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(A) The work done per cycle is the area enclosed by the $P-V$ diagram:$W = 	ext{Area} = (2V_0 - V_0) 	imes (3P_0 - P_0) = V_0 	imes 2P_0 = 2P_0V_0$

Vedclass · Accepted Answer

$19$

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(A) The work done per cycle is the area enclosed by the $P-V$ diagram:$W = 	ext{Area} = (2V_0 - V_0) 	imes (3P_0 - P_0) = V_0 	imes 2P_0 = 2P_0V_0$.Heat is absorbed during processes $AB$ and $BC$:For process $AB$ (isochoric): $Q_{AB} = nC_V \Delta T = n \left( \frac{3R}{2} ight) \left( \frac{P_B V_A}{nR} - \frac{P_A V_A}{nR} ight) = \frac{3}{2} V_A (P_B - P_A) = \frac{3}{2} V_0 (3P_0 - P_0) = 3P_0V_0$.For process $BC$ (isobaric): $Q_{BC} = nC_P \Delta T = n \left( \frac{5R}{2} ight) \left( \frac{P_B V_C}{nR} - \frac{P_B V_B}{nR} ight) = \frac{5}{2} P_B (V_C - V_B) = \frac{5}{2} (3P_0) (2V_0 - V_0) = \frac{15}{2} P_0V_0$.Total heat absorbed $Q_{	ext{in}} = Q_{AB} + Q_{BC} = 3P_0V_0 + 7.5P_0V_0 = 10.5P_0V_0 = \frac{21}{2} P_0V_0$.Efficiency $\eta = \frac{W}{Q_{	ext{in}}} 	imes 100 = \frac{2P_0V_0}{(21/2)P_0V_0} 	imes 100 = \frac{4}{21} 	imes 100 \approx 19.04\%$.Thus,the efficiency is close to $19\%$.

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to $.......\%$

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