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Question Number 34891 by arnabmaiti550@gmail.com last updated on 12/May/18

Answered by ajfour last updated on 12/May/18

$$Workdone=Areaunderellipse$$$$\Rightarrow W=\pi P_0{V}_0$$$$Heatreceived=Areaunder$$$$upperellipsecurveandVaxes$$$$betweenV=\frac{V_0}{2}andV=\frac{3V_0}{2}$$$$+△(Internalenergywhile$$$$expansion)$$$$\Rightarrow Q=P_0{V}_0+\frac{\pi P_0{V}_0}{2}+\frac{3}{2}{P}_0(\frac{3V_0}{2}-\frac{V_0}{2})$$$$efficiency=\frac{workdonebygas}{heatreceived}$$$$\Rightarrow \eta =\frac{\pi P_0{V}_0}{\frac{5}{2}{P}_0{V}_0+\frac{\pi P_0{V}_0}{2}}=\frac{2\pi }{\pi +5}.$$$$So\left(a\right),and\left(c\right).$$

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