KINETIC THEORY OF GASES

MULTIPLE CHOICE QUESTIONS

1. At what temperature will the root mean square velocity of

oxygen molecules be sufficient so as the escape from the
earth?
(A) 1.6 × 105 K (B) 16 × 105K
(C) .16 × 105 K (D) 160 × 105 K
Explanation: (A)
 32 KT = 12 mve2
Where ve = escape velocity of earth = 11.1 km/sec
m = mass of 1 molecule of oxygen = 5.34 × 10–26
 mve2 
T=  
 3K 
 

 5.32 10 26  (11.1103 ) 

T = 
 3  (1.38 10  23 )


 

T = 1.6 × 105 K

2. The first excited state of hydrogen atom is 10.2 eV above its

ground state. What temperature is needed to excite hydrogen
atoms to first excited level –
(A) 7.88 × 104K (B) .788 × 104 K
(C) 78.8 × 104 K (D) 788 × 104 K
Explanation: (A)
K.E. Per atom = 3/2 KT

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 K.E. of the hydrogen atom = 10.2 eV
 10.2 eV = 10.2 × (1.6 × 10–19) Joule
T = 23 × K.E. perK atom
T = 23 × 10.2 1.6 10 19
1.38 10 23

Where, k = 1.38 × 10–23 J/mole. ºK

 T = 7.88 × 104 K

3. At what temperature does the average translational kinetic

energy of a molecule in a gas equal to the kinetic energy of
an electron accelerated from rest through a potential
difference of 5 volt.
(A) 386.5 × 103K (B) 3.865 × 103K
(C) .38 × 103K (D) 38.65 × 103 K
Explanation: (D)
K.E. of the electron is
5 eV = 5 × 1.6 × 10–19J
But K.E. = 3/2 KT
 5 × 1.6 × 10–19 = 3/2 (1.38 × 10–23) × T
T= 5 1.6 10 19  2
3 1.38 10  23

 T = 38. 65 × 103 K

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4. Two ideal gases at temperature T1 and T2 are mixed. There
is no loss of energy. If the masses of molecules of the two
gases are m1 and m2 and number of their molecules are n1
and n2 respectively, the temperature of the mixture will be
(A) nT  Tn
1 2
(B) nT  nT 1 2

1 2 1 2

(C) n nT  nn T
2 1 1 2
(D) n Tn  nn T
1 1 2 2

1 2 1 2

Explanation: (D)
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Total energy of molecules of first gas = 2
n1 K T1
Total energy of molecules of second gas = 32 n2 KT2
Total energy of molecules of mixture = 32 K (n1T1 + n2T2)
 32 (n1 + n2) KT = 32 K(n1T1 + n2T2)
 T = n(Tn  nn T)1 1 2 2

1 2

5. The atomic weight of iodine is 127. A standing wave in a

tube filled with iodine gas at 400 K has nodes that are 6.77cm
apart when the frequency is 1000 vib/sec. iodine is
(A) Monoatomic (B) Diatomic
(C) Triatomic (D) None of these
Explanation: (B)
 = 2 × 6.77 cm = 13. 54cm
= n = 1000 × 13.54 = 1.354 × 104cm/sec.
We know that
= VR
T/M
where molecular weight

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 M = Ax with x = 1 if iodine is monoatomic x = 2 it diatomic
and A is atomic weight
  = Ax  2
RT
= 0.7x
Where x = 2 as iodine is diatomic
 = 1.4 (right value of diatomic gas)

6. Certain perfect gas is found obey PV3/2 = const. during

adiabatic process. If such a gas at initial temperature T is
adiabatically compressed to half the initial volume, in final
temperature will be –
(A) 2 T (B) 2T
(C) 2 2 T (D) 4T
Explanation: (A)
PV3/2 = constant (given)
Put
P = nRT
V

  nRT
V 

 (V
3/2
) = constant
When V changes to V/2 the temperature becomes 2 T.

