JEE Physics Kinetic Theory MCQs Set 09

Physics Objective Questions and Answers: Kinetic Theory of gases

Review structured MCQ sets for JEE Physics Kinetic Theory of gases. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the molecules
(a) is decreased
(b) is increased
(c) remains unchanged
(d) increases or decreases according to the nature of the gas
Show Answer & Explanation

Answer: (b) is increased

Explanation:
1. Removing gas lowers the number density.
2. The mean free path λ ∝ 1/n therefore increases.

Teacher's Note:
1. Fewer molecules, longer path.

Question 2: P, V and T respectively denote the pressure, volume and temperature of two identical gas samples. On mixing them, the new temperature and volume are T and V respectively. The final pressure of the mixture is
(a) P
(b) 2P
(c) zero
(d) 3P
Show Answer & Explanation

Answer: (b) 2P

Explanation:
1. The amount of gas doubles while T and V are unchanged.
2. Since P ∝ n at fixed T and V, the pressure doubles to 2P.

Teacher's Note:
1. P ∝ n.

Question 3:
Assertion (A): When we place a gas cylinder on a moving train, its internal energy increases.
Reason (R): Its temperature remains constant.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. The internal energy depends on the random motion only, which does not change, so the assertion is false.
2. The temperature does remain constant, so the reason is true.

Teacher's Note:
1. Bulk motion is not internal energy.

Question 4: Consider the following statements. I. The ratio C_P/C_V for a diatomic gas is more than that of a monoatomic gas. II. The ratio C_P/C_V is more for helium gas than for hydrogen.
(a) I only
(b) II only
(c) I and II
(d) none of these
Show Answer & Explanation

Answer: (b) II only

Explanation:
1. γ = 7/5 for diatomic and 5/3 for monoatomic.
2. So I is false and II (helium is monoatomic) is true.

Teacher's Note:
1. Fewer degrees of freedom, larger γ.

Question 5: As per the law of equipartition of energy, each vibrational mode gives how many degrees of freedom?
(a) 1
(b) 2
(c) 3
(d) 0
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. A vibrational mode has both kinetic and potential energy terms.
2. So it contributes 2 degrees of freedom, with energy kT.

Teacher's Note:
1. Vibration counts as 2.

Question 6: Choose the false statement(s). I. C_P − C_V = R is true for monoatomic gases only. II. The specific heat of a gas at constant pressure is greater than the specific heat at constant volume. III. The mean free path of molecules of a gas decreases with increase in number density of the molecules.
(a) I only
(b) II only
(c) I and II
(d) I, II and III
Show Answer & Explanation

Answer: (a) I only

Explanation:
1. Mayer's relation holds for every ideal gas, so I is false.
2. II is true, and III is true since λ ∝ 1/n.

Teacher's Note:
1. Mayer's relation is general.

Question 7: Boyle's law is applicable for an
(a) adiabatic process
(b) isothermal process
(c) isobaric process
(d) isochoric process
Show Answer & Explanation

Answer: (b) isothermal process

Explanation:
1. PV = constant needs a constant temperature.
2. That is an isothermal process.

Teacher's Note:
1. Boyle: isothermal.

Question 8: The kinetic theory of gases provides a base for
(a) Charles' law
(b) Boyle's law
(c) both Charles' law and Boyle's law
(d) none of these
Show Answer & Explanation

Answer: (c) both Charles' law and Boyle's law

Explanation:
1. From P = (1/3)ρ⟨v²⟩ and the relation of ⟨v²⟩ with T, both PV = constant at fixed T and V ∝ T at fixed P follow.
2. So both laws are explained.

Teacher's Note:
1. Kinetic theory explains the gas laws.

Question 9: A polyatomic gas with n degrees of freedom has a mean energy per molecule given by
(a) \( \frac{nkT}{N} \)
(b) \( \frac{nkT}{2N} \)
(c) \( \frac{nkT}{2} \)
(d) \( \frac{3kT}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{nkT}{2} \)

Explanation:
1. Each degree of freedom carries (1/2)kT.
2. With n degrees of freedom the mean energy is nkT/2.

Teacher's Note:
1. E = nkT/2.

