12

P-186 Physics

Kinetic Theory
Kinetic Theory of an Ideal nRT é 1 1 ù nRT é l1 - l2 ù
TOPIC 1 (c) ê + ú (d) ê ú
Gas and Gas Laws g ë l2 l1 û g ë l1 l2 û

1. Initially a gas of diatomic molecules is contained in a 5. The temperature of an open room of volume 30 m3 increases
cylinder of volume V1 at a pressure P1 and temperature from 17°C to 27°C due to sunshine. The atmospheric pressure
250 K. Assuming that 25% of the molecules get in the room remains 1 × 105 Pa. If ni and nf are the number of
dissociated causing a change in number of moles. The molecules in the room before and after heating, then nf – ni
pressure of the resulting gas at temperature 2000 K, will be : [2017]
when contained in a volume 2V1 is given by P2. The ratio (a) 2.5 × 1025 (b) –2.5 × 1025
P2/P1 is ______. [NA Sep. 06, 2020 (I)] (c) –1.61 × 1023 (d) 1.38 × 1023
6. For the P-V diagram given for an ideal gas,
2. The change in the magnitude of the volume of an ideal gas
when a small additional pressure DP is applied at a constant 1

temperature, is the same as the change when the P

Constant
P=
temperature is reduced by a small quantity DT at constant V
pressure. The initial temperature and pressure of the gas
were 300 K and 2 atm. respectively. If | DT |= C | DP | ,
then value of C in (K/atm.) is __________. 2
[NA Sep. 04, 2020 (II)] V
3. The number density of molecules of a gas depends on out of the following which one correctly represents the
their distance r from the origin as , n(r) = n0e–ar4. Then T-P diagram ? [Online April 9, 2017]
the total number of molecules is proportional to : 2
[12 April 2019 II] 2
T
(a) n0a –3/4 (b) n0 a1/2 (a) T (b)
(c) n0 a1/4 (d) n0 a–3 1 1
4. A vertical closed cylinder is separated into two parts by a P
P
frictionless piston of mass m and of negligible thickness.
The piston is free to move along the length of the cylinder. T T
The length of the cylinder above the piston is l1, and that 2 1
1 2
below the piston is l2, such that l1 > l2. Each part of the
cylinder contains n moles of an ideal gas at equal temperature (c) (d)
T. If the piston is stationary, its mass, m, will be given by:
P P
(R is universal gas constant and g is the acceleration due to
7. Chamber I Chamber II
gravity) [12 Jan. 2019 II]
ideal real
RT é l1 - 3l2 ù RT é 2l1 + l2 ù gas gas
(a) ê ú (b) g ê l I ú
ng ë l1 I 2 û ë 1 2 û 1 2 3 4
 Kinetic Theory P-187

There are two identical chambers, completely thermally (a) 104 N/m2 (b) 108 N/m2
insulated from surroundings. Both chambers have a 3
(c) 10 N/m 2 (d) 1016 N/m2
partition wall dividing the chambers in two compartments. 13. The temperature, at which the root mean square velocity
Compartment 1 is filled with an ideal gas and of hydrogen molecules equals their escape velocity from
Compartment 3 is filled with a real gas. Compartments 2 the earth, is closest to : [8 April 2019 II]
and 4 are vacuum. A small hole (orifice) is made in the [Boltzmann Constant kB = 1.38 × 10 J/K–23
partition walls and the gases are allowed to expand in
vacuum. Avogadro Number NA = 6.02 × 1026 /kg
Statement-1: No change in the temperature of the gas Radius of Earth : 6.4 × 106 m
takes place when ideal gas expands in vacuum. However, Gravitational acceleration on Earth = 10 ms–2]
the temperature of real gas goes down (cooling) when it (a) 800 K (b) 3 × 105 K
expands in vacuum. 4
Statement-2: The internal energy of an ideal gas is only (c) 10 K (d) 650 K
kinetic. The internal energy of a real gas is kinetic as 14. A mixture of 2 moles of helium gas (atomic mass = 4u), and
well as potential. [Online April 9, 2013] 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in
(a) Statement-1 is false and Statement-2 is true. a container. The ratio of their rms speeds
(b) Statement-1 and Statement-2 both are true.
Statement-2 is the correct explanation of Statement-1. é Vrms ( helium ) ù
ê ú is close to : [9 Jan. 2019 I]
(c) Statement-1 is true and Statement-2 is false. ë Vrms ( argon ) û
(d) Statement-1 and Statement-2 both are true.
(a) 3.16 (b) 0.32
Statement-2 is not correct explanation of Statement-1. (c) 0.45 (d) 2.24
8. Cooking gas containers are kept in a lorry moving with 15. N moles of a diatomic gas in a cylinder are at a temperature
uniform speed. The temperature of the gas molecules T. Heat is supplied to the cylinder such that the tempera-
inside will [2002] ture remains constant but n moles of the diatomic gas get
(a) increase converted into monoatomic gas. What is the change in
(b) decrease the total kinetic energy of the gas ?
(c) remain same [Online April 9, 2017]
(d) decrease for some, while increase for others
1
(a) nRT (b) 0
2
Speed of Gas, Pressure
TOPIC 2 3 5
and Kinetic Energy (c) nRT (d) nRT
2 2
16. In an ideal gas at temperature T, the average force that a
9. Number of molecules in a volume of 4 cm 3 of a perfect molecule applies on the walls of a closed container
monoatomic gas at some temperature T and at a pressure depends on T as Tq. A good estimate for q is:
of 2 cm of mercury is close to? (Given, mean kinetic en- [Online April 10, 2015]
ergy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2,
density of mercury = 13.6 g/cm3) [Sep. 05, 2020 (I)] 1
(a) (b) 2
(a) 4.0 × 1018 (b) 4.0 × 1016 2
16 1
(c) 5.8 × 10 (d) 5.8 × 1018 (c) 1 (d)
10. Nitrogen gas is at 300°C temperature. The temperature 4
17. A gas molecule of mass M at the surface of the Earth has
(in K) at which the rms speed of a H2 molecule would be
kinetic energy equivalent to 0°C. If it were to go up
equal to the rms speed of a nitrogen molecule, is
straight without colliding with any other molecules, how
_____________. (Molar mass of N2 gas 28 g);
high it would rise? Assume that the height attained is much
[NA Sep. 05, 2020 (II)]
less than radius of the earth. (kB is Boltzmann constant).
11. For a given gas at 1 atm pressure, rms speed of the [Online April 19, 2014]
molecules is 200 m/s at 127°C. At 2 atm pressure and at
273k B
227°C, the rms speed of the molecules will be: (a) 0 (b)
[9 April 2019 I] 2Mg
(a) 100 m/s (b) 80 5 m/s 546k B 819k B
(c) (d)
(c) 100 5 m/s (d) 80 m/s 3Mg 2Mg
12. If 1022 gas molecules each of mass 10–26 kg collide with a 18. At room temperature a diatomic gas is found to have an
surface (perpendicular to it) elastically per second over r.m.s. speed of 1930 ms–1. The gas is:
an area 1 m2 with a speed 104 m/s, the pressure exerted [Online April 12, 2014]
by the gas molecules will be of the order of : (a) H2 (b) Cl2
[8 April 2019 I] (c) O2 (d) F2
P-188 Physics