7. In a certain process the pressure of one mole ideal gas varies

with volume according to the relation P =  a  , where a,
2
V
1    
  b  

b are constants, when the volume of gas V = b, then

temperature of the gas will be -

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 (A) 2abR (B) ab/R
(C) ab (D) zero
Explanation: (A)
 T = PV
R
a a
At V = b, P = (1  1)
= 2

T = ab
2R

8. An air bubble of volume V0 is released by a fish at a depth h

in a lake. The bubble rises to the surface. Assume constant
temperature and standard atmospheric pressure above the
lake. The volume of the bubble just before touching the
surface will be (density) of water is 
(A) V0 (B) V0(gh/P)
(D) V0 1  Pgh 
V0
(C)  gh 
1  
 P 

Explanation: (D)
As the bubble rises the pressure gets reduced for constant
temperature, if P is the standard atmospheric pressure, then
(P + gh) V0 = PV
or V = V0 1  Pgh 

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9. Two gases occupy two containers A and B the gas in A, of
volume 0.10m3, exerts a pressure of 1.40 MPa and that in B
of volume 0.15 m3 exerts a pressure 0.7 MPa. The two
containers are united by a tube of negligible volume and the
gases are allowed to intermingle. Then it the temperature
remains constant, the final pressure in the container will be
(in MPa) –
(A) 0.70 (B) 0.98
(C) 1.40 (D) 2.10
Explanation: (B)
We Know that
PAVA = nART, PBVB = nB RT
and Pf (VA + VB) = (nA + nB) RT
Pf (VA + VB) = PAVA + PBVB
 Pf =  PA VA  PB VB 
  = 1.4  0.1  0.7  0.15
0.1  0.15
MPa = 0.98 MPa
 VA  VB 

10. If the pressure of a gas contained in a closed vessel is

increased by 0.5% when heated by 2ºC, then the initial
temperature must be
(A) 127ºC (B) 273ºC
(C) 400ºC (D) 673ºC
Explanation: (A)
Using PV = nRT, we note that
P1V = nRT1
P1 (1.005)V = nR (T1 + 2)

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 (note P = P2 – P1 = 0.005 P1 and
T = 2ºC = 2K
Dividing we get 1.005 = T T 2
1

or 0.005T1 = 2  T1 = 400
Thus in 0ºC, t1 = 400 – 273 = 127ºC.

11. What is the degree of freedom of gas ? If at STP the velocity

of sound in it is 330 m/sec and gas density = 1.3 mg/cm3.
(A) 2 (B) 3
(C) 5 (D) 4
Explanation: (C)
 = P


P = 1.013 × 105 N/m2,  = 1.3kg/m3, v = 330 m/s

= v2P

= 1.4
Let f be the number of degree of freedom then
Cv = f R/2 and Cp= fR/2 + R = R (1 + f/2)
 = CC = 2 f f = 1.4
P

 (f = 5)

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12. The pressure of an ideal gas is written as P = 23 VE . Here E
refers to –
(A) Translational kinetic energy
(B) Rotational kinetic energy
(C) Vibrational kinetic energy
(D) Total kinetic energy
Explanation: (A)
Pressure of the gas
P = 13 mN
V
v 2
rms

Energy E = 1 2
2
v rms

1  v 2  v 22  ...  v 2n 
= mN 1 
2  N 
 
1 1 1
= 2
mv12  mv 22  .... mv 2n
2 2

So energy is basically the sum of energies of all molecules

which represents translational kinetic energy.

13. 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
Explanation: (D)
Kinetic energy according to kinetic theory of gases is given
by,
E = 32 RT

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 R is a universal constant and temperature remains same for
all gases. So kinetic energy is same for molecules of all gases
at a given temperature.

14. A gas is filled in a container at pressure 'P0' . If the mass of

each molecule is halved and speeds are double, then
resulting pressure will be -
(A) 4P0 (B) 2P0
(C) P0 (D) 4P0/ 2
Explanation: (B)
P0 = 13 Nm
V
v (m = mass of each molecule, V = volume of the
2
rms

gas)
Now, m' = 1/2 m
v'rms = 2vrms
2
1 Nm 2
P' = 13 Nm2.4Vv = 2.
rms
3 V
v rms = 2P0

