n mole of a perfect gas and 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 $V_{2} = 2 V_{1}$ 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 :

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$0$

$n R T (\ln 2 + \frac{1}{2})$

$n R T \ln 2$

$n R T (\ln 2 - \frac{1}{2})$

answer is D.

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Detailed Solution

$\begin{array}{l} A t p o i n t A a n d B t e m p e r a t u r e i s s a m e . \\ T_{1} = T_{2} \\ \Rightarrow \frac{P_{1} V_{1}}{n R} = \frac{P_{2} V_{2}}{n R} \\ \Rightarrow P_{2} = \frac{P_{1}}{2} \\ Work done in isothermal process, \\ W_{AB} = n R T \ln (\frac{V_{f}}{V_{i}}) \\ \Rightarrow W_{AB} = P_{1} V_{1} \ln (\frac{2 V_{1}}{V_{1}}) \\ \Rightarrow W_{AB} = P_{1} V_{1} \ln (2) \\ W o r k d o n e i n i s o b a r i c p r o c e s s, \\ W_{B C} = P_{2} (V_{1} - V_{2}) \\ \Rightarrow W_{B C} = P_{2} (V_{1} - 2 V_{1}) \\ \Rightarrow W_{BC} = \frac{P_{1}}{2} (- V_{1}) \\ \Rightarrow W_{BC} = - \frac{P_{1} V_{1}}{2} \\ Work done in isochoric process, W_{CA} = 0 \\ Hence, Net work done, W_{ABCA} = P_{1} V_{1} ℓ n 2 - \frac{P_{1} V_{1}}{2} \\ A s p e r i d e a l g a s e q u a t i o n, P_{1} V_{1} = nRT \\ \Rightarrow W_{ABCA} = n R T (\ln 2 - \frac{1}{2}) \end{array}$