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Question

# Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is:

A
200Rln2
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B
100Rln2
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C
300Rln2
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D
400Rln2
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Solution

## The correct option is D 200Rln2Work done during process A→B(AB) (Isobaric process)WAB=nR(TB−TA) =100nRWork done during process B→C(BC) (Isothernal process)WBC=nRTB ln(PnPc)=nRTB ln(2)Work done during process C→D(CD)WCD=nR(TC−TD)=−100nRWnet=WAB+WDC+WCD+WDA=nR 300 ln(12)+nR 400 ln(2)=100 nR[3ln(12)+4 ln(2)]=100 nR[ln(12)3×(2)4]=100 nR ln(2) Wnet=200 R ln(2) [∵n=2]

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