1. State the three processes

For one mole of ideal gas:

• Initial state $A=(P_1,V_1)$
• $A \to B$: reversible isothermal expansion
• $B \to C$: constant volume cooling
• $C \to A$: reversible adiabatic compression

We need the net work done by the gas: $W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$

2. Process $A \to B$: Isothermal expansion

Since the process is isothermal for one mole of ideal gas, $P_1V_1=RT$

Pressure is reduced to half: $P_B=\frac{P_1}{2}$

For isothermal process, $PV=\text{constant}$, so $P_1V_1=P_BV_B$ $P_1V_1=\frac{P_1}{2}V_B \Rightarrow V_B=2V_1$

Work done in reversible isothermal expansion: $W_{AB}=RT\ln\frac{V_B}{V_A}=RT\ln\frac{2V_1}{V_1}=RT\ln 2$

3. Process $B \to C$: Constant volume cooling

At constant volume, $W_{BC}=0$

Also, since pressure is reduced to one-fourth of initial pressure, $P_C=\frac{P_1}{4}$

And because $B \to C$ is at constant volume, $V_C=V_B=2V_1$

So state $C$ is: $C\left(\frac{P_1}{4},2V_1\right)$

4. Process $C \to A$: Reversible adiabatic compression

For a reversible adiabatic process, $TV^{\gamma-1}=\text{constant}$

Let temperature at $A$ be $T_A=T$.

At state $C$, $T_C=\frac{P_CV_C}{R}=\frac{\left(\frac{P_1}{4}\right)(2V_1)}{R}$ Using $P_1V_1=RT$, $T_C=\frac{1}{2}\cdot \frac{P_1V_1}{R}=\frac{T}{2}$

Now compute work in adiabatic compression. For one mole, $W_{CA}=\frac{R(T_C-T_A)}{\gamma-1}$ This gives work done by the gas during compression, which is negative.

Substitute $T_C=\frac{T}{2}$ and $T_A=T$: $W_{CA}=\frac{R\left(\frac{T}{2}-T\right)}{\gamma-1}$ $W_{CA}=-\frac{RT}{2(\gamma-1)}$

5. Net work done

$W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$ $W_{\text{net}}=RT\ln 2+0-\frac{RT}{2(\gamma-1)}$

Therefore, $W_{\text{net}}=RT\left(\ln 2-\frac{1}{2(\gamma-1)}\right)$

6. Match with options

This matches:

Option B $RT\left(\ln 2-\frac{1}{2(\gamma-1)}\right)$

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So, the answer agrees with the stored correct answer.