An ideal monoatomic gas is taken round the cycle ABCA as shown in the figure. The work done during the cycle is

zero

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4 PV

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6 PV

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9 PV

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Solution

The correct option is B 4 PV
Work done = area enclosed by PV diagram representing the cycle.
$Δ W = \frac{1}{2} \times B C \times A C = \frac{1}{2} \times 4 P \times 2 V = 4 P V$ .

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