Kinetic Theory

Class 11

10 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Ideal gas equation and kinetic interpretation of pressure

Exam focus: At a fixed temperature, rms speed depends only on molar mass — lighter molecules move faster, not slower, which is why hydrogen effuses faster than oxygen at the same temperature.

Kinetic Theory of Gases: Ideal Gas Equation and Pressure

Physics Wallah - Alakh Pandey

Covers the ideal gas equation, Boyle's/Charles' laws and the kinetic interpretation of pressure; also bundles in mean free path, used separately for that subtopic below.

PhysicsNEET 2025show ▾

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheres at a temperature of

2 7^{\circ}

C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given

R = \frac{100}{12} J mo l^{- 1} K^{- 1}

, molecular mass of

O_{2}

= 32, 1 atm pressure

= 1.01 \times 1 0^{5} N/ m^{2}

]

A.0.125 kg
B.0.144 kg
C.0.116 kg✓
D.0.156 kg

Solution

Final moles $n = \frac{P V}{R T} = \frac{12 \times 1.01 \times 1 0 ^{5} \times 0.030}{( 100/12 ) \times 300} \approx 14.5$ mol (gauge 11 atm means absolute 12 atm). Withdrawn $\approx 18.2 - 14.5 = 3.7$ mol $\approx 0.116$ kg.

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PhysicsNEET 2025show ▾

A container has two chambers of volumes

V_{1} = 2

litres and

V_{2} = 3

litres separated by a partition made of a thermal insulator. The chambers contain

n_{1} = 5

and

n_{2} = 4

moles of ideal gas at pressures

p_{1} = 1

atm and

p_{2} = 2

atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

A.1.3 atm
B.1.6 atm✓
C.1.4 atm
D.1.8 atm

Solution

Temperature is the same throughout (no heat exchange with outside, ideal gases). Final $P = \frac{n _{1} + n _{2}}{V _{1} + V _{2}} R T$ and $R T = \frac{p _{1} V _{1}}{n _{1}} = \frac{p _{2} V _{2}}{n _{2}}$ . Using $p_{1} V_{1} + p_{2} V_{2} = (n_{1} + n_{2}) R T$ : $P = \frac{1 \times 2 + 2 \times 3}{5} = 1.6$ atm.

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PhysicsNEET 2024show ▾

The following graph represents the

T

–

V

curves of an ideal gas (where

T

is the temperature and

V

the volume) at three pressures

P_{1}

P_{2}

and

P_{3}

, compared with those of Charles's law represented as dotted lines. Then the correct relation is: Question 47 as printed in the NEET 2024 booklet

A. $P_{1} > P_{3} > P_{2}$
B. $P_{2} > P_{1} > P_{3}$
C. $P_{1} > P_{2} > P_{3}$ ✓
D. $P_{3} > P_{2} > P_{1}$

Solution

For an ideal gas at fixed pressure, $V = \frac{n R}{P} T$ : the $T$ – $V$ line through the origin has slope $\propto \frac{1}{P}$ , so the steepest line corresponds to the lowest pressure. Reading the slopes in the figure gives $P_{1} > P_{2} > P_{3}$ .

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PhysicsNEET 2023show ▾

The temperature of a gas is

- 5 0^{\circ}

C. To what temperature should the gas be heated so that the rms speed is increased by 3 times?

A.669 °C
B.3295 °C✓
C.3097 K
D.223 K

The worked solution is part of a paid pack.

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PhysicsNEET 2022show ▾

The volume occupied by the molecules contained in 4.5 kg of water at STP, if the intermolecular forces vanish away, is:

A. $5.6 m^{3}$ ✓
B. $5.6 \times 1 0^{6} m^{3}$
C. $5.6 \times 1 0^{3} m^{3}$
D. $5.6 \times 1 0^{- 3} m^{3}$

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PhysicsNEET 2021show ▾

Match Column-I and Column-II and choose the correct match from the given choices.

Column-I	Column-II
(A) Root mean square speed of gas molecules	(P) $\frac{1}{3} nm \overset{v}{ˉ}^{2}$
(B) Pressure exerted by ideal gas	(Q) $\frac{3 R T}{M}$
(C) Average kinetic energy of a molecule	(R) $\frac{5}{2} R T$
(D) Total internal energy of 1 mole of a diatomic gas	(S) $\frac{3}{2} k_{B} T$

A.(A)-(R), (B)-(P), (C)-(S), (D)-(Q)
B.(A)-(Q), (B)-(R), (C)-(S), (D)-(P)
C.(A)-(Q), (B)-(P), (C)-(S), (D)-(R)✓
D.(A)-(R), (B)-(Q), (C)-(P), (D)-(S)

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PhysicsNEET 2020show ▾

A cylinder contains hydrogen gas at a pressure of 249 kPa and temperature 27 °C. Its density is: (

R = 8.3 J mo l^{- 1} K^{- 1}

)

A. $0.5 kg/ m^{3}$
B. $0.2 kg/ m^{3}$ ✓
C. $0.1 kg/ m^{3}$
D. $0.02 kg/ m^{3}$

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2. Degrees of freedom, law of equipartition, specific heats

Exam focus: Each degree of freedom contributes ½kT, so Cv = (f/2)R: monatomic gas has f=3 (Cv=3/2R), diatomic has f=5 at ordinary temperatures (Cv=5/2R), and only gains its two vibrational modes — becoming f=7 — at high temperature.

Degrees of Freedom and the Law of Equipartition of Energy

Magnet Brains

Focused single-concept lecture on degrees of freedom and equipartition, which is used to derive specific heats of gases.

3. Mean free path

Exam focus: Mean free path depends on the number density of molecules, hence inversely on pressure at fixed temperature, not on molecular speed — doubling pressure at constant T halves it, but raising T at constant volume leaves it unchanged.

Mean Free Path

Magnet Brains

Dedicated single-concept lecture on mean free path.

Kinetic Theory

1. Ideal gas equation and kinetic interpretation of pressure

Ideal gas equation and kinetic interpretation of pressure

2. Degrees of freedom, law of equipartition, specific heats

Degrees of freedom, law of equipartition and specific heats

3. Mean free path

Mean free path