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PT diagram of one mole of an ideal monoatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is
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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 2.5RT
let A=1,B=2,C=3,D=4
now P1=P4=P & P2=P3=2P

so by ideal gas equation we have
V1=3RTP V2=RT2P

now,
1) for process 1-2
W12=P1V1P2V2γ1
so W12=2PRT2PP3RTP0.4=5RT

2)for process 3-4
W34=2P2RTPP5RTP0.4=2.5RT

3) for process 4-1
W41=P1(V4V1)
W41=P[5RTP3RTP]=2RT

4) for process 2-3
W23=2P[2RTPRT2P]=3RT

now adding[1,2,3,4] we get the net work done
Wtotal=W12+W23+W34+W41
=5RT+3RT+2.5RT+2RT=2.5RT

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