maxwell-Boltzmann statistics The mean, rms and most probable velocities
§
As mentioned in 1.2, there are three dynamic quantities that can be drawn from a given distribution function. These quantities are the mean, rms and most probable values. For example, we will use the MB speed distribution function given in (2.21) to derive the mean, rms and most probable velocities of an ideal gas in a closed container having total volume V and N molecules. It is expected that maximum information can be obtained by calculating such values. Here we calculate the mean velocity vave, which is by definition written as
∞
vave =
0
v f (v) dv = 4π
m 2πkT
3/2
∞
0
e−
1 2
mv2 /kT
v 3 dv.
(2.27)
2.3. Applications of Maxwell-Boltzmann statistics
11
Using (2.9) and a simple relation, Γ(n+1) = n!, the above equation yields
8kT , πm
vave =
(2.28)
where m and T are the mass and temperature of such a gas, respectively. The rms velocity can be derived in a similar manner to that used for the mean velocity. Thus, we have
∞
2
(v )ave =
2
v f (v) dv = 4π
0
m 2πkT
3/2
∞
e−
1 2
mv2 /kT
v 4 dv.
(2.29)
0
Again, applying (2.9) and Γ(n + 1) = n Γ(n) to (2.29) yields 3kT (v 2 )ave = , m
vrms =
(v 2)ave =
3kT . m
(2.30)
The above result for the rms velocity is in line with the sum of the equations in (2.12). To make this clear, we here rewrite it as 1 1 1 1 3 m (v 2 )ave = m (vx2 )ave + m (vy2)ave + m (vz2)ave = kT . 2 2 2 2 2
(2.31)
The most probable velocity occurs when the the speed distribution function f (v) defined in (2.21) is maximum, i.e., the first derivative of this function is zero. Thus, we have df (v) = 4π dv
m 2πkT
3/2
for which we obtain
2v
−
mv 3 kT
e−
1 2
mv2 /kT
= 0,
(2.32)
2kT . (2.33) m The results for the mean, rms and most probable velocities confirm that they are in general different. The rms velocity is the one, which is directly related to the mean energy of a system having expression for energy in a special form that will be discussed below. vmax =
2.3.2
Equipartition principle of energy
In this section, we discuss the equipartition principle for systems having energy in the form of quadratic terms. Such systems can be an ideal gas with the same probability of translational movements in all directions, in which molecular interactions are negligible, or a harmonic oscillator where potential interactions between individual components exist. For convenience, we first examine the application of the principle to an ideal gas, and later extend the result to a harmonic oscillator.
12
2. Maxwell-Boltzmann Statistics
The mean energy of a perfectly ideal gas having energy between and + d can be, with the help of (2.26), calculated as
∞
ave =
f () d = 2π
0
1 πkT
3/2
∞
e−/kT 3/2 d.
(2.34)
0
Using a functional approach of Γ function below,
∞
xn e−ax dx =
0
we obtain ave = 2π
1
1 πkT
Γ(n + 1),
an+1
3/2
kT
5/2
√
3 3 π = kT . 4 2
(2.35)
(2.36)
When the gas is in equilibrium, the total mean energy will be equally distributed among the translational kinetic energies associated with momentum in the x, y and z directions. This is well described by (2.12) and (2.31). Thus, we have 1 ave = m (v 2 )ave = 3 2
× 12 m (v )
2 x ave
3 = kT , 2
(2.37)
and hence
1 1 m (vx2 )ave = kT . (2.38) 2 2 The result shows that each component of momentum that appears as a squared term in the expression for the energy will contribute 12 kT per molecule. When an equilibrium system is in rotational and vibrational motions as well, the total energy of the system is the sum of all possible forms of the energy associated with types of motions. Potential interaction between the individual particles may lead to vibrational motion, where the corresponding potential energy is in the form of a quadratic term. For example, the energy of a one-dimensional harmonic oscillator can be written as p2x 1 = + µx2 , 2m 2
(2.39)
where µ denotes the restoring force per unit displacement. Now we introduce a symbol which is actually the same as ( )ave used before. Thus, we have
x
=
p2x 2m
+
1 2 µx 2
1 1 = kT + kT = kT . 2 2
,
(2.40)
Notice that the LHS of (2.40) describes the mean energy corresponding to the x-component
2.3. Applications of Maxwell-Boltzmann statistics
13
of momentum of 1-D harmonic oscillator. For 3-D harmonic oscillator, we have
xyz
2.3.3
p2y p2z 1 2 1 2 = + + + µx + µy + 2m 2m 2 2 1 1 1 1 1 1 = kT + kT + kT + kT + kT + kT = 3kT . 2 2 2 2 2 2 p2x 2m
1 2 µz 2
(2.41)
Specific heats
In this section, we derive a useful quantity, namely specific heat , by defining the total energy of a classical system of N particles as E = N . The specific heat discussed here is that of perfectly ideal gases. For the ideal gas with total N molecules and the mean energy , as given in (2.36), we have
3 3 3 E = N = N kT = nN o kT = nRT, 2 2 2
(2.42)
where n denotes the number of moles, N o = 6.02 1023 mole−1 is the Avogrado’s number and R = 8.314 Jmole−1 K−1 is the universal gas constant. We generalise this result for a given system with total f degrees of freedom at temperature T to calculate the total energy for the system as follows, 1 1 E = f N kT = f nRT. (2.43) 2 2
×
×
×
It is useful to introduce the molar specific heat at constant volume V , defined as the heat capacity per unit mole. Unlike the heat capacity, the molar specific heat depends only on the nature of the system under consideration. The specific heat at constant volume cV is thus written as 1 ∂E 1 1 cV = = f N o k = f R. (2.44) n ∂T V 2 2
×
×
For monoatomic molecules, cV = 32 R, as directly measured from classical thermodynamics. For diatomic molecules, there are two additional degrees of freedom regarding rotational motions. Therefore, the diatomic gases have a total of 5 degrees of freedom, for which their specific heat at constant volume cV = 52 R. Another quantity, the molar specific heat at constant pressure P is defined as
1 ∂E cP = n ∂T
= cV + R,
(2.45)
P
by which cP is always greater than cV . The above relation is of practical importance as calculations by statistical physics are usually performed for a fixed volume, while experimental measurements are carried out under conditions of constant pressure. Therefore, cV is a theoretically calculated value and
14
2. Maxwell-Boltzmann Statistics
cP is an experimentally measured parameter. The determination of these quantities provides information about internal energy, and hence molecular structure. When heat is added to a gas at constant volume, there will be no work done and so the amount of heat added is converted to increase internal energy of the gas. When the heat is added at constant pressure, work is done. Hence, less amount of heat is used to increase the internal energy, resulting in a smaller increase in temperature change, compared with that at constant volume.
2.4
The Boltzmann partition function
In this section, we introduce a special function that links statistical expression for microscopic systems to the corresponding thermodynamic functions. The detailed properties of this function, together with the discussion of the Helmholtz free energy, will be later discussed in Chapter 5. But for now, it is sufficient to state that the Boltzmann partition function Z describes how energy is distributed among systems in the classical assembly. This function is defined as ∞ Z = g() e−/kT d, (2.46)
0
where g() d is the number of states having energy between and + d, and e−/kT is known as the Boltzmann factor . Based on physical argument, g() d can be expressed as the product of the density of states B and element volume dΓ in phasa space, or simply g() d = B dΓ. Thus, (2.46) can be written as
×
∞
Z =
0
= BV
∞
Be
−/kT
dΓ =
2
− p
e
2
2
−∞
/2mkT
= BV 2πmkT
2
B e−( px + py + pz )/2mkT dx dy dz d px d py d pz
∞
2
p sin θ dθ dφ d p = 4πBV
2
e− p
/2mkT 2
p d p
(2.47)
0
3/2
.
The above partition function can be used to calculate pressure P , a measured quantity of thermodynamic functions. With the help of this function, the total pressure of a classical system containing N particles is written as P = N kT
N kT = . V
∂nZ ∂V
T
= NkT
d n V dV
(2.48)
The last expression for P is well-known as the equation of state of a perfectly ideal gas, P V = N kT = nRT , as frequently found in standard books of fundamental physics.
2.5. Exercise
2.5
15
Exercise
1. Diketahui fungsi distribusi untuk sistem kontinu sebagai berikut: f (x) =
4x − x e λ λ2 2
for x
≥0
(a) Tunjukkan bahwa total peluang adalah 1. (b) Hitunglah nilai x rata-rata, rms dan yang paling mungkin dari sistem tersebut. 2. Hitunglah kecepatan rms untuk atom gas Helium pada suhu 300 K. 3. Hitunglah pada suhu berapa molekul gas Hidrogen memiliki kecepatan yang sama. 4. Diketahui fungsi distribusi untuk sistem kontinu sebagai berikut: f (v) = Av 2 e−
1 2
mv2 /kT
,
A adalah konstanta
(a) Tentukan A untuk atom yang tiap mol-nya memiliki massa sebesar 36 gram pada suhu 0◦ Celcius. (b) Tentukan laju rata-rata partikel. (c) Tentukan waktu bebas rata-rata, bila diketahui lintasan bebas rata-rata partikel ˚ adalah 60 A. ˚ Pada suhu 27◦ Celcius, 5. Satu mol gas Neon bermassa 20,15 gram dan berjari-jari 1,25 A. tekanannya 1,013 104 Nm−2 .
×
(a) Tentukan jarak rata-rata antara 2 atom Neon. (b) Tentukan jumlah tumbukan rata-rata per detik yang dialami oleh satu atom Neon. ˚ bermassa 4 gram 6. Diketahui satu mol gas beratom tunggal berjari-jari atom 1,05 A pada volume 17,5 liter, tekanan 1,01 105 Nm−2 dan suhu 27◦Celcius.
×
(a) Tentukan jarak rata-rata antar atom. (b) Tentukan jarak bebas rata-rata. (c) Tentukan waktu bebas rata-rata (d) Jika terdapat lubang seluas 10−2 mm2 , tentukan jumlah atom persatuan waktu yang menerobos keluar dari lubang tersebut. 7. Gas ideal terperangkap dalam ruang tertutup berupa silinder homogen dengan luas penampang A dan tinggi L (searah terhadap sumbu z ). Tentukan energi rata-rata molekul gas ideal tersebut. Bagaimana bila energi potensialnya diperhitungkan?
