Kinetic Theory of Gases

Chapter # 24

Kinetic Theory of Gases

SOLVED EXAMPLES

Example 24.1. Calculate the rms speed of nitrogen at STP (pressure = 1 atm and temperature = 0ºC). The density of nitrogen in these conditions is 1.25 kg/m3. Sol. At STP, the pressure is 1.0 × 105 N/m2. The rms speed is vrms =

3p p

3  10 5 N / m 2 =

1.25 kg / m 3

= 490 m/s. Example 24.2 If the rms speed of nitrogen molecules is 490 m/s at 273 K, find the rms speed of hydrogen molecules at the same temperature. Sol. The molecular weight of nitrogen is 28 g/mole and that of hydrogen is 2 g/mole. Let m1, m2 be the masses and v1, v2 be the rms speeds of a nitrogen molecule and a hydrogen molecule respectively. Then m1 = 14 m2. Using equation (24.6),

1 1 m v 2 = m2v22 2 1 1 2 m1 v2 = v1 m = 490 m/s × 2

or,

14 = 1830 m/s.

Example 24.3 Calculate the number of molecules in each cubic metre of a gas at 1 atm and 27ºC. Sol. We gave pV = NkT or,

pV kT

N=

(1.0  10 5 N / m 2 ) (1m 3 ) =

(1.38  10  23 J / K ) (300 K )

= 2.4 × 10 25. Example 24.4 Find the rms speed of oxygen molecules in a gas at 300 K. Sol.

vrms =

=

3RT M0

3  (8.3 J / mol  K )  (300 K ) 32 g / mol

=

3  8.3  300 m/s = 483 m/s. 0.032

Example 24.5 At what external pressure will water boil at 140ºC? Use table (24.1) for vapour pressure data and express the answer in atm. Sol. The saturation vapour pressure of water at 140ºC is 2710 mm of Hg. Thus, water will boil at 140ºC at this pressure. Now 760 mm of Hg = 1 atm. 2710 atm = 3.56 atm. 760 The pressure inside a pressure cooker is of this order when it whistles. So, the temperature inside is of the order of 140ºC which helps in cooking the food much faster.

Thus, 2710 mm of Hg =

Example 24.6 The vapour pressure of air at 20ºC is found to be 12 mm of Hg on a particular day. Find the relative humidity. Use the data of table (24.1) manishkumarphysics.in

Page # 1

Chapter # 24 Kinetic Theory of Gases Sol. The saturation vapour pressure of water at 20ºC is 17.5 mm of Hg. Thus, the relative humidity is

vapour pressure of air SVP at the same temperature 12 mm of Hg = 17.5 mm of Hg = 0.69 that is, 69% Example 24.7 In an experiment with Regnault’s hygrometer, dew appears at 10ºC when the atmospheric temperature is 40ºC. Using table (24.1), find the relative humidity. Sol. The dew point is 10ºC. The saturation vapour pressure at this temperature is 8.94 mm of Hg from table (24.1). Also, the saturation vapour pressure of air at 40ºC is 55.1 mm of Hg. The relative humidity expressed in percentage =

vapour pressure at the dew point × 100% SVP at the air temperature

=

8.94 × 100% = 16.2%. 55.1

QUESTIONS FOR SHORT Objective - I

ANSWER

1.

Which of the following parameters is the same for molecules of all gases at a given temperature? (A) mass (B) speed (C) momentum (D*) kinetic energy fdlh fn;s x;s rki ij leLr xSlksa ds fy;s fuEu esa ls dkSuls dkjd (parameters) ,d leku gS (A) nzO;eku (B) pky (C) laosx (D*) xfrt ÅtkZ

2.

A gas behaves more closely as an ideal gas at

xSlsa vkn'kZ xSl tSlk O;ogkj djrh gS &

3.

(A) low pressure and low temperature (C) high pressure and low temperature (A) de nkc vkSj de rki ij (C) mPp nkc vkSj de rki ij

(B*) low pressure and high temperature (D) high pressure and high temperature (B*) de nkc vkSj mPp rki ij (D) mPp nkc vkSj mPp rki ij

The pressure of an ideal gas is written as P =

2E . Here E refers to 3V

fdlh vkn'kZ xSl ds fy, nkc P = (A*) translatioal kinetic energy (C) vibrational kinetic energy (A*) LFkkukUrfj; xfrt ÅtkZ (C) dkEifud xfrt ÅtkZ

2E 3V

gksrk gS ;gk¡ E dk vFkZ gS & (B) rotational kinetic energy (D) total kinetic energy (B) ?kw.khZ; xfrt ÅtkZ (D) dqy xfrt ÅtkZ

4.

The energy of a given sample of an ideal gas depends only on its (A) volume (B) pressure (C) density (D*) temperature vkn'kZ xSl ds fn;s x;s uewus dh ÅtkZ fuHkZj djrh gS] blds (A) vk;ru ij (B) nkc ij (C) ?kuRo ij (D*) rki ij

5.

Which of the following gases has maximum rms speed at a given temperature ? (A*) hydrogen (B) nitrogen (C) oxygen (D) carbon dioxide fdlh fn;s x;s rki ij fuEu esa ls fdl xSl dh oxZ ek/; ewy] ewy pky lokZf/kd gksxh (A*) gkbMªkstu (B) ukbVª k st u (C) vkWDlhtu (D) dkcZu MkbZ vkDlkbM

6.

Fig. shows graphs of pressuer vs. density for an ideal gas at two temperature T1 and T2. fn[kk;s x;s fp=k esa fdlh vkn'kZ xSl ds fy, nkc vkSj ?kuRo esa nks rkiksa T1 o T2 ij xzkQ [khapk x;k

manishkumarphysics.in

gSA

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Chapter # 24

(A*) T1 > T2 (A*) T1 > T2


(B) T1 = T2 (B) T1 = T2

(C) T1 < T2 (C) T1 < T2

(D) any of the three is possible (D) rhuksa eas ls dksbZ Hkh lEHko

7.

The mean square speed of the molecules of a gas at absolute temperature T is proportional to ijerki T ij fdlh xSl ds v.kqvksa dh oxZek/; pkyw lekuqikrh gksrh gS (A) 1/T (B) T (C*) T (D) T2

8.

Suppose a container is evacuated to leave just one molecule of a gas in it. Let a and rms represent the average speed and the rms speed of the gas. eku yhft, fdlh ik=k dks fuokZfrr djds blesa dsoy ,d v.kq NksM+ fn;k tkrk gS vkSj a rFkk rms vkSlr pky rFkk oxZ

ek/; ewy osx O;Dr djrk gS rks & (A) a > rms (A) a > rms

(B) a < rms (B) a < rms

(C*) a = rms (C*) a = rms

(D) rms is undefined (D) rms vifjHkkf"kr gSA

9.

The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at the same temperature is about dejs ds rki ij vkWDlhtu dh oxZ ek/;ewy pky yxHkx 500 eh@ls gSA leku rki ij gkbMªkstu dh oxZ ek/;ewy pky yxHkx gksxh (A) 125 m/s (B*) 200 m/s (C) 800 m/s (D) 31 m/s

10.

The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be lerkih crZu esa j[ks xSl dk nkc 200 kPa gSA ;fn blesa ls vk/kh xSl dks fudky fn;k tk; rks nkc nkc gksxkA (A*) 100 kPa (B) 200 kPa (C) 400 kPa (D) 800 kPa

11.

The rms speed of oxygen molecules in a gas is . If the temperature is doubled and the O2 molecule dissociate into oxygen atoms, the rms speed will become fdlh xSl esa vkWDlhtu v.kqvksa dk oxZek/; ewy osx  gSA ;fn rki dks nqxuq k dj fn;k tk; rFkk vkWDlhtu v.kq vkWDlhtu

ijek.kqvksa esa VwV tk; rks oxZek/; ewy pky gksxhA (A) 

12.

13.

The quantity pV/kT represents jkf'k pV/kT O;Dr djrh gS & (A) mass of the gas (C) number of moles of the gas (A) xSl dk nzO;ekuA (C) xSl ds eksyksa dh la[;kA

(C*) 2 

(D) 4

(B) kinetic energy of the gas (D*) number of molecules in the gas (B) xSl dh xfrt ÅtkZA (D*) xSl esa v.kqvksa dh la[;kA

The process on an ideal gas, shown in fig. is fp=k esa n'kkZ;k x;k] ,d vkn'kZ xSl dk izØe gS -

(A) isothermal (A) lerkih; 14.

(B) 2

(B) isobaric (B) lenkch;

(C*) isochoric (C*) levk;rfud

(D) none of these (D) buesa ls dksbZ ugha

There is some liquid in a closed bottle. The amount of liquid is continuously decreasing. The vapour in the remaining part (A) must be saturated (B) must be unsaturated (C*) may be unsaturated (D) there will be no vapour manishkumarphysics.in

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Chapter # 24


,d can ckWVy esa dksbZ nzo Hkjk gqvk gSA nzo dh ek=kk fujUrj de gks jgh gSA 'ks"k cps Hkkx esa ok"i (vapour) (A) larÌr gksxhA (B) vlarÌr gksxhA (C*) larÌr gks ldrh gSA (D) ogk¡ dksbZ ok"i ugha gksxhA 15.

There is some liquid in a closed bottle. The amount of liquid is remains constant as time passses. The vapour in the remaining part (A*) must be saturated (B) must be unsaturated (D) may be unsaturated (D) there will be no vapour ,d can ckWVy esa dksbZ nzo Hkjk gqvk gSA le; xqtjus ds lkFk nzo dh ek=kk fu;r jgrh gSA 'ks"k cps gq, Hkkx esa ok"i (vapour)(A*) larÌr gksxh (B) vlarÌr gksxhA (D) larÌr gks ldrh gS (D) ogk¡ dksbZ ok"i ugha gksxhA

16.

Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel (A) increases continuously (B) decreases continuously (C) first increases and then decreases (D*) first increases and then becomes constant izkjEHk esa fuokZfrr ,d can esa fu;r nj ls ok"i izfo"V djokbZ tkrh gSA ik=k dk nkc (A) lrr~ :i ls c<+sxkA (B) lrr~ :i ls de gksxkA (C) igys c<+x s k rFkk fQj de gksxk (D*) igys c<+sxk rFkk fQj fu;r gks tk;sxkA

17.

