A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in the process is (given, γ=3/2) :

Solution

Physics sir · 4 steps

Find the final temperature from TV^(γ-1) first, and the adiabatic work formula does the rest.

1. Final temperature. In an adiabatic process TV^(γ-1) stays constant. Here γ - 1 = 1/2 and the volume doubles, 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. Take one mole, as the options assume. Then W = R(T - T/√2)/(1/2) = 2RT(1 - 1/√2) = RT(2 - √2).
3. Check the sign. 2 - √2 ≈ 0.59 is positive. That fits, because a gas does positive work when it expands.
4. So. The work done by the gas is W = RT[2 - √2], which is option (d).

The idea to keep. For adiabatic work, find T₂ from TV^(γ-1) = constant, then use W = nR(T₁ - T₂)/(γ - 1). Work done by an expanding gas must come out positive.

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