Helium gas undergoes a cycle ABCDA (consistin | vedclass

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(B) For a monoatomic gas like Helium, $C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$.$1$. Work done $(W)$ in the cycle is the area enclosed by the rect

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$15.4$

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(B) For a monoatomic gas like Helium, $C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$.$1$. Work done $(W)$ in the cycle is the area enclosed by the rectangle $ABCD$:$W = (2V_0 - V_0) 	imes (2P_0 - P_0) = V_0 P_0$.$2$. Heat is supplied during processes $AB$ (isochoric) and $BC$ (isobaric):Heat supplied $(Q_{in})$ = $Q_{AB} + Q_{BC}$.$Q_{AB} = n C_V \Delta T = n (\frac{3}{2}R) (T_B - T_A) = \frac{3}{2} (P_B V_0 - P_A V_0) = \frac{3}{2} (2P_0 V_0 - P_0 V_0) = \frac{3}{2} P_0 V_0$.$Q_{BC} = n C_P \Delta T = n (\frac{5}{2}R) (T_C - T_B) = \frac{5}{2} (P_C V_C - P_B V_B) = \frac{5}{2} (2P_0 	imes 2V_0 - 2P_0 	imes V_0) = \frac{5}{2} (2P_0 V_0) = 5 P_0 V_0$.$3$. Total heat supplied $(Q_{in})$ = $\frac{3}{2} P_0 V_0 + 5 P_0 V_0 = \frac{13}{2} P_0 V_0 = 6.5 P_0 V_0$.$4$. Efficiency $(\eta)$ = $\frac{W}{Q_{in}} 	imes 100 = \frac{P_0 V_0}{6.5 P_0 V_0} 	imes 100 = \frac{1}{6.5} 	imes 100 \approx 15.38 \% \approx 15.4 \%$.

Helium gas undergoes a cycle $ABCDA$ (consisting of two isochoric and two isobaric lines) as shown in the figure. The efficiency of this cycle is nearly ..... $\%$ (Assume the gas to be close to an ideal gas).

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