A vessel A has volume V and a vessel B has volume 2V. Both contain some water which has constant volume. The pressure in the space above water is Pa for vessel A and Pb for vessel B : ,d ik=k A dk vk;ru V gS rFkk ik=k B dk vk;ru 2V gSA nksuksa esa dqN ikuh gS] ftldk vk;ru fu;r gSA ikuh ds Åij nkc dk eku ik=k A esa Pa rFkk ik=k B esa Pb gS rks : (A*) Pa = Pb (B) Pa = 2Pb (C) Pb = 2Pa (D) Pb = 4Pa

Objective - II 1.

Consider a collision between an oxygen molecule and a hydrogen molecules in a misture of oxygen and hudrogen kept at room temperature. Which of the following are possible ?

eku yhft, dejs ds rki ij j[ks vkWDlhtu vkSj gkbMªkt s u ds feJ.k esa ,d vkWDlhtu v.kq vkSj ,d gkbMªkt s u v.kq es VDdj gksrh gSA fuEu esa dkSu ls lEHko gS & (A) The kinetic energies of both the molecules increase.

nksuksa v.kqvksa dh xfrt ÅtkZ,a c<+ tk;sxhA (B) The kinetic energies of both the molecules decrease

nksuksa v.kqvksa dh xfrt ÅtkZ,a ?kV tk;sxhA (C*) The kinetic energy of the oxygen molecule increases and that of the hydrogen molecules decresases.

vkWDlhtu v.kq dh xfrt ÅtkZ c<+sxh rFkk gkbMªkt s u v.kq dh xfrt ÅtkZ ?kVsxhA (D*) The kinetic energy of the hydrogen molecules increases and that of the oxyzen molecule decreases.

gkbMªkstu v.kq dh xfrt ÅtkZ c<+sxh rFkk vkWDlhtu v.kq dh xfrt ÅtkZ ?kVsxhA 2.

Consider a mixture of oxygen and hydrogen kept at room tempertaure. As compared to a hydrogen molecule an oxygen molecule hits the wall

dejs ds rki ij vkWDlhtu vkSj gkbMªkt s u vkSj gkbMªkt s u dk feJ.k ekfu,A gkbMªkt s u v.kqvksa dh rqyuk esa vkWDlhtu v.kq nhokj ls VDdj ekjsxa s & (A) With greater average speed (C) with greater average kinetic energy (A) vf/kd vkSlr pky lsA (C) vf/kd vkSlr xfrt ÅtkZ lsA 3.

(B*) with smaller average speed (D) with smaller average kinetic energy. (B*) de vkSlr pky lsA (D) de vkSlr xfrt ÅtkZ lsA

Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium ?

fdlh vkn'kZ xSl ds v.kq lkE;koLFkk esa gS]a fuEu esa ls dkSulh jkf'k dk vkSlr 'kwU; gksxk & (A) kinetic energy (A) xfrt ÅtkZ 4.

(B*) momentum (B*) laosx

(C) density (C) ?kuRo

(D) speed (D) pky

Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gas ?

eksyksa dh la[;k] vk;ru rki dks leku j[kus ij çR;sd vkn'kZ xSl ds fy, fuEu esa ls dkSulh jkf'k leku gksxh& (A) rms speed of a milecule

(B) density manishkumarphysics.in

(C*) pressure Page # 4

Chapter # 24 Kinetic Theory of Gases (D) average magnitude of momentum. (A) v.kq dh oxZek/; ewy pky (B) ?kuRo (C*) nkc 5.

(D) laox s

dk vkSlr ifjek.kA

The average momentum of a molecule in a sample of an ideal gas depends on

vkn'kZ xSl ds uewus esa ,d v.kq dk vkSlr laox fuHkZj djrk gS & (A) temperature (A) rki 6.

(B) number of moles (B) eksyksa dh la[;k

(C) volume (C) vk;ru

(D*) none of these (D*) buesa ls dksbZ ughaA

Which of the following quantities is the same for all ideal gases at the same temperature ?

çR;sd vkn'kZ xSl ds fy, leku rki ij] fuEu esa dkSu lh jkf'k leku gksxh & (A*) the kinetic energy of 1 mole (C*) the number of molecules in 1 mole (A*) ,d eksy dh xfrt ÅtkZA (C*) ,d eksy esa v.kqvksa dh la[;kA 7.

(B) the kinetic energy of 1 g (D) the number of molecules in 1 g (B) ,d xzke dh xfrt ÅtkZA (D) ,d xzke esa v.kqvksa dh la[;kA

Consider the quantity MkT / pV of an ideal gas where M is the mass of the gas. It depends on the eku yhft, fd jkf'k MkT / pV ,d vkn'kZ xSl dh gSA tgk¡ M xSl dk nzO;eku gSA ;g fuHkZj djrk gSA (A) temperature of the gas (B) volume of the gas (C) pressure of the gas (D*) nature of the gas (A) xSl ds rki ijA (B) xSl ds vk;ru ijA (C) xSl ds nkc ijA (D*) xSl ds çd`fr ijA

WORKED OUT EXAMPLES 1.

A vessel of volume 8.0 x 10 -3 m 3 contains an ideal gas at 300 K and 200 kPa. The gas is allowed to leak till the pressure falls to 125 kPa. Calculate the amount of the gas (in moles) leaked assuming that the temperature remains constant.

Solution : As the gas leaks out, the volume and the temperature of the remaining gas do not change. The number of moles of the gas in the vessel is given by n

n1 

pV . The number of moles in the vessel before the leakage is RT

V V (p  p 2 ) V p1 p and that after the leakage is n 2  2 . Thus, the amount leaked is n1 -n2 = 1 RT RT RT

(200  125 ) x 10 3 N / m 2 x 8.0 x 10 3 m3  0.24 mol . = (8.3J / mol  K ) x (300 K ) 2.

A Vessel of volume 2000 cm3 contains 0.1 mole of oxygen and 0.2 mole of carbon dioxide. If the temperature of the mixture is 300 K, find its pressure.

Sol.

We have p =

nRT . V The pressure due to oxygen is (0.1mol ) (8.3 J / mol  K ) (300 K )

p1 =

( 2000 x 10

6

3

m )

 1.25 x 10 5 Pa

Similarly, the pressure due to carbon dioxide is p2 = 2.50 x 105 Pa. The total pressure in the vessel is p = p1 + p2 =(1.25 + 2.50 ) x 105 Pa = 3.75 x 105 Pa. 3. Sol. :

A mixture of hydrogen and oxygen has volume 2000 cm3, temperature 300K, pressure 100 kPa and mass 0.76 g. Calculate the masses of hydrogen and oxygen in the mixture. Suppose there aare n1 moles of hydrogen and n2 moles of oxygen in the mixture. The pressure of the mixture will be. p=

n1 RT n 2 RT RT   (n1  n 2 ) V V V manishkumarphysics.in

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Chapter # 24


or, 100 x 103 Pa = (n1 + n2 )

(8.3J / mol  K (300 K ) 2000 x 10 6 m 3

(n1 + n2 ) = 0.08 mol. ...(i) The mass of the mixture is n1 x 2g/mol + n2 x 32g/mol = 0.76 g or, n1 + 16n2 = 0.38 mol. ....(ii) From (i) and (ii), n1 -0.06 mol and n2 = 0.02 mol. The mass of hydrogen = 0.06 x 2 g = 0.12 g and the mass of oxygen = 0.02 x 32 g =0.64 g. 4.

Sol.:

A mercury monometer ( figure 24-W1) consits of two unequal arms of equal cross-section 1 cm2 and lengths 100 cm and 50 cm. The two open ends are sealed with air in the tube at a pressure of 80 cm of mervury. Some amount of mercury is now introduced in the manometer through the stopcock connected to it. If mercury rises in the shorter tube to a length 10 cm in steady state, find the length of the mercury column risen in the longer tube. Let p 1 and p2 be the pressures in centicetre of mercury in the two arms after introducing mercury in the tube. Suppose the mercury column rises in the second arm to lo cm.

P2

P1 0 cm 10cm

Using pV = constant for the shorter arm, (80 cm) (50 cm) = p1 (50 cm – 10 cm) or, p1 = 100 cm. .... (i) Using pV = constant for the longer arm, (80 cm) (100 cm) = p2(100 – 0) cm From the figure, p1 = p2 + (0 – 10)cm Thus by (i), 100 cm = p2 + (0 – 10)cm. or, p2 = 110 cm – 0 cm. Putting in (ii), (110 – 0) (100 – 0) = 8000 or, 02 – 210 0 + 3000 = 0 or, 0 = 15.5. The required length is 15.5 cm. 5.

Sol.

....(ii)

An ideal gas has pressure p0, volume V0 and temperature T0. It is taken through an isochoric process till its pressure is doubled. It is now isothermally expanded to get the original pressure. Finally, the gas is isobarically compressed to its original volume V0. (a) Show the process on a p–V diagram. (b) What is the temperature in the isothermal part of the process? (c) What is the volume at the end of the isothermal part of the process? (a) The process is shown in a p–V diagram in figure. The process starts from A and goes through ABCA.

2P0

P0

C

A v0

(b) Applying pV = nRT at A and B, p0V0 = nRT0 manishkumarphysics.in

Page # 6

Chapter # 24 Kinetic Theory of Gases and (2p0)V0 = nRTB. Thus, TB = 2T0 This is the temepratrure in the isothermal part BC. (c) As the process BC is isothermal, TC = TB = 2T0. Applying pV = nRT at A and C, p0V0 = nRT0 and p0VC = nR(2T0) or, VC = 2V0. 6.

A cylic process ABCA shown in the V–T diagram (figure) is performed with a constant mass of an ideal gas. Show the same process on a p–V diagram. In the figure. CA is parallel to the V-axis and BC is parallel to the T-axis.

