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In the cyclic process given in the $\mathrm{P}-\mathrm{V}$ diagram, the work done is
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Verified Answer
The correct answer is:
$\frac{\pi}{4}\left(P_2-P_1\right)\left(V_2-V_1\right)$
From diagram,
$\begin{aligned}
& \text { major axis, } 2 \mathrm{~b}=\mathrm{V}_2-\mathrm{V}_1 \Rightarrow \mathrm{b}=\frac{\mathrm{V}_2-\mathrm{V}_1}{2} \\
& \text { minor axis, } 2 \mathrm{a}=\mathrm{P}_2-\mathrm{P}_1 \Rightarrow \mathrm{a}=\frac{\mathrm{P}_2-\mathrm{P}_1}{2} \\
& \text { Work done }=\text { Area of cyclic work }=\pi \mathrm{ab} \\
& =\pi \times\left(\frac{\mathrm{P}_2-\mathrm{P}_1}{2}\right) \times\left(\frac{\mathrm{V}_2-\mathrm{V}_1}{2}\right) \\
& =\frac{\pi}{4}\left(\mathrm{P}_2-\mathrm{P}_1\right)\left(\mathrm{V}_2-\mathrm{V}_1\right)
\end{aligned}$
$\begin{aligned}
& \text { major axis, } 2 \mathrm{~b}=\mathrm{V}_2-\mathrm{V}_1 \Rightarrow \mathrm{b}=\frac{\mathrm{V}_2-\mathrm{V}_1}{2} \\
& \text { minor axis, } 2 \mathrm{a}=\mathrm{P}_2-\mathrm{P}_1 \Rightarrow \mathrm{a}=\frac{\mathrm{P}_2-\mathrm{P}_1}{2} \\
& \text { Work done }=\text { Area of cyclic work }=\pi \mathrm{ab} \\
& =\pi \times\left(\frac{\mathrm{P}_2-\mathrm{P}_1}{2}\right) \times\left(\frac{\mathrm{V}_2-\mathrm{V}_1}{2}\right) \\
& =\frac{\pi}{4}\left(\mathrm{P}_2-\mathrm{P}_1\right)\left(\mathrm{V}_2-\mathrm{V}_1\right)
\end{aligned}$
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