1. State the processes and key state variables

For the ideal gas at state $A$: $P_1V_1=nRT$

Since $A \to B$ is isothermal at temperature $T$ and volume doubles: $V_2=2V_1$

Using $PV=nRT$ for the same temperature, $P_2V_2=P_1V_1=nRT$ So, $P_2=\frac{P_1V_1}{V_2}=\frac{P_1}{2}$

Thus the states are:

• $A:(P_1,V_1)$
• $B:(P_2,2V_1)$ with $P_2=\frac{P_1}{2}$
• $C:(P_2,V_1)$

2. Work done in process $A \to B$ (isothermal expansion)

For isothermal expansion of an ideal gas, $W_{AB}=nRT\ln\left(\frac{V_2}{V_1}\right)$ Since $V_2=2V_1$, $W_{AB}=nRT\ln 2$

3. Work done in process $B \to C$ (isobaric compression)

At constant pressure $P_2$, volume changes from $2V_1$ to $V_1$: $W_{BC}=P_2(V_C-V_B)=P_2(V_1-2V_1)=-P_2V_1$ Now, $P_2=\frac{P_1}{2}, \qquad P_1V_1=nRT$ Hence, $P_2V_1=\frac{P_1V_1}{2}=\frac{nRT}{2}$ So, $W_{BC}=-\frac{nRT}{2}$

4. Work done in process $C \to A$ (isochoric)

In an isochoric process, $dV=0$, so $W_{CA}=0$

5. Total work done in the complete cycle

$W_{\text{cycle}}=W_{AB}+W_{BC}+W_{CA}$ $W_{\text{cycle}}=nRT\ln 2-\frac{nRT}{2}+0$ $W_{\text{cycle}}=nRT\left(\ln 2-\frac{1}{2}\right)$

6. Match with options

The correct option is: $\boxed{\text{C: } nRT\left(\ln 2-\frac{1}{2}\right)}$