V C

B A T

Sol.

pV = nR, we T see that the pressure p is constant in this part. This is represented by the part A’B’ in the p-V diagram. During The p–V diagram is shown in figure. During the part AB of figure, V is proportional to T. Thus,

p is constant. As the temperature decreases, pressure also T decreases. This is represented by the part B’C’ in the p-V diagram. During the part CA, the temperature remains constant so that pV = constant. Thus, p is inversely proportional to V. This is represented by the port C’A’ in the p-V diagram. the part, BC, volume is constant. Thus,

p A'

B' C' v

7.

Sol.

Two closed vessels of equal volume contain air at 105 kPa, 300 K and are connected through a narrow tube. If one of the vessels is now maintained at 300 K and the other at 400 K, what will be the pressure in the vessels ? Let the initial pressure, volume and temperature in each vessel be p0(=105 kPa), V0 and T0(=300 K). Let the number of moles is each vessel be n. When the first vessel is maintained at temperature T0 and the other is maintained at T’ = 400 K, the pressure change. Let the common pressure becomes p’ and the number of moles in the two vessels become n1 and n2. We have

p,'n1

p,'n2

T0=300K p0V0 = nRT0 ....(i) p’V0 = n1RT0 ....(ii) p’V0 = n2RT’ ....(iii) and n1 + n2 = 2n ....(iv) Putting n, n1 and n2 from (i), (ii) and (iii) in (iv), p V p' VB p' V0  2 0 0 RT0 RT' RT0 or,

 T'T0 p   T0 T'

 2p 0   T 0 

or,

2p 0 T' 2  105kPa  400K p’ = T' T = = 120 kPa. 400 K  300 K 0 manishkumarphysics.in

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Chapter # 24 8.

Sol.


A vessel contains 14 g of hydrogen and 96 g of oxygen of STP. (a) Find the volume of the vessel. (b) Chemical reaction is induced by passing electric spark in the vessel till one of the gases is consumed. The temperature is brought back to its starting value 273 K. Find the pressure in the vessel. (a) The number of moles of hydrogen = 14g / 2g = 7 and the number of moles of oxygen = 96 g / 32 g = 3. The total number of moles in the vessel = 7 + 3 = 10. The pressure is 1 atm = 1.0 × 105 N/m2 and the temperature = 273 K. Now pV = nRT .... (i) or,

V=

nRT p

(10 mol)  (8.3 J / mol  K )  ( 273 K )

=

1.0  10 6 N / m 2

= 0.23 m3 . (b) When electric spark is passed, hydrogen reacts with oxygen to form water (H2O). Each gram of hydrogen reacts with eight grams of oxygen. Thus,. 96 g of oxygen will be totally consumed together with 12 g of hydrogen. The gas left in the vessel will be 2 g of hydrogen which is n’ = 1 mole. Neglecting the volume of the water formed, p’V = n’RT. ...(ii) From (i) and (ii),

p' n' 1   p n 10 or,

9.

Sol.

10.

Sol.

p’ = p × 0.10 = 0.10 atm.

A borometer reads 75 cm of mercury. When 2.0 cm3 of air at atmospheric pressure is introduced into the space above the mercury level, the volume of this space becomes 50 cm3. Find the length by which the mercury column descends. Let the pressure of the air in the barometer by p. We have, p × 50 cm3 = (75 cm of mercury) × (2.0 cm3) or p = 3.0 cm of mercury. The atmospheric pressure is equal to the pressure due to the mercury column plus the pressure due to the air inside. Thus, the mercury column descends by 3.0 cm. A barometer tube is 1 m long and 2 cm2 in cross-section. Mercury stands to a height of 75 cm in the tube. When a small amount of oxygen is introduced in the space above the mercury level, the leve falls by 5 cm. Calculate the mass of the oxygen intorduced. Room temperature = 27ºC, g = 10 m/s2 and density of mercury = 13600 kg/m3. The pressure of oxygen in the space above the mercury level = 5 cm of mercury = 0.05 m × 13600 kg/m3 × 10 m/s2 = 6800 N/m2. The volume of oxygen = (2 cm2) × (25 cm + 5 cm) = 60 cm3 = 6 × 10–5 m3. The temperature = (273 + 27) K = 300 K. The amount of oxygen is n=

=

pV RT

(6800 N / m2 )  6  10 5 m3 (8.3 J / mol  K )  (300 K )

= 16.4 × 10–5 mol. The mass of oxygen is (16.4 × 10–5 mol) × (32 g/mol) = 5.24 × 10–3 g. 11.

Figure shows a vertical cylindrical vessel separated in two parts by a frictionless pistion free to move along the length of the vessel. The length of the cylinder is 90 cm and the piston divides the cylinder in the ratio of 5 : 4. Each of the two parts of the vessel contains o.1 mole of an ideal gas. The temperature of the gas is 300 K is each part. Calculate the mass of the piston. manishkumarphysics.in

Page # 8

Chapter # 24


90cm

Sol.

Let 1 and 2 be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, 1 = 50 cm and 2 = 40 cm. Let the pressure in the upper and lower parts be p1 and p2 respectively. Let the area of cross-section of the cylinder be A. The temperature in both parts is T = 300 K. Consider the equilibrium of the pistion. The forces acting onthe piston are (a) its weight mg (b) p1 A downward, by the upper part of the gas and (c) p2 A upwards, by the lower part of the gas. Thus, p2 A = p1 A + mg .... (i) Using pV = nRT for the upper and the lower parts p1L1A = nRT .....(ii) and p2L2A = nRT. .....(iii) Putting p1A and p2A from (ii) and (iii) into (i).

nRT nRT   mg 2 1 Thus,

m=

=

nRT  1 1    g   2 1 

(0.1 mol) (8.3 J / mol  K ) (300 K )  1 1     0 . 4 m 0 . 5 m 9 .8 m / s 2 

= 12.7 kg. 12.

Figure shows a cylindrical tube of volume V0 divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gases are filled in the two parts. When the separation is kept in the middle, the pressure are p1 and p2 in the left part and the right part respectively. The separation is slowly slid and is released at a position where it can stay in equilibrium. Find the volume of the two parts.

p1 Sol.

p2

As the separation is conducting, the temperatures in the two parts will be the same. Suppose the common temperature is T when the separation is in the middle. Let n1 and n2 be the number of moles of the gas in the left part and the right part respectively. Using ideal gas equation,

and

Thus,

p1

V0  n1RT 2

p2

V0  n 2RT 2 n1 p1  n2 p 2

.....(i)

The separator will stay in equilibrium at a position where the pressures on the two sides are equal. Suppose the volume of the left part is V1 and of the right part is V2 in this situation. Let the common pressure be p’. Also, let the common temperature in this situation be T’. Using ideal gas equation, p’V1 = n1RT’ and p’V2 = n2RT or, Also,

V1 n1 p1  = V2 n2 p 2 .

[using (i)]

V1 + V2 = V0


Page # 9

Chapter # 24


Thus, 13.

Sol.

V1 =

p 2 V0 p1V0 and V2 = p  p 1 2 p1  p 2

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing nercury and the parts on its two sides containing air at the same pressure p. When the tube is held at an angle of 60º with the vertical, the length of the air column above and below the mercury pallet are 46 cm and 44.5 cm respectively. Calculate the pressure p in centimetres of mercury. The temperature of the system is kept at 30ºC. When the tube is kept inclined to the vertical, the length of the upper part is 1 = 46 cm and that of the lower part is 2 = 44.5 cm. When the tube lies horizontally,the length on each side is

1   2 46 cm  44.5 cm = = 45.25 cm. 2 2 Let p1 and p2 be the pressure in the upper and the lower parts when the tube is kept inclined. As the temperature is constant throughout, we can apply Boyle’s law. For the upper part, p11 A = p0A 0 =

p 0 p1 =  1

or,

.....(i)

Similarly, for the lower part,

p 0 p2 =  2

.....(ii)

1

60º 2

Now consider the equilibrium of the mercury pallet when the tube is kept in inclined position. Let m be the mass of the mercury. The forces along the length of the tube are (a) p1 A down the tube (b) p2 A up the tube and (c) mg cos 60º down the tube.

mg cos 60º A Putting form (i) and (ii), Thus,

p2 = p1 +

p 0 p 0 mg = + 2 1 2A or,

 1 1 mg p0      = 2A 1  2

or,

p=

mg  1 1 2A 0      1  2

If the pressure p is equal to a height h of mercury, p = hg. Also, m = (5 cm) Ap so that

or,

hg =

h=

(5 cm) Ag  1 1 2A 0      1  2

(5 cm)  1 1   2( 45.25 cm)   44 . 5 cm 46 cm  

= 75.39 cm. The pressure p is equal to 75.39 cm of mercury.


Page # 10

//////////////////////

Chapter # 24 Kinetic Theory of Gases 14. A ideal monatomic gas is confined in a cylinder by a spring-loaded piston of cross-section 8.0 × 10–3 m2. Initially the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed state (figure). The gas is heated by a small heater until the pistion moves out slowly by 0.1 m. Calculate the final temperature of the gas. The force constant of the spring is 8000 N/m, and the atmospheric pressure is 1.0 × 10 5 N/m2. The cylinder and the piston are thermally insulated. The piston and the spring are masseless and there is no friction between the piston and the cylinder. Neglect any heat-loss through the lead wires of the heater. The piston and the spring are massless and there is no friction between the piston and the cylinder. Neglect any heat-loss through the lead wires of the heater. The heat capacity of the heater coil is neglibible.

Sol.

Initially the spring is in its relaxed state. So, the pressure of the gas equals the atmospheric pressure. Initial pressure = p1 = 1.0 × 10 5 N/m2. Final pressure

= p2 = p1 +

= 1.0 × 105 N/m2 +

kx A (8000 N / m) (0.1m) 8.0  10 3 m 2

= 2.0 × 10 5 N/m2. Final volume = V2 = V1 + Ax = 2.4 × 10–3 m3 + 8.0 × 10–3 m2 × 0.1 m = 3.2 × 10–3 m3 .

p1V1 T1

Using

T2

=

p 2 V2 T2 , (2.0  10 5 N / m 2 ) (3.2  10 3 m 3 )

p 2 V2 = p V T1 1 1

=

(1.0  10 5 N / m 2 ) (2.4  10 2 m 3 )

× 300 K

= 800 K. 15.

