JEE Physics Kinetic Theory MCQs Set 18

Answer: (d) \( \frac75 \)

Explanation:
1. For a rigid diatomic gas C_P = 7R/2 and C_V = 5R/2.
2. The ratio is 7/5.

Teacher's Note:
1. γ = 7/5.

Answer: (b) temperature of the gas

Explanation:
1. The average kinetic energy is (3/2)kT.
2. It depends only on the temperature, not on the nature or volume of the gas.

Teacher's Note:
1. Only T matters.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A single atom has 3 translational degrees of freedom.
2. The reason is true by definition but does not explain the assertion.

Teacher's Note:
1. Atom: 3 degrees.

Answer: (a) the average kinetic energy of the gaseous molecules

Explanation:
1. For an ideal gas the average translational kinetic energy per molecule is (3/2)kT.
2. So T measures the average kinetic energy.

Teacher's Note:
1. Temperature measures mean kinetic energy.

Answer: (c) I, II and III

Explanation:
1. v_rms ∝ √T.
2. At constant volume P ∝ T, so raising P raises T.
3. At constant pressure raising T also raises v_rms. All three work.

Teacher's Note:
1. v_rms depends on T.

Answer: (d) I, II and III

Explanation:
1. The average translational kinetic energy is (3/2)kT.
2. It depends only on T, so all three statements are true.

Teacher's Note:
1. KE depends only on T.

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

Explanation:
1. C_V/C_P = 3/5 for monoatomic and 5/7 for diatomic, so the assertion is true.
2. A monoatomic gas has fewer degrees of freedom, so the reason is false.

Teacher's Note:
1. Monoatomic: 3 degrees only.

Answer: (a) I only

Explanation:
1. Both gases are at the same temperature in equilibrium.
2. The mean kinetic energy per molecule is (3/2)kT, equal for both.

Teacher's Note:
1. Same T, same mean KE.

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

Explanation:
1. At low density the molecules are far apart, so the ideal gas equation applies well.
2. The reason explains the assertion.

Teacher's Note:
1. Low density, ideal behaviour.

Answer: (d) 12.42 × 10⁻²¹ J, 684 m/s

Explanation:
1. The mean energy is proportional to T, so it doubles to 12.42 × 10⁻²¹ J.
2. v_rms ∝ √T, so it becomes 484 × √2 ≈ 684 m/s.

Teacher's Note:
1. KE ∝ T and v ∝ √T.

Answer: (b) II and III

Explanation:
1. I is correct for translational energy.
2. An ideal gas has no potential energy, so II is wrong.
3. The energy is shared equally per degree of freedom, not equally between the translational and rotational parts, so III is wrong.

Teacher's Note:
1. Equal share per degree of freedom.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Driving warms the tyre air, so P ∝ T rises and the assertion is true.
2. The reason is a true statement but is unrelated to the explanation.

Teacher's Note:
1. Pressure follows temperature.

Answer: (a) 1 : 1

Explanation:
1. Both gases are at the same temperature.
2. The average translational kinetic energy per molecule is (3/2)kT for each, so the ratio is 1 : 1.

Teacher's Note:
1. Mean KE depends only on T.

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

Explanation:
1. An ideal gas has no intermolecular potential energy.
2. Its internal energy depends on T only, so it is independent of volume.

Teacher's Note:
1. U = U(T).

Answer: (b) low pressure and high temperature

Explanation:
1. At low pressure the molecules are far apart and the volume of the molecules is negligible.
2. At high temperature the molecular kinetic energy is large compared with intermolecular attraction.

Teacher's Note:
1. Low P and high T.

Answer: (d) 2 : 1

Explanation:
1. N ∝ PV/T.
2. N_A/N_B = (2 × 2)/2 = 2, that is 2 : 1.

Teacher's Note:
1. N = PV/kT.

Answer: (c) mean kinetic energy

Explanation:
1. Average translational kinetic energy is (3/2)kT, independent of the gas.
2. Momentum, rms speed and mean free path depend on the molecular mass or size.

Teacher's Note:
1. Mean KE depends only on T.

Answer: (c) \( \frac52RT \)

Explanation:
1. A rigid diatomic molecule has 5 degrees of freedom (3 translational and 2 rotational).
2. U = 5 × (1/2)RT = (5/2)RT per mole.

Teacher's Note:
1. U = fRT/2.

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

Explanation:
1. A gas has many specific heats (C_P, C_V and others) depending on the process, so the assertion is false.
2. The definition given is correct.

Teacher's Note:
1. Specific heat depends on the process.

Answer: (a) inversely proportional to the number of molecules per unit volume

Explanation:
1. λ = 1/(√2 π d² n).
2. It is inversely proportional to the number density n (and to d², not d).

Teacher's Note:
1. λ ∝ 1/n.

Answer: (d) I, II and III

Explanation:
1. Near room temperature the specific heats of gases are practically constant.
2. At low temperature some degrees of freedom freeze out and the specific heat tends to zero, so all three are true.

Teacher's Note:
1. Quantum freezing of degrees of freedom.

Answer: (d) have the same mass but negligible volume

Explanation:
1. Molecules of one gas are identical in mass.
2. Their size is negligible compared with the distances between them.

Teacher's Note:
1. Point-like identical molecules.

Answer: (c) three

Explanation:
1. A point-like particle moving freely in space has 3 translational degrees of freedom.

Teacher's Note:
1. Free motion in 3D.

Answer: (b) the number of collisions per unit time will increase

Explanation:
1. The molecules move faster as T rises.
2. They strike the walls more often and with greater momentum, so the number of collisions per unit time increases.

Teacher's Note:
1. Hotter gas, more frequent collisions.

Answer: (b) \( 5\times10^4\ \text{J} \)

Explanation:
1. V = m/ρ = 0.25 m³.
2. Thermal energy of a diatomic gas = (5/2)PV = 2.5 × 8 × 10⁴ × 0.25 = 5 × 10⁴ J.

Teacher's Note:
1. U = (5/2)PV.