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Degrees of freedom and equipartition of energy: NEET MCQs with answers

52 questions on this topic of Kinetic Theory of Gases (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. According to the law of equipartition of energy, the average kinetic energy associated with each degree of freedom per molecule of any ideal gas at absolute temperature T is:

  1. (3/2) k_B T
  2. k_B T
  3. (1/2) k_B T
  4. (5/2) k_B T
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Answer: (C) (1/2) k_B T

The law of equipartition of energy states that for any dynamical system in thermal equilibrium, the total energy is distributed equally among its various degrees of freedom, and the energy associated with each degree of freedom per molecule is equal to (1/2) k_B T.

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2. Two non-reactive ideal gases, 1 mole of a monatomic gas and 1 mole of a rigid diatomic gas, are enclosed in a container at temperature T. The ratio of the average translational kinetic energy of a monatomic molecule to that of a diatomic molecule is:

  1. 3 : 5
  2. 3 : 2
  3. 5 : 3
  4. 1 : 1
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Answer: (D) 1 : 1

Translational kinetic energy depends purely on temperature and is equal to (3/2) k_B T per molecule for all gases, regardless of whether the gas is monatomic, diatomic, or polyatomic. Thus, at the same temperature T, the ratio is 1 : 1.

hard

3. A gas consisting of non-linear triatomic rigid molecules is heated from 300 K to 400 K at constant volume. The change in internal energy of 2 moles of this gas is: (R = 8.31 J/(mol·K))

  1. 2493 J
  2. 4986 J
  3. 6232 J
  4. 7479 J
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Answer: (B) 4986 J

A non-linear rigid triatomic molecule has f = 6 degrees of freedom (3 translational + 3 rotational). Therefore, Cv = (6/2)R = 3R. The change in internal energy is ΔU = n Cv ΔT = 2 × 3R × (400 - 300) = 6 × 8.31 × 100 = 4986 J.

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4. A triatomic molecule is linear (such as carbon dioxide, CO₂) and is treated as rigid without any vibrational modes. The molar heat capacity at constant pressure (Cp) for this gas is:

  1. (5/2) R
  2. (9/2) R
  3. 4 R
  4. (7/2) R
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Answer: (D) (7/2) R

A linear molecule has 3 translational and 2 rotational degrees of freedom, totaling f = 5 degrees of freedom when rigid. Therefore, Cv = (5/2) R and Cp = Cv + R = (5/2) R + R = (7/2) R.

easy

5. Which of the following graphs correctly represents the variation of the molar heat capacity at constant volume (C_v) with temperature T for a real diatomic gas like H₂ over a wide temperature range (from very low to very high temperatures)?

  1. C_v remains constant at (5/2)R throughout all temperatures.
  2. C_v decreases continuously from (7/2)R to (3/2)R as temperature increases.
  3. C_v increases in discrete plateaus from (3/2)R to (5/2)R, and finally to (7/2)R.
  4. C_v increases linearly with temperature from zero without any plateaus.
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Answer: (C) C_v increases in discrete plateaus from (3/2)R to (5/2)R, and finally to (7/2)R.

At very low temperatures, only translational modes are active (C_v = 3/2 R). At moderate temperatures, rotational modes become active (C_v = 5/2 R). At very high temperatures, vibrational modes are also excited, adding R to give C_v = 7/2 R. This creates a characteristic stepped plateau curve.

hard

6. If 2 moles of a gas with adiabatic index γ = 7/5 is mixed with 1 mole of another gas with adiabatic index γ = 5/3, what is the effective adiabatic index (γ_mix) of the mixture?

  1. 1.47
  2. 1.40
  3. 1.53
  4. 1.60
Show answer and explanation

Answer: (A) 1.47

For the first gas (γ₁ = 7/5), Cv₁ = R/(γ₁ - 1) = R/(2/5) = 2.5R = 5R/2. For the second gas (γ₂ = 5/3), Cv₂ = R/(γ₂ - 1) = R/(2/3) = 1.5R = 3R/2. The effective Cv of the mixture is Cv_mix = (n₁Cv₁ + n₂Cv₂) / (n₁ + n₂) = [2(5R/2) + 1(3R/2)] / 3 = (5R + 1.5R)/3 = 6.5R / 3 = 13R/6. Then Cp_mix = Cv_mix + R = 19R/6. Thus, γ_mix = Cp_mix / Cv_mix = 19/13 ≈ 1.4615 ≈ 1.47.

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7. For a gas of rigid linear triatomic molecules at moderate temperature, the ratio of the molar heat capacity at constant pressure to the molar heat capacity at constant volume (γ = C_p / C_v) is:

  1. 4/3
  2. 9/7
  3. 5/3
  4. 7/5
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Answer: (D) 7/5

A linear triatomic molecule (like CO₂ without vibration) has 3 translational and 2 rotational degrees of freedom, making f = 5 degrees of freedom. Therefore, C_v = (5/2)R and C_p = C_v + R = (7/2)R, giving γ = C_p / C_v = 7/5.

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8. An ideal gas has an adiabatic exponent γ = 1.25. Assuming each vibrational degree of freedom contributes two quadratic energy terms, what is the total number of degrees of freedom f for this gas?

  1. 6
  2. 7
  3. 8
  4. 10
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Answer: (C) 8

The adiabatic index is related to degrees of freedom by γ = 1 + 2/f. Given γ = 1.25 = 5/4, we have 1 + 2/f = 5/4 ⇒ 2/f = 1/4 ⇒ f = 8.

easy

9. Treating helium as an ideal monatomic gas at room temperature (no vibrational modes), the ratio Cp/Cv is:

  1. 5/3
  2. 7/5
  3. 9/7
  4. 4/3
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Answer: (A) 5/3

A monatomic molecule has f = 3 degrees of freedom; γ = 1 + 2/f = 5/3.

easy

10. Treating hydrogen as an ideal rigid diatomic gas at room temperature (no vibrational modes), the ratio Cp/Cv is:

  1. 7/5
  2. 4/3
  3. 5/3
  4. 9/7
Show answer and explanation

Answer: (A) 7/5

A rigid diatomic molecule has f = 5 degrees of freedom; γ = 1 + 2/f = 7/5.

easy

11. Treating argon as an ideal monatomic gas at room temperature (no vibrational modes), the ratio Cp/Cv is:

  1. 5/3
  2. 7/5
  3. 9/7
  4. 4/3
Show answer and explanation

Answer: (A) 5/3

A monatomic molecule has f = 3 degrees of freedom; γ = 1 + 2/f = 5/3.

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12. The root-mean-square speed of a gas molecule of mass m at absolute temperature T is v_rms. The contribution to its total translational kinetic energy along any single chosen Cartesian axis (x-axis) is:

  1. m v_rms²
  2. (1/3) m v_rms²
  3. (1/2) m v_rms²
  4. (1/6) m v_rms²
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Answer: (D) (1/6) m v_rms²

By isotropy of gas molecules, the mean square speeds along the axes are equal: v_x² = v_y² = v_z² = (1/3) v_rms². The kinetic energy along the x-axis is (1/2) m v_x² = (1/2) m ((1/3) v_rms²) = (1/6) m v_rms².

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