Work for the following process ABCD on a monoatomic gas is :

 

At A and D the temperatures of the gas will be equal, so 

$\mathrm{\Delta E}=0,\mathrm{\Delta H}=0$

$\begin{array}{l}\text{ Now }\mathrm{W}={\mathrm{W}}_{\mathrm{AB}}+{\mathrm{W}}_{\mathrm{BC}}+{\mathrm{W}}_{\mathrm{CD}}=-{\mathrm{P}}_0{\mathrm{V}}_0-2{\mathrm{P}}_0{\mathrm{V}}_0\ln 2+{\mathrm{P}}_0{\mathrm{V}}_0\\ =-2{\mathrm{P}}_0{\mathrm{V}}_0\ln 2\end{array}$