16
2. Maxwell-Boltzmann Statistics
8. A flux of 1012 neutrons/m2 emerges each second from a port of nuclear reactor. If these neutrons have a Maxwell-Boltzmann energy distribution corresponding to temperature T = 300 K, calculate the density of neutrons in the beam (taken from Beiser, 1988, Ch.15, problem 7).
∞
9. By direct integration of 0 n() d with n() taken from equation 2.59, show that the total energy of the assembly is 32 RT where R = N k (taken from Pointon, 1978, Ch.2, problem 2). 10. Calculate the mean values of vx and vx2 (taken from Pointon, 1978, Ch.3, problem 2). 11. Show that the integration of equation 3.10 over all polar angles leads to distribution in equation 3.8 (taken from Pointon, 1978, Ch.3, problem 3). 12. Derive the variation of pressure with height in a column of a gas at temperature T (a) using the fact that the change in pressure over a height dh is -ρg dh, where ρ is the density of the gas (b) using the Boltzmann factor to give the concentration gradient of the molecules (taken from Pointon, 1978, Ch.3, problem 1). 13. By what factor must the absolute temperature of a gas be increased to double the rms speed of its molecules? (taken from Tipler, 1999, Ch.18, p.563, problem 35). 14. A mole of He molecules is in one container and a mole of CH4 molecules is in a second container. Both are in standard conditions. Which molecules have the greater mean free path? (taken from Tipler, 1999, Ch.18, p.563, problem 37). 15. The molecules of a monoatomic ideal gas are escaping by effusion through a small hole in a wall of an enclosure maintained at absolute temperature T. (a) By physical reasoning (without actual calculation) do you expect the mean kinetic energy o of a molecule in the effusing beam to be equal to, or greater than, or less than the mean kinetic energy i of a molecule within the enclosure?
(b) Calculate o for a molecule in the effusing beam (taken from Reif , 1985, Ch.7, p.287, problem 7.30).
Chapter 3 Bose-Einstein Statistics In this chapter, we discuss Bose-Einstein (BE) statistics. This statistics is used to describe the statistical behaviour of bosons - quantum systems having integer or zero spin. As the spin classification of particles determines the nature of the energy distribution over the available energy states, the bosons do not obey the Pauli exclusion principle. Thus, there is no restriction for how many bosons are allowed to occupy the same quantum state. This chapter is outlined as follows. The concept of weight of configuration, or the number of ways of distributing particles over the available states, is first introduced in 4.1. The number of particles occupying a certain state is derived in 3.2, followed by the discussion of the condition under which a gas could be considered as a classical gas, or as a Bose-Einstein gas in 3.3. As real examples, the applications of the BE statistics in boson systems, such as electromagnetic radiation in space and elastic waves in solid, are discussed. The classical Planck’s radiation formula is derived in 3.4 while the Debye theory for specific heats of solid is examined in 3.5.
§
§
§
§
3.1
§
Weight of configuration
There are two principal differences between classical and boson systems. In the MB statistics, the basic assumption made is that particles are distinguishable. Thus, the interchange of two classical particles leads to a new arrangement. Whilst, in the BE statistics the assembly consists of indistinguishable particles, hence the interchange of two identical particles gives no new configuration. Another difference between these systems arises from the nature of the available energy states. Rather than a continuous form of energy, bosons are distributed over discrete levels of energy. Before further discussing bosons in detail, it is necessary to calculate the number of ways of arranging particles within the available energy states. Let us consider two classical particles, labelled as a and b, occupying two energy states. Based on the MB statistics, there 17
18
3. Bose-Einstein Statistics
will be four possible arrangements for these particles,
| | | | | | | | | | | | a
b
b
arrangement 1
a
ab
arrangement 2
ab
arrangement 3
arrangement 4
However, the first two arrangements are, according to the BE statistics, completely identical. Thus, they are considered to be united, resulting in only three arrangements for the system.
| | | | | | | | | a
a
arrangement 1
aa
aa
arrangement 2
arrangement 3
How about if such particles are distributed over three energy states. You will find easily that 9 configurations are possible for classical arrangements while only 6 are found for boson arrangements. The details of these are left for students for exercise. This difference in the number of arrangements between classical and boson systems is associated with different statistical formulations for calculation of possible configurations within each assembly. For classical statistics, the number of possible configurations W (also known as the weight of configuration ) is given by giN i W = N ! , (3.1) N ! i i
where gi and N i are the number of states and the number of particles in the ith sheet , respectively, and N is the total number of particles. For boson statistics, this quantity is written as (N i + gi 1) ! W = . (3.2) N ! (g 1) ! i i i
− −
We can then calculate the number of ways of arranging particles using the above formulations.
3.2
Population of particles
Having discussed the weights of configurations for both the MB and BE statistics, it is now necessary to find an expression for the number of particles distributed over a particular sheet. We begin with the classical case by writing (3.1) and taking a natural logarithm of it,
n W = n N ! + n
i
= n N ! +
i
= n N ! +
giN i N i !
giN i n N i ! N i n gi
i
(3.3)
− n N ! i
3.2. Population of particles
19
≈ N n N − N to the above equation.
It is useful to apply Stirling’s approximation : n N ! Equation (3.3) becomes n W = N n N
− N +
= N n N +
i
i
N i n gi
i
where we have used
− N n N + N
N i n gi
i
− N n N i
i
i
,
N i = N .
The most probable configuration corresponds to the condition under which n W reaches maximum. This is simply written as d n W = 0, or 0 = d(N n N ) +
d N i n gi
i
dN = n N dN + N + N
− n gi
=
− N n N i
n gi
i
i
× dN − n N × dN − N × i
i
i
i
dN i N i
(3.4)
n N i dN i ,
i
where dN i = d N i = dN = 0 because N is constant (see also below). It is important to mention here that the above condition goes together with the imposed limitations on the values of total particle N and total energy E , respectively, for which the followings hold, N i = N = constant
i
N i i = E = constant
i
dN i = d
i
N i = dN = 0
i
d(N i i ) = d
i
N i i = dE = 0
(3.5)
i
Thus we can write d n W + α dN + β dE = 0,
(3.6)
where α and β are Lagrange multipliers that are to be later determined. Now we have all ingredients needed to develop Lagrangian method using (3.4), (3.5) and (3.6). Equation (3.6) becomes n gi n N i + α + βi dN i = 0, (3.7)
i
which requires
n gi
−
− n N + α + β i
i
= 0.
After several simple steps, we obtain the particle population N i in the ith sheet, N i =
gi e−(α+βi )
.
(3.8)
20
3. Bose-Einstein Statistics
§
From the discussion of the Boltzmann partition function in 2.4 we have Z in a continuous form,
∞
Z =
g() e−/kT d.
0
Rewriting this Boltzmann partition function into a discrete form gives Z =
gi e−i /kT .
(3.9)
i
Dividing both sides of (3.9) by N results in
gi e−i /kT . N i
Z = N
(3.10)
In order to derive the number of particles occupying a particular sheet, we here define N Z
≡e
α
(3.11)
and substitute it into (3.10) to obtain N i =
gi e−(α−i /kT )
.
(3.12)
It is clear from (3.8) and (3.12) that β =
1 . − kT
(3.13)
The same result can also be obtained from thermodynamics. It is important to keep it in mind that either (3.8) or (3.12) is frequently referred to as classical distribution (also known as classical population number ), which describes population of particles within each sheet in the classical assembly. In a similar manner to the above steps, we can derive the number of bosons, or boson population number, occupying available quantum states by first taking a natural logarithm of (3.2),
n W = n
i
=
N i + gi
i
− −
(N i + gi 1) ! N i ! (gi 1) !
−
1 n N i + gi
− − 1
(3.14) N i n N i
− (g − 1) n (g − 1), i
i
where we have applied Stirling approximation to the above derivation. The most probable configuration for boson distribution is obtained when (3.14) reaches
3.3. The Bose-Einstein gas
21
maximum, or d n W = 0. Again, using (3.5) and (3.6) we can write 0 = d n W + α dN + β dE =
− 1) − n N + α + β
n (N i + gi
i
i
i
dN i .
(3.15)
The solution of (3.15) requires
− 1) − n N + α + β (3.16) = n (N + g ) − n N + α + β , where we have again used Stirling approximation, i.e., N 1 to simplify calculation. After 0 = n (N i + gi i
i
i
i
i
i
i
simple algebra, we obtain the population number of bosons in the form of N i =
gi e−(α−i /kT )
− 1,
(3.17)
−
where we have used β = 1/kT . Equation (3.17) is frequently referred to as quantum distribution for bosons (also known as boson population number), which describes population of bosons within each sheet in the boson system. It can then be shown easily that (3.17) approaches to (3.12) when gi N i . It follows that the number of states is much larger than the population of particles. Note that α has not been yet determined, and that it is the purpose of the next section to derive the actual form of α.
3.3
The Bose-Einstein gas
In this section, we examine conditions under which a gas may obey either the MB or BE statistics. In doing so, we need to assume that each available energy state in the quantum assembly occupies a definite volume in phasa space. From the discussion of the Boltzmann partition function in 2.4, it is defined that g() d = B dΓ, where g() d denotes the number of available energy states having energy in the range to + d and B is a unique constant associated with the definite volume, which will be determined below.
§
For a single allowed energy state (i.e., g = 1) occupying element volume dΓ in phasa space, we can write 1 = B dΓ = B dx dy dz d px d py d pz = B dx d px dy d py dz d pz . Assume that the changes in position and momentum in each direction are finite, the above
22
3. Bose-Einstein Statistics
expression can then be written as 1 = B ∆x ∆ px ∆y ∆ py ∆z ∆ pz
h
h
h
= B h3 ,
for which B = 1/h3 . Here we have taken ∆x ∆ px = h from the Heisenberg’s uncertainty principle. Similar expressions for the other two directions also hold. Thus, equation (2.47) can be written as V 3/2 (3.18) Z = 3 2πmkT . h Hence, equation (3.11) becomes
N N h3 = Z V 2πmkT
3/2
= eα = A,
(3.19)
where A is constant. The necessary condition under which the gas can be considered as a classical gas is closely related to the value of A. In most cases, this value is sufficiently small (i.e., A−1 1) such that A−1 e/kT 1 A−1 e/kT . If this condition is satisfied then boson population (3.17) approaches to the classical distribution (3.12).
