In the cyclic process given in the P-V diagra | vedclass

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(D) From the given $P-V$ diagram,the cyclic process forms an ellipse.The length of the axis along the $V$-axis is $2b = V_2 - V_1$,so $b = \frac{V_2 -

Vedclass · Accepted Answer

$\frac{\pi}{4}\left(P_2-P_1\right)\left(V_2-V_1\right)$

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(D) From the given $P-V$ diagram,the cyclic process forms an ellipse.The length of the axis along the $V$-axis is $2b = V_2 - V_1$,so $b = \frac{V_2 - V_1}{2}$.The length of the axis along the $P$-axis is $2a = P_2 - P_1$,so $a = \frac{P_2 - P_1}{2}$.The work done in a cyclic process is equal to the area enclosed by the loop in the $P-V$ diagram.Area of an ellipse = $\pi ab$.Therefore,Work done = $\pi 	imes \left(\frac{P_2 - P_1}{2}ight) 	imes \left(\frac{V_2 - V_1}{2}ight) = \frac{\pi}{4}(P_2 - P_1)(V_2 - V_1)$.

In the cyclic process given in the $P-V$ diagram,the work done is

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