Sol.

Assume that the temperature remains essentially constant in the upper part of the atmosphere. Obtain an expression for the variation in pressure in the upperatmosphere with height. The mean molecular weight of air is M. Suppose the pressure at height h is p and that at h + dh is p + dp. Then dp = – g dh, ....(i) Now considering any small volume V of air of mass m, pV = nRT = or,

p=

or,

r=

m RT M

m RT RT = V M M M p. RT

Putting in (i), dp = – p

or,



p0

M pg dh RT h

dp M   g dh p RT

 0

p Mgh ln p = – RT 0 where p0 is the pressure at h = 0. or,

Thus,

p = p0 e



Mgh RT


Page # 11

Chapter # 24 16.


A horizontal tube of length  closed at both ends contains an ideal gas of molecules weight M. The tube is rotated at a constant angular velocity  about a vertical axis passing through an end. Assuming the temperature to be uniform and constant, show that 

Sol.

M2  2 2RT

p2 = p1 e , where p2 and p1 denote the pressure at the free end and the fixed end respectively. Consider an element of the gas between the cross-sections at distances x and x + dx from the fixed end (figure). If p be the pressure at x and p + dp at x + dx, the force acting on the element towards the centre is Adp, where A is the cross-sectional area. As this element is going in a circle of radius x,

x  Adp = (dm)2 x ....(i) where dm = mass of the element. Using pV = nRT on this element, pAdx = or,

dm =

dm RT M

MpA dx. RT

Putting in (i), Adp = p2

or,



p1

or,

or, 17.

Sol.

18. Sol.

MpA 2  x dx RT 1

dp M2  x dx RT RT

 0

p2 M2  2 ln p = 1 2RT p2 = p1 e



M2  2 2RT

.

A barometer tube contains a mixture of air and saturated water vapour in the space about the mercury column. It reads 70 cm when the actual atmospheric pressure is 76 cm of mercury. The saturation vapour pressure at room temperature is 1.0 cm of mercury. The tube is now lowered in the reaservoir till the space above the mercury column is reduced to half its original volume. Find the reading of the barometer. Assume that the temperature remains constant. The pressure due to the air + vapour is 76 cm – 70 cm = 6 cm of mercury. The vapour is saturated and the pressure due to it is 1 cm of mercury. The pressure due to the air is, therefore, 5 cm of mercury. As the tube is lowered and the volume above the mercury is decreased, some of the vapour will condense. The remaining vapour will again exert a pressure of 1 cm of mercury. The pressure due to air is doubled as the volume is halved. Thus, pair = 2 × 5 cm = 10 cm of mercury. The pressure due to the air + vapour = 10 cm + 1 cm = 11 cm of mercury. The barometer reading is 76 cm – 11 cm = 65 cm. Find the mass of water vapour per cubic metre of air at temperature 300 K and relative humidily 50%. The saturation vapour pressure at 300 K is 3.6 kPa and the gas constant R = 8.3 J/mol -K. At 300 K, the saturation vapour pressure = 3.6 kPa. Considering 1 m3 of volume,

m RT M where m = mass of vapour and M = molecular weight of water. pV = nRT =


Page # 12

Chapter # 24 Thus,

Kinetic Theory of Gases Mp V

m=

RT

(18 g / mol) (3.6  103 Pa) (1m3 ) = = 26 g (8.3 J / mol  K ) (300 K ) As the relative humidity is 50%, the amount of vapour present in 1 m3 is 26 g × 0.50 = 13 g. 19.

Sol.

The temperature and the relative humidity of air are 20ºC and 80% on a certain day. Find the fraction of the mass of water vapour that will condence if the temperature falls to 5ºC. Saturation vapour pressures at 20ºC and 5ºC are 17.5 mm and 6.5 mm of mercury respectively. The relative humidity is

vapour pressure of the air SVP at the same temperature Thus, the vapour pressure at 20ºC = 0.8 × 17.5 mm of Hg = 14 mm of Hg. Consider a volume V of air. If the vapour pressure is p and the temperature is T, the mass m of the vapour present is given by

m RT M

pV =

MV p . .... (i) R T The mass present at 20ºC is or,

m=

m1 =

MV 14 mm of Hg R 293 K

When the air is cooled to 5ºC, some vapour condenses and the air gets saturated with the remaining vapour. The vapour pressure at 5ºC is, therefore, 6.5 mm of mercury. The mass of vapour present at 5ºC is, therefore, m2 =

MV 6.5 mm of Hg R 278 K

The fraction condensed = 20.

Sol.

m1  m 2 m  1 2 m1 m1

=1–

6.5 293 × = 0.51. 278 14

A vessel containing wate is put in a dry sealed room of volume 76m3 at a temperature of 15ºC. The saturation vapour pressure of water at 15ºC is 15 mm of mercury. How much water will evaporate before the water is in equilibrium with the vapour? Water will be in equilibrium with its vapour when the vapour gets saturated vapour pressure = 15 mm of mercury = (15 × 10–3 m) (13600 kg/m2) (9.8 m/s2) = 2000 N/m2. Using gas law, pV =

m=

Mp V

m RT M

(18g / mol)(2000 N / m 2 ) (76 m3 ) = (8.3J / mol  K ) (288 K )

RT Thus, 1.14 kg of water will evaporate.

21.

Sol.

A jar contains a gas and a few drops of water at absolute temperature T1. The pressure in the jar is 830 mm of mercury. The temperature of the jar is reduced by 1%. The saturation vapour pressures of water at the two temperatures are 30 mm of mercury and 25 mm of mercury. Calculate the new pressure in the jar. At temperature T1, the total pressure is 830 mm of mercury. Out of this, 30 mm of mercury is due to the vapour and 800 mm of mercury is due to the gas. As the temperature decreases, the pressure due to the gas decreases according to the gas law. here the volume is constant, so,

p 2 p1  T2 T1 manishkumarphysics.in

Page # 13

Chapter # 24


T2 p2 = T p1. 1

or,

As T2 is 1 % less than T1 T2 = 0.99 T1 and hence, p2 = 0.99 p1 0.99 × 800 mm of mercury = 792 mm of mercury. The vapour is still saturated and hence, its pressure is 25 mm of mercury. The total pressure at the reduced temperature is p = (792 + 25) mm of mercury. = 817 mm of mercury. 22.

Calculate the mass of 1 litre of moist air at 27ºC when the barometer reads 753.6 mm of mercury and the dew point is 16.1ºC. Saturation vapour pressure of water at = 0.001293 g/cc, density of saturated water vapour at STP = 0.000808 g/cc.

Sol.

We have pV =

m RT M

m Mp = ... (i) V RT The dew point is 16.1ºC and the saturation vapour pressure is 13.6 mm of mercury at the dew point. This means that the present vapour pressure is 13.6 mm of mercury. At this pressure and temperature, the density of vapour will be or

=

=

=

Mp RT

(18 g / mol) (13.6  10 3 m)(13600 kg / m3 ) (9.8 m / s 2 ) (8.3 J / mol  K ) (300 K )

= 13.1 g/m3. Thus, 1 litre of moist air at 27ºC contains 0.0131 g of vapour. The pressure of dry air at 27ºC is 753.6 mm – 13.6 mm = 740 mm of mercury. The density of air at STP is 0.001293 g/cc. the density at 27ºC is given by equation (i),

1 p1 / T1  1 p 2 / T2 or,

p 2 T1 2 = T p 1 2 1

=

740  273 × 0.001293 g/cc. 300  760

= 0.001457 g/cc. Thus, 1 litre of moist air contains 1.145 g of dry air. The mass of 1 litre of moist air is 1.1457 g + 0.0131 g  1.159 g.

EXERCISE 1. Ans. 2. Ans. 3. Ans. 4. Ans.

Calculate the volume of 1 mole of an ideal gas at STP. ekud rki o nkc (STP) ij 1 eksy vkn'kZ xSl ds vk;ru dh 2.24 × 10–2 m3

x.kuk dhft;sA

Find the number of molecules of an ideal gas in a volume of 1.000 cm3 at STP. ekukd rki o nkc (STP) ij 1.000 lseh3 vk;ru ds fy;s vkn'kZ xSl ds v.kqvksa dh la[;k 2.685 × 1019

Kkr dhft;sA

Find the number of molecules in 1 cm3 of an ideal gas at 0ºC and at a pressure of 10–5 mm of mercury. 0°C rki ij ,oa 10–5 feeh ikjn~ nkc ij vkn'kZ xSl ds 1 lseh3 vk;ru esa v.kqvksa dh la[;k Kkr dhft;sA 3.53 × 1011 Calculate the mass of 1 cm 3 of oxygen kept at STP. STP ij j[kh x;h vkWDlhtu ds 1 lseh3 vk;ru ds nzO;eku 1.43 mg

dh x.kuk dhft;sA


Page # 14

Chapter # 24 5.

Ans. 6.


Equal masses of air are sealed in two vessels, one of volume V0 and the other of volume 2V0. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels. ok;q dk leku nzO;eku nks ik=kksa esa can fd;k x;k gS] ,d dk vk;ru V0 rFkk nwljs dk 2V0 gSA ;fn izFke ik=k dk rki 300 K rFkk nwljs dk rki 600 K j[kk tkrk gS] nksuksa ik=kksa esa nkcksa dk vuqikr Kkr dhft;sA 1:1 An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10–3 mm of mercury at 27ºC. Complete the number of air molecules contained in the bulb. Avogadro constant = 6 × 10 23 per mol, density of mercury = 136000 kg/m3 and g = 10 m/s2. 250 ?ku lseh vk;ru okyk ,d fo|qr&cYc cukrs le; 10–3 feeh ikjn~ nkc ,oa 27ºC rki ij can fd;k x;k FkkA cYc

esa ifjc) ok;q ds v.kqvksa dh la[;k dh x.kuk dhft;sA vkokxknzks fu;rkad dh la[;k dh x.kuk dhft;sA vkokxknzks fu;rkad = 6 × 10 23 izfr eksy] ikjs dk ?kuRo = 136000 fdxzk/eh3 rFkk g = 10 m/s2 Ans.