− ≈
Consider here, for example, Helium gas with molar mass of 4 gram per mole−1 at standard atmospheric pressure of 1 atm. The value of A for the He gas is approximately 3 10−6 at T = 300 K, drops to 0.15 at T = 4 K. Thus, the He gas at room temperature obeys classical distribution, and even at low temperature of T = 4 K, the He gas will behave, to some degree, as a classical gas.
×
3.4
Black-body radiation
Here we consider electromagnetic radiation within an enclosure of volume V with enclosure walls being maintained at temperature T . The radiation can be considered as a collection of photons, and is assumed to obey the BE statistics. When the photons set in motion within the container, some of them are continuously absorbed and reemitted by the walls. In this context, the number of photons inside the enclosure is not conserved. Because (3.6) must hold, it follows that α must be zero, leading to A = 1. Thus, (3.17) can be rewritten in the form of g() d N () d = /kT , (3.20) e 1
−
where we have invoked e−α = A−1 = 1. N () d and g() d denote the number of photons and the number of available states with energy in the range to + d, respectively.
3.4. Black-body radiation
23
Note that the energy of a photon depends on the frequency ν or equivalently on the wavelength λ of the radiation, i.e., = hν = hc/λ, and that it is radiated in space over the whole values of either frequency or wavelength. Thus, equation (3.20) can be written in two forms, g(ν ) dν ehν/kT 1 g(λ) dλ N (λ) dλ = hc/λkT , 1 e N (ν ) dν =
− −
(3.21)
where N (ν ) dν and g(ν ) dν denote the number of photons and the number of available states with frequency in the range ν to ν + dν , respectively. Similar definitions hold for N (λ) dλ and g(λ) dλ in the second equation of (3.21). The energy distribution of the photons can be obtained by rewriting the number of available energy states g as g=2
dΓ dx dy dz d px d py d pz V 4πp 2 d p = 2 = 2 , h3 h3 h3
(3.22)
where the multiplier “ 2 ” in front of element volume dΓ arises from the fact that there are two independent directions of spin orientation, that is parallel and anti-parallel to the photon linear momentum. Another physical argument based on which the multiplier 2 is derived is that according to the particle-wave duality, a photon with a definite spin state corresponds to a propagating electromagnetic wave having two independent directions of polarisation, with each direction being perpendicular to the wave propagation. For simplicity, it is assumed that photons are linearly polarised. The number of available states having wavelength in the range λ to λ + dλ can be derived with the help of the De Broglie’s hypothesis , p = h/λ. Substituting this relation into (3.22) yields 8π g(λ) dλ = V 4 dλ. (3.23) λ Here we define the density of states having wavelength between λ and λ + dλ as the number of available states g(λ) dλ per unit volume V such that (3.23) becomes
G
dλ. G (λ) dλ = 8π λ 4
(3.24)
The photons are distributed over the available states, in which the associated energy density in the range of wavelength between λ and λ + dλ is defined as the product of the photon energy and the density of photons (i.e., the number of photons per unit volume, N/V )
E
N
24
3. Bose-Einstein Statistics
as follows,
E (λ) dλ = × N hc G (λ) dλ = × λ e −1 hc/λkT
=
8πhc dλ λ5 ehc/λkT
(3.25)
− 1.
The above expression is known as the Planck’s radiation formula for the spectral distribution of radiant energy within a thermal equilibrium enclosure at temperature T . Two possible, interesting approximations can be directly derived from the last expression of (3.25). At sufficiently long wavelengths, or when hc/λ kT , we have
ehc/λkT
hc + · · · · · · −1 − 1 = 1 + λkT hc ≈ 1 + λkT
ehc/λkT for which (3.25) becomes
dλ, E (λ) dλ = 8πkT λ
(3.26)
4
which is known as the classical Rayleigh-Jeans’ formula for the spectral distribution of radiant energy within an enclosure. On the other extreme, at sufficiently short wavelengths, or when hc/λ kT , we have ehc/λkT 1 ehc/λkT
− ≈
for which (3.25) becomes
dλ E (λ) dλ = 8πhc λ e 5
hc/λkT
,
(3.27)
which is known as the classical Wien’s formula for the distribution of radiant energy within an enclosure. Thus, it has been shown that the Planck’s formula (3.25) for black-body radiation is valid over the whole spectrum of wavelengths, or accordingly frequencies. The formula covers the empirically observed findings of the spectral distribution of electromagnetic radiation that are previously unexplained by classical electrodynamic and thermodynamic theories. Let us now assume that there is a small hole in the enclosure wall, giving a way for the photons to radiate out. The number of photons emitted per unit time per unit area of the hole (i.e., the flux of photons crossing the hole) is written as
N
rad (λ) dλ
=
1 c 4
× N (λ) dλ,
because the photon is always in move with the velocity of light c in free space.
(3.28)
3.4. Black-body radiation
25
By analogy, the energy radiated out per unit time per unit area of the hole is then given by
1 c (λ) dλ. 4 Substituting the Planck’s formula (3.25) into (3.29) results in
E
rad (λ) dλ
× E
=
2πhc 2 dλ (λ) dλ = rad λ5 ehc/λkT
E
(3.29)
− 1.
(3.30)
Integrating both sides of (3.30) gives
E ∞
∞
rad (λ) dλ
=
0
2πhc 2 dλ λ5 ehc/λkT
0
− 1,
and arranging the integral leads to E rad
2πk 4 T 4 = h3 c2
∞
0
(hc/λkT )3 hc d , ehc/λkT 1 λkT
−
where E rad denotes the total energy radiated out of the hole per unit time per unit area. Now let us introduce x = hc/λkT as a variable of integration such that E rad
∞
2πk 4 T 4 = h3 c2
0
x3 ex
− 1 dx.
(3.31)
The integral can be evaluated with the help of some steps (see Gasiorowicz, 1996, p.5),
∞
0
x3 ex
− 1 dx =
∞
∞
3
x dx e
0
−x
∞
−nx
e
n=0
=
n=0
1 (n + 1)4
∞
∞
0
3
−y
y e
dy = 6
1
1 π4 = . n4 15
Equation (3.31) then becomes E rad =
2π5 k4 15h3 c2
T 4 = σ T 4 ,
(3.32)
where σ is called the Stefan’s constant (5.7 10−8 Watt m−2 K−4 ). The above equation is known as the Stefan-Boltzmann law for the energy density of electromagnetic radiation at temperature T . It is found empirically that an additionally constant e, the emissivity of a substance should be added to the RHS of (3.32). The magnitude of this quantity depends on the characteristics (or surface) of the object, which is valued between 0 and 1. Objects that are grouped into perfect heat absorbers (or heat radiators) have unit emissivity. Conversely, imperfect heat absorbers (or heat radiators) correspond to the value of emissivity between 0 and 1. Black holes in the universe are considered as perfect black-body.
×
26
3.5
3. Bose-Einstein Statistics
The specific heat of solid
§
As previously discussed in 3.4, electromagnetic radiation is quantised as photons. In this section, the energy of elastic waves propagated in solids is quantised as phonons, which obey the BE statistics. The energy of a phonon also depends upon frequency. For convenience, we use the phonon frequency ν as a dynamic variable. Thus, the phonon population having frequency between ν and ν + dν is written as N (ν ) dν =
g(ν ) dν . ehν/kT 1
(3.33)
−
Here we use Debye approximation , g(ν ) dν = Cν 2 dν for ν ν m , where C is a constant and ν m is the maximum frequency corresponding to three modes of vibration, that is one longitudinal polarisation and two transverse relative to the direction of propagation of the waves. It should be noted that the former polarisation is associated with a compressional wave and the latter corresponds to shear waves. Hence, for a solid having N atoms, there exists a total of 3N allowed states. It follows that
≤
ν m
3N =
ν m
g(ν ) dν = C
0
ν 2 dν,
0
3 for which C = 9N/ν m . Thus, equation (3.33) becomes
9N ν 2 dν N (ν ) dν = 3 hν/kT ν m e 1
for ν
−
≤ ν .
(3.34)
m
The total energy of phonons in solid is determined by the energy of a phonon = hν and the number of phonons N (ν ) dν with frequency in the range ν to ν + dν ,
ν m
E =
0
9N h N (ν ) dν = 3 ν m
ν m
0
ν 3 dν ehν/kT
− 1.
(3.35)
In the same manner as the definition in (2.44), the specific heat of solid can be derived with the help of (3.35),
1 ∂E cV = n ∂T
V
9N o h2 1 = 3 ν m kT 2
ν m
0
ν 4 ehν/kT dν ehν/kT
−1
2.
(3.36)
We introduce a new variable, x = hν/kT and a new parameter, namely Debye temperature Θ = hν m /k to replace (3.36) by cV = 9N o k
− T Θ
3
Θ/T
0
x4 ex dx ex
1
2.
(3.37)
3.6. Exercise
27
There will be two regions of temperature for which (3.37) is evaluated. For a region of high temperatures, it is understood that Θ/T 1 and ex 1 + x 1. In this condition, equation (3.37) becomes
cV = 9N o k
≈
T Θ
3
Θ/T
≈
x2 dx = 3R,
(3.38)
0
where R replaces N o k. The above result is the same as that derived from the classical equipartition of energy. At low temperatures, however, Θ/T Thus, equation (3.37) becomes
1 hold and the upper limit approaches to infinity.
− − T Θ
cV = 9R
3
∞
x4 ex dx ex
0
1
2.
(3.39)
The integral can be evaluated with the help of some steps below, ∞
0
x4 ex dx ex
1
2
∞
1 4π 4 = . n4 15
= 24
1
Substituting the above expression into (3.39) results in 3
T Θ
12 cV = π4 R 5
.
(3.40)
Note that the above results for high and low temperatures derived from (3.37) are in good agreement with experimental measurements of the specific heat of solids. The equation is therefore usually referred to as the Debye theory for the specific heat of solid, and is applicable to most metals.
3.6
Exercise
1. Jelaskan dengan dua cara pendekatan dalam hal apa Distribusi Bose-Einstein (3.17) menuju ke Distribusi Maxwell-Boltzmann (3.12). 2. Persamaan radiasi elektromagnetik Planck untuk foton dalam ruang tertutup pada temperatur T dinyatakan oleh
E (λ) dλ = 8πhc λ e 5
dλ
hc/λkT
−1
(a) Carilah bentuk persamaan radiasi Planck dalam variabel frekuensi ν .
