Degree of Freedom and Law of Equipartition of Energy – Degrees of freedom: NEET MCQs with Answers

10 free questions on Degree of Freedom and Law of Equipartition of Energy – Degrees of freedom for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Degrees of freedom. Law of equipartition of energy and applications to specific heat capacities of gases

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Question 1

Previous-year question
According to the law of equipartition of energy, the number of vibrational modes of a polyatomic gas of constant γ = Cₚ/Cᵥ is [Re-NEET 2024]
  1. (4 + 3γ)/(γ – 1)
  2. (3 + 4γ)/(γ – 1)
  3. (4 – 3γ)/(γ – 1)
  4. (3 – 4γ)/(γ – 1)
Show answer
Answer: (C) (4 – 3γ)/(γ – 1)
Explanation: A non-linear polyatomic molecule has f = 3 + 3 + 2v degrees of freedom, where each vibrational mode counts twice. Since f = 2/(γ – 1), v = 1/(γ – 1) – 3 = (4 – 3γ)/(γ – 1).

Question 2

Previous-year question
The average thermal energy of a molecule of a monatomic gas is (k_B is the Boltzmann constant and T the absolute temperature) [NEET 2020]
  1. (3/2)k_BT
  2. (5/2)k_BT
  3. (7/2)k_BT
  4. (1/2)k_BT
Show answer
Answer: (A) (3/2)k_BT
Explanation: A monatomic molecule has only 3 translational degrees of freedom, each contributing (1/2)k_BT, so its average thermal energy is (3/2)k_BT.

Question 3

Previous-year question
A gaseous mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is [NEET 2017]
  1. 15 RT
  2. 9 RT
  3. 11 RT
  4. 4 RT
Show answer
Answer: (C) 11 RT
Explanation: U = μ₁(5/2)RT + μ₂(3/2)RT = 2 × (5/2)RT + 4 × (3/2)RT = 5RT + 6RT = 11RT.

Question 4

Previous-year question
A gas mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is: [NEET 2017]
  1. 15 RT
  2. 9 RT
  3. 11 RT
  4. 4 RT
Show answer
Answer: (C) 11 RT
Explanation: O₂ is diatomic (f = 5) and Ar is monoatomic (f = 3). U = 2 × (5/2)RT + 4 × (3/2)RT = 5RT + 6RT = 11RT.

Question 5

Previous-year question
4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 J/K. If the speed of sound in this gas at NTP is 952 m/s, then the heat capacity at constant pressure is: (Take gas constant, R = 8.3 J K⁻¹ mol⁻¹) [Re-AIPMT 2015]
  1. 8.5 J K⁻¹ mol⁻¹
  2. 8.0 J K⁻¹ mol⁻¹
  3. 7.5 J K⁻¹ mol⁻¹
  4. 7.0 J K⁻¹ mol⁻¹
Show answer
Answer: (B) 8.0 J K⁻¹ mol⁻¹
Explanation: 22.4 L at NTP is 1 mole, so M = 4 g = 4 × 10⁻³ kg. From v = √(γRT/M): γ = v²M/(RT) = (952² × 4 × 10⁻³)/(8.3 × 273) ≈ 1.6. Then C_p = γC_v = 1.6 × 5.0 = 8.0 J K⁻¹ mol⁻¹.

Question 6

Previous-year question
The ratio of specific heats C_p/C_v = γ in terms of degrees of freedom (n) is given by: [AIPMT 2015]
  1. 1 + 2/n
  2. 1 + n/2
  3. 1 + 1/2
  4. 1 + n/3
Show answer
Answer: (A) 1 + 2/n
Explanation: C_v = (n/2)R and C_p = (n/2 + 1)R. Therefore γ = C_p/C_v = (n + 2)/n = 1 + 2/n.

Question 7

Previous-year question
If C_p and C_v denote the specific heats (per unit mass) of an ideal gas of molecular weight M, then: [AIPMT 2010]
  1. C_p − C_v = R
  2. C_p − C_v = R/M
  3. C_p − C_v = MR
  4. C_p − C_v = R/M²
Show answer
Answer: (B) C_p − C_v = R/M
Explanation: Mayer’s relation for molar specific heats is C_p(molar) − C_v(molar) = R. Dividing throughout by the molecular weight M converts them to specific heats per unit mass, giving C_p − C_v = R/M.
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Question 8

Previous-year question
The translational kinetic energy of molecules of one mole of a monoatomic gas is U = 3NkT/2. The value of the atomic specific heat of the gas under constant pressure will be: [AIPMT 2006]
  1. (3/2) R
  2. (5/2) R
  3. (7/2) R
  4. (9/2) R
Show answer
Answer: (B) (5/2) R
Explanation: For a monoatomic gas, C_v = dU/dT = (3/2)Nk = (3/2)R. Then C_p = C_v + R = (3/2)R + R = (5/2)R.

Question 9

Previous-year question
The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant pressure to that at constant volume is: [AIPMT 2006]
  1. 7/5
  2. 8/7
  3. 5/7
  4. 9/7
Show answer
Answer: (A) 7/5
Explanation: C_v = C_p − R = (7/2)R − R = (5/2)R. Therefore γ = C_p/C_v = (7/2)/(5/2) = 7/5.

Question 10

Each atom of a metal is treated as a three-dimensional oscillator obeying equipartition of energy. Using R = 8.31 J mol⁻¹ K⁻¹, what is the estimated specific heat capacity of a metal of molar mass 60 g mol⁻¹ at room temperature, assuming its volume change on heating is negligible?
  1. 139 J kg⁻¹ K⁻¹
  2. 693 J kg⁻¹ K⁻¹
  3. 416 J kg⁻¹ K⁻¹
  4. 277 J kg⁻¹ K⁻¹
Show answer
Answer: (C) 416 J kg⁻¹ K⁻¹
Explanation: An oscillator in one dimension has average energy 2 × ½k_BT = k_BT, so in three dimensions it has 3k_BT and one mole has U = 3RT. For a solid ΔV is negligible, so ΔQ ≈ ΔU and the molar heat capacity is C = 3R = 24.93 J mol⁻¹ K⁻¹. Dividing by the molar mass of 0.060 kg mol⁻¹ gives about 416 J kg⁻¹ K⁻¹.

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