10. KTG S

5555 KINETIC THEOKINETIC THEOKINETIC THEOKINETIC THEORY OF GASESRY OF GASESRY OF GASESRY OF GASES

CET WORKSHEETS SOLUTIONS

Kinetic Theory Of Gases

1. d) 0273 C−

Molecular motion ceases at absolute

zero temperature = 0273 C−

2. c) 00 C

3. b) column of mercury 76 mm high at 00 C under stranded gravity

As STP pressure is equivalent to the

pressure exerted by column of

mercury 76 m high at 00 C or 273 K

under standard gravity

4. c) torr

mm of Hg is also called torr ∴1mm of Hg = 1 torr

5. b) 10 : 1

S.I. unit = 2/N m

C.G.S. unit = 2/dyne cm

2

2

1 /

1 /

N mratio

dyne cm=

1 N = 510 dyne 2 4 21 10m cm=

5 4 2

2

10 /10

1 /

dyne cmratio

dyne cm∴ =

2

2

10 /

1 /

dyne cm

dyne cm=

= 10 : 1

6. b) N

n

0

NN

n=

7. b) 0

Mm

N=

8. c) fundamental constant and equal to 236.023 10×

Avogadro’s number is a fundamental

constant and equal to 236.023 10×

9. a) mN

Mn

=

10. c) A -3, B -2, C - 4

11. b) A B

B A

P P

V V=

According to Boyle’s law

1P

V∝

or PV = constant

A A B BP V P V∴ =

A B

B A

P V

P V∴ =

12. b) 1 2

1 2

V V

T T= at constant pressure

According to Charle’s law

V T∝ at constant pressure

V

T∴ = constant

1 2

1 2

V V

T T∴ =

13. c)

PV = constant

14. a)

∝V T

15. d)


44

tan=

Vcons t

T

16. a)

∝P T

17. d)

tan=P

cons tT

18. b) 52 10 Pa×

By Boyle’s law

1P

V∝

1 1 2 2PV PV=

1 12

2

PVP

V∴ =

1

2

2

2 1 1

V L

V L

=

= − =

510 2

1

×=

52 10 Pa= ×

19. c) 51

102

Pa×

By Boyle’s law

1P

V∝

1 1 2 2PV PV=

1 12

2

PVP

V∴ =

1

2

2

2 2 4

V L

V L

=

= + =

510 2

4

×=

5110

2Pa= ×

20. d) 3 P / 2

( ) ( )

22

3

VV V

initial Final

→ →

1P

V∝

f i

i f

P V

P V∴ =

2

3

f i

VP P

V= ×

3 3

2 2i

PP= × =

21. b) 0.04 3m

By Boyle’s law

1P

V∝

1 1 2 2PV PV=

1 12

2

PVV

P∴ =

21

22

5 /

2.5 /

P N m

P N m

=

=

5 0.02

2.5

×= 2 0.02= ×

= 0.04 3m

22. c) 0.02/3 3m

By Boyle’s law

1 1 2 2PV PV=

1 12

2

PVV

P∴ =

21

22

5 /

15 /

P N m

P N m

=

=


45

5 0.02

15

×=

20.02/

3N m=

23. c) 4P/3, 3V/4

Initial state : P V T

3V V→

1P

V∝

3

PP∴ →

Second state : 3

P,3V,T

4

3 3

P P→

1P

V∝

3

4

VV∴ →

Final state : 4 3

, ,3 4

P VT

24. d) 25 %

Volume is decreased by 20% ∴final volume

2

20

100V V V= −

= 0.8 V

By Boyle’s law

1P

V∝

PV = Constant

1 1 2 2PV PV∴ =

2 0.8PV P V= ×

20.8

PP =

Increase in pressure =0.8

PP−

=1 0.8

0.8P

−

0.2

0.8P= ×

∴% increase0.2

1000.8

= ×1

1004

= ×

= 25%

25. c) 9V /11

( ) ( )

9 9 /11T T T

Initial Final

→ →

By Boyle’s law

V T∝

f f

i i

V T

V T=

f

f i

i

TV V

T= ×

9 / 11T

VT

= ×

9

11

V=

26. b) 1200 K

By Boyle’s law

V T∝ (T = absolute temperature)

2 2

1 1

T V

T V=

22 1

1

VT T

V= ×

1 2

1

1 , 3 1 3

127 273 400

V L V L

T K

= = × =

= + =

3

4001

= ×

= 1200K.

