Consider the cyclic process `ABCA` on a sample of `2.0 mol` of an ideal gas as shown in the figure. Temperature of the gas at `A` and `B` are `300 K` and `500 K` respectively. A total of `1200 J` heat is withdrawn from the sample in the process. Find the work done by the gas in part `BC`. take `R=8.3 JK^(-1) mol^(-1)`

Explanation

The change in internal energy during the cyclic process is zero. Hence, the heat supplied to the gasis equal to the work done by it. Hence
`W _(AB) + W_(BC) + W_(CA) =-1200J." "(i)`

The word doen during the process AB is
`W _(AB) =P_(A) (V_(B) - V _(A))`
=nR(T_(B) -T_(A))`
`= (2.0 mol) (8.3 J/mol-K) (200 K) = 3320 J `The work done by the gas during the process CA is zero as the volume remains constant. From (i)
`3320J + W_(BC) =-120 J`
or `W _(BC) =-4520J`
`=-4520J`