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Question

Six moles of an ideal gas performs a cycle shown in figure. The temperatures are TA=600 K,TB=800 K,TC=2200 K and TD=1200 K. The work done by the cycle ABCDA is
678377_8485fab9708c46e9ad8a835b81d86c7f.png

A
20 kJ
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B
30 kJ
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C
40 kJ
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D
60 kJ
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Solution

The correct option is B 40 kJ
Given, n=6
AB: isochoric process, ΔWAB=0
BC: isobaric process, ΔWBC=nRΔT
=nR(TcTa)
=6R(2200800)
=8400R
CD:isochoric process, ΔWCD=0
DA: isobaric process, ΔWDA=nRΔT
=nR(TATD)
=6R(6001200)
=3600R
ΔWcyclic=ΔWAa+ΔWac+ΔWCo+ΔWDA
=8400R3600R=4800R
=4800×8.3
=39840J40kJ.

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