Question

# Pâˆ’T diagram of one mole of an ideal monoatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is

A
2.5RT
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B
2RT
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C
1.5RT
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D
3.5RT
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Solution

## The correct option is B 2.5RTlet A=1,B=2,C=3,D=4now P1=P4=P & P2=P3=2Pso by ideal gas equation we haveV1=3RTP V2=RT2Pnow,1) for process 1-2W1−2=P1V1−P2V2γ−1so W1−2=2PRT2P−P3RTP0.4=−5RT2)for process 3-4W3−4=2P2RTP−P5RTP0.4=−2.5RT3) for process 4-1W4−1=P1(V4−V1)W4−1=P[5RTP−3RTP]=2RT4) for process 2-3W2−3=2P[2RTP−RT2P]=3RTnow adding[1,2,3,4] we get the net work doneWtotal=W1−2+W2−3+W3−4+W4−1=−5RT+3RT+−2.5RT+2RT=−2.5RT

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