19. In the isothermal expansion of 10g of gas from volume The total internal energy, U of a mole of this gas, and the
V to 2V the work done by the gas is 575J. What is the æ Cp ö
root mean square speed of the molecules of the gas at value of g ç = ÷ are given, respectively, by:
è Cv ø
that temperature? [Online April 25, 2013] [Sep. 06, 2020 (I)]
(a) 398m/s (b) 520m/s
(c) 499m/s (d) 532m/s 5 6 7
20. A perfect gas at 27°C is heated at constant pressure so as (a) U = RT and g = (b) U = 5RT and g =
2 5 5
to double its volume. The final temperature of the gas will
be, close to [Online May 7, 2012] 5 7 6
(c) U = RT and g = (d) U = 5RT and g =
(a) 327°C (b) 200°C 2 5 5
(c) 54°C (d) 300°C 27. In a dilute gas at pressure P and temperature T, the mean
21. A thermally insulated vessel contains an ideal gas of time between successive collisions of a molecule varies
molecular mass M and ratio of specific heats g. It is
moving with speed v and it's suddenly brought to rest. with T is : [Sep. 06, 2020 (II)]
Assuming no heat is lost to the surroundings, its 1
temperature increases by: [2011] (a) T (b)
T
( g - 1) Mv 2 K gM 2v
(a) (b) K
2 gR 2R 1
(c) (d) T
( g - 1) ( g - 1) 2 T
(c) Mv 2 K (d) 2( g + 1) R Mv K
2R 28. Match the Cp/Cv ratio for ideal gases with different type
22. Three perfect gases at absolute temperatures T1, T2 and T3 of molecules : [Sep. 04, 2020 (I)]
are mixed. The masses of molecules are m1, m2 and m3 and Column-I Column-II
the number of molecules are n1, n2 and n3 respectively. Molecule Type Cp/Cv
Assuming no loss of energy, the final temperature of the (A) Monatomic (I) 7/5
mixture is : [2011] (B) Diatomic rigid molecules (II) 9/7
n1T1 + n2T2 + n3T3 n1T12 + n2T22 + n3T32 (C) Diatomic non-rigid molecules(III) 4/3
(a) (b)
n1 + n2 + n3 n1T1 + n2T2 + n3T3 (D) Triatomic rigid molecules (IV) 5/3
n12T12 + n22T22 + n32T32 (T1 + T2 + T3 ) (a) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(c) (d) (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
n1T1 + n2T2 + n3T3 3
23. One kg of a diatomic gas is at a pressure of (c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
8 × 104N/m2. The density of the gas is 4kg/m3. What is (d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
the energy of the gas due to its thermal motion?[2009] 29. A closed vessel contains 0.1 mole of a monatomic ideal
(a) 5 × 104 J (b) 6 × 104 J gas at 200 K. If 0.05 mole of the same gas at 400 K is
4
(c) 7 × 10 J (d) 3 × 104 J added to it, the final equilibrium temperature (in K) of
24. The speed of sound in oxygen (O2) at a certain temperature the gas in the vessel will be close to _________.
is 460 ms–1. The speed of sound in helium (He) at the same [NA Sep. 04, 2020 (I)]
temperature will be (assume both gases to be ideal)
[2008] 30.
(a) 1421 ms –1 (b) 500 ms –1