28
3. Bose-Einstein Statistics
(b) Jika λm menyatakan besarnya panjang gelombang radiasi pada kerapatan energi maksimum, tunjukkan dari persamaan radiasi Planck bahwa λm T = 2, 9 10−3 mK.
×
(c) Kerapatan energi maksimum cahaya matahari terjadi pada panjang gelombang radiasi 4,84 10−7 m. Dengan menganggap matahari sebagai benda hitam, berapakah temperatur permukaan matahari ?
×
3. Using equation (4.20) of Pointon (1978), or alternatively the equation below
N (λ) dλ = 8π λ e 4
dλ
hc/λkT
−1
show that the total number of photons per unit volume of an enclosure at temperature T is given by ∞ kT 3 1 16π hc n3 n=1
where the n’s are positive integers (taken from Pointon, 1978, Ch.4, problem 1).
Chapter 4 Fermi-Dirac Statistics In this chapter, we examine a group of quantum particles having half-integer spin which obey the Pauli exclusion principle. Such particles are called fermions and the quantum statistics used to describe the fermions is commonly known as Fermi-Dirac (FD) statistics. This chapter is outlined as follows. The number of ways of arranging fermions or the weight of configuration is shortly introduced in 4.1, followed by the discussion of the fermion population in 4.2. The discussion of these sections is mostly similar to that for the BE statistics, and thus many details are left for students. The concepts of the Fermi energy and Fermi function, as well as their properties under ideal conditions, are examined in 4.3. Finally, the application of the Fermi distribution to real cases is discussed. Because of the high concentration of conduction electrons in a metal, an electron gas cannot be treated as a classical gas. It will be later shown in 4.4 that the FD statistics is appropriate for the gas in a metal.
§
§
§
§
4.1
Weight of configuration
Let us consider an isolated, quantum assembly containing N non-interacting fermions with total energy E . These fermions are distributed over energy sheets, which include all available states gi . As with bosons, fermions are indistinguishable particles, hence any rearrangement of these particles within the available states will make no configuration. Unlike bosons, the number of fermions occupying a single state is limited by the exclusion principle. It follows that there exists only one particle or no particle per unit state. Thus, the number of possible configuration for fermions can be written as W =
i
gi ! , N i ! (gi N i ) !
−
(4.1)
where N i is the number of particles in the sheet i, which will be derived in the next section. 29
30
4.2
4. Fermi-Dirac Statistics
Population of fermions
Recall again that in the Lagrange method (3.6), we have d n W + α dN + β dE = 0, where α and β are the Lagrange multipliers. In a manner similar to the steps discussed in Chapter 3, we derive the fermion population by first taking the natural logarithm of (4.1), and then applying (3.5) to the above expression. After several simple algebra, we obtain N i =
gi e−(α+βi ) + 1
.
(4.2)
The above expression is called the Fermi-Dirac distribution or fermion population number, which may be written in the form of N () d =
g() d = f () g() d, e−(α+β) + 1
where f () is f () =
1 e−(α+β) + 1
.
(4.3)
(4.4)
−
Here β = 1/kT as before, but this time α is defined as F /kT (different from that for bosons), where F is known as the Fermi energy. With this new definition for α, we can rewrite (4.4) as 1 f () = (− )/kT , (4.5) e F +1 which is generally called as the Fermi function. This function provides the probability of finding fermions with energy in the given sheets.
4.3
The Fermi-Dirac gas
It is widely known that the absolute zero of temperature has never been achieved, and so the condition under which the Fermi-Dirac gas behaves at T = 0 K is considered to be an ideal case. Therefore, we here examine the behaviour of the Fermi function and derive the Fermi energy at this temperature for comparison. We begin with analysing f () at T = 0 K as follows. 1 f () = −∞ =1 for < F (0) e +1 and 1 f () = ∞ =0 for > F (0) e +1
4.3. The Fermi-Dirac gas
31
The above values for f () at T = 0 K imply that all states with energy smaller than F (0) are occupied while those having energy greater than F (0) are empty. These values are used to evaluate the Fermi energy at T = 0 K. In doing so, we first write explicitly the number of available states for fermions and to relate it to the volume occupied by these states in phasa space. Recall again that the relation between the number of states g() d and element volume in phasa space dΓ is given by g() d = B dΓ dx dy dz d px d py d pz =2 h3 V 4πp 2 d p =2 , h3 where the multiplier “ 2 ” arises from the fact that for fermions there are two independent directions of spin orientation. Based on classical linear momentum-kinetic energy relation p2 = 2m, the above expression can be rewritten as g() d =
4π V (2m)3/2 1/2 d. 3 h
(4.6)
Now integrating both sides of (4.3) and inserting (4.6) yields
∞
∞
N () d =
0
f () g() d,
0
where the LHS of the above integral denotes the total number of fermions and the RHS can be separated into two parts as follows.
F (0)
N =
∞
f () g() d +
0
f () g() d
F (0)
Substituting the values of Fermi function f () at T = 0 K and equation (4.6) to the above expression results in 4π V N = 3 (2m)3/2 h = 4π V
F (0)
0 3/2
2m h2
1/2 d
2 F (0)3/2 , 3
from which we obtain the Fermi energy at T = 0 K below, F (0) =
3N 8πV
2/3
h2 2m
(4.7)
32
4. Fermi-Dirac Statistics
F (0) in eV
Fermi-Dirac gas Helium-3 atoms electron gas in Lithium electron gas in Potassium
0.94
× 10
−3
4.7 2.1
Table 4.1:
The values of the Fermi energies at some Fermi-Dirac gases.
T
T F in K 10 5.4 2.4
× 10 × 10
3 3
= 0 K and Fermi temperatures for
We may equate the Fermi energy at T = 0 K to thermal energy kT to derive a Fermi temperature T F as follows, 2/3 3N h2 T F = . (4.8) 8πV 2mk
We here provide examples of the values of F (0) and T F for some different Fermi-Dirac gases in Table 4.1. It is clear from Table 4.1 that molecular gases, such as He-3 atoms which consist of fermions, have a low Fermi temperature compared with room temperature.
§
However, it has been already shown in 3.3 that the He-4 gas behaves as a classical gas at T = 300 K. It means that the energy distribution of the gas molecules at ordinary (room) temperature will approximate to the classical, MB statistics. The condition under which molecular gases obey classical distribution does not hold for an electron gas in a metal. It is therefore necessary to discuss the electron gas in the next section.
4.4
The electron gas
The fact that the Fermi temperature for the electron gas in a metal is high (see Table 4.1) leads to a fundamental idea: an increase in temperature from absolute zero to a value around room temperature may only affect the electrons with the energy near the Fermi energy. This can be shown below, for = F
− kT
f () =
1 = 0.73 e−1 + 1
1 = 0.50 e0 + 1 1 for = F + kT f () = 1 = 0.27 e +1 The mean values of dynamics properties of the electrons can then be evaluated easily. For example, the mean energy of the electron gas at T = 0 K is given by for = F
f () =
∞
(0) =
0
N () d , N () d
∞
0
(4.9)
4.5. Exercise
33
where N () d is given in (4.3 (4.3). ). After After several several simple simple steps, we obtain obtain 35 F (0) for the mean energy at T = 0 K. In a similar manner to the above steps, the mean velocity of the electron at T = 0 K can also be evaluated using
∞
0
v (0) =
v N ( N () d , N () d
(4.10)
∞
0
and the result is 34 vF , where vF is the velocity of the electron at the Fermi energy. Note that equations (4.7 (4.7)) and (4.9 (4.9)) hold for T = 0 K only. only. In fact, the Fermi Fermi energy energy and the mean energy of the gas electrons are a function of temperature. Thus, it is necessary to find the general form of the Fermi energy F and the mean energy of the electrons at temperature T . T . The details of calculations are left for students (see Appendix 7 of Pointon, 1978 for help), but the main results are here presented. The Fermi energy is
−
F = F (0) 1
π2 T 12 T F F
2
(4.11)
and the mean energy of the electrons is
=
F (0)
3 π2 T + 5 4 T F F
2
.
(4.12)
The second terms in the bracket of both (4.11 (4.11)) and (4.12 (4.12)) are considered as the correction terms that are in fact relatively small compared with the first terms in each equation.
4.5
Exercise
1. Jelask Jelaskan an dengan dengan dua cara pendek pendekatan dalam dalam hal apa Distri Distribusi busi Fermi Fermi-Di -Dirac rac (4.2) 4.2) menuju menuju ke Distribusi Distribusi Maxwell-Bo Maxwell-Boltzmann ltzmann (3.12 (3.12). ). 2. Jelaskan Jelaskan perbedaan antara antara Statistik Statistik Bose-Einstein Bose-Einstein and Statistik Statistik Fermi-Dirac. ermi-Dirac. 3. Gas isotop Helium-3 mengikuti Statistik Statistik Fermi-Dirac Fermi-Dirac dengan fungsi distribusi sebagai berikut gi N i = (− )/kT +1 e F (a) Turunkan energi Fermi pada temperatur 0 K. (b) Tunjukkan bahwa energi rata-rata elektron pada temperatur 0 K adalah 35 F (0). (c) Hitunglah energi rata-rata elektron pada temperatur 0 K.
34
4. Fermi-Dirac Statistics
4. Perak mempunyai rapat massa 10,5 103 kgm−3 dan pada temperatur 0 K, 1 mol atomny atomnyaa mengandu mengandung ng 107,88 gram. Elektro Elektron n logam logam perak p erak tunduk tunduk pad padaa Statist Statistik ik Fermi-Dirac.
×
(a) Hitunglah kerapatan elektron logam perak pada temperatur 0 K. (b) Hitunglah energi Fermi pada temperatur 0 K. (c) Hitunglah energi rata-rata elektron pada temperatur 0 K. (d) Hitunglah energi Fermi pada temperatur 300 K. 5. Show that the mean velocity of an electron in an electron gas at the absolute zero of temperature is 34 vF , where vF is the velocity of an electron at the Fermi energy (taken from Pointon, 1978, Ch.5, problem 1). 6. Show that, for a gas in which the molecules behaves behaves as fermions, the value value of the Fermi energy is approximately F
≈
N h3 kT n V (2 V (2πmkT πmkT ))3/2
for reasonably high temperatures (taken from Pointon, 1978, Ch.5, problem 3).
