If one mole of an ideal gas at $$(P_1, V_1)$$ 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 $$(B \to C)$$. Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to:

For the complete cycle,

$$W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$$

For A→B, expansion is isothermal.

For one mole,

Since at A,

and

$$V_B=2V_1$$

so

$$W_{AB}=P_1V_1\ln2$$

For B→CB \to CB→C, volume is constant, so

$$W_{BC}=0$$

For C→A, process is adiabatic compression.

For adiabatic process,

Q=0

and

For compression from C to A,

$$W_{CA}=\frac{P_CV_C-P_AV_A}{\gamma-1}$$

Now

and

$$P_AV_A=P_1V_1$$

Thus

So net work done is

$$W_{\text{net}}=RT\ln2-\frac{P_1V_1}{2(\gamma-1)}$$