8.0 × 1015

7.

A gas cylinder has walls that can bear a maximum pressure of 1.0 × 10 6 Pa. It contains a gas at 8.0 × 10 5 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break. ,d xSl flys.Mj dh nhokjsa vf/kdre 1.0 × 10 6 ikLdy nkc lgu dj ldrh gSA blesa 8.0 × 105 nkc rFkk 300 K rki

ij ,d xSl Hkjh gqbZ gSA flys.Mj dks /khjs&/khjs xeZ fd;k tkrk gSA vk;ru esa ifjorZu dks ux.; ekudj] og rki Kkr dhft;sA ftl ij flys.Mj QV tk;sxkA Ans.

375 K

8.

2 g of hydrogen is sealed in a vessel of volume 0.02 m 3 and is maintained at 300 K. Calculate the pressure in the vessel. 2 xzke gkbMªkt s u 0.02 eh3 vk;ru okys ik=k esa Hkjh gqbZ gS] ftldk rki 300 K gSA ik=k esa nkc dh x.kuk dhft;sA 5 1.24 × 10 Pa

Ans. 9. Ans. 10.

The density of an ideal gas is 1.25 × 10–3 g/cm3 at STP. Calculate the molecular weight of the gas. STP ij ,d vkn'kZ xSl dk ?kuRo 1.25 × 10–3 xzke/lseh3 gSA xSl ds vkof.kd Hkkj dh x.kuk dhft;sA 28.3 g/mol The temperature and pressure at Simla are 15.0ºC and 72.0 cm of mercury and at Kalka these are 35ºC and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla. f'keyk esa rki ,o na kc 15.0ºC ,oa 72.0 ikjn~ LrEHk gS rFkk dkydk esa buds eku 35ºC ,oa 76.0 lseh ikjn~LrEHk gSA dkydk

,oa f'keyk esa ok;q ?kuRoksa dk vuqikr Kkr dhft;sA Ans.

0.987

11.

Figure shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected in the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1 : 3. Find the ratio of the pressures in the two parts of the vessel.

fp=k esa :nks"e nhokjksa okyh ,d csyukdkj uyh iznf'kZr gS ftlesa ,d Å"ekjks/kh i`Fkd~dkjh yxk gqvk gSA i`Fkd~dkjh dks cká ;kaf=kd O;oLFkk }kjk uyh esa f[kldk;k tk ldrk gSA nksuksa vksj ,d vkn'kZ xSl leku nkcksa ,oa lekurkiksa ij izfo"V djk nh tkrh gSA i`Fkddkjh lkE;koLFkk esa e/; esa cuk jgrk gSA bldk vc ,d ,slh fLFkfr esa f[kldk;k tkrk gS fd ;g uyh dks 1 : 3 esa foHkkftr djrk gSA ik=k esa nksuksa Hkkxksa esa nkcksa dk vuqikr Kkr dhft;sA

Ans.

3:1

12.

Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which the rms speed is double the speed calculated in the previous part. 300 K rki okys gkbMªkstu ds uewus esa gkbMªkstu v.kqvksa dh oxZ ek/; ewy pky Kkr dhft;sA og rki Kkr dhft;s ftl

ij oxZ ek/; ewy pky fiNys Hkkx esa x.kuk ls izkIr eku dh nqxuh gksxhA Ans.

1930 m/s, 1200 K

13.

A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules. ekud rki o nkc (STP) ij vkn'kZ xSl dk 0.177 xzke izfrn'kZ 1000 lseh3 vk;ru ?ksjrk gSA xSl ds v.kqvksa dh oxZek/;ewy manishkumarphysics.in

Page # 15

Chapter # 24


pky dh x.kuk dhft;sA Ans.

1300 m/s

14.

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10–19 J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10–23 J/K. ok;q ds v.kqvksa dh vkSlr LFkkukarj.k xfrt ÅtkZ 0.040 eV (1 eV = 1.6 × 10–19 twy) gSA ok;q ds rki dh x.kuk dhft;sA cksYV~eku fu;rkad k = 1.38 × 10–23 twy/dsfYou 310 K

Ans. 15.

Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth. 300 K rki ij vkWDlhtu ds ,d izfrn'kZ ij fopkj dhft;sA xSl ds v.kq dks i`Foh ds O;kl ds cjkcj nwjh r; djus esa yxk

vkSlr le; Kkr dhft;sA Ans. 16.

Ans. 17.

8.0 hour Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0ºC. Mass of a helium molecule = 6.64 × 10–27 kg and Boltzmann constant = 1.38 × 10 –23 J/K. 0ºC rki ij ghfy;e xSl ds uewus esa ,d ghfy;e v.kq ds jSf[kd laox s dk vkSlr ifjek.k Kkr dhft;sA ghfy;e v.kq dk nzO;eku = 6.64 × 10–27 fdxzk rFkk cksYV~eku fu;rkad = 1.38 × 10–23 twy@dsfYou 8.0 × 10–24 kg–m/s The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

,d gkbMªkt s u izfrn'kZ esa v.kqvksa dh vkSlr pky] ghfy;e izfrn'kZ esa v.kqvksa dh vkSlr pky ds cjkcj gSA gkbMªkt s u izfrn'kZ rFkk ghfy;e izfrn'kZ ds rkiksa ds vuqikr dh x.kuk dhft;sA Ans.

1:2

18.

At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth?

fdl rki ij gkbMªkt s u xSl ds v.kqvksa dh ek/; pky] i`Foh ds iyk;u pky ds cjkcj gks tk;sxh\ Ans.

11800 K

19.

Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

nks xSlksa ds feJ.k esa gkbMªkstu v.kqvksa rFkk ukbVªkstu v.kqvksa dh ek/; pkyksa dk vuqikr Kkr dhft;sA Ans.

3.74

20.

Figure shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part of the mass of molecule in the right part.

fp=k esa fLFkj Å"ekjks/kh i`Fkddkjh ls foHkkftr ,d ik=k iznf'kZr fd;k x;k gSA nksuksa Hkkxksa esa vyx&vyx vkn'kZ xSlas Hkjh gqbZ gSA cka;s Hkkx ds v.kqvksa dh oxZ ek/; ewy pky nk;sa Hkkx ds v.kqvksa dh ek/; pky ds cjkcj gSA ck;sa Hkkx ds v.kqvksa ds nzO;eku rFkk nk;sa Hkkx ds v.kqvksa ds nzO;eku ds vuqikr dh x.kuk dhft;sA

Ans.

1.18

21.

Estimate the number of collisions per second suffered by a molecules in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10–5 cm. STP ij gkbMªkt s u izfrn'kZ esa ,d v.kq }kjk izfr lsd.M dh x;h VDdjksa dh x.kuk dhft;sA ek/; eqDr iFk ¼nks Øekxr VDdjksa ds chp r; dh x;h vkSlr nwjh½ = 1.38 × 10–5 lsehA 1.23 × 10 10

Ans. 22.

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45º with it. How many molecules strike each square metre of the wall per second? ,d can ik=k esa 1 ok;qe.Myh; nkc (100 kPa) rFkk 300 K rki ij gkbMªkt s u xSl Hkjh gqbZ gSA (a) v.kqvksa dh ek/; pky dh x.kuk dhft;sA (b) eku yhft;s fd v.kq nhokj ls vkSlru 45º dks.k cukrs gq, nhokj ls Vdjkrs gSAa nhokj ds izfr oxZ ehVj

{ks=kQy ls izfr lsd.M fdrus v.kq Vdjk;saxs\


Page # 16

Chapter # 24 Ans. (a) 1780 m/s 23.

Ans. 24.

Kinetic Theory of Gases (b) 1.2 × 10 28

Air is pumped into an automobile tyre’s tube upto a pressure of 200 kPa in the morning when the air temperature is 20ºC. During the day the temperature rises to 40ºC and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature. izkr%dky tc ok;q dk rki 20°C gS] ,d eksVjxkM+h ds Vk;j esa 200 kPa nkc rd iEi ls gok Hkjh tkrh gSA fnu esa rki 40ºC rd c<+ tkrk gS rFkk V~;wc 2% izlkfjr gks tkrh gSA bl rki ij V~;c w esa gok ds nkc dh x.kuk dhft;sA 209 kPa Oxygen is filled in a closed metal jar of volume 1.0 × 10–3 m–3 at a pressure of 1.5 × 10 5 Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 105 Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding. 1.0 × 10–3 eh–3 vk;ru okys /kkrq ds ,d can tkj esa 1.5 × 10 5 ikWLdy nkc rFkk 400 K rki ij vkWDlhtu Hkjh gqbZ gSA tkj esa FkksM+k lk fjlko gSA ok;qe.Myh; nkc 1.0 × 105 ikLdy rFkk ok;qe.Myh; rki 300 K gSA tc tkj ds vUnj rki

o nkc] cká okrkoj.k ds cjkcj gks tkrk gS] tkj ls fjl dj ckgj fudyh xSl dk nzO;eku Kkr dhft;sA Ans.

0.16 g

25.

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 × 10 5 Pa and density of water = 1000 kg/m3. ,d 3.3 eh- xgjh unh ds iSna as esa 2.0 feeh f=kT;k okyk ok;q dk cqycqyk mRiUu gksrk gSA tc ;g lrg ij vk tkrk gS] bldh f=kT;k dh x.kuk dhft;sA ok;qe.Myh; nkc = 1.0 × 10 5 ikLdy rFkk ty dk ?kuRo = 1000 fdxzk@eh3 2.2 mm

Ans. 26.

Ans. 27.

Ans. 28.

Ans. 29.

Ans. 30.

Ans. 31.

Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m3. One of the tubes gets punctured and the volume of the tube reduces to 0.0005 m3. How many moles of air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas. lkbfdy fjD'kk ds Vk;jksa esa iEi ls 2 ok;qe.Myh; nkc rd gok Hkjh x;h gSA bl nkc ij izR;sd V~;cw dk vk;ru 0.002 eh3 gSA buesa ls ,d V~;cw iaDpj gks tkrh gS rFkk bldk vk;ru 0.0005 eh3 rd de gks tkrk gSA ok;q ds fdrus eksy fjl dj ckgj fudy x;s\ eku yhft;s fd rki 300 K fu;r jgrk gS rFkk ok;q vkn'kZ xSl dh Hkkafr O;ogkj djrh gSA 0.14 0.040 g of He is kept in a closed container initially at 100.0ºC. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J. k 0.040 g of He is kept in a closed container initially at 100.0ºC. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J. k 196ºC During an experiment, an ideal gas is found to obey an additional law pV2 = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V. k During an experiment, an ideal gas is found to obey an additional law pV2 = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V. k T/2 A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture. k A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture. k 2250 N/m2 A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston. 25 cm Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy

p 1  p A pB      T 2  TA TB  manishkumarphysics.in

Page # 17

Chapter # 24


when equilibrium is achieved. fp=k esa n`<+ nhokjksa okys nks ik=kksa A o B esa vkn'kZ xSlas Hkjh gqbZ gSA ik=k A esa nkc rki rFkk vk;ru pA, TA, V ,oa ik=k B esa pB, TB, V gSA ik=kksa dks vc ,d NksVh uyh ls tksM+ fn;k tkrk gSA O;Dr dhft;s fd tc lkE;koLFkk izkIr gks tk;sxh rks nkc p rFkk rki T fuEu lEcU/k dks larq"V djsxsa :

p 1  p A pB      T 2  TA TB  when equilibrium is achieved. k

A 32.

B

A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0ºC). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100ºC). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached. 50 ?ku lseh vk;ru okys ik=k esa ok;q ¼ek/; v.kqHkkj = 28.8 xzke) Hkjh gqbZ gS ,oa bldks ok;qe.My esa [kksy fn;k tkrk gS] tgk¡ nkc 100 kPa gSA ik=k dks fi?kyrs gq, cQZ (0ºC) dh uko esa j[kk tkrk gSA (a) m"eh; lkE;koLFkk izkIr djus ij ik=k esa gok dk nzO;eku Kkr dhft;sA (b) ik=k dks vc ,d uko esa j[k fn;k tkrk gS] ftlesa mcyrk gqvk ikuh (100ºC) gSA ik=k esa gok dk nzO;eku Kkr dhft;sA (c) vc ik=k dks can djds fi?kyrs gq, cQZ okyh uko esa j[k fn;k tkrk gSA Å"eh; lkE;koLFkk

izkIr gksus ij gok dk nkc Kkr dhft;sA Ans.

(a) 0.058 g

(b) 0.0468 g

(c) 73.0 kPa

33.

A uniform tube closed at one end, contains a pallet of mercury 10 cm long. When the tube is kept vertically with the closed end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure = 75 cm of mercury. dkap dh ,d le:i uyh] ,d fljs ij can gSA blesa 10 lseh yEch ikjs dh xksyh gSA tc uyh dks bl izdkj m/okZ/kj j[kk tkrk gS fd bldk can fljk Åij dh vksj jgrk gS] rks ifjc) ok;q LrEHk dh yEckbZ 20 lseh gSA ;fn bldks uhps dh vksj

bl izdkj mYVk dj fn;k tk;s fd bldk can fljk uhps dh vksj jgs rks blesa ifjoc) ok;q LrEHk dh yEckbZ Kkr dhft;sA ok;qe.Myh; nkc = 75 lseh ikjn~ LrEHkA Ans.

15 cm

34.

A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27ºC and at a pressure 76 cm of mercury. The air column on one side is maintained at 0ºC and the other side is maintained at 127ºC. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass. nksuksa fljksa ij can dh gqbZ dkap dh uyh dh yEckbZ 100 lseh gSA ;g {kSfrt j[kh gqbZ gS rFkk blds e/; esa 10 lseh yEckbZ esa ikjk gSA uyh ds nksuksa fljksa esa 27ºC rki ,oa 76 lseh ikjs ds nkc ij ok;q Hkjh gqbZ gSA uyh ls ,d vksj ds ok;q LrEHk dk rki 0ºC j[kk tkrk gS rFkk nwljh vksj dk 127ºC j[kk tkrk gSA B.Mh vksj ds ok;q LrEHk dh yEckbZ dh x.kuk dhft;sA ikjs

rFkk dkap ds vk;ru esa ifjorZu dks ux.; eku yhft;sA Ans.

36.5 cm

35.

An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60º? Assume the temperature to remain constant.

le:i vk/kkj okyh ,d ladjh m/oZ uyh ds can fljs rFkk blesa Hkjs gq, ikjn~ LrEHk ds chp esa ,d vkn'kZ xSl ifjc) gSA uyh dk Åij fljk ok;qe.My esa [kqyk gqvk gSA ok;qe.Myh; nkc 76 lseh- ikjn~ LrEHk ds rqY; gSA uyh esa ikjn~ LrEHk rFkk ifjc) ok;q LrEHk dh yEckbZ;k¡ Øe'k% 20 lseh- ,oa 43 lseh- gSA ;fn uyh dks m/okZ/kj ry esa /khjs&/khjs 60º dks.k ls >qdk;k tkrk gS rks uyh esa ok;q LrEHk dh yEckbZ fdruh gksxh \ ;g eku fyft;s fd rki fu;r jgrk gSA Ans.

48 cm

36.

Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of manishkumarphysics.in

Page # 18

Chapter # 24 Kinetic Theory of Gases the separator, reached after a long times. fp=k esa 30 lseh- yEch ,d csyukdkj uyh iznf'kZr gSA ftls chp esa ,dne lgh

cSBus okyk inkZ yxkdj foHkkftr fd;k x;k gSA inkZ Å"ek dk nqcy Z pkyd gS rFkk uyh esa vklkuh ls f[kld ldrk gSA ik=k ds nksuksa Hkkxksa esa vkn'kZ xSl Hkjh gqbZ gSA izkjEHk esa Hkkxksa A o B ds rki Øe'k% 400 K ,oa 100 K gSA inkZ fp=k esa iznf'kZr fLFkfr ls {kf.kd :i ls lkE;oLFkk dh fLFkfr rd f[kld ldrk gSA yEcs le; i'pkr~ insZ }kjk izkIr dh x;h lkE; fLFkfr Kkr dhft;sA 20 cm

10 cm

400 K A

100 K B

Ans.

10 cm from the left end

37.

A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume – rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out. V0 vk;ru okys ,d ik=k esa nkc p0 ,oa rki T ij ,d vkn'kZ xSl Hkjh x;h gSA iEi dh lgk;rk ls xSl dks rki fu;r j[krs gq, bl ik=k ls fu;r vk;ru nj dV/dt = r ls ckgj fudkyk tkrk gSA ckgj fudkyh x;h xSl dk nkc] ik=k es vUnj okys nkc ds cjkcj gSA Kkr dhft;s % (a) le; ds Qyu :i esa xSl dk nkc] (b) iEi ls izkjfEHkd xSl dh vk/kh xSl ckgj fudkyus

esa yxk le;A

V0 ln 2 

Ans.

(a) p = p0 e  t / V0

38.

One mole of an ideal gas undergoes a process p =

(b)

p0 1  ( V / V0 )2

, where p0 and V0 are constants. Find the

temperature of the gas when V = V0. p0

,d eksy vkn'kZ xSl fuEu izØe ls xqtjrh gS % p = 1  ( V / V )2 , tgk¡ p0 ,oa V0 fLFkjkad gSA tc V = V0 gks rks xSl dk 0 rki Kkr dhft;sA Ans. 39.

p 0 V0 –1 2 R mol . Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows etc. O;Dr dhft;s fd fnu vkSj jkr ds rki ifjorZu es]a fdlh dejs esa fLFkr ok;q (vkn'kZ xSl dh Hkkafr ekudj) dh vkarfjd ÅtkZ]

fu;r jgrh gSA eku yhft;s fd ok;qe.Myh; nkc fu;r jgrk gS rFkk dejs dh ok;q dk f[kM+fd;ksa vkfn ds }kjk okrkoj.k ls vknku&iznku gksus ds dkj.k nkc dk ;g eku Hkh fu;r jgrk gSA 40.

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl, of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate

dN . dl

fp=k esa 5 lseh- f=kT;k rFkk 20 lseh- yEckbZ dh ,d csyukdkj uyh n'kkZ;h x;h gSA ;g ,d dkdZ ls dldj can dh x;h gSA dkWdZ rFkk uyh ds chp ?k"kZ.k xq.kkad 0.20 gSA uyh esa 1 ok;qe.Myh; nkc rFkk 300 K rki ij ,d vkn'kZ xSl Hkjh gqbZ gSA uyh dks /khjs&/khjs xeZ fd;k tkrk gS rfkk ;g izfs {kr gksrk gS fd 600 K rki ij dkWdZ mNydj ckgj vk tkrk gSA ekukfd dkEdZ dh ifjf/k ds vuqfn'k vR;Yi yEckbZ (fp=k) ij vfHkyEcor~ lEidZ cy dk ifjek.k dN gSA ;g ekurs gq, fd fdlh Hkh {k.k ij iwjh uyh dk rki le:i gS] Ans.

dN dl

dh x.kuk dhft;sA

1.25 × 10 4 N/m manishkumarphysics.in

Page # 19

Chapter # 24 Kinetic Theory of Gases 41. Figure shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T0 and its pressure is p0 which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T0 ? [HCV_Chp_24_Ex_41] fp=k esa A vuqiLz Fk dkV {ks=kQy okyh ,d csyukdkj uyh iznf'kZr gSA ftlds nksuksa fljksa ij nks fiLVu yxs gq, gS] ftudks ,d /kkfRod rkj ls tksMk+ x;k gSA izkjEHk esa xSl dk rki T0 rFkk bldk nkc p0 gS tks ok;qe.Myh; nkc ds cjkcj gSA (a) rkj esa ruko dk eku fdruk gS \ (b) ;fn rki 2T0 rd c<+k fn;k tk;s rks ruko dk eku fdruk gksxk ?

