NEETAceToday's NEET Daily

NEET Kinetic Theory of Gases MCQs with answers

289 questions on Kinetic Theory of Gases for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

Practise all 289 questions freePlay today's NEET Daily

Free with a Google sign-in; NEET Daily needs no sign-in at all.

Practise Kinetic Theory of Gases topic by topic

15 sample questions

easy

1. A rigid container holds a gas composed of N molecules, each of mass m, confined in a volume V. If the mass of each molecule is doubled while their root-mean-square speed is halved, the new pressure exerted by the gas becomes:

  1. Unchanged
  2. Doubled
  3. Halved
  4. One-fourth of the initial pressure
Show answer and explanation

Answer: (C) Halved

The pressure exerted by an ideal gas is given by P = (1/3) ρ (v_rms)² = (1/3) (N m / V) (v_rms)². If the mass m is doubled (m' = 2m) and the speed v_rms is halved (v'_rms = v_rms/2), the new pressure is P' = (2)(1/2)² P = (2/4) P = P/2. Thus, the pressure is halved.

medium

2. The rms speed of the molecules of a gas at 300 K is 400 m/s. Its rms speed at 2700 K is:

  1. 3600 m/s
  2. 32400 m/s
  3. 133 m/s
  4. 1200 m/s
Show answer and explanation

Answer: (D) 1200 m/s

v_rms ∝ √T, so v = 400 × √(2700/300) = 1200 m/s.

hard

3. Two identical containers A and B each hold 1 mole of the same ideal gas at temperature T₀ and pressure P₀. Container A is compressed isothermally to half its original volume. Container B is compressed at constant pressure to half its original volume. If W_A and W_B are the magnitudes of work done on the gas in containers A and B respectively, the ratio W_A / W_B is:

  1. 2 ln 2
  2. ln 2
  3. (ln 2) / 2
  4. 4 ln 2
Show answer and explanation

Answer: (A) 2 ln 2

For container A (isothermal compression from V₀ to V₀/2): work done on the gas is W_A = RT₀ ln(V₀ / (V₀/2)) = RT₀ ln 2. For container B (isobaric compression from V₀ to V₀/2 at pressure P₀): work done on the gas is W_B = P₀(V₀ - V₀/2) = P₀V₀ / 2 = RT₀ / 2. Therefore, the ratio is W_A / W_B = (RT₀ ln 2) / (RT₀ / 2) = 2 ln 2.

medium

4. An ideal gas is kept in a sealed container of fixed volume. When the temperature of the gas is increased from 300 K to 600 K, how do the mean free path (λ) and the collision frequency (ν) of the molecules change?

  1. λ decreases by a factor of 2, ν increases by a factor of √2
  2. λ doubles, ν remains unchanged
  3. λ remains unchanged, ν doubles
  4. λ remains unchanged, ν increases by a factor of √2
Show answer and explanation

Answer: (D) λ remains unchanged, ν increases by a factor of √2

The mean free path is given by λ = 1 / (√2 π n d²), where n = N/V is the number density. Since the container has fixed volume and fixed number of molecules, n is constant, so λ remains unchanged. The collision frequency is ν = v_avg / λ ∝ √T / λ. As T increases from 300 K to 600 K (by a factor of 2), v_avg and hence ν increase by a factor of √2.

easy

5. At a given temperature and pressure, equal volumes of two different ideal gases must contain the same:

  1. Root-mean-square speed
  2. Mass of gas
  3. Number of molecules
  4. Total kinetic energy per gram
Show answer and explanation

Answer: (C) Number of molecules

According to Avogadro's hypothesis, which directly follows from the ideal gas law PV = N kᵦ T, equal volumes of all ideal gases at identical temperature and pressure contain an identical number of molecules (N). Mass and molecular speeds depend on the molar mass of the specific gas.

hard

6. Equal masses of helium (He, molar mass = 4 g/mol) and oxygen (O₂, molar mass = 32 g/mol) are kept in separate vessels at the same temperature. Treating oxygen as a rigid diatomic gas, the ratio of the total internal energy of helium to that of oxygen is:

  1. 12 : 5
  2. 5 : 12
  3. 3 : 5
  4. 24 : 5
Show answer and explanation

Answer: (A) 12 : 5

Let the mass of each gas be m. The moles are n_He = m/4 and n_O2 = m/32. The internal energy is U = n × (f/2) R T. For helium, f = 3, so U_He = (m/4) × (3/2) R T = (3m/8) R T. For oxygen, f = 5, so U_O2 = (m/32) × (5/2) R T = (5m/64) R T. The ratio U_He / U_O2 = (3/8) / (5/64) = (3 × 64) / (8 × 5) = 24 / 10 = 12 / 5.

medium

7. The rms speed of the molecules of a gas at 300 K is 400 m/s. Its rms speed at 75 K is:

  1. 800 m/s
  2. 200 m/s
  3. 100 m/s
  4. 25 m/s
Show answer and explanation

Answer: (B) 200 m/s

v_rms ∝ √T, so v = 400 × √(75/300) = 200 m/s.

hard

8. The diameter of a gas molecule is 0.20 nm. If the number density of the gas is 2.5 × 10²⁵ molecules/m³, what is the order of magnitude of its mean free path? (Take π ≈ 3.14, √2 ≈ 1.41)

