A ideal monoatomic gas is carried around the cycle $ABCDA$ as shown in the fig. The efficiency of the gas cycle is
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Heat is absorbed only during processes $AB$ and $BC$
Gas is monoatomic.Hence $C_{v}=\frac{3}{2} R, C_{p}=\frac{5}{2} R$
Heat absorbed during $\mathrm{AB}, \Delta Q_{A B}=n C_{v} \Delta T=\frac{3}{2} n R \Delta T=\frac{3}{2} V \Delta P=3 P_{o} V_{o}$
Heat absorbed during $\mathrm{BC}, \Delta Q_{B C}=n C_{p} \Delta T=\frac{5}{2} n R \Delta T=\frac{5}{2} P \Delta V=\frac{15}{2} P_{o} V_{o}$
Hence, net heat absorbed $=\Delta Q=\Delta Q_{A B}+\Delta Q_{B C}=\frac{21}{2} P_{o} V_{o}$
Also, net work done $=\Delta W$ =area under $P-V$ graph $=2 P_{o} V_{o}$
Hence efficiency $=\frac{\Delta W}{\Delta Q}=\frac{4}{21}$
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