An ideal gas is taken through the cycle `AtoBtoCtoA,` as shown in the figure, If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CtoA is

Explanation

`DeltaW_(AB)=PDeltaV=(10)(2-1)=10J`
`DeltaW_(BC)=0` (as V=constant)
From first law of thermodynamics
`DeltaQ=DeltaW+DeltaU=0` (process ABCA is cyclic)
`therefore DeltaQ=DeltaW_(AB)+DeltaW_(BC)+DeltaW_(CA)`
`therefore DeltaW_(CA)=DeltaQ-DeltaW_(AB)-DeltaW_(BC)=5-10-0=5J`