27. d) V/2, T/2

Initial state : P V T

4T T→

By Boyle’s law

V T∝

4V V∴ →

Second state : P,4T

4

8 2

V VV → =

By Boyle’s law

V T∝

4

8 2

T TT∴ → =

Final state : P, 2

V,

2

T

28. a) decrease by 15%

By Boyle’s law

V T∝

∴As volume decreases by 15% , then

temperature also decreases by 15%


46

29. c) 3.33 N / 3m

By Gay Lussac’s Law

P T∝

1 1

2 2

P T

P T∴ =

22 1

1

TP P

T= ×

1

1

327 273 600

127 273 400

T K

T K

= + =

= + =

400 2

5 5600 3

= × = ×

= 3.33 N / 3m

30. b) T /5

( ) ( )

2

10 10 5

P P PP

Initial Final

→ → =

By Boyle’s law

f f

i i

P T

P T=

f

f i

i

PT T

P= ×

/ 5P

TP

= ×

5

T=

31. a) 25 %

By Gay Lussac’s Law

P T∝ ∴if T increases by 25%

P increases by 25%

32. a) 2313 10×

According to Avogadro’s law at the

same temperature and pressure, equal

volume of gases contain equal

number of molecules.

∴Number of molecules in cylinder B 2313 10= ×

33. c) 1 2P P>

As a constant temperatures T we get

the two values

1 2V and V corresponding to the two

Clearly 2 1V V>

As constant temperature

1P

V∝ 1 2P P∴ >

34. d) 1 2t t>> so that 2t is negligible

35. d) remains constant

36. b) rate of change of momentum

imparted to walls per unit time per

unit area

Rate of change of momentum

imparted to walls per unit time per

unit area is the pressure exerted by a

gas on the wall of container.

37. d) –2 mc

38. d) both ‘b’ and ‘c’

39. c) viscosity

40. c) both ‘a’ and ‘b’

41. d) both ‘a’ and ‘b’

,

PVPV nRT

T= = constant

42. d) neither attract nor repel each other

43. c) translational kinetic energy of

molecules

44. c) K.E.

45. d) neither ‘a’ nor ‘b’

Ideal gas can neither be liquefied nor

solidified.


S.I. unit of R : N/m mole K,J/mole K

47. a) [R] = 1 2 2 1 1[ ]M L T mol− − −θ

48. c) 4.157 J / gm K

1 mole 2 2H gm=

8.314

2R∴ = J/gm K

= 4.157 J/gm K

49. a) J/ K

S.I. unit of K is J/K

50. c) 0927 C


47

We know that

1 1 2 2

1 2

PV PV

T T=

2 12P P= and 2 12V V= ,then

2 2 2

1 1 1

T PV

T PV=

1 1

1 1

2 24

P V

PV

×= =

2 14T T=

( )4 27 273= +

4 300= ×

1200K=

01200 273 C= −

0970 C=

51. d) 2.75 atm

1 1 2 2

1 2

=PV P V

T T

21 30 5.2

300 143

× ×∴ =

P

2

30 143

300 5.2∴ = ×P

= 2.75atm

52. a) 12

PV = nRT

PVn

RT=

6 31.8 10 16.62 10

8.31 300

−× × ×=

×

5 3

2

18 16.62 10 10

8.31 3 10

−× ×= ×

×

6 2 12= × =

53. d) 4 : 1

For cylinder A :

A A A

mP V nRT RT

M= =

(m = mass of gas)

' 'AA

m N RT m RTPV N

M M= = × ..(i)

( 'm = mass of each molecule )

For cylinder B :

B B B B

mP V nRT RT

M= =

( )' 22

4

Bm N R TVP

M× =

'

4

BPV m N RT

M=

'

B

m RTN

M= × …..(ii)

4

1

A

B

N

N=

. . : 4 :1A Bi e N N =

54. a) 5 22.49 10 /N m×

By Dalton’s law

0 NP P P= +

We , know

PV = nRT

m RTP

M V=

oo

o

m RTP

M V= ×

NN

N

m RTP

M V= ×

o N

o N

m m RTP

M M V

∴ = + ×

3

16 14 8.3 300

32 28 10 10−

× = + ×

×

31 1249 10

2 2

= + × ×

51 2.49 10P = × × 5 22.49 10 /P N m∴ = ×

55. d) 8.3 atm

By Dalton’s law

0 2N COP P P P= + +

PV = nRT

m RTP

M V= ×

oo

o

m RTP

M V= × , N

N

N

m RTP

M V= × ,

2

2

2

CO

CO

CO

m RTP

M V= ×

2

2

COo N

o N CO

mm m RTP

M M M V

∴ = + + ×

3

16 7 11 8.3 300

32 28 44 3 10−

× = + + ×

×


48

3

1 1 1 8.3 300

2 4 4 10−

× = + + ×

51 18.3 10

2 2

= + × ×

58.3 10= ×

8.3 .P atm∴ ≈

56. c) 158 10× molecules

We have

PV = nRT

m

RTM

= × (m = mass of gas)

