A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiab constant for the gas is γ=3/2, then the work done by the gas in the process is :

Solution

Physics sir · 4 steps

Find the final temperature from TV^(γ-1) first, then work is just the drop in internal energy.

1. Final temperature. For an adiabatic process TV^(γ-1) stays constant. Here γ - 1 = 1/2, so T₂ = T(V/2V)^(1/2) = T/√2.
2. Work formula. No heat enters, so the work done by the gas equals the fall in its internal energy. For n moles that gives the formula below.

   W = nR(T₁ - T₂)/(γ - 1)

3. Put in numbers. With n = 1 and γ - 1 = 1/2: W = 2R(T - T/√2) = 2RT - √2 RT, since 2/√2 = √2.
4. So. W = RT[2 - √2], about 0.59 RT, which is option (b).

The idea to keep. In an adiabatic change, get the new temperature from TV^(γ-1) = constant, then use W = nR(T₁ - T₂)/(γ-1). Check that the answer has units of RT, which is energy.

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