Question 10: The relation PV = RT can describe the behaviour of a real gas at
(a) high temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) low temperature and high pressure
Show Answer & Explanation

Answer: (b) high temperature and low pressure

Explanation:
1. A real gas behaves ideally when intermolecular forces and molecular volume are negligible.
2. That happens at high temperature and low pressure.

Teacher's Note:
1. Ideal limit: high T, low P.

Question 11: The number of degrees of freedom for each atom of a monoatomic gas is
(a) 3
(b) 5
(c) 6
(d) 1
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. A single atom can translate along x, y and z only.
2. So it has 3 degrees of freedom.

Teacher's Note:
1. Monoatomic: 3.

Question 12: Half a mole of helium gas is contained at STP. The heat energy needed to double the pressure of the gas keeping the volume constant (specific heat of the gas = 3 J/g K) is
(a) 3276 J
(b) 1638 J
(c) 819 J
(d) 409.5 J
Show Answer & Explanation

Answer: (b) 1638 J

Explanation:
1. Mass = 0.5 × 4 = 2 g, and T doubles from 273 K to 546 K, so ΔT = 273 K.
2. Q = 2 × 3 × 273 = 1638 J.

Teacher's Note:
1. Q = m c ΔT.

Question 13: One mole of an ideal diatomic gas goes from state A (P = 5 kPa, V = 4 m³) to state B (P = 2 kPa, V = 6 m³). The change in internal energy of the gas is
(a) −20 kJ
(b) 20 J
(c) −12 kJ
(d) 20 kJ
Show Answer & Explanation

Answer: (a) −20 kJ

Explanation:
1. ΔU = (5/2)(P_B V_B − P_A V_A) for a diatomic gas.
2. P_A V_A = 20 kJ and P_B V_B = 12 kJ, so ΔU = 2.5 × (−8) = −20 kJ.

Teacher's Note:
1. ΔU = (f/2)Δ(PV).

Question 14: Consider the following statements about an ideal gas at any given temperature T. I. The coefficient of volume expansion at constant pressure is the same for all ideal gases. II. The average translational kinetic energy per molecule of oxygen gas is 3KT (K being the Boltzmann constant). III. In a gaseous mixture, the average translational kinetic energy of the molecules of each component is the same. IV. The mean free path of molecules increases with decrease in pressure. Select the correct ones.
(a) I, II and III
(b) III and IV only
(c) I, II and IV
(d) I, III and IV
Show Answer & Explanation

Answer: (d) I, III and IV

Explanation:
1. I is true (γ = 1/T for all ideal gases).
2. II is false since the energy is (3/2)KT.
3. III is true as KE depends on T only, and IV is true as λ ∝ 1/n.

Teacher's Note:
1. Mean translational KE = 3kT/2.

Question 15: Match Column I (physical variables) with Column II (expressions). (n = number of gas molecules per unit volume, k = Boltzmann constant, T = absolute temperature, m = mass of the particle)
Column IColumn II
(A) Most probable velocity(1) nkT
(B) Energy per degree of freedom(2) √(3kT/m)
(C) Pressure(3) √(2kT/m)
(D) R.M.S. velocity(4) kT/2

(a) A-3, B-1, C-4, D-2
(b) A-4, B-3, C-1, D-2
(c) A-3, B-4, C-1, D-2
(d) A-3, B-2, C-1, D-4
Show Answer & Explanation

Answer: (c) A-3, B-4, C-1, D-2

Explanation:
1. v_mp = √(2kT/m) (A-3).
2. Energy per degree of freedom is kT/2 (B-4).
3. P = nkT (C-1).
4. v_rms = √(3kT/m) (D-2).

Teacher's Note:
1. Learn the three characteristic speeds.

Question 16:
Assertion (A): The internal energy of a real gas is a function of both temperature and volume.
Reason (R): Internal kinetic energy depends on temperature and internal potential energy depends on volume.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A real gas has intermolecular forces, so its potential energy depends on the average separation, that is on volume.
2. The kinetic part depends on T, so the reason explains the assertion.

Teacher's Note:
1. U = KE(T) + PE(V).

Question 17: The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of the specific heat at constant pressure to that at constant volume is
(a) \( \frac57 \)
(b) \( \frac97 \)
(c) \( \frac75 \)
(d) \( \frac87 \)
Show Answer & Explanation

Answer: (c) \( \frac75 \)

Explanation:
1. C_V = C_P − R = (5/2)R.
2. γ = (7/2)/(5/2) = 7/5.

