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In a cyclic process shown in the figure ...

In a cyclic process shown in the figure an ideal gas is adiabatically taken from A to B the work done on the gas during the process `AtoB` is 80 J, when the gas is taken from B to a the heat absorbed by the gas is 40 J. What will be work done by the gas in process `BtoA`?

Text Solution

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`DeltaQ_(AB)=0,DeltaW_(AB)=-80J`
`DeltaQ_(BA)=40J, DeltaW_(BA)=?`
From the first law of thermodynamics
`(sumQ)_("cycle")=(sumW)_("cycle")`
`DeltaQ_(AB)=DeltaQ_(BA)=DeltaW_(AB)+DeltaW_(BA)`
`0+40=-80+DeltaW_(BA)`
`DeltaW_(BA)=120J`
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