JEE Main 2024 · 4 April · evening · Single correct · +4 / −1

A sample of gas at temperature T is adiabatically expanded to double its volume. Adiabatic constant for the gas is γ=3 / 2. The work done by the gas in the process is:
(μ=1 mole )

  1. (a) R T[2 √2-1]
  2. (b) R T[2-√2]
  3. (c) R T[1-2 √2]
  4. (d) R T[√2-2]
Answer and solution

Answer (b) R T[2-√2]

Solution

Physics sir · 3 steps

Get the final temperature from TV^(γ-1) first, then the adiabatic work is just a temperature drop.

  1. Final temperature. In an adiabatic process TV^(γ-1) stays constant, and here γ - 1 = 1/2. So T₂ = T(V/2V)^(1/2) = T/√2. The gas cools as it expands.

    W = nR(T₁ - T₂)/(γ - 1)

  2. Put in numbers. With n = 1 and γ - 1 = 1/2, W = 2R(T - T/√2) = 2RT - √2 RT, using 2/√2 = √2.
  3. So. W = RT(2 - √2) ≈ 0.59 RT. This is positive, as it should be for an expanding gas, so the answer is (b).

The idea to keep. For adiabatic work, find T₂ from TV^(γ-1) = constant and use W = nR(T₁ - T₂)/(γ - 1). A gas that expands does positive work, so check the sign.

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 2024.