JEE Main 2020 · 3 September · evening · Single correct · +4 / −1

To raise the temperature of a certain mass of gas by 50ᵒC at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100ᵒC at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

  1. (a) 6
  2. (b) 7
  3. (c) 5
  4. (d) 3
Answer and solution

Answer (a) 6

Solution

Physics sir · 4 steps

Turn each heat into heat per degree, so the two heat capacities give you R and then f.

  1. Constant pressure. Qₚ = nCₚ Δ T, so nCₚ = 160/50 = 3.2 cal K⁻¹.
  2. Constant volume. Qᵥ = nCᵥ Δ T, so nCᵥ = 240/100 = 2.4 cal K⁻¹. The temperature changes are different, so divide each heat by its own Δ T.
  3. Find nR. Mayer's relation Cₚ - Cᵥ = R gives nR = 3.2 - 2.4 = 0.8 cal K⁻¹.

    nCᵥ = f/2 nR ⇒ 2.4 = f/2 × 0.8 ⇒ f = 6

  4. So. As a check, γ = 3.2/2.4 = 4/3 and f = 2/(γ - 1) = 6. Each molecule has 6 degrees of freedom, option (a).

The idea to keep. Get Cₚ and Cᵥ per degree first. Their difference is R, and Cᵥ = f/2R gives f directly, with no need to know the mass or the number of moles.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Kinetic Theory of Gases

Asked about 130 times in JEE Main since 2020. This question is from 2020.