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A thermodynamic system is taken through the cycle `ABCD` as shown in the figure. Heat rejected by the gas during the cycle is
image
A. `PV`
B. `2PV`
C. `4PV`
D. `1/2 PV`

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Correct Answer - B
From the first law of thermodynamics
`DeltaQ = DeltaU + DeltaW`
For cycle process `DeltaU=0`
Hence, `Q = W`
Work done `W` = area under `p-V` disgram
`implies Q = -2PV`.

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