Ans.

(a) zero

(b) p0 A

42.

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h2 of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

fp=k esa ikuh ls Hkjk gqvk cgqr cM+k ,d csyukdkj can Vsd a iznf'kZr gSA izkjEHk esa ikuh dh lrg ds Åij ifjc) gok dh Å¡pkbZ h0 gS rFkk gok dk nkc 2p0 gS tgk¡ p0 ok;qe.Myh; nkc gSA Vsd a ds Åijh fljs ls h1 uhps bldh nhokj esa ,d fNnz gS] ftlls ikuh ckgj vk jgk gSA ,d m/okZ/kj yEch uyh fp=kkuqlkj Vsd a ls tksMh+ x;h gSA (a) izkjEHk esa yEch uyh esa Vsd a ds Åij ry ls ikuh dh Å¡pkbZ h2 Kkr dhft;sA (b) fNnz ls ckgj fudyrs gq, ikuh dh pky Kkr dhft;sA (c) tc ikuh fNnz ls ckgj fudyuk can gks tkrk gS] yEch uyh esa] Vsad ds Åijh ry ls ikuh dh Å¡pkbZ Kkr dhft;sA Ans. 43.

p0 (a) g  h 0

2  (b)  p 0  g(h1  h 0 )   

1/ 2

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg (figure). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process. ,d yEcs csyukdkj ik=k esa ,d ?k"kZ.k jfgr fiLVu yxk gqvk gSA ftldk vuqiLz Fk dkV {ks=kQy 10 lseh-2 rFkk Hkkj 1 fdxzkgS] blesa ,d vkn'kZ xSl Hkjh gqbZ gS (fp=k)A ;g csyukdkj ik=k ,d vU; c<+s d{k esa j[kk gqvk gS] ftlesa 100 fdyks ikWLdy ok;qe.Myh; nkc ij ok;q Hkjh gqbZ gSA xSl LrEHk dh yEckbZ 20 lseh- gSA ;fn d{k dks ,d fuokZr iEi dh lgk;rk ls iwjk

fuokZfrr dj fn;k tkrk gS] xSl LrEHk dh yEckbZ fdruh gksxh \ eku fyft;s fd lEiw.kZ izØe esa rki fu;r jgrk gSA Ans.

2.2 m

44.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder. ,d yEcs csyukdkj ik=k esa ,d ?k"kZ.k jfgr fiLVu yxk gqvk gSA ftldk vuqiLz Fk dkV {ks=kQy 10 lseh-2 rFkk Hkkj 1 fdxzkgS] blesa ,d vkn'kZ xSl Hkjh gqbZ gSA ik=k esa xSl LrEHk dh yEckbZ 20 lseh- gSA ok;qe.Myh; nkc 100 fdyks ikWLdy gSA vc

ik=k dks ,d varfj{k ;ku esa ys tk;k tkrk gS] tks i`Foh ds pkjksa vksj mixzg dh Hkkafr ifjØek dj jgk gSA varfj{k ;ku esa ok;q nkc 100 fdyks ikWLdy j[kk x;k gSA flys.Mj esa xSl LrEHk dh yEckbZ Kkr dhft;sA Ans.

22 cm

45.

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62ºC. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible. leku vk;ru okys dkap ds nks cYc ,d ladjh uyh ls tqM+s gq, gS] buesa 0ºC rki rFkk 76 lseh- ikjs ds nkc ij ,d xSl Hkjh gqbZ gSA vc buesa ls ,d cYc dks fio?kyrs gq, cQZ dh ukn esa rFkk nwljs dks 62ºC fLFkj rki okyh ikuh dh ukn esa j[k

fn;k tkrk gSA cYcksa ds vUnj nkc dk u;k eku fdruk gS \ tksM+us okyh uyh dk vk;ru ux.; gSA Ans.

84 cm of mercury

46.

The weather report reads, “Temperature 20ºC : Relative humidity 100%”. What is the dew point? eksle dh fjiksVZ gS ^^ rki 20ºC : vkisf{kd vknzrk 100%** vkslkad (dewpoint) dk eku fdruk gS \ 20ºC

Ans.


Page # 20

Chapter # 24 47.

Ans. 48.


The condition of air in a closed room is described as follows. Temperature = 25ºC, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25ºC = 3.2 kPa. fdlh can dejs esa fLFkr gok dh voLFkk vxzkuqlkj O;ä dh x;h gSA rki = 25ºC , vkisf{kd vknzrk = 60%, nkc = 104 kPaA ;fn rki vifjofrZr j[krs gq, dejs esa ls lEiw.kZ ty ok"i gVk yh tk;s rks u;k nkc fdruk gksxk \25ºC ij larI` r ok"i nkc = 3.2 kPaA 102 kPa The temperature and the dew point in an open room are 20ºC and 10ºC. If the room temperature drops to 15ºC, what will be the new dew point? ,d [kqys gq, dejs esa rki rFkk vkslkad (dewpoint) Øe'k% 20ºC rFkk 10ºC gSA ;fn dejs dk rki fxjdj 15ºC gks tk;s

rks vkslkad dk u;k eku fdruk gksxk \ Ans.

10ºC

49.

Pure water vapour is trapped in a vessel of volume 10 cm3. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing. 10 lseh-3 vk;ru okys ik=k esa 'kq) ty ok"i ifjc) gSA vkisf{kd vknzrk 40% gSA ok"i dks /khjs&/khjs ,oa lerkih; :i ls

lEihfM+r fd;k tkrk gSA ok"i ds vk;ru dk og eku Kkr dhft;s ftl ij ;g la?kfur gksuk 'kq: gks tk;sxhA Ans.

4.0 cm3

50.

A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapour pressure at the room temperature is 1.0 cm. ,d ok;qnkc ekih (barometer) dh uyh 80 lseh yEch gSA fdlh fof'k"V fnol ij bldk ikB~;kad 76 lseh gSA uyh esa ikuh dh FkksM+h lh ek=kk izfo"V djk nh tkrh gS ,oa ikB~;kad 75.4 lseh rd de gks tkrk gSA ;fn dejs ds rki ij larI` r ok"i nkc 1.0 lseh gSA ;fn dejs ds rki ij larI` r ok"i nkc 1.0 lseh gS rks ikjn~ LrEHk ds Åij okys LFkkuesa vkisf{kd vknzrk

Kkr dhft;sA Ans.

60%

51.

Using figure, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm. fuEu fp=k dk mi;ksx djds ok;qe.Myh; nkc ,oa 0.5 ok;qe.Myh; nkc ij feFkkby ,Ydksgky dk DoFkukad Kkr

dhft;sA

Ans.

65ºC, 48ºC

52.

The human body has an average temperature of 98º F. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure for the vapour pressures. ekuo 'kjhj dk vkSlr rki 98º F gksrk gSA eku yhft;s fd :f/kj ufydkvksa esa jDr dk ok"i nkc] 'kq) ty ds ok"i nkc

dh Hkkafr O;ogkj djrk gSA :f/kd dks mcyus ls jksdus ds fy;s vko';d U;wure ok;qe.Myh; nkc Kkr dhft;sA

Ans.

50 mm of mercury

53.

A glass contains some water at room temperature 20ºC. Refrigerated water is added to it slowly. When the temperature of the glass reaches 10ºC, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20ºC and at 8.9 mm of mercury it is 10ºC. ,d fxykl esa dejs ds rki 20ºC ij ikuh Hkjk gqvk gSA blesa /khjs&/khjs 'khrfyr ty feyk;k tkrk gSA tc fxykl dk rki de gksdj 10ºC jg tkrk gS] bldh cká lrg ij NksVh cwna as la?kfur gks tkrh gSA dejs esa vkisf{kd vknzrk dh x.kuk dhft;sA 17.5 feeh ikjs ds nkc ij ty dk DoFkukad 20ºC rFkk 8.9 feeh ikjs ds nkc ij 10ºC gSA 51%

Ans. 54.

Ans.

50 m3 of saturated vapour is cooled down from 30ºC to 20ºC. Find the mass of the water condensed. The absolute humidity of saturated water vapour is 30 g/m3 at 30ºC and 16 g/m3 at 20ºC. 50 eh3 larI` r ok"i 30ºC ls 20ºC rd B.Mh dh tkrh gSA la?kfur ty dh ek=kk Kkr dhft;sA ty ok"i dh ije vknzrk (absolute humidity) dk eku 30° C ij 30 xzke@eh3 rFkk 20ºC ij 16 xzke@eh3 700 g manishkumarphysics.in

Page # 21

Chapter # 24 Kinetic Theory of Gases 55. A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapour pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value. ,d cSjksehVj dk lgh ikB~;kad 76 lseh ikjs ds cjkcj gSA cSjksehVj dh uyh esa ,d MªkiW j dh lgk;rk ls ikuh dh cwna s izfo"V

djk nh tkrh gSA ikjs ds LrEHk dh Å¡pkbZ igys de gksrh gS] rRi'pkr~ fu;r gks tkrh gSA ;fn ok;qe.My ds rki ij larI` r ok"i nkc 0.80 lseh ikjs ds cjkcj gks] ikjn~ LrEHk dh Å¡pkbZ rc Kkr dhft;s tc bldk eku U;wure gks tk;sA

Ans.

75.2 cm

56.

50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27ºC. The water level in the jar is same as the level outside. The saturation vapour pressure at 27ºC is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar. ikuh ds Åij mYVs tkj esa 50 ?ku lseh vkWDlhtu ,df=kr dh x;h gSA ok;qe.Myh; nkc 99.4 kPa rFkk dejs dk rki 27ºC gSA tkj esa ikuh dk ry] blds ckgj okys ikuh ds ry ds cjkcj gSA 27ºC ij larI` r ok"i nkc 3.4 kPa gSA tkj esa ,df=kr

vkWDlhtu esa eksy la[;k dh x.kuk dhft;sA Ans.

1.93 × 10–3

57.

