An ideal mono atomic gas is carried around the cycle ABCDA as shown in the fig. The efficiency of the gas cycle is

Heat is absorbed only during process AB and BC Here $C_r=\frac{3}{2}R{C}_P=\frac{5}{2}R$

$\mathrm{\Delta }{Q}_{AB}=n{C}_V\mathrm{\Delta }T=\frac{3}{2}nR\mathrm{\Delta }T=\frac{3}{2}V\mathrm{\Delta }P=3P_0{V}_0$  

Heat absorbed during BC $\mathrm{\Delta }{Q}_{BC}=n{C}_P\mathrm{\Delta }T=\frac{5}{2}nR\mathrm{\Delta }T=\frac{5}{2}P\mathrm{\Delta }V=\frac{15}{2}{P}_0{V}_0$

Hence net heat absorbed

$\mathrm{\Delta }Q=\mathrm{\Delta }{Q}_{AB}+\mathrm{\Delta }{Q}_{BC}=\frac{15}{2}{P}_0{V}_0+3P_0{V}_0$ $=\frac{21}{2}{P}_0{V}_0$

Also net work done W= area under P-V  Graph  $=2P_0{V}_0$   

Hence efficiency $n=\frac{\mathrm{\Delta }W}{\mathrm{\Delta }Q}=\frac{4}{21}$