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Find the work done during the perfectly ...

Find the work done during the perfectly circular cyclic process as shown in the diagram.

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The area of a circle is `pi R^2`. But here you have two radii. The horizonatal radius `R_1 =((V_2-V_1))/2 ` and the vertical radius `R_2 = ((P_1 -P_2))/2`. You know that the area of an ellipe is `piR_1R_2`. Where `R_1 &R_2` are somi-major and semi minor axis respectively. If `R_1 = R_2` the ellipse becomes a circle. So, here the work done = area `= piR_1R_2 = pi((V_2V_1))/2,((P_1-P_2))/2`. The +ve sign is due to the fact that the cycle is clockwise.
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