(c) 650 ms–1 (d) 330 ms–1

25. At what temperature is the r.m.s velocity of a hydrogen
molecule equal to that of an oxygen molecule at 47°C? Consider a gas of triatomic molecules. The molecules are
[2002] assumed to be triangular and made of massless rigid rods
whose vertices are occupied by atoms. The internal energy
(a) 80 K (b) –73 K
(c) 3 K (d) 20 K of a mole of the gas at temperature T is :
[Sep. 03, 2020 (I)]
Degree of Freedom, Specific 5 3
TOPIC 3 Heat Capacity, and Mean (a) RT (b) RT
2 2
Free Path
9
26. Molecules of an ideal gas are known to have three (c) RT (d) 3RT
2
translational degrees of freedom and two rotational 31. To raise the temperature of a certain mass of gas by 50°C
degrees of freedom. The gas is maintained at a at a constant pressure, 160 calories of heat is required.
temperature of T. When the same mass of gas is cooled by 100°C at constant
 Kinetic Theory P-189

volume, 240 calories of heat is released. How many 37. Consider a mixture of n moles of helium gas and 2n
degrees of freedom does each molecule of this gas have moles of oxygen gas (molecules taken to be rigid) as an
(assume gas to be ideal)? [Sep. 03, 2020 (II)] ideal gas. Its CP/CV value will be: [8 Jan. 2020 II]
(a) 5 (b) 6 (a) 19/13 (b) 67/45
(c) 3 (d) 7 (c) 40/27 (d) 23/15
32. A gas mixture consists of 3 moles of oxygen and 5 moles of Cp 5
argon at temperature T. Assuming the gases to be ideal 38. Two moles of an ideal gas with C = are mixed with 3
V 3
and the oxygen bond to be rigid, the total internal energy
(in units of RT) of the mixture is : [Sep. 02, 2020 (I)] Cp 4
(a) 15 (b) 13 moles of another ideal gas with C = . The value of
V 3
(c) 20 (d) 11 Cp
33. An ideal gas in a closed container is slowly heated. As its for the mixture is: [7 Jan. 2020 I]
temperature increases, which of the following statements CV
are true? [Sep. 02, 2020 (II)] (a) 1. 45 (b) 1.50
(1) The mean free path of the molecules decreases (c) 1.47 (d) 1.42
(2) The mean collision time between the molecules 39. Two moles of helium gas is mixed with three moles of
decreases hydrogen molecules (taken to be rigid). What is the molar
(3) The mean free path remains unchanged specific heat of mixture at constant volume?
(4) The mean collision time remains unchanged (R = 8.3 J/mol K) [12 April 2019 I]
(a) (2) and (3) (b) (1) and (2) (a) 19.7 J/mol L (b) 15.7 J/mol K
(c) (3) and (4) (d) (1) and (4) (c) 17.4 J/mol K (d) 21.6 J/mol K
34. Consider two ideal diatomic gases A and B at some 40. A diatomic gas with rigid molecules does 10 J of work
temperature T. Molecules of the gas A are rigid, and have when expanded at constant pressure. What would be the
a mass m. Molecules of the gas B have an additional heat energy absorbed by the gas, in this process ?
m [12 April 2019 II]
vibrational mode, and have a mass . The ratio of the (a) 25 J (b) 35 J
4
(c) 30 J (d) 40 J
specific heats (CA B
V and CV ) of gas A and B, respectively 41. A 25×10 – 3 m3 volume cylinder is filled with 1 mol of O2
is: [9 Jan 2020 I]
gas at room temperature (300 K) . The molecular diameter
(a) 7 : 9 (b) 5 : 9
of O2, and its root mean square speed, are found to be 0.3
(c) 3 : 5 (d) 5 : 7
nm and 200 m/s, respectively. What is the average collision
35. Two gases-argon (atomic radius 0.07 nm, atomic weight
rate (per second) for an O2 molecule?
40) and xenon (atomic radius 0.1 nm, atomic weight 140)
[10 April 2019 I]
have the same number density and are at the same
(a) ~1012 (b) ~1011
temperature. The ratio of their respective mean free
(c) ~1010 (d) ~1013
times is closest to: [9 Jan 2020 II]
42. When heat Q is supplied to a diatomic gas of rigid
(a) 3.67 (b) 1.83
molecules, at constant volume its temperature increases
(c) 2.3 (d) 4.67
by DT. The heat required to produce the same change in
36. The plot that depicts the behavior of the mean free time t
temperature, at a constant pressure is :
(time between two successive collisions) for the molecules
[10 April 2019 II]
of an ideal gas, as a function of temperature (T),
qualitatively, is: (Graphs are schematic and not drawn to 2 5
(a) Q (b) Q
scale) [8 Jan. 2020 I] 3 3
7 3
(c) Q (d) Q
5 2
t t 43. An HCl molecule has rotational, translational and
(a) (b) vibrational motions. If the rms velocity of HCl molecules
in its gaseous phase is v , m is its mass and k B is
1
T T Boltzmann constant, then its temperature will be:
[9 April 2019 I]
mv 2 mv 2
t (a) (b)
t 6kB 3k B
(c) (d)
mv 2 mv 2
1 (c) (d)
T T 7k B 5k B
P-190 Physics