Chapter 5 Termodinamika Gas Distribusi statistik klasik (MB) maupun statistik kuantum (BE dan FE) melibatkan dua besaran makroskopik, yaitu energi total E dan temperatur mutlak T . T . Pembahasan Pembahasan kedua kedua macam distribusi statistik tersebut (klasik dan kuantum) mengasumsikan bahwa tidak ada interaksi antara sistem dan lingkungan, sedemikian sehingga jumlah total partikel N dan energi total E dapat dianggap tetap, jadi berlaku dN dN = 0 dan dE dE = 0. Nam Namun sepert sepertii telah dibahas pada Bab 3 untuk radiasi elektomagnetik yang menembus lubang pada ruang tertutup tertutup,, jumlah jumlah partik partikel el (foton) (foton) berubah berubah setiap setiap saat, saat, jadi jadi dN = 0 tidak tidak lagi lagi berla berlaku ku.. Pembahasan lebih jauh memberikan kemungkinan interaksi antara sistem dan lingkungan, sedemikian sehingga bisa saja berlaku dN dN = 0 tetapi dE dE = 0, atau bisa juga berlaku dN dN = 0 tetapi dE dE = 0, atau keduanya dN dN = 0 dan dE dE = 0.
Rincian topik bahasan Bab 5 ini adalah sebagai berikut. Konsep entropi entropi untuk sistem tertutup diperkenalkan pada pasal 5.1, 5.1, diikuti dengan pembahasan penerapan entropi pada gas klasik pada pasal 5.2. 5.2. Salah satu contoh gejala fisis yang tidak dapat dijelaskan dengan konsep statistik gas klasik adalah peristiwa yang dikenal sebagai Paradoks Gibbs, yang akan dibahas pada pasal 5.3. Konsep statistik statistik gas semi klasik akan akan dibahas pada pasal 5.4 sebagai alternatif solusi untuk mengatasi masalah Paradoks Gibbs.
5.1 5.1
Kons Konsep ep entr entrop opii
Satu dari dua pengali Lagrange, yaitu β telah diturunkan (lihat kembali persamaan 3.13), 3.13), dimana diperoleh β = 1/kT . /kT . Besaran Besaran ini digunak digunakan an un untuk tuk merumusk merumuskan an ulang ulang bentuk bentuk persamaan diferensial Lagrange (3.6 (3.6)) yang diterapkan untuk sistem tertutup, dimana jumlah partikel N dan volume V dapat dianggap tetap, jadi berlaku dN dN = 0 dan dV dV = 0. Hukum pertama Termodinamika dalam bentuk diferensial menyatakan bahwa
−
dQ = dE + E + P d P dV, 35
(5.1)
36
5. Termodinamika Gas
dimana dQ adalah perubahan panas yang terjadi pada sistem, dE adalah perubahan energi total sistem (energi total E juga dikenal sebagai energi dalam U dalam ilmu termodinamika) dan P dV adalah kerja luar akibat adanya perubahan volume dV . Mengingat tidak ada perubahan volume (dV = 0) pada sistem tertutup, maka berlaku dQ = dE . Jadi persamaan diferensial Lagange (3.6) berikut, d n W + α dN + β dE = 0, berubah menjadi d n W + β dQ = 0.
(5.2)
Pada tahap ini, konsep entropi diperkenalkan sebagai derajat keteraturan sistem. Seperti halnya temperatur mutlak T dan energi total E , maka entropi sistem S juga merupakan fungsi keadaan yang nilainya secara umum berbeda untuk keadaan yang berbeda. Dalam hal ini, perubahan entropi dS merupakan besaran penting yang menggambarkan dinamika sistem, dimana nilainya hanya ditentukan oleh keadaan awal dan keadaan akhir (dikenal sebagai diferensial eksak). Perubahan entropi dS dari satu keadaan ke keadaan yang lain dirumuskan sebagai dQ dS = T
atau
∆S = S akhir
− S
awal
=
dQ T
(5.3)
Pembahasan detil tentang perubahan entropi dapat dipelajari dalam kuliah termodinamika, dimana dibahas perubahan entropi berbagai macam proses, baik proses reversibel maupun non-reversibel. Dengan bantuan definisi β dan persamaan diferensial (5.3) di atas, maka persamaan (5.2) dapat diubah menjadi 1 d n W dS = 0, k dengan mana kita bisa memperoleh
−
S = knW.
(5.4)
Perumusan entropi di atas sering dikenal sebagai relasi Boltzmann, yang menghubungkan peluang termodinamik W (besaran mikro) dan fungsi termodinamik entropi S (besaran makro). Pada kuliah ini, pembahasan ditekankan pada kegunaan (5.4) untuk menjelaskan beberapa gejala makroskopis dari kacamata mikroskopis. Akan ditunjukkan bahwa (5.4) memberikan hasil yang sama dengan pengamatan makroskopis untuk sistem gas klasik. Bagaimanapun keberhasilan persamaan ini bergantung pada bentuk perumusan peluang termodinamik W yang dibangun berdasarkan asumsi-asumsi dasar tertentu.
5.2. Gas klasik
5.2
37
Gas klasik
Pada bagian ini, akan ditunjukkan bahwa persamaan keadaan gas ideal dapat diturunkan dengan bantuan perumusan peluang termodinamika klasik (3.1), W = N !
i
giN i , N i !
yang disubstitusikan ke dalam persamaan (5.4). Kita akan peroleh bentuk seperti di bawah ini,
− − − − −
S = k n N !
i
=k
NnN
giN i N i !
=k
i
N +
i
= k NnN
n N ! + n
N +
giN i n N i !
= k NnN
N +
n giN i
i
N i n gi
N i n N i +
i
= k NnN +
giN i N i !
i
N i n gi
− n N i !
i
N i
i
N i n N i .
i
i
Kemudian kita substitusikan persamaan (3.10) berikut N i =
N gi e−i/kT Z
untuk memunculkan secara eksplisit fungsi partisi Z , energi total (dalam) E dan temperatur mutlak T , sedemikian sehingga perumusan entropi di atas menjadi
S = k NnN +
N i n gi
i
= k NnN +
i
=k
− − − − − N i n
i
N i n gi
N i n N +
i
N i n gi + N n Z
i
= k N n Z +
i
N gi e−i /kT Z
i
i
N i n e−i /kT
N i n gi
i
N i n gi e−i /kT
N i n Z
i
N i i /kT = k N n Z + E/kT .
Adalah sangat jelas dari bentuk terakhir di atas bahwa S = N k n Z +
E . T
(5.5)
38
5. Termodinamika Gas
Sekarang kita siap untuk menurunkan persamaan keadaan gas ideal. Langkah pertama adalah mengalikan kedua ruas persamaan (5.5) dengan T , kemudian mengatur suku-suku persamaan tersebut sedemikian sehingga diperoleh bentuk berikut ini, E
− T S = −NkT nZ.
Dalam ilmu termodinamika, ruas kiri dari bentuk perumusan di atas dikenal sebagai fungsi bebas Helmholtz F. Jadi kita sekarang punya bentuk berikut, F =
−NkT nZ.
(5.6)
3/2
Dengan mensubstitusikan fungsi partisi Boltzmann (2.47), Z = BV 2πmkT , ke dalam persamaan (5.6) di atas, kita akan peroleh bentuk eksplisit dari energi bebas Helmholtz, yaitu 3/2 F = NkT nBV 2πmkT . (5.7)
−
Persamaan keadaan gas ideal dapat diperoleh dengan memanfaatkan salah satu definisi yang diturunkan dari hukum pertama Termodinamika, yaitu
−
∂F ∂nZ P = = NkT ∂V T ∂V d n V = NkT dV NkT = . V
T
Bentuk ungkapan di atas tidak lain merupakan persamaan keadaan gas ideal, P V = N kT . Selain persamaan keadaan gas ideal tersebut di atas, kita bisa juga menurunkan ungkapan energi total (dalam) gas ideal E dengan memanfaatkan definisi berikut,
∂F/T E = T ∂T 3 N kT 2 = 2 T 3 = NkT. 2
−
2
V
2
= NkT
∂nZ ∂T
V
Sekali lagi kita bisa membuktikan keampuhan perumusan entropi sistem S pada (5.5) yang menjelma menjadi energi bebas Helmholtz F pada (5.7), sebelum digunakan untuk menurunkan beberapa kuantitas gas ideal. Namun demikian, perumusan entropi pada (5.5) tidak dapat digunakan untuk menjelaskan gejala pencampuran gas dalam ruang tertutup bersekat. Gejala fisis ini akan kita bahas pada pasal berikutnya.
5.3. Paradoks Gibbs
5.3
39
Paradoks Gibbs
Untuk menguji lebih lanjut apakah perumusan entropi sistem pada (5.5) berlaku umum, maka pada bagian ini dibahas situasi fisis dimana dua gas klasik dengan jumlah partikel masing-masing N 1 dan N 2 , massa total masing-masing m1 dan m2 mula-mula ditempatkan pada dua buah ruang terpisah bersekat yang masing-masing volumenya adalah V 1 dan V 2 . Misalkan bahwa temperatur T kedua ruang dijaga tetap. Adalah menarik untuk dipelajari apa yang akan terjadi bila sekat pemisah kedua ruang dibuka, dengan demikian memberikan kemungkinan interaksi antara partikel-partikel dari kedua gas tersebut. Dalam konteks ini, pembicaraan ditekankan pada nilai entropi sistem dua gas tersebut sebelum dan sesudah sekat dibuka. Marilah kita mulai pembahasan dengan menghitung nilai entropi total sebelum sekat 3/2 dibuka. Dengan mensubstitusikan fungsi partisi Boltzmann pada (2.47), Z = BV 2πmkT , ke dalam persamaan (5.5), kita akan peroleh bentuk eksplisit entropi sistem, yaitu
S = N k n B + N k n V + Nkn 2πmkT
3/2
+
E . T
(5.8)
Berdasarkan persamaan (5.8) di atas, kita bisa tuliskan entropi masing-masing gas sebelum sekat dibuka, yaitu
E 1 T E 2 3/2 S 2 = N 2 k n B + N 2 k n V2 + N 2k n 2πm 2 kT + . T S 1 = N 1 k n B + N 1 k n V1 + N 1k n 2πm 1 kT
3/2
+
(5.9)
Dengan demikian entropi total sistem dua gas tersebut sebelum sekat dibuka adalah
3/2
S = S 1 + S 2 = (N 1 + N 2 ) k n B + (N 1 + N 2 ) k n V + N 1 k n 2πm 1 kT E 1 + E 2 3/2 + N 2 k n 2πm 2 kT + , T dimana telah diasumsikan bahwa V 1 = V 2 = V .