'm N

RTM

=

'

.

m mass of each molecule

N no of molecules

=

=

'm RT

NM

=

∴PV = NKT ……….(i)

where

'm R

M= constant = K

At STP

0 0 0 0PV N KT= ……….(ii)

where

0N = Avogadro’s no.

equation (i) ÷ equation (ii)

0 0 0 0

PV N T

PV N T= ×

00

0 0

TP VN N

P V T= × × ×

4

23 10 250 2736 10

76 22400 300

−

= × × × ×

164095010

51072= ×

160.8018 10= × 158 10N∴ ≈ × molecules

57. a) 3 units

Mean free path

= 1 2 3 .......... nd d d d

n

+ + +

= 1 3 2 4 2 1 5 6

8

+ + + + + + +

=

24

8

= 3 units

58. a) zero

59. a) becomes four times of its pervious

value

We have

C T∝

2C T∝

22 2

211

C T

TC=

2

22 1

1

CT T

C

= ×

( )2

2 12T T=

2 14T T=

60. a) 00 C

We know

2

5

7He HC C=

2

2

33 5

7

HHe

He H

RTRT

M M=

2

2

5

7

HHe

He H

TT

M M=

5 273

4 7 2

HeT=

25 273

449 2

HeT = × ×

050273 273 0

49K C= × ≈ =

61. d) 3.0 atm

21

3P C= ρ

218.99 10 3180 3180

3

−= × × × ×

90910476

3=

5

30303492

1.01 10=

×

5300034 10−= ×

= 3.0 atm


49

62. a) 614.25 K

2 1

3

2C C=

we know

C T∝

2 2

1 1

C T

C T=

12

1

3

2

273

CT

C=

23

2 273

T=

29

4 273

T∴ =

2

9273

4T = ×

2T = 614.25 K

63. b) H HeC C=

We have

3 HH

H

P VC

M=

3 HeHe

He

P VC

M=

HeH H

He He H

MC P

C P M= ×

1 4

2 2= ×

H HeC C∴ =

64. a) 400

3

We know velocity depends on

temperature. So,

2 2

1 1

=C T

C T

22 1

1

= ×T

C CT

400

200300

= ×

200 2

3

×=

400

/3

= m s

65. a) 2

1

ρ

ρ

3PC =

ρ

1C ∝

ρ

1 2

2 1

C

C

ρ∴ =

ρ

66. b) 0.732 × initial velocity

∝C T

2 2

1 1

=C T

C T

1

1

3=

T

T

3=

2 13= ×C C

Increase in velocity 2 1= −C C

1 13= −C C

( ) 13 1= − × C

( ) 11.732 1= − × C

10.732= × C

0.732= × initial velocity.

67. a) : 3λ

sound

PV

γ=

ρ and

3rms

PC =

ρ

/

3 /

sound

rms

V P

C P

γ ρ∴ =

ρ 3

γ=

: : 3sound rmsV C∴ = γ

68. b) 22.16 10 /m s×

. . .r m sC =

( ) ( ) ( )2

2 2 21 10 2 10 3 10

3

× + × + ×

= ( ) ( ) ( )4 4 41 10 4 10 9 10

3

× + × + ×


50

= ( ) 41 4 9 10

3

+ + ×

= 41410

3×

= 44.667 10×

= 22.16 10 /m s×

69. b) 7 T

C BC C=

3 3C B

C B

RT RT

M M∴ =

C B

C B

T T

M M∴ =

28 4

CT T∴ =

28 7 .4

C

TT T= × =

70. b) 3 : 2

3=

ρrms

PC

1 1 2

2 2 1

ρ= ×

ρ

C P

C P

3 3 3

2 2 2= × =

= 3 : 2

71. a)

21

3

mNC

V

P = 21

3

MC

V=

21

3

NmC

V

72. d) two third of mean kinetic energy per

unit volume

= 2

2 1

3 2

NmC

V

21. .

