3 moles of an ideal monoatomic gas perfoms a cycle shown in the figure. The gas temperature at A,B,C,D are TA=400K,TB=800K,TC=2400K,TD=1200K. Find the work done by the gas.
A
12 kJ
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B
20 kJ
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C
120 kJ
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D
24 kJ
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Solution
The correct option is B20 kJ As process AB,CD pass through origin on P vs T graphs, we can say that P∝T⇒V=Constant. Hence, Isochoric. WAB=WCD=0
While process BC,DA are isobaric in nature
WBC=3R(TC−TB) WDA=3R(TA−TD) ∴ Total work done WAB+WCD+WBC+WDA=3R(TA+TC−TB−TD) =3R(400+2400−800−1200) =2400R=20kJ