JEE Main · 2012 · 29 Apr 2012 · Offline · Heat & Thermodynamics

Over one loop the work done equals the rectangular area enclosed on the $P$--$V$ diagram, $P_0{V}_0$. Heat enters the gas only on the two legs where its temperature rises: on the constant-volume leg, $Q=n{C}_V\Delta T=\frac{3}{2}{P}_0{V}_0$ for this monatomic helium, and on the constant-pressure leg, $Q=n{C}_P\Delta T=\frac{5}{2}(2P_0{V}_0)=5P_0{V}_0$, so the total heat supplied is $\frac{13}{2}{P}_0{V}_0$. The efficiency is the work divided by the heat taken in, $\eta =\frac{P_0{V}_0}{\frac{13}{2}{P}_0{V}_0}=\frac{2}{13}\approx 0.154$, that is about $15.4\%$.