44. The specific heats, C p and Cv of a gas of diatomic 51. Two moles of an ideal monoatomic gas occupies a volume
molecules, A, are given (in units of J mol–1 k–1) by 29 and V at 27°C. The gas expands adiabatically to a volume 2 V.
22, respectively. Another gas of diatomic molecules, B, Calculate (1) the final temperature of the gas and (2) change
has the corresponding values 30 and 21. If they are treated in its internal energy. [2018]
as ideal gases, then: [9 April 2019 II] (a) (1) 189 K (2) 2.7 kJ
(a) A is rigid but B has a vibrational mode. (b) (1) 195 K (2) –2.7 kJ
(b) A has a vibrational mode but B has none. (c) (1) 189 K (2) –2.7 kJ
(c) A has one vibrational mode and B has two. (d) (1) 195 K (2) 2.7 kJ
(d) Both A and B have a vibrational mode each. 52. Two moles of helium are mixed with n with moles of
45. An ideal gas occupies a volume of 2 m 3 at a pressure of 3 C 3
× 106 Pa. The energy of the gas: [12 Jan. 2019 I] hydrogen. If P = for the mixture, then the value of n
CV 2
(a) 9 × 106 J (b) 6 × 104 J is [Online April 16, 2018]
(c) 108 J (d) 3 × 102 J (a) 3/2 (b) 2
46. An ideal gas is enclosed in a cylinder at pressure of 2 atm (c) 1 (d) 3
and temperature, 300 K. The mean time between two 53. Cp and Cv are specific heats at constant pressure and
successive collisions is 6 × 10–8 s. If the pressure is doubled constant volume respectively. It is observed that
and temperature is increased to 500 K, the mean time Cp – Cv = a for hydrogen gas
between two successive collisions wiil be close to: Cp – Cv = b for nitrogen gas
[12 Jan. 2019 II] The correct relation between a and b is : [2017]
–7
(a) 2 × 10 s (b) 4 × 10 s –8 (a) a = 14 b (b) a = 28 b
(c) 0.5 × 10–8 s (d) 3 × 10–6 s 1
(c) a = b (d) a = b
14
47. A gas mixture consists of 3 moles of oxygen and 5 moles 54. An ideal gas has molecules with 5 degrees of freedom.
of argon at temperature T. Considering only translational The ratio of specific heats at constant pressure (C p) and
and rotational modes, the total internal energy of the at constant volume (Cv) is : [Online April 8, 2017]
system is : [11 Jan. 2019 I] 7
(a) 6 (b)
(a) 15 RT (b) 12 RT 2
(c) 4 RT (d) 20 RT 5 7
(c) (d)
48. In a process, temperature and volume of one mole of an 2 5
ideal monoatomic gas are varied according to the relation 55. An ideal gas undergoes a quasi static, reversible process
VT = K, where K is a constant. In this process the in which its molar heat capacity C remains constant. If
temperature of the gas is increased by DT. The amount of during this process the relation of pressure P and volume
heat absorbed by gas is (R is gas constant) : V is given by PVn = constant, then n is given by (Here CP
[11 Jan. 2019 II] and CV are molar specific heat at constant pressure and
constant volume, respectively) : [2016]
1 1
(a) RΔT (b) KRΔT CP - C C - CV
2 2 (a) n = (b) n =
2K C - CV C - CP
3
(c) RΔT (d) ΔT CP C – CP
2 3 (c) n = (d) n =
49. Two kg of a monoatomic gas is at a pressure of 4 × 104 CV C – CV
N/m2. The density of the gas is 8 kg/m3. What is the 56. Using equipartition of energy, the specific heat
order of energy of the gas due to its thermal motion? (in J kg–1 K–1) of aluminium at room temperature can
be estimated to be (atomic weight of aluminium = 27)
[10 Jan 2019 II] [Online April 11, 2015]
3 5
(a) 10 J (b) 10 J (a) 410 (b) 25
(c) 104 J (d) 106 J (c) 1850 (d) 925
50. A 15 g mass of nitrogen gas is enclosed in a vessel at 57. Modern vacuum pumps can evacuate a vessel down to a
a temperature 27°C. Amount of heat transferred to the pressure of 4.0 × 10–15 atm. at room temperature (300
K). Taking R = 8.0 JK–1 mole–1, 1 atm = 105 Pa and
gas, so that rms velocity of molecules is doubled, is NAvogadro = 6 × 1023 mole–1, the mean distance between
about: [Take R = 8.3 J/K mole] [9 Jan. 2019 II] molecules of gas in an evacuated vessel will be of the
(a) 0.9 kJ (b) 6 kJ order of: [Online April 9, 2014]
(c) 10 kJ (d) 14 kJ (a) 0.2 mm (b) 0.2 mm
(c) 0.2 cm (d) 0.2 nm
 Kinetic Theory P-191

