$n$ mole of a perfect gas undergoes a cyclic process $ABCA$ (see figure) consisting of the following processes.
$A \to B$: Isothermal expansion at temperature $T$ so that the volume is doubled from $V_1$ to $V_2 = 2V_1$ and pressure changes from $P_1$ to $P_2$.
$B \to C$: Isobaric compression at pressure $P_2$ to initial volume $V_1$.
$C \to A$: Isochoric change leading to change of pressure from $P_2$ to $P_1$.
Total work done in the complete cycle $ABCA$ is:
Answer: (A) $nRT\left(\ln2 - \frac12\right)$
$W_{AB} = nRT\ln\dfrac{V_2}{V_1} = nRT\ln 2$
$W_{BC} = P_2(V_1 - 2V_1) = -P_2V_1$. At B, $P_2(2V_1) = nRT$, so $P_2V_1 = \dfrac{nRT}{2}$ and $W_{BC} = -\dfrac{nRT}{2}$.
$W_{CA} = 0$ (constant volume).
$$W = nRT\left(\ln 2 - \frac12\right)$$
Solution by Sreeraj P, M.Sc Physics