A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir. ,d =kqfV;qDr cSjksehVj esa dqN ok;q rFkk larÌr ty ok"i gSA tc ok;qe.Myh; nkc 76.0 lseh ikjs ds cjkcj gS rks bldk ikB~;kad 74.0 lseh gS rFkk tc ok;qe.Myh; nkc 74.0 lseh gS rFkk tc ok;qe.Myh; nkc 74.0 lseh ikjs ds cjkcj gS rks bldk ikB~;kad 72.10 lseh gSA ok;q ds rki ij larI` r ok"i nkc = 1.0 lseh ikjkA ikjs dh ukn esa ikjs ds ry ds Åij cSjksehVj

dh uyh dh yEckbZ Kkr dhft;sA Ans.

91.1 cm

58.

On a winter day, the outside temperature is 0ºC and relative humidity 40%. The air from outside comes into a room and is heated to 20ºC. What is the relative humidity in the room? The saturation vapour pressure at 0ºC is 4.6 mm of mercury and at 20ºC it is 18 mm of mercury. lnhZ ds fnu es]a ckgj dk rki 0ºC rFkk vkisf{kd vknzrk 40% dejs esa ckgj ls gok vUnj vkrh gS rFkk 20ºC rki rd xeZ gks tkrh gSA dejs esa vkisf{kd vknzrk fdruh gS\ 0ºC rki ij larI` r ok"i nkc 4.6 feeh ikjk rFkk 20ºC rki ij 18 feeh

ikjk gSA Ans.

9.5%

59.

The temperature and humidity of air are 27ºC and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at 27ºC = 3600 Pa. fdlh fof'k"V fnol ij ok;q dk rki rFkk vkisf{kd vknzrk 27ºC ,oa 50% gSA 1 ?ku ehVj ok;q dk larI` r djus ds fy;s vko';d ok"i dh ek=kk dh x.kuk dhft;sA 27ºC ij larI` r ok"i nkc = 3600 ikWLdy 13 g

Ans. 60.

Ans. 61.

Ans.

The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m3. The saturation vapour pressure at 300 K is 3.3 kPa. Calculate the mass of the water vapour present in the room. fdlh dejs dk rki rFkk vkisf{kd vknzrk Øe'k% 300 K rFkk 20% gSA dejs dk vk;ru 50 eh3 gSA 300 K ij larI` r ok"i nkc 3.3 kPa gSA dejs mifLFkr ty ok"i ds nzO;eku dh x.kuk dhft;sA 238 g The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m3. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa. 50 eh3 vk;ru okys dejs dk rki ,oa vkisf{kd vknzrk Øe'k% 300 K o 20% gSA blds Q'kZ dks ikuh ls /kks;k tkrk gS] 500 xzke ikuh Q'kZ ls fpidk jg tkrk gSA ;g ekurs gq, fd okrkoj.k ls dksbZ vkokxeu ugha gks jgk gS] tc Q'kZ iwjk lw[k tkrk gS] vkisf{kd vknzrk Kkr dhft;sA rki ,oa nkc ds ifjorZu ux.; ekus tk ldrs gSaA 300 K ij larÌr ok"i nkc = 3.3 kPa 62%


Page # 22

Chapter # 24 Kinetic Theory of Gases 62. A bucket full of water is placed in a room at 15ºC with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5ºC how much more water will evaporate? The saturation vapour pressure of water at 15ºC and 20ºC are 1.6 kPa and 2.4 kPa respectively. ,d dejs esa 15ºC rki ij izkkjfEHkd vkisf{kd vknzrk 40% gSA blesa ikuh ls Hkjh gqbZ ,d ckYVh j[k nh tkrh gSA dejs ds vk;ru 50 eh3 gSA (a) fdruk ikuh okf"ir gks tk;sxk\ (b) ;fn dejs dk rki 5ºC c<+k fn;k tk;s rks fdruk ikuh vkSj okf"ir gksxk\ 15ºC rFkk 20ºC ij larÌr ok"i nkc Øe'k% 1.6 kPa rFkk 2.4 kPa gSA Ans. (a) 361 g (b) 296 g

EXTRA QUESTIONS 2.

Find the number of molecules of an ideal gas in a volume of 1.000 cm3 at STP. Ans. 2.685 × 1019

9.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder. [3] Ans. 22 cm

10.

Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth. [3] Ans. 8.0 hr.

11.

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0ºC. Mass of a helium molecule = 6.64 × 10–27 kg and Boltazmann constant = 1.38 × 10–23 J/K. [3] Ans. 8.0 × 10–24 kg-m/s

12.

During an experiment, an ideal gas is found to obey an additional law pV2 = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V. Ans. T/ 2 [3]

13.

Calculate the heat absorbed by a system in going through the cyclic process shown in figure.

[3]

Ans. 31.4 J 14.

Consider the cyclic process ABCA, shown in figure, performed on a sample of 2.0 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process Find the work done by the gas during the part BC. [3]

Ans. –4520 J 15.

A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.


Page # 23

Chapter # 24


[4]

Ans. 4.19 J/cal. 16.

A gas is taken along the path AB as shown in figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

[4]

Ans. –241 J 17.

Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10–5 cm. [4] Ans. 1.23 × 1010

18.

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45º with it. How many molecules strike each square metre of the wall per second ? [4] Ans. (a) 1780 m/s (b) 1.2 × 1028 A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston. [4] Ans. 25 cm

19.

21.

Figure shows a cylindrical tube of volume V0 divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gases are filled in the two parts. When the separator is kept in the middle, the pressures are p1 and p2 in the left part and the right part respectively. The separator is slowly slid and is released at a position where it can stay in equilibrium. Find the volumes of the two parts. [4]

p 2 V0 Ans. V2 = p  p 1 2 22.

A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out. [5] Ans. (a) p = p 0 e  t / V0 (b)

23.

Assume that the temperature remains essentially constant in the upper part of the atmosphere. Obtain an expression for the variation in pressure in the upper atmosphere with height. The mean molecular weight of air is M. [5] 

24.

V0 n2 

Mgh

Ans. p 0 e RT A horizontal tube of length  closed at both ends contains an ideal gas of molecular weight M. The tube is rotated at a constant angular velocity  about a vertical axis passing through an end. Assuming the temperature to be uniform and constant, show that manishkumarphysics.in

Page # 24

Chapter # 24

Kinetic Theory of Gases M  2RT

2 2

p2 = p1e

where p2 and p1 denote the pressure at the free end and the fixed end respectively. Ans. p2 = 25.

[5]

M2  2 p1e 2RT

Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How [5]

Ans. (a) zero (b)

p1T2 (p1  p 2 ) p 2 T1(p1  p 2 ) on the left and on the right  

T1T2 (p1  p 2 ) 3p1p 2 ( T2  T1 )V (d) where  = p1T2 + p2T1  4 A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic e n ergy of the block as it stops slipping past the belt. (c) Find the work done in this frame by the external force holding the belt. [6] Ans. (a) 200 J (b) 200 J (c) 400 J

(c)

26.

36.

Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducing and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperature in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time. [4]

Ans. 10 cm from the left end 39.

Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows etc. [3]

40.

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate

dN . d

d dN

Heat


Page # 25

Chapter # 24 Ans. 1.25 × 10 4 N/m


41.

Figure shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T0 and its pressure is p0 which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T0?

Ans.

(a) zero (b) p0A

42.

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h2 of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole. hg 2p0

h0 h1

Ans.

p0 (a) g  h 0

2  p0  h 0  (p 0  g(h1  h0 ) (b) g   

1/ 2

43.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg (figure). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.

Ans.

2.2 m

44.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder. 22 cm

Ans. 45.

Ans.

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water both maintained at 62ºC. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible. 84 cm of mercury

46. Ans.

The weather report reads. "Temperature 20ºC; Relative humidity 100%. What is the dew point? 20ºC

47.

The condition of air in a closed room is described as follows. Temperature = 25ºC, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25ºC = 3.2 kPa. 102 kPa

Ans.


Page # 26

Chapter # 24 48. Ans. 49. Ans. 50.


The temperature and the dew point in an open room are 20ºC and 10ºC. If the room temperature drops to 15ºC, what will be the new dew point? 10ºC Pure water vapour is trapped in a vessel of volume 10 cm3. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing. 4.0 cm3

Ans.

A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapour pressure at the room temperature is 1.0 cm. 60%

51. Ans.

Using figure of the text, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm. 65ºC, 48ºC

52.

The human body has an average temperature of 98ºCF. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure of the text for the vapour pressures. 50 mm of mercury

Ans. 53.

Ans. 54. Ans. 55.

Ans. 56.

Ans. 57.

Ans. 58.

Ans. 59. Ans. 60.

Ans.

A glass contains some water at room temperature 20ºC. Refrigerated water is added to it slowly. When the temperature of the glass reaches 10ºC, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20ºC and at 8.9 mm of mercury it is 10ºC. 51% 50 m3 of saturated vapour is cooled down from 30ºC to 20ºC. Find the mass of the water condensed. The absolute humidity of saturated water vapour is 30 g/m3 at 30ºC and 16 g/m3 at 20ºC. 700 g A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If saturation vapour pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value. 75.2 cm 50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27ºC. The water level in the jar is same as the level outside. The saturation vapour pressure at 27ºC is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar. 1.93 × 10–3 A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir. 1.93 × 10–3 On a winter day, the outside temperature is 0º and relative humidity 40%. The air from outside comes into a room and is heated to 20ºC. What is the relative humidity in the room? The saturation vapour pressure at 0ºC is 4.6 mm of mercury and at 20ºC it is 18 mm of mercury. 9.5% The temperature and humidity of air are 27ºC and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at 27ºC = 3600 Pa. 13 g The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m3. The saturation vapour pressure at 300 K is 3.3 kPa. Calculate the mass of the water vapour present in the room. 238 g


Page # 27

Chapter # 24 Kinetic Theory of Gases 61. The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m3. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa. Ans. 62% 62.

Ans.

A bucket full of water is placed in a room at 15ºC with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5ºC how much more water will evaporate? The saturation vapour pressure of water at 15ºC and 20ºC are 1.6 kPa and 2.4 kPa respectively. (a) 361 g (b) 296 g


Page # 28

Kinetic Theory of Gases

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