consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in the 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 A total of 1200 J heat is withdrawn from the sample in the process.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 cuclic process is zero. Hence, the heat supplied to the gas is equel 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)-V_(A)`
=nR(T(B)-T_(A))`
`(2.0 mol)(8.3 JK^(-1) mol^(-1))(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`.
`~~-4500 J`.