15. The ratio of number of collisions per second at the walls of

containers by H2 and Ne gas molecules kept at same volume
and temperature is –
(A) 10 : 1 (B) 1 : 10
(C) 1 : 10 (D) 10 : 1
Explanation: (D)
Let's consider the wall perpendiculars to x-axis number of
collisions per second are given by
vx
2L
Now, vx = vrms/ 3
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 'L' and 'T' is same of both. Then, Ratio of speed given by
v rms,H 2 M Ne 20
v rms, Ne
= M H2
= 2
= 10 :1

16. Two containers of equal volumes contain H2 and O2 at same

temperature. If the number of molecules of these gases is
also equal then the ratio of pressure exerted by these will be
-
(A) 1: 1 (B) 4: 1
(C) 8: 1 (D) 16: 1
Explanation: (A)
Pressure of the gas
2 1 nM 3RT
P = 13 nM
V
v rms = 3 V
. M
= nRT
V

Given, nH2 = nO2, TH2 = TO2

VH2 = VO2
 PH2 = PO2

17. The density of carbon dioxide gas at 0ºC and at a pressure of

1.0 × 105 newton/metre2 is 1.98kg/m3. Find the root mean
square velocity of its molecules at 0ºC. Pressure is constant
(A) 39 metre/sec (B) 3.09 metre/sec
(C) 389 metre/sec (D) 38.9 metre/sec
Explanation: (C)
We know that
P = 13  v 2

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 / 2

 vrms =  3P 
 
  

Given that  = 1.98kg/m3 and

P = 1.0 × 105 N/m2
 vrms = 3  1.0  10 5
1.98

 vrms = 389 m/s

18.. 70 J of heat is required to raise the temperature of an ideal

diatomic gas at constant pressure from 30C to 35C. The amount
of heat required (in joules) to raise the temperature of the same
gas through the same range (30C to 35C) at constant volume is
(A) 30 (B) 50
(C) 70 (D) 90
Explanation: (B)
Heat required at constant pressure to raise the temperature by
T : Q1 = nCpT
Heat required at constant volume for the same rise in
temperature: Q2 = nCvT
 QQ  nC nC T C
2

T C
1 5
 v
 7
v

1 p p

  Q2 
5
7
 Q1 
5
7
 70  50 J .

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19. Molar specific heat at constant pressure for a diatomic gas is
(A) 52 R (B) 72 R
(C) 4 R (D) 32 R

Explanation: (B)

20. A given quantity of gas can be taken from a state p , V to a state 1 1

p , V by two different processes. Let Q and W represent the heat

2 2

supplied to the gas and the work done by the gas respectively.
Which of the following must be a constant for both processes?
(A) Q  W (B) Q
(C) W (D) Q  W
Explanation: (D)
 As initial and final temperatures are fixed, U =
constant
Q = W + U U = Q – W = constant
 21.A monatomic gas initially at 17 C is suddenly 

compressed to one eighth of its original volume. The

temperature after compression is
(A) 887 K (B) 36.25 K (C) 2320 K (D) 1160 K
 Sudden compression is adiabatic for which T V T V 1 1
 1
2 2
 1

 1
 = 290  8 = 290  4 = 1160 K
2
T2 = T1  VV
1
 3

 2

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22. Two identical containers A and B with frictionless pistons
contain the same ideal gas at the same temperature and the
same volume V. The mass of gas A is mA and that in B is
mB. The gas in each cylinder is now allowed to expand
isothermally to the same final volume 2V. The change in the
pressure in A and B are found to be P and 1.5P
respectively. Then
(A) 4 mA = 9mB (B) 2mA = 3mB
(C) 3mA = 2mB (D) 9mA = 4mB
Explanation: For gas in A, P   RT m
M V
1
A

  1

 RT  mA
P2   
 M  V2
 RT   1 1
 P  P1  P2    mA  V  V 
 M  1 2 

Putting V1  V and V2  2V
RT mA
We get P 
M 2V
 RT  mB
Similarly for Gas in B, 1.5 P   
 M  2V
From eq. (I) and (II) we get 2mB = 3mA
Hence, (C) is correct.

23. In a given process on an ideal gas dW = 0 and dQ < 0, then

for the gas
(A) the temperature will decreases.
(B) the volume will increases.
(C) the pressure will remain constant.
(D) the temperature will increase
Explanation:
dQ < 0
dU < 0 (or dW = 0)
T2< T1

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 Hence, (A) is correct.

