Six moles of an ideal gas perfomrs a cycle shown in figure. If the temperature are ${\mathrm{T}}_{\mathrm{A}} = 600 \mathrm{K}, {\mathrm{T}}_{\mathrm{B}} = 800 \mathrm{K}, {\mathrm{T}}_{\mathrm{C}}= 2200 \mathrm{K} \mathrm{and} {\mathrm{T}}_{\mathrm{D}} = 1200 \mathrm{K},$ the work done per cycle is

Processes A to B and C to D are parts of straight line graphs of the form y = mx

Also $\mathrm{P} = \frac{\mathrm{\mu R}}{\mathrm{V}}\mathrm{T} (\mathrm{\mu } = 6)$

$\Rightarrow \mathrm{P} \propto \mathrm{T}.$ So volume remains constant for the graphs AB and CD

So no work is done during processes for A to B and C to D i.e., ${\mathrm{W}}_{\mathrm{AB}} = {\mathrm{W}}_{\mathrm{CD}} = 0 \mathrm{and} {\mathrm{W}}_{\mathrm{BC}} = {\mathrm{P}}_2({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}) = \mathrm{\mu R}\left({\mathrm{T}}_{\mathrm{C}}-{\mathrm{T}}_{\mathrm{B}}\right)$

 = 6R(2200-800) = 6R$\times 1400 \mathrm{J}$

Also ${\mathrm{W}}_{\mathrm{DA}} = {\mathrm{P}}_1({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}}) = \mathrm{\mu R}({\mathrm{T}}_{\mathrm{A}}-{\mathrm{T}}_D)$

                  = $6\mathrm{R}(600-1200) = -6\mathrm{R}\times 600 \mathrm{J}$

Hence work done in complete cycle

$\mathrm{W} = {\mathrm{W}}_{\mathrm{AB}}+{\mathrm{W}}_{\mathrm{BC}}++{\mathrm{W}}_{\mathrm{CD}}+{\mathrm{W}}_{\mathrm{DA}}$

     = 0+6Rx1400+(-6Rx600)

     = 6Rx800 = 6x8.3x800 = 40 kJ