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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. 2PV
B. 4PV
C. `(1)/(2) PV`
D. PV

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Correct Answer - A
In cyclic process `DeltaU=0`
So heat absorbed
`DeltaQ=W` =Area under the curve
`=-(2V)(P)=-2PV`
So heat rejected =2PV

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