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An ideal gas is taken path as shown in figure. The net work done in the whole cycle is
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Verified Answer
The correct answer is:
$-3 \mathrm{p}_{1} \mathrm{~V}_{1}$
The net work done in the whole cycle is
$$
=\frac{1}{2} \times\left(3 \mathrm{p}_{1}\right) \times\left(2 \mathrm{~V}_{1}\right)=3 \mathrm{p}_{1} \mathrm{~V}_{1}
$$
If the cycle is anticlockwise work done is negative $\therefore \quad \mathrm{W}_{\text {cycle }}=-3 \mathrm{p}_{1} \mathrm{~V}_{1}$
$$
=\frac{1}{2} \times\left(3 \mathrm{p}_{1}\right) \times\left(2 \mathrm{~V}_{1}\right)=3 \mathrm{p}_{1} \mathrm{~V}_{1}
$$
If the cycle is anticlockwise work done is negative $\therefore \quad \mathrm{W}_{\text {cycle }}=-3 \mathrm{p}_{1} \mathrm{~V}_{1}$
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