58. Figure shows the variation in temperature (DT) with the 60. If CP and CV denote the specific heats of nitrogen per unit
amount of heat supplied (Q) in an isobaric process mass at constant pressure and constant volume
corresponding to a monoatomic (M), diatomic (D) and a respectively, then [2007]
polyatomic (P) gas. The initial state of all the gases are the (a) CP – CV = 28R (b) CP – CV = R/28
same and the scales for the two axes coincide. Ignoring (c) CP – CV = R/14 (d) CP – CV = R
vibrational degrees of freedom, the lines a, b and c 61. A gaseous mixture consists of 16 g of helium and 16 g of
respectively correspond to : [Online April 9, 2013]
Cp
a oxygen. The ratio of the mixture is [2005]
Cv
Q b (a) 1.62 (b) 1.59
(c) 1.54 (d) 1.4
c
62. One mole of ideal monatomic gas (g = 5/3) is mixed with
one mole of diatomic gas (g = 7/5). What is g for the
DT mixture? g Denotes the ratio of specific heat at constant
(a) P, M and D (b) M, D and P pressure, to that at constant volume [2004]
(a) 35/23 (b) 23/15
(c) P, D and M (d) D, M and P
(c) 3/2 (d) 4/3
Cp 63. During an adiabatic process, the pressure of a gas is found
59. A given ideal gas with g = = 1.5 at a temperature T. If
Cv to be proportional to the cube of its absolute temperature.
the gas is compressed adiabatically to one-fourth of its The ratio CP/CV for the gas is [2003]
initial volume, the final temperature will be 4
(a) (b) 2
[Online May 12, 2012] 3
(a) 2 2T (b) 4 T 5 3
(c) (d)
(c) 2 T (d) 8 T 3 2
P-192 Physics

1. (5) Using ideal gas equation, PV = nRT æ 1 1ö

Þ nRT ç l – l ÷ = mg
Þ PV
1 1 = nR ´ 250 [Q T1 = 250 K] ...(i) è 2 1ø

5n nRT æ l1 – l 2 ö
P2 (2V1 = R ´ 2000 [Q T2 = 2000 K] ...(ii) \ m = g ç l ×l ÷
4 è 1 2 ø
Dividing eq. (i) by (ii), 5. (b) Given: Temperature Ti = 17 + 273 = 290 K
P1 4 ´ 250 P 1 Temperature Tf = 27 + 273 = 300 K
= Þ 1 =
2 P2 5 ´ 2000 P2 5 Atmospheric pressure, P0 = 1 × 105 Pa
P2 Volume of room, V0 = 30 m3
\ = 5. Difference in number of molecules, nf – ni = ?
P1
2. (150) In first case, Using ideal gas equation, PV = nRT(N0),
From ideal gas equation N0 = Avogadro's number
PV = nRT PV
Þ n= (N )
P D V + V DP = 0 (As temperature is constant) RT 0

DV = -
DP
V
P0V0 æ 1 1 ö
...(i) \ nf – ni = ç - ÷N
P R è T f Ti ø 0
In second case, using ideal gas equation again
P DV = -nR DT 1 ´ 105 ´ 30 æ 1 1 ö
= ´ 6.023 ´ 10 23 ç - ÷
nR DT 8.314 è 300 290 ø
DV = - ...(ii) = – 2.5 × 1025
P
Equating (i) and (ii), we get 6. (c) From P-V graph,

nR DT DP V 1
=- V Þ DT = DP Pµ , T = constant and Pressure is increasing from 2
P P nR V
to 1 so option (3) represents correct T-P graph.
Comparing the above equation with | DT | = C | DP | , we 7. (a) In ideal gases the molecules are considered as point
have particles and for point particles, there is no internal
V DT 300 K excitation, no vibration and no rotation. For an ideal gas
C= = = = 150 K/atm the internal energy can only be translational kinetic energy
nR DP 2 atm
and for real gas both kinetic as well as potential energy.
8. (c) The centre of mass of gas molecules also moves with
3. (a) N = ò r(dv)
lorry with uniform speed. As there is no relative motion of
r
4
r
4 gas molecule. So, kinetic energy and hence temperature
-ar
= ò n0 e ´ 4pr 2 dr = 4p n òr
2
(e -ar )dr remain same.
0
0 0
3
µ n0 a–3/4 9. (c) Given : K.E.mean = kT = 4 ´ 10 -14
4. (d) Clearly from figure, 2
P2A = P1A + mg P = 2 cm of Hg, V = 4 cm3
nRT × A nRT × A PV PrgV 2 ´ 13.6 ´ 980 ´ 4
or, + mg ; 4 ´ 1018
Al 2 = Al1 N= =
8
KT KT
´ 10 -14
3
n
l1 T P1A 10. (41) Room mean square speed is given by

3RT
vrms =
l2 n P2A mg M
T
 Kinetic Theory P-193

Here, M = Molar mass of gas molecule Now, change in total kinetic energy of the gas
T = temperature of the gas molecule 1
DU = Q = nRT
We have given vN2 = vH2 2
1 mN 2
3RTN2 3RTH 2 16 . (c) Pressure, P = V rm s
3 V
\ =
M N2 M H2 (mN )T
or, P =
V
TH2 573 If the gas mass and temperature are constant then
Þ = Þ TH2 = 41 K
2 28 P µ (Vrms)2 µ T
So, force µ (Vrms)2 µ T
3RT i.e., Value of q = 1
11. (c) Vrms =
M 17. (d) Kinetic energy of each molecule,
v1 T1 (273 + 127) 400 4 2 3
= = = = = K.E. = K B T
v2 T2 (273 + 237) 500 5 5 2
In the given problem,
5 5 Temperature, T = 0°C = 273 K
\ v2 = v1 = ´ 200 = 100 5 m/s. Height attained by the gas molecule, h = ?
2 2
12. (Bouns) Rate of change of momentum during collision 3 819K B
K.E. = K B ( 273) =
mv – (– mv ) 2mv 2 2
= = N K.E. = P.E.
Dt Dt
819K B
Þ = Mgh
so pressure P = N ´ (2mv) 2
Dt ´ A 819K B
or h =
1022 ´ 2 ´ 10–26 ´ 104 2Mg
= = 2 N / m2
1´ 1 3RT
18. (a) Q C=
13. (c) vrms = ve M
3 ´ 8.314 ´ 300
3RT
= 11.2 ´ 103 (1930 ) =
2