2NmC K E

=

∵

=2 . .

3

K E

V

73. b) directly proportional to the mean

square velocity

2

∝P C

74. d) equal in all the three

1 2

/ 2, ,

/ 2

M M M

V V Vρ = ρ = =

3

/ 4

/ 4

M M

V Vρ = =

1 2 3∴ρ = ρ = ρ

also 1 2 3C C C∴ = =

21

3P C rms∴ = ρ

1 2 3. .i e P P P= =

75. d) 16 : 1

RTP

M

ρ=

1P

M∝

1 2

2 1

P M

P M=

32 16

2 1=

1 2: 16 :1P P =

76. b) 6 31.5 10 /erg cm×

( ). . 3

, 12

K EP At NTP P atm

Volume= =

6 21.013 10 /dyne cm= ×

631.013 10

2= × ×

63.03910

2= ×

6 31 5 10= ×. erg / cm

77. c)

9

2times kinetic energy of gas

molecules in cylinder A

E T∝

B B

A A

E T

E T=

1800

400=

1

2

127 273 400

1527 273 1800

T K

T K

= + =

= + =

9

2B AE E=

78. c) 16 : 5

( )

( )

( )( )

( )( )

21

1

22

1

1. . 2

1. .5

2

m CK E

K Em C

=

2

1

2

1

5

C

C

=


51

( )21 16

45 5

= =

( ) ( )1 2

. . : . . 16 : 5K E K E∴ =

79. a) Q

SmdT

=

80. d) 0/cal gm C

81. b) 0 2 2 1[ ]M L T − −θ

82. b) quantity of heat required to raise

temperature of 1 mole of substance

through 1K

83. b) principal specific heat

84. d) P VC dT C dT dW+ =

85. c) P V

RC C

MJ− =

86. c) 1

R

γ −

87. d)

1

Rγ

γ −

88. d) atomicity of gas

89. c) triatomic

90. b)

5

2VC R=

91. c) 8333 J/ kg K

1.6∴ −P VC C

5000P VC C∴ − =

1.6 5000− =V VC C

0.6 5000=VC

5000

0.6VC =

= 8333 J/ kg K

92. c) 4.2 J/cal

JdQ = du + dW

200 J = 502 + 338

502 338 840

200 200J

+= =

= 4.2 J/cal

93. d) 4197 J/Kcal

P

V

C

Cγ =

0.2391.4

VC=

0.2391.171

1.4VC = =

P V

RC C

mJ− =

P V

RTC C

mJT− =

P V

PVC C

mJT− = 1V

m

=

ρ ∵

∴ − =ρ

P V

PC C

JT

51.013 100.239 0.171

1.3 273J

×− =

× ×

51.01310

24.1332= ×

50.04197 10= ×

= 4197 J/Kcal

94. a) 1.984 K cal/K mole K

P

V

C

Cγ =

1.44.96

PC=

1.4 4.96PC∴ = ×

6.944PC =

P VC C R∴ − =

6.944 4.96 R− = R = 1.984 K cal/K mole K

95. b)

=VC K

96. a) PC = PMC

97. c) V PC C>

Due to anomalous behavior of water

from 00 C to 04 C the volume decreases

and than increases.

98. b) 124 J

207PdQ C dT= =

VdQ C dT′ =

P VC dT C dT PdV− =

P VC dT C dT RdT− =

( )207 8.3 10VC dT− =

207 83 124VC dT J dQ′= − = =

99. a) 5

7

For 1 mole of ideal gas

PQ C dT=

Vdu C dT=


52

V

P

Cdu

Q C∴ =

1 10 5

1.4 14 7= = =V

P

CBut

C

5

7

du

Q∴ =

100. c) 36 m/s

When vessel is moved with velocity

v, all gas molecules will have

additional translational velocity.

∴KE. Of molecules = 21

2mv=

K.E. of one mole = 21

2Mv=

When it is suddenly stopped its K.E is

converted in to heat.

