T-V curve of cyclic process is shown below, number of moles of the gas are n find the total work done during the cycle.

Since path AB and CD arc isochoric therefore work done is zero during path AB and CD. Process BC and DA are isothermal, therefore
${\mathrm{W}}_{\mathrm{BC}} = \mathrm{nR}2{\mathrm{T}}_0\ln \frac{{\mathrm{V}}_{\mathrm{C}}}{{\mathrm{V}}_{\mathrm{B}}} = 2{\mathrm{nRT}}_0\ln 2$

${\mathrm{W}}_{\mathrm{DA}} = {\mathrm{nRT}}_0\ln \frac{{\mathrm{V}}_{\mathrm{A} }}{{\mathrm{V}}_{\mathrm{D}}} = -{\mathrm{nRT}}_0\ln 2$
Total work done

${\mathrm{W}}_{\mathrm{BC}}+ {\mathrm{W}}_{\mathrm{DA}} = 2{\mathrm{nRT}}_0\ln 2 - {\mathrm{nRT}}_0 \ln 2 = {\mathrm{nRT}}_0 \ln 2$