M M
3 ´ 8.314 ´ 300
M= » 2 ´10 -3 kg
3kT 1930 ´1930
or = 11.2 ´ 103
m The gas is H2.
3rv
3 ´ 1.38 ´ 10-23 T 3 19. (c) v rms =
or = 11.2 ´ 10 \ v = 104 K mass of the gas
2 ´ 10-3 20. (a) Given, V1 = V
V1rms M2 V2 = 2V
14. (a) Using V = M1 T1 = 27° + 273 = 300 K
2rms
T2 = ?
Vrms ( He ) M Ar 40
= From charle’s law
Vrms ( Ar ) M He = 4 = 3.16
V1 V2
=
T1 T2 (
15. (a) Energy associated with N moles of diatomic gas, Q Pressure is constant )
5
Ui = N RT V 2V
2 or, 300 = T
Energy associated with n moles of monoatomic gas 2
3 \ T2 = 600 K = 600 – 273 = 327°C
= n RT 21. (c) As, work done is zero.
2
Total energy when n moles of diatomic gas converted into So, loss in kinetic energy = heat gain by the gas
1 2 R
3 5 mv = nCv DT = n DT
monoatomic (Uf) = 2n RT + (N - n) RT 2 g -1
2 2
1 2 m R
1 5 mv = DT
= nRT + NRT 2 M g -1
2 2
P-194 Physics

Mv 2 ( g - 1) Cp 2 2 7
\ DT = K And g = = 1+ = 1+ =
2R Cv f 5 5
n1
22. (a) Number of moles of first gas = N 1
A
n2 27. (b) Mean free path, l =
Number of moles of second gas = N 2pnd 2
A where, d = diameter of the molecule
n3
Number of moles of third gas = N n = number of molecules per unit volume
A
If there is no loss of energy then l
But, mean time of collision, t =
P1V1 + P2V2 + P3V3 = PV vrms
n1 n n
RT1 + 2 RT2 + 3 RT3 3kT
NA NA NA But vrms =
R
n1 + n2 + n3
= RTmix
NA l 1
\t = Þtµ
n1T1 + n2T2 + n3T3 3kT T
Tmix = n1 + n2 + n3 m
23. (a) Given, mass = 1 kg 28. (c) As we know,
Density = 4 kg m–3
Cp 2
mass 1 g= = 1+ , where f = degree of freedom
Volume = = m3 Cv f
density 4
Internal energy of the diatomic gas (A) Monatomic, f = 3
5 5 1 2 5
= PV = ´ 8 ´ 104 ´ = 5 ´ 10 4 J \ g = 1+ =
2 2 4 3 3
Alternatively: (B) Diatomic rigid molecules, f = 5
5 5m 5 m PM 2 7
K.E = nRT = RT = ´ [Q PM = dRT ] \ g = 1+ =
2 2M 2M d 5 5
4
5 mP 5 1 ´ 8 ´ 10 (C) Diatomic non-rigid molecules, f = 7
= = ´ = 5 ´ 104 J
2 d 2 4 2 9
\ g = 1+ =
gRT 7 7
24. (a) The speed of sound in a gas is given by v =
M (D) Triatomic rigid molecules, f = 6
g 2 4
\ vµ [As R and T is constant] \ g = 1+ =
M 6 3
vO2 g O2 M He 29. (266.67) Here work done on gas and heat supplied to the
\ = ´ gas are zero.
vHe M O2 g He
Let T be the final equilibrium temperature of the gas in the
1.4 4 vessel.
= ´ = 0.3237 Total internal energy of gases remain same.
32 1.67
vO 2 460 i.e., u1 + u2 = u '1 + u '2
\ vHe = = = 1421 m / s
0.3237 0.3237 or, n1Cv DT1 + n2Cv DT2 = (n1 + n2 )CvT
25. (d) RMS velocity of a gas molecule is given by Þ (0.1)Cv (200) + (0.05)Cv (400) = (0.15)CvT
3RT 800
Vrms = \T = = 266.67 K
M 3
Let T be the temperature at which the velocity of hydrogen
30. (d) Here degree of freedom, f = 3 + 3 = 6 for triatomic non-
molecule is equal to the velocity of oxygen molecule.
linear molecule.
3RT 3R ´ (273 + 47) Internal energy of a mole of the gas at temperature T,
\ =
2 32 f 6
Þ T = 20K U= nRT = RT = 3RT
2 2
26. (c) Total degree of freedom f = 3 + 2 = 5 31. (b) Let Cp and Cv be the specific heat capacity of the gas
Total energy, U =
nfRT 5RT at constant pressure and volume.
=
2 2
 Kinetic Theory P-195