(5.10)
Sekarang kita akan menghitung nilai entropi sistem pada saat sekat telah dibuka. Perlu diketahui bahwa sesudah sekat dibuka, volume total menjadi V 1 + V 2 = 2V , jumlah partikel masing-masing gas tetap N 1 = N 1 dan N 2 = N 2 , massa total masing-masing gas tetap m1 = m1 dan m2 = m2 , energi dalam masing-masing gas tetap E 1 = E 1 dan E 2 = E 2 . Dengan demikian entropi total sistem dua gas sesudah sekat dibuka adalah
3/2
S = (N 1 + N 2 ) k n B + (N 1 + N 2 ) k n 2V + N 1 k n 2πm 1 kT E 1 + E 2 3/2 + N 2 k n 2πm 2 kT + . T
(5.11)
40
5. Termodinamika Gas
Dari kedua persamaan (5.10) dan (5.11) tersebut di atas jelas bahwa perubahan entropi total sistem adalah ∆S = S S = (N 1 + N 2 ) k (n 2V n V ). (5.12)
−
−
Misalkan bahwa kedua gas yang kita bicarakan adalah berjenis sama (dengan kerapatan jenis ρ1 = ρ2 = ρ), maka perubahan entropi total sistem adalah ∆S = 2Nkn 2. Hasil ini sangat menarik mengingat tidak sesuai dengan akal sehat yang menyatakan bahwa derajat keteraturan suatu sistem semestinya sama berapapun volume ruang yang ditempati oleh sistem tersebut. Dengan kata lain, jika kedua jenis gas yang mengisi masing-masing ruang adalah sama maka semestinya tidak boleh ada perubahan entropi pada saat sekat dibuka, atau secara matematis diharapkan ∆S = 0. Tetapi bila hal ini diterima, maka timbul pemikiran bahwa seolah-olah entropi sistem bukan saja bergantung pada keadaan awal dan akhir, melainkan juga bergantung pada ‘riwayat’ sistem tersebut. Dalam hal ini, apakah sistem fisis yang ditinjau terdiri atas gas-gas berjenis sama ataukah tidak. Hal inilah yang merupakan paradoks, karena bertentangan dengan prinsip dasar termodinamika yang menyatakan bahwa entropi adalah fungsi keadaan saja.
5.4
Gas semi klasik
Gibbs adalah orang pertama yang mengamati paradoksial entropi sistem dua gas dalam ruang tertutup bersekat, dan mengajukan usulan perumusan ulang peluang termodinamika untuk mengatasi masalah paradoks tersebut. Untuk lebih jelasnya, kita ingat kembali bahwa perumusan entropi (5.5) E S = N k n Z + T diperoleh dari perumusan peluang termodinamika klasik (3.1) sebagai berikut W = N !
i
giN i . N i !
Nampaknya perumusan peluang termodinamika klasik tersebut di atas yang menjadi biang keladi masalah. Perumusan tersebut berdasarkan asumsi dalam Distribusi Statistik Klasik yang menyatakan bahwa gas klasik terdiri dari partikel-partikel yang dapat dibedakan. Beliau mengusulkan jumlah cara untuk mengatur sistem ke dalam tingkat-tingkat energi yang tersedia, atau yang dikenal sebagai peluang termodinamik W seharusnya N ! kali lebih kecil dari yang diberikan oleh persamaan (3.1). Jadi seharusnya W =
i
giN i . N i !
(5.13)
5.4. Gas semi klasik
41
Hal ini hanya akan bisa dipenuhi bila gas terdiri atas partikel-partikel yang tak terbedakan. Gas semacam ini yang memiliki peluang termodinamik yang mirip dengan persamaan (3.1) kemudian dikenal dengan istilah gas semi klasik dan persamaan (5.13) disebut dengan peluang termodinamik semi klasik . Sekarang saatnya untuk menghitung entropi sistem dengan menggunakan perumusan peluang termodinamik semi klasik (5.13). Perlu diketahui bahwa relasi Boltzmann (5.4) yang menghubungkan besaran makro entropi sistem S dan besaran mikro peluang termodinamik W adalah bentuk dasar perumusan entropi sistem, dan oleh karena itu manjur digunakan untuk semua situasi baik jenis gas yang ditinjau adalah gas klasik maupun semi klasik. Substitusi persamaan (5.13) ke dalam persamaan (5.4) menghasilkan S = k n
i
giN i N i !
= Nk + N k n B + Nkn = Nk
V + Nkn 2πmkT N
3/2
− N k n N + N k n B + N k n V + Nkn
+
E T
(5.14)
2πmkT
3/2
E . T
+
Jelas sekali terlihat bahwa dua suku pertama pada ruas kanan persamaan (5.14) merupakan “pembeda” yang membedakan persamaan entropi semi klasik (5.14) dari persamaan entropi klasik (5.8). Berdasarkan persamaan (5.14) tersebut di atas, maka entropi masing-masing gas sebelum sekat dibuka adalah
− N k n N + N k n B + N k n V + N k n S = N k − N k n N + N k n B + N k n V + N k n S 1 = N 1 k 2
2
1
2
1
2
1
2
1
2
1
2
1
2
E 1 T E 2 3/2 2πm 2kT + T 2πm 1kT
3/2
+
(5.15)
Dengan demikian entropi total sistem gas semi klasik sebelum sekat dibuka adalah S = 2Nk
− 2N k n N + 2N k n B + 2N k n V + 2Nkn
2πmkT
3/2
+
E 1 + E 2 , T
(5.16)
dimana sekali lagi telah diasumsikan V 1 = V 2 = V dan N 1 = N 2 = N serta m1 = m2 = m (untuk jenis gas yang sama). Analogi dengan gas klasik, sekarang kita menghitung entropi total sistem gas semi klasik sesudah sekat dibuka sebagai berikut, S = 2N k
− 2N k n N + 2N k n B + 2N k n V + 2Nkn
2πmkT
3/2
+
E 1 + E 2 . (5.17) T
Jelas sekali terlihat dari kedua persamaan (5.16) dan (5.17) tersebut di atas bahwa tidak
42
5. Termodinamika Gas
ada perubahan entropi sistem sebelum dan sesudah sekat dibuka. Secara matematis, hal ini dituliskan sebagai ∆S = 0. Tidak adanya perubahan entropi total sistem gas semi klasik pada saat sekat dibuka dikarenakan “specific volume” V /N sistem tersebut bernilai sama sebelum dan sesudah sekat dibuka. Menarik untuk dibicarakan di sini bahwa perumusan entropi sistem gas semi klasik pada (5.14) dapat dituliskan dalam bentuk
V 2πmkT S = N k n Nh3
3/2
+
5 . 2
(5.18)
Perumusan di atas dikenal sebagai persamaan Sackur-Tetrode, yang diperoleh dengan jalan mensubstitusikan B = 1/h3 dan E = 32 N kT ke dalam persamaan (5.14). Juga menarik untuk dikaji bahwa tidak ada cara untuk memahami secara klasik mengapa kita harus membagi perumusan peluang termodinamik klasik (3.1) dengan N ! agar diperoleh perhitungan entropi total sistem dua gas dengan benar. Bagaimanapun secara kuantum, hal tersebut dapat dimengerti mengingat atom-atom gas tersusun oleh partikel-partikel takterbedakan yang dibentuk oleh fungsi gelombang yang simetri dan anti-simetri terhadap pertukaran dua partikel. Detil matematis pembahasan masalah ini adalah di luar cakupan kuliah Fisika Statistik.
5.5
Latihan soal
1. Gas yang terdiri atas N buah partikel dengan massa per partikelnya adalah m memiliki temperatur T dan menempati ruang dengan volume V . Partikel-partikel gas tersebut mengikuti hukum Distribusi Semi Kasik. (a) Carilah bentuk fungsi energi bebas Helmholtz. (b) Turunkan bentuk persamaan keadaan gas tersebut. (c) Turunkan ungkapan entropi gas tersebut. (d) Turunkan ungkapan kapasitas kalor gas tersebut. 2. (a) Jelaskan perbedaan antara Statistik Klasik (Maxwell-Boltzmann), Statistik Semi Klasik dan Statistik Kuantum (Bose-Einstein dan Fermi-Dirac). (b) Jelaskan secara fisis apa yang dimaksud dengan Paradoks Gibbs, dan bagaimana cara mengatasi masalah tersebut. 3. Show that, when two different gases A and B of volumes V A and V B and total molecules of N A and N B , respectively, are mixed together at constant temperature T to form a
5.5. Latihan soal
43
total volume of V A + V B , there is an increase in the total entropy given by the mixing term as follows
k (N A + N B ) n (V A + V B )
− N
A
n V A
− N
B
n V B .
Hence deduce that, when V A = V B and N A = N B = N , the mixing term is 2Nkn 2 (taken from Pointon, 1978, Ch.7, problem 1). 4. Derive the equation of state and the total energy of a semi classical perfect gas from the expression for the entropy for such a gas. Explain why the results obtained are the same as those for a classical perfect gas (taken from Pointon, 1978, Ch.7, problem 3). 5. Show that the total partition function derived from equation (7.45) of Pointon (1978) on the assumption that the energy of the N sub-systems is p2x1 p2y1 p2z1 + + + E = 2m 2m 2m
······
p2xN p2yN p2zN + + + 2m 2m 2m
agrees with that obtained by substituting (7.28) into (7.31) of Pointon (1978), i.e., V N Z = 2πmkT N ! h3N
(taken from Pointon, 1978, Ch.7, problem 5).
3N/2
44
5. Termodinamika Gas
Chapter 6 Aplikasi Distribusi Statistik Bab ini merupakan bab terakhir dari rangkaian pembahasan pada kuliah Fisika Statistik, yang meliputi aplikasi fisika statistik pada beberapa sistem fisis sederhana, dimana jumlah partikel dan energi totalnya dapat dianggap tetap (dN = 0 dan dE = 0). Sistem fisis dengan partikel-partikel yang saling berinteraksi tidak ikut dibahas pada bab ini. Namun demikian, hasil penurunan matematis untuk sistem fisis yang ditinjau akan berlaku secara umum. Adapun rincian topik bahasan Bab 6 adalah sebagai berikut. Model fisis dengan dua tingkat energi akan diperkenalkan terlebih dahulu pada pasal 6.1, kemudian diikuti dengan pengenalan konsep osilator harmonik menurut distribusi statistik kuantum pada pasal 6.2. Pembahasan molekul diatom dengan melibatkan gabungan gerak translasi, rotasi dan vibrasi akan diberikan pada pasal 6.3.