  1. 10⁻⁵ m
  2. 10⁻⁹ m
  3. 10⁻⁷ m
  4. 10⁻³ m
Show answer and explanation

Answer: (C) 10⁻⁷ m

Mean free path λ = 1 / (√2 π n d²). Here d = 2 × 10⁻¹⁰ m, so d² = 4 × 10⁻²⁰ m². Thus, λ = 1 / (1.41 × 3.14 × 2.5 × 10²⁵ × 4 × 10⁻²⁰) = 1 / (4.43 × 10⁶) ≈ 2.26 × 10⁻⁷ m. The order of magnitude is 10⁻⁷ m.

medium

9. A gas cylinder contains neon gas (atomic mass = 20 u) at 300 K. What is the average translational kinetic energy of a single neon atom? (Boltzmann constant k_B = 1.38 × 10⁻²³ J/K)

  1. 8.28 × 10⁻²¹ J
  2. 4.14 × 10⁻²¹ J
  3. 2.07 × 10⁻²¹ J
  4. 6.21 × 10⁻²¹ J
Show answer and explanation

Answer: (D) 6.21 × 10⁻²¹ J

The average translational kinetic energy per molecule (or atom) is E = (3/2) k_B T = 1.5 × (1.38 × 10⁻²³ J/K) × 300 K = 6.21 × 10⁻²¹ J.

easy

10. Two vessels of volumes V₁ and V₂ contain the same ideal gas at the same temperature. If the pressure in the first vessel is three times that in the second, the ratio of their number densities (n₁ : n₂) is:

  1. 1 : 3
  2. 9 : 1
  3. 3 : 1
  4. 1 : 9
Show answer and explanation

Answer: (C) 3 : 1

From the kinetic interpretation of pressure, P = n k_B T. Since the temperature is the same for both vessels, P is directly proportional to the number density n. Thus, n₁/n₂ = P₁/P₂ = 3/1.

medium

11. In a sample of an ideal gas, the velocity components of five molecules along the x-axis are measured to be -300 m/s, -100 m/s, 0 m/s, +200 m/s, and +400 m/s. What is the contribution of these molecules to the mean-square speed along the x-direction, (vₓ)ₐᵥ²?

  1. 40,000 m²/s²
  2. 200 m²/s²
  3. 50,000 m²/s²
  4. 60,000 m²/s²
Show answer and explanation

Answer: (D) 60,000 m²/s²

The mean-square speed along the x-direction is the arithmetic average of the squares of the individual x-velocity components: (vₓ)ₐᵥ² = [(-300)² + (-100)² + 0² + (200)² + (400)²] / 5 = [90,000 + 10,000 + 0 + 40,000 + 160,000] / 5 = 300,000 / 5 = 60,000 m²/s².

easy

12. Consider the following statements regarding degrees of freedom and equipartition of energy: Statement I: The translational degrees of freedom of all gas molecules, regardless of their atomicity, is always 3. Statement II: Each vibrational degree of freedom contributes k_B T to the average energy of a molecule because it involves both kinetic and potential energy modes. Choose the correct option:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct.

Gas molecules can move independently along three mutually perpendicular spatial directions (x, y, z), so translational degrees of freedom are always 3. A vibrational mode involves both kinetic energy (motion of atoms) and potential energy (interatomic bond stretching/compression), contributing (1/2)k_B T + (1/2)k_B T = k_B T to internal energy.

medium

13. A hypothetical gas molecule consists of 4 point masses placed at the four vertices of a rigid regular tetrahedron. The number of translational and rotational degrees of freedom for this molecule are, respectively:

  1. 3 translational and 2 rotational
  2. 3 translational and 3 rotational
  3. 2 translational and 3 rotational
  4. 4 translational and 6 rotational
Show answer and explanation

Answer: (B) 3 translational and 3 rotational

Any three-dimensional non-linear rigid body has 6 degrees of freedom in free space: 3 mutually independent translational degrees of freedom (along the x, y, and z axes) and 3 independent rotational degrees of freedom about three mutually perpendicular principal axes.

medium

14. The work done on an ideal gas when it is compressed isothermally from volume V1 to V2 has magnitude:

  1. Zero
  2. nRT ln(V1/V2)
  3. nR(T2 - T1)
  4. P(V1 - V2) at varying pressure
Show answer and explanation

Answer: (B) nRT ln(V1/V2)

For an isothermal process W_by gas = nRT ln(V2/V1). For compression V2 < V1, so work done on the gas is nRT ln(V1/V2).

NCERT Class 11

medium

15. One mole of an ideal gas is compressed at a constant temperature T from an initial pressure P to a final pressure 4P. The work done on the gas during this isothermal compression is:

  1. 2RT ln 2
  2. RT ln 2
  3. 4RT ln 2
  4. (RT/2) ln 2
Show answer and explanation

Answer: (A) 2RT ln 2

Work done on the gas during isothermal compression is W_on = RT ln(P_final / P_initial) = RT ln(4P / P) = RT ln(4) = 2RT ln 2.

Want more on Kinetic Theory of Gases? Practise every question with instant answers, mark your mistakes for revision and see your accuracy by topic. Free to start with a Google sign-in, no card needed.
Start practising Kinetic Theory of Gases

More NEET Physics chapters