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(A) The gas is monoatomic,so $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$.Heat is absorbed only during processes $AB$ and $BC$.For process $AB$ (isoc

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$\frac{4}{21}$

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(A) The gas is monoatomic,so $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$.Heat is absorbed only during processes $AB$ and $BC$.For process $AB$ (isochoric): $\Delta Q_{AB} = n C_v \Delta T = \frac{3}{2} n R \Delta T = \frac{3}{2} V_0 (3P_0 - P_0) = \frac{3}{2} V_0 (2P_0) = 3P_0 V_0$.For process $BC$ (isobaric): $\Delta Q_{BC} = n C_p \Delta T = \frac{5}{2} n R \Delta T = \frac{5}{2} P_0 \Delta V = \frac{5}{2} (3P_0) (2V_0 - V_0) = \frac{15}{2} P_0 V_0$.Total heat absorbed $Q_{in} = \Delta Q_{AB} + \Delta Q_{BC} = 3P_0 V_0 + \frac{15}{2} P_0 V_0 = \frac{21}{2} P_0 V_0$.Net work done $W = 	ext{Area of rectangle } ABCD = (2V_0 - V_0) 	imes (3P_0 - P_0) = V_0 	imes 2P_0 = 2P_0 V_0$.Efficiency $\eta = \frac{W}{Q_{in}} = \frac{2P_0 V_0}{\frac{21}{2} P_0 V_0} = \frac{4}{21}$.

Column-$1$	Column-$2$
$P$: Process-$I$	$A$: Adiabatic
$Q$: Process-$II$	$B$: Isobaric
$R$: Process-$III$	$C$: Isochoric
$S$: Process-$IV$	$D$: Isothermal

An ideal monoatomic gas is carried around the cycle $ABCDA$ as shown in the figure. The efficiency of the gas cycle is

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