At constant pressure, heat required

5
DQ1 = nC p DT R
CvA
5
B 7 = 2 =
Þ 160 = nC p × 50 \ Cv = R Hence C B 7 7
...(i) 2 v R
2
At constant volume, heat required
35. (Bonus) Mean free path of a gas molecule is given by
DQ2 = nCv DT
1
l=
Þ 240 = nCv ×100 ...(ii) 2pd 2 n
Dividing (i) by (ii), we get Here, n = number of collisions per unit volume
d = diameter of the molecule
160 C p 50 Cp 4
= × Þ = If average speed of molecule is v then
240 Cv 100 Cv 3
l
Cp 4 2 Mean free time, t =
g= = = 1+ (Here, f = degree of freedom) v
Cv 3 f 1 1 M
Þ t= =
Þ f = 6. 2
2pnd v 2pnd 2 3RT
32. (a) Total energy of the gas mixture, æ 3RT ö
f1n1 RT1 f 2 n2 RT2 çQ v = M ÷
Emix = + è ø
2 2 2
5 5 M \ t1 = M1 ´ d2
= 3 ´ RT + ´ 3RT = 15RT \ tµ
d2 t2 d12 M2
2 2
33. (a) As we know mean free path 40 æ 0.1 ö
2
= ´ç ÷ = 1.09
1 140 è 0.07 ø
l=
æ Nö mean free path
2 ç ÷ pd 2 1
èV ø 36. (c) Relaxation time (t ) µ Þ tµ
speed v
Here, N = no. of molecule and, v µ T
V = volume of container
1
d = diameter of molecule \tµ
T
But PV = nRT = nNKT
1
N P Hence graph between t v/s is a straight line which
Þ = =n T
V KT is correctly depicted by graph shown in option (c).
1 KT 37. (a) Helium is a monoatomic gas and Oxygen is a diatomic
l= gas.
2 pd 2 P
For constant volume and hence constant number density 3 5
For helium, CV1 = R and CP1 = R
2 2
P
n of gas molecules is constant.
T 5 7
For oxygen, CV2 = R and CP = R
So mean free path remains same. 2 2 2
As temperature increases no. of collision increases so N1CP1 + N 2 CP2
relaxation time decreases. g=
34. (d) Specific heat of gas at constant volume N1CV1 + N 2 CV2

1 5 7
Cv = fR; f = degree of freedom n. R + 2n. R
2 g= 2 2 = 19nR ´ 2
For gas A (diatomic) Þ 3 5 2(13nR )
n. R + 2n. R
f = 5 (3 translational + 2 rotational) 2 2
5 æC ö 19
\C A = R \ç P ÷ =
v 2 è V ø mixture 13
C
For gas B (diatomic) in addition to (3 translational + 2
rotational) 2 vibrational degree of freedom.
P-196 Physics

n1C p + n2C p 1
1 2 43. (a) mv 2 = 3k BT
38. (d) Using, gmixture= 2
n1Cv + n2Cv
1 2
mv 2
n1 n n +n or T =
Þ + 2 = 1 2 6k B
g 1 –1 g 2 –1 g m –1 CP 29
3 2 5 44. (b) gA = = = 1.32 < 1.4 (diatomic)
Þ + = Cv 22
4 5 g m –1
–1 –1 30 10
3 3 and g B = = = 1.43 > 1.4
21 7
9 2´3 5 5 Gas A has more than 5-degrees of freedom.
Þ + = Þ g m –1 =
1 2 g m –1 12 45. (a) Energy of the gas, E
17 f f
Þ gm = = 1.42 = nRT = PV
12 2 2
n1[Cv1 ] + n2 [Cv2 ] f
= (3 ´ 106 )(2) = f ´ 3 ´ 106
39. (c) [Cv]min = 2
n1 + n2
Considering gas is monoatomic i.e., f = 3
é 3R 5R ù Energy, E = 9 × 106 J
ê2´ 2 + 3´ 2 ú 1
46. (b) Using, t=
= êê 2+3
ú
ú 2npd 2 Vavg
ë û
= 2.1 R = 2.1 × 8.3 = 17.4 J/mol–k T é no.of molecules ù
\t µ êë\n = úû
P Volume
Cv 1 1 5
40. (b) F = C = r = (7 / 5) = t1 500 P
p 7 or, = ´ » 4´10–8
6 ´10 –8
2P 300
W 5 2
or = 1- = f1 f
Q 7 7 47. (a) U = n1RT + 2 n 2RT
2 2
7 7 ´ 10 Considering translational and rotational modes, degrees
or Q = W = = 35 J
2 2 of freedom f1 = 5 and f2 = 3
41. (c) V = 25 × 10–3 m3, N = 1 mole of O2
T = 300 K 5 3
\ u = (3RT) + ´ 5RT
Vrms = 200 m/s 2 2
1 U = 15RT
\ l= 48. (a) According to question VT = K
2Npr 2
1 <V> we also know that PV = nRT
Average time = = 200.Npr 2 . 2
t l æ PV ö
ÞT = ç
23 è nR ÷ø
2 ´ 200 ´ 6.023 ´ 10
= .p ´ 10-18 ´ 0.09
25 ´ 10 -3 æ PV ö
ÞVç ÷ = k Þ PV = K
2

The closest value in the given option is = 1010 è nR ø

42. (c) Amount of heat required (Q) to raise the temperature R
at constant volume Q C= + CV (For polytropic process)
1– x
Q = nCvDT ...(i)
R 3R R
Amount of heat required (Q1) at constant pressure C= + =
Q1 = nCPDT ...(ii) 1– 2 2 2
Dividing equation (ii) by (i), we get \DQ = nC DT
Q1 C p R
\ = = ´DT [here, n = 1 mole]
Q Cv 2
49. (c) Thermal energy of N molecule
æ 7ö æ C p 7ö
æ3 ö
Þ Q1 = (Q) ç ÷
è 5ø çQ g = C = 5 ÷ = N ç kT ÷
è v ø è2 ø
 Kinetic Theory P-197