101. d) 1 : 32

7PV = constant

7

2 1

1 2

P V

P V

=

5/3

2 1

1 18

P V

P V

=

5/3 5

1 1 1

8 2 32

= = =

2 1: 1: 32P P =

102. a) 3 : 2

For adiabatic gases

1P T

−γ γ = constant; ( )1

consatntT

P

γ

−γ=

( )1T P

− −γγ ∝ , ( )1T P

γ −γ ∝ , ( )/ 1T P

γ−γ ∝ ( )/ 1

P Tγ−γ∝

But , 3P T∝

31

γ=

γ −

3 3γ = γ −

2 3γ =

3

2γ =

: 3 : 2P VC C∴ =

103. b) nRdT

Energy supplied =PdQ nC dT=

Rise in internal energy du =

Vdu nC dT=

External work done dW = −dQ du

P VnC dT nC dT= −

( )P VndT C C= −

= nRdT

104. d) ( )Pn C R dT−

Rise in internal energy = dQ – dw

PnC dT nRdT= −

( )PndT C R= −

( )Pn C R dT= −

105. c) 1400 J

Heat absorbed = 1000J VnC dT=

VnC dT= = 1000

5 10 1000VC× × =

1000 10020

5 10 5VC = = =

×

But

P VC C R− =

20 8PC − =

8 20 28PC = + = (amount of heat absorbed at constant

pressure ) VnC dT=

5 28 10= × ×

=1400J.

106. d) 3

2

R

For perfect gas

3.

2 2V

fC R R= =

107. d) 5

2

R

51 .

2 2V

fC R

= + =

108. a) the body at lower temperature absorbs

heat

109. d) 1 2 2 1[ ]M L T

− −θ 110. c) product of specific hat and mass of

substance

Water equivalent = mass of substance × specific heat

111. b) A-III, B-II, C-VI, D-V, E-IV, F-I

112. b) A,B,E only

Processes that involve absorption of

heat – fusion, vaporization,

sublimation.

113. b) 34.184 10 :1×

Heat unit : Mechanical unit

:Cal J

gm Kg=

3

4.186:

10

J J

Kgkg−

=


53

34.186 10

:J J J

kg Kg

×=

34.184 10 :1= ×

114. d) hidden heat 115. c) 540 Kcal / kg

4536

2

QL

m= = = 2268 J/Kg

2268

4.2=

= 540 Kcal / kg 116. c) 2

4.00 10 Kcal×

QL

m=

Q = Lm

80 5= ×

= 400 K cal

24 10 .Kcal= ×

117. a) 4 Kg

QL

m=

3

3

200 10

50 10

Qm

L

×= =

×

= 4 Kg.


( )2 1−

= = =P V VPdV PdV

Lem mJ mJ

119. c) 499.73 Kcal / kg

L = Li + Le

Li = L – Le

= 540 – 40.27

= 499.73 Kcal / kg 120. b) 499.73, 40.27

5 6

3

1.013 10 3340 10

2 10 42

−

−

× × ×= =

× ×

PdVLe

mJ

21.013 167010

4200

×= ×

20.4027 10= ×

40.27 /Kcal Kg=

i eL L L= +

i eL L L= −

540 40.27= −

= 499.73 Kcal / kg

121. c) 80.0024119 Kcal/kg

( ) 3 3 3 31 1.1 10 0.1 10dV m m− −= − × = − ×

e

PdVL

mJ=

( )5 31.013 10 0.1 10

1 4200

−× × − ×=

×

5 4

1.013 10 10

4200

−× ×= −

1.013 10

4200

×= −

0.0024119 / .Kcal Kg= −

i eL L L= +

i eL L L∴ = −

( )80 0.0024119= − −

= 80 + 0.0024119

= 80.0024119 Kcal/Kg.

122. a) 190.5 Kcal

External work done = dW= mPdV

J

55 10 1.6

4200

× ×=

= 190.5 Kcal 123. b) 79.99 Kcal / kg , -32.64 10 cal/kg×

( )2 1P V VLe

J

−=

2 1

1 1P

LeJ

−

ρ ρ =

( )1 2

1 2

ρ − ρ=

ρ ρ

P

J

( )510 1000 900

1000 900 4200

× −=

× ×

7

7

10

10 9 42=

× ×

1

378=

32.64 10 /Kcal kg−= ×

i eL L L= +

i eL L L= −

380 2.64 10

−= − ×

79.99 /Kcal Kg=


54

124. b) 533.3 cal / gm

Heat produced by heater in 15 min =

amount of heat required to boil 100H m= ×

∴Heat produced in 1 min = 100

15

m ×

∴Heat produced in 80 min =

8100

15

m × ×

Heat required to convert water boiling

into steam = heat produced in 80 min

80100

15

mmL

×=

80100

15L

×=

= 533.3 cal / gm

10. KTG S

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