Consider the cyclic process ABCA, shown in figure, performed on a sample of `2.0 mol` of an ideal gas. A total of `1200 J` of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

Explanation

In a cyclic process
`Delta U = 0`
From the first law of thermodynamics, for the cyclic process
`Q = Delta U + W`
`W = Q - Delta U = - 1200 - 0`
`= - 1200 J`
From `C` to `A`, `Delta v = 0`
`:. W_(CA) = 0`
For the whole cycle
`W_(AB) + W_(BC) + W_(CA) = W`
`= 0 1200 J`
As `W_(CA) = 0`
`:. W_(AB) + W_(BC) = - 1200 J`
Work done form `A` to `B` :
In the process `V poro T`, so pressure remains constant
we know that `PV = n RT`
or `P Delta V = nR Delta T`
`:. W_(AB) = P Delta V = nR Delta T`
`= 2 xx 8.31 xx (500 - 300) = 3324 J`
Subsituting this value in Eq. (i) we get
`3324 + W_(BC) = - 1200`
`W_(BC) = - 4524 J`