A sample of an ideal gas has pressure $p_0$, volume ${\mathrm{V}}_0$ and temperature ${\mathrm{T}}_0$. It is isothermally expanded to twice its original volume. It is then compressed at constant pressure to have the original volume ${\mathrm{V}}_0$. Finally, the gas is heated at constant volume to get the original temperature. The heat absorbed in the process is

The process is cyclic so that the change in internal energy is zero. The heat supplied is, therefore, equal to the work done by the gas. The work done during ab is
${\mathrm{W}}_1 = {\mathrm{nRT}}_0\ln \frac{2{\mathrm{V}}_0}{{\mathrm{V}}_0} ={\mathrm{nRT}}_0\ln 2 = {\mathrm{p}}_0{\mathrm{V}}_0\ln 2$

Also from the ideal gas equation,
$p_a {\mathrm{V}}_{\mathrm{a}} = {\mathrm{p}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}} \mathrm{or} {\mathrm{p}}_{\mathrm{b}} = \frac{{\mathrm{p}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}}{{\mathrm{V}}_{\mathrm{b}}} = \frac{{\mathrm{p}}_0{\mathrm{V}}_0}{2{\mathrm{V}}_0} = \frac{{\mathrm{p}}_0}{2}$

In the step bc,the pressure remains constant. Hence the work done is, ${\mathrm{W}}_2= \frac{{\mathrm{p}}_0}{2}({\mathrm{V}}_0-2{\mathrm{V}}_0) = -\frac{{\mathrm{p}}_0{\mathrm{V}}_0}{2}$
In the step ca,the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is
$\mathrm{W} = {\mathrm{W}}_1 + {\mathrm{W}}_2 = {\mathrm{p}}_0{\mathrm{V}}_0[\ln 2-\frac{1}{2}]$

Hence, the heat supplied to the gas is ${\mathrm{p}}_0{\mathrm{V}}_0[\ln 2- \frac{1}{2}]$