6.1
Sistem dua tingkat energi
Bagian ini membahas sistem N partikel yang terdistribusi ke dalam dua tingkat energi 1 dan 2 yang tak-terdegenerasi (g1 = g2 = 1), dimana 2 - 1 = . Misalkan bahwa tingkat energi 1 ditetapkan berisi N 1 partikel dan tingkat energi 2 ditetapkan berisi N 2 partikel, maka dapat dituliskan N 1 + N 2 = N . Berikut akan ditunjukkan bahwa distribusi partikel pada masing-masing tingkat energi merupakan fungsi dari e/kT . Langkah-langkah untuk menurunkan fungsi distribusi partikel tersebut adalah sebagai berikut. Mula-mula kita mengasumsikan bahwa partikel-partikel yang menghuni kedua tingkat energi adalah partikel-partikel yang terbedakan, agar N 1 dan N 2 bersifat tetap. Jadi, distribusi statistik yang digunakan untuk memecahkan masalah ini adalah Distribusi Maxwell-Boltzmann, dimana jumlah cara untuk mengatur partikel ke dalam tingkat-tingkat energi yang tersedia dinyatakan menurut persamaan (3.1). Karakteristik termodinamik sistem tersebut di atas digambarkan oleh fungsi entropi 45
46
6. Aplikasi Distribusi Statistik
melalui relasi Boltzmann (5.4) sebagai berikut, S = k n W = k n
N ! . N 1 ! N 2 !
(6.1)
Dengan memanfaatkan pendekatan Stirling, persamaan (6.1) berubah menjadi
S = k N n N
− N n N − N n N 1
1
2
2
.
(6.2)
Seperti telah dibahas pada pasal 5.2, fungsi energi bebas Helmholtz dituliskan sebagai F = E T S . Dengan demikian energi bebas Helmholtz sistem tersebut adalah
−
− kT N n N − N n N − N n N = N + N − kT N n N − N n N − N n N
F = E
1 1
2 2
1
1
2
2
1
1
2
2
(6.3)
.
Agar sistem dalam keadaan setimbang, maka nilai energi bebas Helmhotz haruslah minimum. Dengan kata lain, turunan parsial dari energi bebas Helmholtz (6.3) terhadap N 1 atau N 2 di atas haruslah berharga nol. Jadi, ∂F =0 ∂N 1
∂F = 0. ∂N 2
atau
(6.4)
Substitusi persamaan (6.3) ke dalam salah satu persamaan (6.4) menghasilkan 0 = 2 = 2
−
− − kT − − kT 1 1
n N 1 n
n N 2
N 1 . N 2
Dengan mengatur suku-suku pada persamaan di atas, maka akan diperoleh
−
N 1 2 1 = N 2 kT N 1 n = n e( − )/kT N 2 N 1 = e( − )/kT = e/kT . N 2 n
2
2
1
1
Bentuk perumusan di atas adalah seperti yang diharapkan pada temperatur yang cukup tinggi, dimana partikel cenderung mengisi tingkat energi yang lebih tinggi (N 1 N 2). Substitusi N 1 = N N 2 ke dalam perumusan tersebut di atas dan mengatur suku-suku sehingga akan diperoleh populasi partikel pada tingkat energi 2 sebagai berikut
−
N 2 =
N . 1 + e/kT
(6.5)
6.2. Osilator harmonik
47
Energi total sistem dapat diturunkan dengan mudah dengan bantuan (6.5), yaitu E = N 1 1 + N 2 2
− N ) + N = N + ( − ) N = (N
2
1
1
2
= N1 +
2 2
1
2
(6.6)
N . 1 + e/kT
Berbekal persamaan (6.6) dapat ditunjukkan dengan mudah bahwa kapasitas panas zat yang terdiri atas sistem dua tingkat energi pada volume tetap adalah C V =
6.2
∂E ∂T
V
N2 e/kT = . kT 2 (1 + e/kT )2
(6.7)
Osilator harmonik
Banyak sekali masalah fisika yang dapat dipelajari dengan pemodelan sistem fisis yang memanfaatkan osilator harmonik untuk menggambarkan sifat dinamik dari sistem yang bersangkutan. Perlu diketahui di sini bahwa pendekatan osilator harmonik dapat dilakukan dengan dua cara, yaitu mekanika klasik dan mekanika kuantum. Ingat kembali pembahasan ekipartisi energi pada sub-pasal 2.3.2, dimana ditemukan energi rata-rata osilator harmonik satu dimensi adalah kT (persamaan 2.40). Hal ini berarti energi kinetik osilator dapat berharga berapa saja (terdistribusi secara kontinu) bergantung pada besarnya temperatur. Sebaliknya, pendekatan mekanika kuantum memberikan batasan harga untuk energi osilator yaitu n =
n+
1 2
hν,
(6.8)
dimana n = 0, 1, 2, ......, dan ν adalah frekuensi osilator. Jadi, menurut mekanika kuantum energi osilator terdistribusi secara diskrit dalam tingkat-tingkat energi yang diperbolehkan (merupakan kelipatan setengah dari hν ). Tingkat energi terendah diperoleh untuk n = 0, yaitu = 12 hν . Tingkat energi terendah ini disebut dengan tingkat energi keadaan dasar . Tingkat-tingkat energi lain yang lebih tinggi berkaitan dengan harga n > 0 disebut dengan tingkat energi keadaan eksitasi . Dalam paragraf berikut akan dicoba diturunkan energi rata-rata osilator harmonik satu dimensi menurut mekanika kuantum dengan memanfaatkan persamaan (6.8) di atas. Hasil yang diperoleh akan dibandingkan dengan hasil menurut mekanika klasik. Langkah pertama
48
6. Aplikasi Distribusi Statistik
adalah menuliskan kembali persamaan (3.9), Z =
gi e−i /kT .
(6.9)
i
Mengingat tingkat-tingkat energi osilator harmonik adalah tak-terdegenerasi (g = 1) dan dengan bantuan persamaan (6.8), maka persamaan (6.9) di atas dapat dituliskan ulang dalam bentuk ∞ ∞
Z =
e−(n+1/2)hν/kT = e−
1 2
hν/kT
n=0
e−nhν/kT .
(6.10)
n=0
Substitusi deret berikut xn = (1 di atas akan menghasilkan
− x)
−1
dimana x = e−hν/kT ke dalam persamaan (6.10) e−
Z =
1 2
hν/kT
. (6.11) 1 e−hν/kT Energi rata-rata osilator harmonik dapat dengan mudah dihitung melalui perumusan yang telah diperkenalkan sebelumnya pada bagian akhir pembahasan pasal 5.2,
−
− − − − 1
∂nZ e− hν/kT 2 ∂ = kT = kT n ∂T V ∂T 1 e−hν/kT ∂ = kT 2 n e− hν/kT n 1 e−hν/kT ∂T 1 hν hν = kT 2 + 2 kT 2 kT 2 (ehν/kT 1) 1 hν = hν + hν/kT . 2 e 1
2
2
1 2
(6.12)
− Pada temperatur cukup tinggi berlaku hν kT , sedemikian sehingga hν/kT
e
≈
hν 1 1+ + kT 2
hν kT
2
sampai pada suku order kedua. Substitusi deret tersebut di atas ke dalam persamaan (6.12) akan menghasilkan
= hν
− 1 kT + 1 2 hν
hν kT
= kT .
(6.13)
Hasil tersebut di atas menunjukkan bahwa perumusan energi rata-rata osilator harmonik menurut mekanika kuantum pada temperatur tinggi akan sama dengan perumusan menurut mekanika klasik.
6.3. Gas diatomik
6.3
49
Gas diatomik
Model fisis lain yang dapat digunakan untuk menggambarkan dinamika sistem adalah model molekul diatomik. Dalam model ini, energi total sistem terdiri dari energi yang berasal dari gerak translasi, rotasi, vibrasi, orbital dan spin. Gerak orbital yang dimaksud di sini adalah gerakan elektron disekeliling inti atom, sedangkan gerak spin adalah gerak berputar inti atom terhadap porosnya. Gerak translasi dapat didekati dengan mekanika klasik, namun keempat gerak yang lain secara umum mensyaratkan adanya kuantisasi (diskritisasi) energi. Fungsi partisi total sistem molekul diatomik dapat dituliskan sebagai Z = Z t
× Z × Z × Z × Z , r
v
e
(6.14)
n
dimana indeks t, r, v, e dan n masing-masing untuk gerak translasi, rotasi, vibrasi, elektron dan nuklir. Kita akan turunkan satu per satu kelima bentuk fungsi partisi tersebut dimulai dari yang paling mudah. Fungsi partisi untuk gerak translasi adalah yang paling mudah mengingat kita sudah menurunkan sebelumnya pada persamaan (2.47),
Z t = BV 2πmkT
3/2
=
V
3
2πmkT
3/2
.
(6.15)
Berikutnya yang juga mudah adalah fungsi partisi untuk gerak vibrasi karena gerak vibrasi molekul diatomik sebagai satu kesatuan dapat dipandang sebagai osilator harmonik satu dimensi. Jadi fungsi partisi gerak vibrasi dapat didekati oleh persamaan (6.11), Z v =
e− 1
1 2
hν/kT −hν/kT
−e
,
(6.16)
dimana ν menyatakan frekuensi karakteristik vibrasi molekul diatomik. Kemudian fungsi partisi dari kontribusi gerak orbital elektron dapat ditentukan langsung dari persamaan (6.9) berikut, Z e =
i
gi e−i /kT
≈g
o
+ g1 e−
1
/kT
,
(6.17)
dimana telah diambil dua suku pertama dari deret fungsi partisi yang bersangkutan sebagai pendekatan yang cukup baik. Dalam hal ini go menyatakan derajat degenerasi keadaan dasar molekul diatomik, g1 menyatakan derajat degenerasi keadaan tereksitasi pertama, dan 1 menyatakan tingkat energi eksitasi pertama. Pada umumnya tingkat-tingkat energi tereksitasi berharga jauh lebih besar dari energi termal kT , kecuali untuk temperatur tinggi, dimana energi termal tersebut menjadi dominan.