INTEGER TYPE QUESTIONS

24. The specific heat of argon at constant volume is 0.075
kcal/kg K. Calculate its atomic weight, [R = 2cal/mol K]
Explanation:As argon is monatomic, its molar specific heat at
constant volume will be
3 3
C  R  2  3 cal/mol K
V
2 2

Now as CV = McV with CV = 0.075 cal/gK

3 = M × 0.075
3
Mw = 0.075 = 40 gm/mole

25. At the top of a mountain a thermometer reads 7ºC and

barometer reads 70 cm of Hg. At the bottom of the mountain
they read 27ºC and 76 cm of Hg respectively. Compare the
density of the air at the top with that at the bottom.
Explanation:
M  M 
PV = Mw
RT as   
 Mw 

P R  M 
or 
T M w as V  
 

Now as M and R are same for top and bottom

 P   P   P T

T
  
T
 i.e., 

P

T
T T B

 T  B B B T

 T 70 300 75
So   
 B 76 280 76
= 0.9868

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26. A closed container of volume 0.02 m3 contains a mixture of
neon and argon gases at temperature of 27ºC and pressure of
1 × 105 Nm2. The total mass of the mixture is 28 g. If the
gram molecular weights of neon and argon are 20 and 40
respectively, find the masses of the individual gases in the
container, assuming them to be ideal. Given R = 8.314
J/mol/K.
Explanation: Let m g be the mass of neon. Then, the mass
of argon is (28 – m)g.
Total number of moles of the mixture,
= 20m  2840 m  2840 m …(A)
1 10  0.02
5
Now,= PV   0.8 …(B)
RT 8.314  300
28  m
Equating (A) and (B), 40
= 0.8
or 28 + m = 32
or m = 4g mass of argon = (28 – 4)g = 24 g

27. A gas is taken through a v (cm3)

C
cyclic process ABCA as 850
shown in the Figure. If 3.6
650 B
calories of heat is given in the A

process, one calorie is 250 400  (kPa)

equivalent to

15
Explanation:
 Heat given = Work done in the process
= 21  (400 – 250)  103 (850 – 650)  10–6
= 21  150  200  10–3 = 15 J  3.6 cal.
 1 cal = 15
3 .6
= 4.17 J.

28. The average translational kinetic energy and the rms speed
of molecules in a sample of oxygen at 300K are 6.21 10-21J
and 484 m/s respectively. The corresponding values at 600K
area nearly (assuming ideal gas behavior).

Explanation:
3
KE  kT ,
2
3RT
v rms 
M
KE2 T
i.e. KE1
 2 2
T1

 KE2 = 2KE1 = 2  6.21  1021 = 12.42  1021J

v rms,2 T2
  2
v rms,1 T1

 v rms,2  2  v rms,1 = 684 m/s.

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29. Two moles of helium gas ( = 5/3) are initially at 27°C and
occupy a volume of 20 litres. The gas is first expanded at
constant pressure until the volume is doubled. Then it
undergoes an adiabatic change until the temperature returns
to its initial value. What is the work done by the gas?
Explanation:
Work done by the gas in the part ab
= Pa (Vb – Va)
= nRT2 – nRT1
= 2 mol × (8.3 J/mol-K) × (600 K – 300 K) = 4980 J
Pb V b  PcVc
The work done in the adiabatic part bc =  1

nR (T2 T1) 4980 J

=  1
= 5/3 1
= 7470 J

The net work done by the gas

= 4980 J + 7470 J
= 12450 J
30. Two cylinders A and B fitted with pistons contain equal
amounts of an ideal diatomic gas at 300K. The piston A is
free to move, while that of B is held fixed. The same amount
of heat is given to the gas in each cylinder. If the rise in
temperature of the gas in A is 30K, then the rise in
temperature of the gas in B is

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Explanation:
For cylinder A. For cylinder B
dQ = nCpdT dQ = nCvdT
nCPdT= nCV dT
 dT 
C  30
P

C
= 30  1.4 = 42 K
v

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