Teacher's Note:
1. γ = C_P/C_V.

Question 18: In the kinetic theory of gases, which of these statements is/are true? I. The pressure of a gas is proportional to the mean speed of the molecules. II. The root mean square speed of the molecules is proportional to the pressure. III. The rate of diffusion is proportional to the mean speed of the molecules. IV. The mean translational kinetic energy of a gas is proportional to its kelvin temperature.
(a) II and III only
(b) I, II and IV
(c) I and III only
(d) III and IV only
Show Answer & Explanation

Answer: (d) III and IV only

Explanation:
1. P = (1/3)ρ⟨v²⟩ depends on the mean square speed, so I and II are false.
2. Diffusion rate ∝ mean speed (III) and mean KE = (3/2)kT (IV) are true.

Teacher's Note:
1. Check each statement with P = ρv²/3 and KE = 3kT/2.

Question 19: The specific heat of a gas
(a) has only two values c_p and c_v
(b) has a unique value at a given temperature
(c) can have any value between 0 and ∞
(d) depends upon the mass of the gas
Show Answer & Explanation

Answer: (c) can have any value between 0 and ∞

Explanation:
1. The heat needed depends on the process followed.
2. For different processes the specific heat can take any value between 0 and ∞.

Teacher's Note:
1. Depends on the process.

Question 20: Molecules of an ideal gas behave like
(a) inelastic rigid spheres
(b) perfectly elastic non-rigid spheres
(c) perfectly elastic rigid spheres
(d) inelastic non-rigid spheres
Show Answer & Explanation

Answer: (c) perfectly elastic rigid spheres

Explanation:
1. The model assumes perfectly elastic collisions with no deformation of the molecules.
2. Hence they behave as perfectly elastic rigid spheres.

Teacher's Note:
1. Elastic hard spheres.

Question 21: A gas at a pressure P₀ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be
(a) \( 4P_0 \)
(b) \( 2P_0 \)
(c) \( P_0 \)
(d) \( \frac{P_0}{2} \)
Show Answer & Explanation

Answer: (b) \( 2P_0 \)

Explanation:
1. P ∝ m v².
2. The new value is (1/2) × 2² = 2 times the old one, so P = 2P₀.

Teacher's Note:
1. P ∝ mv².

Question 22:
Assertion (A): On reducing the volume of a gas at constant temperature, the pressure of the gas increases.
Reason (R): On reducing the volume, the number of molecules per unit volume increases and as a result more collisions with the walls exert greater pressure on the walls.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Fewer volume means more molecules per unit volume at fixed speed.
2. More wall collisions per second raise the pressure, so the reason explains the assertion.

Teacher's Note:
1. Boyle's law from molecular collisions.

Question 23: The ratio of the molar specific heats C_P/C_V for a gas whose molecules have 5 degrees of freedom is
(a) 7/5
(b) 5/3
(c) 3/2
(d) 9/7
Show Answer & Explanation

Answer: (a) 7/5

Explanation:
1. γ = 1 + 2/f = 1 + 2/5.
2. γ = 7/5.

Teacher's Note:
1. γ = 1 + 2/f.

Question 24: The phenomenon of Brownian movement may be taken as evidence of
(a) the kinetic theory of matter
(b) the electromagnetic theory of radiation
(c) the corpuscular theory of light
(d) the photoelectric phenomenon
Show Answer & Explanation

Answer: (a) the kinetic theory of matter

Explanation:
1. Brownian motion is the random jiggling of small particles caused by collisions with molecules.
2. It supports the kinetic theory of matter.

Teacher's Note:
1. Random molecular collisions.

Question 25: If C_P and C_V denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
(a) \( C_P-C_V=28R \)
(b) \( C_P-C_V=R/28 \)
(c) \( C_P-C_V=R/14 \)
(d) \( C_P-C_V=R \)
Show Answer & Explanation

Answer: (b) \( C_P-C_V=R/28 \)

Explanation:
1. Per unit mass, C_P − C_V = R/M.
2. For nitrogen M = 28, so the difference is R/28.

Teacher's Note:
1. Per gram: R/M.

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