If one mole of an ideal gas at (P₁, V₁) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B → C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :

Solution

Physics sir · 3 steps

Go leg by leg: the isotherm gives work, constant volume gives none, the adiabatic takes some back.

1. Temperatures first. A and B are on the same isotherm, so both are at T. From B to C the volume stays 2V₁ while the pressure halves, so the temperature also halves: T_C = T/2.

   Process	What happens	Work by the gas
   A to B	isothermal, V₁ → 2V₁	RT ln 2V₁/V₁ = RT ln 2
   B to C	constant volume	0
   C to A	adiabatic, T/2 → T	R(T_C - T_A)/(γ - 1) = -RT/2(γ - 1)

2. Adiabatic sign. For an adiabatic process the work done by the gas is nR(Tᵢ - T_f)/(γ - 1). In this compression the gas heats up from T/2 to T, so this work is negative.

   Wₙₑₜ = RT ln 2 + 0 - RT/2(γ - 1) = RT(ln 2 - 1/2(γ - 1))

3. So. The net work done by the gas is RT(ln 2 - 1/2(γ - 1)), which is option (b).

The idea to keep. For a cycle, add the work of each leg with its sign. Constant volume gives zero work. Adiabatic work is nR(Tᵢ - T_f)/(γ - 1), so you only need the end temperatures.

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