N 3 3 3 54. (d) The ratio of specific heats at constant pressure (C p)

= RT = ( nRT ) = PV and constant volume (Cv)
NA 2 2 2
Cp æ 2ö
3 æmö 3 æ 2ö = g = ç1 + ÷
= Pç ÷= Pç ÷ Cv è fø
2 è r ø 2 è8ø
where f is degree of freedom
3 2 Cp æ 2 ö 7
= ´ 4´10 ´ = 1.5 ´10 J
4 4

2 8 = ç1 + ÷ =
Cv è 5 ø 5
therefore, order = 104 J
55. (d) For a polytropic process
50. (c) Heat transferred,
R R
Q = nCv DT as gas in closed vessel C = Cv + \ C - Cv =
To double the rms speed, temperature should be 4 times 1- n 1- n
R R
i.e., T' = 4T as vrms = 3RT / M \ 1- n = \ 1- =n
C - Cv C - Cv
15 5 ´ R C - C v - R C - C v - Cp + C v
\Q= ´ ´ ( 4T - T ) \ n= =
28 2 C - Cv C - Cv
é CP 7 ù C - Cp
êë\ CV = γ diatomic = 5 & C p - Cv = R úû = (Q C p - C v = R )
C - Cv
or, Q = 10000 J = 10 kJ
56. (d) Using equipartition of energy, we have
51. (c) In an adiabatic process
TVg–1 = Constant or, T1V1g–1 = T2V2g–1 6
KT = mCT
5 2
For monoatomic gas g =
3 3 ´1.38 ´10 –23 ´ 6.02 ´1023
C=
300 27 ´10–3
(300)V2/3 = T2(2V)2/3 Þ T2 = \ C = 925 J/kgK
(2) 2/3
T2 = 189 K (final temperature) 57. (b)
f 58. (b) On giving same amount of heat at constant pressure,
Change in internal energy DU = n R DT there is no change in temperature for mono, dia and
2
polyatomic.
æ 3 öæ 25 ö
= 2 ç ÷ç ÷ (-111) = -2.7 kJ æ No. of molecules ö
è 2 øè 3 ø ( DQ) P = mC p DT ç m =
52. (b) Using formula, è Avogedro 's no. ÷ø
n1g 1 n 2 g 2 1
+ or DT µ
æ Cp ö g1 - 1 g 2 - 1 no. of molecules
g mixture = ç ÷ =
è Cv ø mix n1 n 59. (c) TV g -1 = constant
+ 2
g1 - 1 g 2 - 1 g -1
T1V1g -1 = T2V2
æ Cp ö 3
Putting the value of n1 = 2, n2 = n, ç ÷ = 1
è C v ø mix 2 1 æV ö 2
Þ T (V ) 2 =T2 ç ÷
è 4ø
5 7
g 1 = , g 2 = and solving we get, n = 2 é Vù
3 5
êëQ g = 1.5, T1 = T ,V1 = V and V2 = 4 úû
53. (a) As we know, Cp – Cv = R where Cp and Cv are molar
specific heat capacities 1
æ 4V ö 2
R \ T2 = ç ÷ T = 2T
or, Cp – Cv = èV ø
M
60. (b) According to Mayer's relationship
R
For hydrogen (M = 2) Cp – Cv = a = CP – CV = R, as per the question (CP – CV) M = R
2
Þ CP – CV = R/28
R
For nitrogen (M = 28) Cp – Cv = b = Here M = 28 = mass of 1 unit of N2
28
61. (a) For mixture of gas specific heat at constant volume
a
\ = 14 or, a = 14b n1Cv1 + n2Cv2
b
Cv =
n1 + n2
P-198 Physics

No. of moles of helium, Cp 47 R 18

mHe 16 Þ
\ = ´ = 1.62
n1 = M = =4 Cv 18 29 R
He 4
Number of moles of oxygen, 5 7
62. (c) g 1 = g =
16 1 3 2 5
n2 = = n1 = 1, n2 = 1
32 2
n1 + n2 n n
3 1 5 5 = 1 + 2
4 ´ R + ´ R 6R + R g -1 g1 - 1 g 2 - 1
= 2 2 2 = 4
\ Cv æ 1ö 9 1 +1 1 1 3 5
çè 4 + ÷ø Þ = + = + =4
2 2 g - 1 5 - 1 7 -1 2 2
3 5
29 R ´ 2 29 R 2 3
= = and \ =4 Þ g=
9´ 4 18
g -1 2
Specific heat at constant pressure
63. (d) P µ T 3 Þ PT -3 = constant ....(i)
5R 1 7 R
n1C p1 + n2C p2 4´ + ´ But for an adiabatic process, the pressure temperature
Cp = = 2 2 2 relationship is given by
(n1 + n2 ) æ 1ö
çè 4 + ÷ø P1-g T g = constant
2
g
7 Þ PT 1-g
= constt. ....(ii)
10 R + R
= 4 = 47 R g 3
9 18 From (i) and (ii) = -3 Þ g = -3 + 3g Þ g =
1- g 2
2