Two moles of helium gas are taken along the path ABCD (as shown in Fig.). The work done by the gas is

 

A $\to$B is an isobaric process, 

                        V $\infty$ T

            So,       $∆W_{AB}$ = nR$∆T$ = 2 × R × (750 – 250) = 1000 R

            B $\to$C is an isochoric process 

                   $∆W_{BC}$ = 0 and 

            C $\to$D is an isothermal process

                        $∆W_{CD}$ = nRT ln$\left(\frac{V_f}{V_i}\right)$

                        = 2 × R × 1000 ln= 2000 R ln$\left(\frac{20}{15}\right)$

            Total work done, $∆$W =  $∆W_{AB}$+$∆W_{BC}$+$∆W_{CD}$