A monoatomic ideal gas goes through a cyclic | vedclass

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(A) The work done in a complete cycle is equal to the area under the closed curve on the $P-V$ diagram.Work done $W = 	ext{Area} = (2V - V) 	imes (3

Vedclass · Accepted Answer

$19.04$

Vedclass · Answer

(A) The work done in a complete cycle is equal to the area under the closed curve on the $P-V$ diagram.Work done $W = 	ext{Area} = (2V - V) 	imes (3P - P) = V 	imes 2P = 2PV$.For a monoatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$ and at constant pressure is $C_P = \frac{5}{2}R$.Heat is absorbed by the system during processes $AB$ and $BC$.For process $AB$ (isochoric): $Q_{AB} = n C_V \Delta T = n \left(\frac{3}{2}Right) \Delta T = \frac{3}{2} \Delta(PV) = \frac{3}{2} V(3P - P) = \frac{3}{2} V(2P) = 3PV$.For process $BC$ (isobaric): $Q_{BC} = n C_P \Delta T = n \left(\frac{5}{2}Right) \Delta T = \frac{5}{2} P \Delta V = \frac{5}{2} (3P)(2V - V) = \frac{15}{2} PV$.Total heat absorbed $Q_{in} = Q_{AB} + Q_{BC} = 3PV + \frac{15}{2} PV = \frac{6PV + 15PV}{2} = \frac{21}{2} PV$.Efficiency $\eta = \frac{W}{Q_{in}} 	imes 100 = \frac{2PV}{\frac{21}{2} PV} 	imes 100 = \frac{4}{21} 	imes 100 \approx 19.04 \%$.

$A$ monoatomic ideal gas goes through a cyclic process as shown in the figure. The efficiency of this process is (in $\%$)

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