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Question

One mole of a vander Waal's gas obeying the equation
(p+aV2)(Vb)=RT
undergoes the quasi-static cyclic process which is shown in the p-V diagram. The net heat absorbed by the gas in this process is:
461994.png

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(p1p2)(V1V2)
Workdone during the complete cycle is equal to the area enclosed by the P-V graph
W=12×(OB)×(OA)=12(p1p2)(V1V2)
Also the change in internal energy in a complete cycle is Zero i.e ΔU=0

Using first law of thermodynamics : Q=ΔU+W
Q=0+W=W
Thus the heat absorbed by the gas in a complete cycle Q=W=12(p1p2)(V1V2)

497050_461994_ans.png

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