3 moles of an ideal monoatomic gas performs ABCDA cyclic process as shown in figure. The gas temperature are  ${\mathrm{T}}_{\mathrm{A}}=400 \mathrm{K},$ ${\mathrm{T}}_{\mathrm{B}}=800 \mathrm{K},$${ \mathrm{T}}_{\mathrm{C}}=2400 \mathrm{K}$ and ${\mathrm{T}}_{\mathrm{D}}=1200\mathrm{K}$ , the work done by the gas is (approximately) (R= 8.314 J/mole.K)

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.

${\mathrm{W}}_{\mathrm{AB}}={\mathrm{W}}_{\mathrm{CD}}=0$

$\text{ and }{\mathrm{W}}_{\mathrm{BC}}={\mathrm{p}}_{\mathrm{B}}\left({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}\right)$

$=\mathrm{nR}\left({\mathrm{T}}_{\mathrm{C}}-{\mathrm{T}}_{\mathrm{B}}\right)==3\mathrm{R}(2400-800)=4800\mathrm{R}$

${\mathrm{W}}_{\mathrm{DA}}={\mathrm{p}}_{\mathrm{A}}\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}}\right)$

$=\mathrm{nR}\left({\mathrm{T}}_{\mathrm{A}}-{\mathrm{T}}_{\mathrm{D}}\right)=3\mathrm{R}(400-1200)=-2400\mathrm{R}$

Total work done $\mathrm{W}={\mathrm{W}}_{\mathrm{AB}}+{\mathrm{W}}_{\mathrm{DA}}$ 

$\mathrm{W}=2400\mathrm{R}=19953.6\mathrm{J}\simeq 20\mathrm{kJ}$