JEE Main 2023 · 31 January · evening · Single correct · +4 / −1

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be :

  1. (a) 572 J
  2. (b) 441 J
  3. (c) 525 J
  4. (d) 735 J
Answer and solution

Answer (c) 525 J

Solution

Physics sir · 3 steps

At constant pressure only Cᵥ/Cₚ of the heat stays in the gas; the rest pushes the gas outward.

  1. Count degrees of freedom. A diatomic molecule that rotates but does not vibrate has 3 translational and 2 rotational degrees of freedom, so f = 5. That gives Cᵥ = 5/2R and Cₚ = Cᵥ + R = 7/2R.
  2. Split the heat. At constant pressure the heat supplied is Q = nCₚ Δ T. The internal energy always changes by Δ U = nCᵥ Δ T, whatever the process. Dividing the two makes n and Δ T cancel.

    (Δ U)/Q = Cᵥ/Cₚ = 5/7

  3. So. Δ U = 5/7 × 735 J = 525 J. The other 210 J is the work done by the expanding gas, so the answer is (c).

The idea to keep. For heat given at constant pressure, Δ U : W : Q = Cᵥ : R : Cₚ. Get f from the molecule's motion and the split follows at once.

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

Study Thermodynamics

Asked about 150 times in JEE Main since 2020. This question is from 2023.