50
6. Aplikasi Distribusi Statistik
Perhitungan fungsi partisi gerak rotasi agak sedikit sulit. Kita bayangkan tiap molekul gas terdiri atas 2 atom bermassa sama m terpisah pada jarak 2a. Momen inersia sistem ini terhadap sumbu yang melewati titik tengah garis hubung kedua atom tersebut adalah = 2ma2, dimana a adalah jarak dari salah satu atom menuju ke titik pusat sistem. Bila sistem ini tidak berada dalam pengaruh medan potensial apapun, maka energi totalnya hanya merupakan energi kinetik rotasi saja. Jadi, kita dapat menuliskan persamaan Hamilton dalam bentuk operator untuk menggambarkan dinamika rotasi sistem diatomik tersebut sebagai berikut, H ψ = (E + V ) ψ = (E + 0) ψ = E ψ, (6.18)
J
J
dimana H adalah operator Hamiltonian yang menyatakan energi total sistem, E menyatakan operator energi kinetik, V menyatakan operator energi potensial, dan ψ menyatakan fungsi gelombang sistem yang bersangkutan. Analogi dengan gerak translasi, energi kinetik gerak rotasi dapat digambarkan oleh L2 E = , (6.19) 2
dimana L adalah operator momentum sudut orbital, yang bekerja terhadap fungsi gelombang menurut L2 ψ = 2 ( + 1) ψ, dimana adalah bilangan kuantum momentum sudut orbital ( = 0, 1, 2, ......). Substitusikan bentuk di atas ke dalam persamaan (6.19) akan diperoleh
Eψ =
2 (
2
+ 1)
J
ψ.
(6.20)
Koreksi kecil dilakukan untuk perumusan momentum sudut. Dengan mempertimbangkan momentum sudut spin, maka bilangan kuantum momentum sudut orbital diganti dengan bilangan kuantum momentum sudut total j. Dengan demikian persamaan 6.20) berubah menjadi 2 j( j + 1) Eψ = ψ. (6.21) 2
J
Secara implisit, persamaan (6.21) di atas menyatakan bahwa tingkat-tingkat energi rotasi yang diijinkan adalah 2 j( j + 1) j = , (6.22) 2
J
dimana j = 12 . Perlu diketahui bahwa untuk setiap harga j akan terdapat (2 j + 1) kemungkinan harga m j (bilangan kuantum magnetik total), yaitu m j = - j,...... 0, ...... j. Kemungkinan sebanyak itu dapat dipandang sebagai akibat degenerasi kuantum. Jadi dalam hal ini derajat degenerasinya adalah g j = (2 j+1) sedemikian sehingga persamaan (6.9) dapat
±
6.3. Gas diatomik
51
dituliskan ulang menjadi ∞
Z r =
∞
−j /kT
g j e
=
j=0
(2 j + 1) e−j /kT
j=0
∞
=
(6.23)
(2 j + 1) e− j( j+1)K/kT ,
j=0
dimana
2
2
K = h /8π J .
Fungsi partisi dari kontribusi gerak orbital elektron dikembangkan dari persamaan (6.9) berikut, Z e = gi e−i /kT go + g1 e− /kT , (6.24)
≈
i
1
dimana telah diambil dua suku pertama dari deret fungsi partisi yang bersangkutan sebagai pendekatan yang cukup baik. Dalam hal ini go menyatakan derajat degenerasi keadaan dasar molekul diatomik, g1 menyatakan derajat degenerasi keadaan tereksitasi pertama, dan 1 menyatakan tingkat energi eksitasi pertama. Pada umumnya tingkat-tingkat energi tereksitasi berharga jauh lebih besar dari energi termal kT , kecuali untuk temperatur tinggi, dimana energi termal tersebut menjadi dominan. Yang terakhir adalah fungsi partisi gerak spin Z n dari inti atom. Fungsi partisi ini bukan merupakan fungsi temperatur, melainkan merupakan konstanta biasa. Tidak begitu penting membicarakan berapa nilai sebenarnya dari fungsi partisi ini. Setelah memperoleh kelima bentuk fungsi partisi (lihat kembali persamaan 6.14), maka fungsi partisi sistem molekul diatomik secara lengkap dapat dituliskan sebagai Z =
V 3
2πmkT
3/2
e−
1 2
hν/kT
×1−e
−hν/kT
∞
×
(2 j + 1) e− j( j+1)K/kT
j=0
×
go + g1 e−
1
/kT
×
Z n. (6.25)
Perlu dipahami di sini bahwa kita meninjau gas semi klasik yang terdiri dari N buah molekul diatomik yang identik satu sama lain. Entropi sistem gas semacam ini dinyatakan oleh persamaan (5.14), yang dapat dituliskan ulang dalam bentuk S = Nk + N k n B + Nkn Z E = Nk + Nkn + . N T
V + Nkn 2πmkT N
3/2
+
E T
Dengan bantuan persamaan (6.26) dan definisi fungsi energi bebas Helmholtz F = E maka kita dapat menurunkan fungsi Helmholtz untuk gas semi klasik dalam bentuk F =
Z N kT n . N !
−
(6.26)
− T S , (6.27)
52
6. Aplikasi Distribusi Statistik
Cobalah bandingkan persamaan (6.27) dengan bentuk fungsi Helmholtz untuk gas klasik pada persamaan (5.6), F = NkT nZ = kT n Z N .
−
−
Jelas terlihat sekali lagi bahwa terdapat perbedaan faktor N ! antara besaran makro energi bebas Helmholtz F menurut distribusi klasik dan semi klasik. Sebelumnya adalah perbedaan (dengan faktor yang sama N !) antara besaran mikro peluang termodinamik W menurut distribusi klasik dan semi klasik (lihat kembali pembahasan pasal 5.4). Dengan membandingkan perumusan energi bebas Helmholtz untuk sistem klasik dan semiklasik, maka dapat didefinisikan besaran fungsi partisi total sistem semi klasik sebagai berikut Z N = . (6.28) N ! Substitusikan persamaan (6.28) ke dalam persamaan (6.27), akan diperoleh
Z
Z
F =
−kT n Z .
(6.29)
Dengan bantuan persamaan (6.29) kita bisa menghitung besarnya energi total gas diatomik semi klasik sebagai berikut,
Z
∂F/T ∂ n ∂nZ E = T 2 = kT 2 = N kT 2 ∂T V ∂T ∂T V V ∂ n Zt ∂ n Zr ∂ n Zv ∂ n Ze ∂ n Zn = NkT 2 + + + + ∂T ∂T ∂T ∂T ∂T
−
.
Suku terakhir ruas kanan jelas berharga nol mengingat Z n adalah konstanta. Keempat suku pertama ruas kanan memberikan hasil berikut ini, 3 ∂ E = NkT + N kT 2 n 2 ∂T
∞
j=0
1 hν + Nhν + hν/kT 2 e 1
−
(2 j + 1) e− j( j+1)K/kT
N g1 1 e− + go + g1 e−
1
/kT
1
/kT
.
(6.30)
Pada tahap ini akan sangat bermanfaat bila didefinisikan suatu besaran karakteristik, dengan mana kita bisa membandingkan kontribusi masing-masing gerak dalam persamaan (6.30) di atas. Besaran karakteristik yang dimaksud adalah temperatur karakteristik rotasi, vibrasi dan orbital elektronik berturut-turut dinyatakan sebagai berikut, θr =
K, k
θv =
hν , k
θe =
1 . k
(6.31)
Pada umumnya nilai ketiga besaran karakteristik tersebut di atas memenuhi hubungan
6.4. Latihan soal
53
berikut θr θv θe . Pada temperatur rendah (T θr ), tiga suku pada ruas kanan persamaan (6.30) menjadi dominan, yaitu suku pertama 32 N kT , suku kontribusi dari energi keadaan dasar gerak vibrasi 12 Nhν , dan suku kontribusi dari energi keadaan dasar gerak orbital elektron Ngo . Dari ketiga suku ini, hanya suku pertama saja yang merupakan fungsi temperatur. Suku ini merupakan kontribusi dari gerak translasi. Jadi kapasitas panas C V gas diatomik pada temperatur rendah dinyatakan sebagai C V =
∂E ∂T
V
3 3 3 = N k = nN o k = nR. 2 2 2
(6.32)
Dengan kata lain, kalor jenis cV gas diatomik pada temperatur rendah dituliskan sebagai cV =
C V 3 = R. n 2
(6.33)
Hasil tersebut di atas sama dengan hasil yang diperoleh melalui perhitungan distribusi klasik. Bagaimanapun, perhitungan semi klasik memungkinkan kita untuk menentukan kapasitas panas ataupun kalor jenis gas diatomik pada temperatur sedang dan tinggi. Untuk sekedar diketahui, pada temperatur sedang kalor jenis gas diatomik adalah 52 R, sedangkan pada temperatur tinggi 72 R. Detil penurunan tidak diberikan dalam kuliah ini, namun demikian mahasiswa diharapkan untuk melakukannya buat olahraga ringan.
6.4
Latihan soal
1. Calculate the maximum specific heat for an assembly of systems having two energy levels separated by an energy hν if the upper level is doubly degenerate (taken from Pointon, 1978, Ch.8, problem 6). 2. Two different groups of n fermions are distributed among m states of the same energy where m > n. Show that the entropy is given by
2k m n m
− n n n − (m − n) n (m − n)
(taken from Pointon, 1978, Ch.8, problem 2).
3. Show that at high temperatures the free energy of an assembly of quantum mechanical, 1D oscillators approximates to kT n (hν/kT ) per oscillator, corresponding to a mean thermal energy of kT (taken from Pointon, 1978, Ch.8, problem 3). 4. Evaluate the partition function at temperature T for a classical 1D harmonic oscillator
54
6. Aplikasi Distribusi Statistik
having an energy of
p2x 1 = + µx2 2m 2 and hence find the mean energy of such an oscillator at this temperature (taken from Pointon, 1978, Ch.8, problem 4). 5. Show that the equation of state of a diatomic gas as derived from (8.27) of Pointon (1978) is the same as that for a monoatomic gas (taken from Pointon, 1978, Ch.8, problem 5). 6. Write out the partition function for a 3D harmonic oscillator assuming that it obeys classical and quantum mechanics. Show that both give a mean energy of 3kT at high temperatures (taken from Pointon, 1978, Ch.8, problem 7).
