Explanation

(c) Work done $\mathrm{\Delta}W = p\mathrm{\Delta}V$
At constant pressure
$\mathrm{\Delta}W = p\left( V_{f} - V_{i} \right) = nR(T_{f} - T_{i})$
At constant temperature
$\mathrm{\Delta}W = nRT\ In\ \left( \frac{V_{f}}{V_{i}} \right) = nRT\ In\ \left( \frac{P_{i}}{P_{f}} \right)$
$\therefore\ \ \ \mathrm{\Delta}W_{AB} = 1 \times R \times (2T - T) = RT$
$\ \ \ \ \ \ \mathrm{\Delta}W_{BC} = 1 \times R \times 2T\ In\frac{2p}{P} = 2RT\ In\ 2$
$\ \ \ \ \ \ \mathrm{\Delta}W_{CD}\text{=}1 \times R \times (T - 2T) = - RT$
$\ \ \ \ \ \ \mathrm{\Delta}W_{DA} = 1 \times R \times T\ In\ \left( \frac{p}{2p} \right) = RT\ In\ (\frac{1}{2})$
Net work done in the complete cycle is
$\mathrm{\Delta}W = \mathrm{\Delta}W_{AB} + \mathrm{\Delta}W_{BC} + \mathrm{\Delta}W_{CD} + \mathrm{\Delta}W_{DA}$
$= RT + 2RT\ In\ 2 - RT + RT\ In\ \left( \frac{1}{2} \right)$
$= 2RT\ In\ 2 + RT\ In\ 1 - RT\ In\ 2$
$= 2RT\ In\ 2 - RT\ In\ 2$
$= RT\ In\ 2$