One mole of a monoatomic ideal gas left(c_{V} | vedclass

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(C) The efficiency of the cycle is given by $\eta = \frac{W_{	ext{net}}}{Q_{	ext{supplied}}}$.The net work done $W_{	ext{net}}$ is the area enclose

Vedclass · Accepted Answer

$\frac{\pi}{32+\pi}$

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(C) The efficiency of the cycle is given by $\eta = \frac{W_{	ext{net}}}{Q_{	ext{supplied}}}$.The net work done $W_{	ext{net}}$ is the area enclosed by the cycle on the $P-V$ diagram. The area of a quarter ellipse with semi-axes $a = \frac{V_{0}}{2}$ and $b = \frac{P_{0}}{2}$ is $W_{	ext{net}} = \frac{1}{4} \pi a b = \frac{1}{4} \pi \left(\frac{V_{0}}{2}ight) \left(\frac{P_{0}}{2}ight) = \frac{\pi P_{0} V_{0}}{16}$.Heat is supplied to the gas only during the process $CA$. For process $CA$,the change in internal energy is $\Delta U_{CA} = n C_{V} \Delta T = \frac{3}{2} R (T_{A} - T_{C})$. Using $PV = nRT$,we have $T_{A} = \frac{(3/2 P_{0}) V_{0}}{R} = \frac{3 P_{0} V_{0}}{2R}$ and $T_{C} = \frac{P_{0} (V_{0}/2)}{R} = \frac{P_{0} V_{0}}{2R}$.Thus,$\Delta U_{CA} = \frac{3}{2} R \left(\frac{3 P_{0} V_{0}}{2R} - \frac{P_{0} V_{0}}{2R}ight) = \frac{3}{2} P_{0} V_{0}$.The work done during the path $CA$ is the area under the curve $CA$,which is the area of the rectangle plus the area of the quarter ellipse: $W_{CA} = P_{0} \left(\frac{V_{0}}{2}ight) + \frac{\pi P_{0} V_{0}}{16} = \frac{P_{0} V_{0}}{2} + \frac{\pi P_{0} V_{0}}{16}$.Total heat supplied $Q_{	ext{supplied}} = W_{CA} + \Delta U_{CA} = \left(\frac{P_{0} V_{0}}{2} + \frac{\pi P_{0} V_{0}}{16}ight) + \frac{3}{2} P_{0} V_{0} = 2 P_{0} V_{0} + \frac{\pi P_{0} V_{0}}{16} = P_{0} V_{0} \left(2 + \frac{\pi}{16}ight) = P_{0} V_{0} \left(\frac{32 + \pi}{16}ight)$.Efficiency $\eta = \frac{\pi P_{0} V_{0} / 16}{P_{0} V_{0} (32 + \pi) / 16} = \frac{\pi}{32 + \pi}$.

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