Kinetic Theory — Class 11 MCQs with Answers

Kinetic Theory — important questions & MCQs with answers (Class 11 Physics)

13 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 53 are timed and scored when you sign in.

1. Q1Easy

   Ideal gas equation:

   A.PV = nRT✓ Correct

   B.PV = T

   C.P = nRT

   D.PV² = nRT

   Answer: A. PV = nRT

   Explanation: The ideal gas law PV = nRT links pressure, volume, moles and temperature via R = 8.314 J/(mol·K). Dropping n (P = RT) or T (PV = T), or squaring V, breaks the proportionality this law defines.

2. Q2Easy

   Ideal gas assumes molecules:

   A.Have size

   B.Point particles with no interaction except elastic collisions✓ Correct

   C.Have viscosity

   D.Stick together

   Answer: B. Point particles with no interaction except elastic collisions

   Explanation: The ideal gas model treats molecules as point particles with negligible size that interact only through instantaneous elastic collisions. Giving them finite size or viscosity describes real gases, which the ideal model ignores.

3. Q3Easy

   Kinetic theory gives P = (1/3) n m v_rms², where n is the number of molecules per unit volume. At constant temperature, doubling n changes the pressure to:

   A.2P✓ Correct

   B.P/2

   C.4P

   D.P (unchanged)

   Answer: A. 2P

   Explanation: v_rms = √(3kT/m) depends only on temperature and molecular mass, so holding T fixed leaves it unchanged and P ∝ n directly: doubling the number density doubles the pressure. Answering 4P wrongly assumes v_rms also scales with n.

4. Q4Easy

   Degrees of freedom for monatomic gas:

   A.3✓ Correct

   B.5

   C.7

   D.Variable

   Answer: A. 3

   Explanation: A monatomic molecule such as He has only 3 translational degrees of freedom (x, y, z motion) and no rotational or vibrational modes. Choosing 5 or 7 wrongly borrows the count used for diatomic or polyatomic gases.

5. Q5Easy

   Equipartition theorem: each degree of freedom gets:

   A.R/2

   B.kT

   C.(1/2)kT per molecule✓ Correct

   D.Variable

   Answer: C. (1/2)kT per molecule

   Explanation: The equipartition theorem assigns exactly (1/2)kT of energy to each degree of freedom, per molecule. Writing kT or R/2 confuses the per-degree-of-freedom share with the total energy or the per-mole gas constant.

6. Q6Medium

   At STP (P = 1 atm, T = 273 K), volume of 1 mole of gas:

   A.22.4 L✓ Correct

   B.2.24 L

   C.224 L

   D.0.224 L

   Answer: A. 22.4 L

   Explanation: At STP (1 atm, 273 K), PV = nRT gives one mole of any ideal gas a fixed molar volume of 22.4 L = 0.0224 m³. Shifting the decimal point to 2.24 L or 224 L is the classic power-of-ten slip in this constant.

7. Q7Medium

   v_rms of nitrogen (M = 28 g/mol) at 300 K:

   A.100 m/s

   B.100

   C.1000

   D.≈ 517 m/s✓ Correct

   Answer: D. ≈ 517 m/s

   Explanation: v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.028 kg/mol) ≈ 517 m/s for nitrogen at 300 K. Forgetting to convert M from 28 g/mol to 0.028 kg/mol would give an answer off by a large power of ten.

8. Q8Medium

   For diatomic gas at very high T, additional degree of freedom is:

   A.Cannot say

   B.Translational

   C.Rotational

   D.Vibrational (2 more)✓ Correct

   Answer: D. Vibrational (2 more)

   Explanation: At very high temperature, the bond of a diatomic molecule starts to vibrate, adding 2 extra degrees of freedom — one kinetic and one potential — on top of the usual 3 translational and 2 rotational modes. Calling this purely rotational misses the potential-energy term.

9. Q9Medium

   Total energy of N molecules of monatomic gas at T:

   A.(5/2)NkT

   B.NkT

   C.(3/2)NkT✓ Correct

   D.Variable

   Answer: C. (3/2)NkT

   Explanation: Each monatomic molecule has 3 translational degrees of freedom contributing (1/2)kT, so N molecules carry total energy (3/2)NkT. Using (5/2)NkT wrongly adds the 2 rotational degrees that only diatomic molecules possess.

10. Q10Medium

    Internal energy of n moles of monatomic gas at T:

    A.(5/2)nRT

    B.(3/2)nRT✓ Correct

    C.nRT

    D.Variable

    Answer: B. (3/2)nRT

    Explanation: Internal energy of n moles of a monatomic ideal gas is U = (3/2)nRT, since each mole has 3 translational degrees of freedom contributing (1/2)RT. Writing (5/2)nRT borrows the diatomic result, which includes rotation.

11. Q11Hard

    Mean free path is:

    A.Speed

    B.Average distance between successive collisions✓ Correct

    C.Force

    D.Pressure

    Answer: B. Average distance between successive collisions

    Explanation: Mean free path λ is defined as the average distance a molecule travels between two successive collisions, λ = 1/(√2 nπd²). It is a distance, not a speed or a force — those describe unrelated quantities.

12. Q12Hard

    Mixture of n₁ moles A and n₂ moles B in volume V at T. Total pressure:

    A.Variable

    B.n₁ RT/V

    C.(n₁ + n₂)RT/V (Dalton's law)✓ Correct

    D.Cannot compute

    Answer: C. (n₁ + n₂)RT/V (Dalton's law)

    Explanation: By Dalton's law of partial pressures, each gas in a mixture acts independently, so total pressure is the sum of what each would exert alone: P_total = (n₁ + n₂)RT/V. Using only n₁RT/V ignores the second gas's contribution entirely.

13. Q13Hard

    At what temperature does v_rms equal 1 km/s for hydrogen (M = 2 g/mol)?

    A.Cannot say

    B.100 K

    C.300

    D.≈ 80 K✓ Correct

    Answer: D. ≈ 80 K

    Explanation: From v_rms = √(3RT/M), T = Mv_rms²/(3R) = (0.002 kg/mol)(1000 m/s)²/(3 × 8.314) ≈ 80 K for hydrogen. Forgetting to convert M from 2 g/mol to 0.002 kg/mol would inflate this answer by a factor of 1000.

Kinetic Theory — FAQs

More Class 11 Physics chapters

• 01Physical World
• 02Units and Measurements
• 03Motion in a Straight Line
• 04Motion in a Plane
• 05Laws of Motion
• 06Work, Energy and Power

Class 11 — other subjects

• Class 11 Biology chapters
• Class 11 Chemistry chapters
• Class 11 Computer Science chapters
• Class 11 English chapters
• Class 11 Mathematics chapters