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Question

# One mole of a van der Waals gas obeying the equation (P+aV2)(Vâˆ’B)=RT undergoes the quasi-static cyclic process which is shown in the Pâˆ’V diagram. The net heat absorbed by the gas in this process is

A
12(P1P2)(V1V2)
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B
12(P1+P2)(V1V2)
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C
12(P1+av21P2av22)(V1V2)
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D
12(P1+av21+P2+av22)(V1V2)
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Solution

## The correct option is A 12(P1−P2)(V1−V2)For a cyclic process, the heat absorbed by the gas is equal to the work done in one complete cycle.Work done in a cyclic process is equal to the area enclosed by the PV curve.Hence net heat absorbed=12×Base×Height=12×(V1−V2)×(P1−P2)=12(P1−P2)(V1−V2)

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