Question

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 A total of `1200 J` heat is withdrawn from the sample in the process.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)`

Answer 1

The change in internal energy during the cyclic process is zero. Hence, the heat supplied to the gas is equal to the work done by it Hence,
`W_(AB) + W_(BC) + W_(CA) = - 1200 J`
The work done during the process AB is
`W_(AB) = P_(A) (V_(B) - B_(A))`
`nRT (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).
`3320 J + W_(BC) = - 1200 J`
or `W_(BC) = - 4520 J - - 4520`