Using the given P-V diagram, the work done by | vedclass

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(D) The work done in a $P-V$ diagram is equal to the area under the curve. The path is $	ext{ABCD}$.For path $	ext{AB}$ (isobaric expansion): $	ext

Vedclass · Accepted Answer

$-3$

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(D) The work done in a $P-V$ diagram is equal to the area under the curve. The path is $	ext{ABCD}$.For path $	ext{AB}$ (isobaric expansion): $	ext{W}_{	ext{AB}} = 	ext{P}_0(3	ext{V}_0 - 2	ext{V}_0) = 	ext{P}_0	ext{V}_0$.For path $	ext{BC}$ (isochoric process): $	ext{W}_{	ext{BC}} = 0$ (since $\Delta	ext{V} = 0$).For path $	ext{CD}$ (isobaric compression): $	ext{W}_{	ext{CD}} = 2	ext{P}_0(	ext{V}_0 - 3	ext{V}_0) = 2	ext{P}_0(-2	ext{V}_0) = -4	ext{P}_0	ext{V}_0$.The total work done is $	ext{W}_{	ext{ABCD}} = 	ext{W}_{	ext{AB}} + 	ext{W}_{	ext{BC}} + 	ext{W}_{	ext{CD}} = 	ext{P}_0	ext{V}_0 + 0 - 4	ext{P}_0	ext{V}_0 = -3	ext{P}_0	ext{V}_0$.

Using the given $P-V$ diagram, the work done by an ideal gas along the path $\text{ABCD}$ is $-$ (in $\text{P}_0 \text{V}_0$)

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