Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram

The net work done on the gas in the cycle ABCDA is :

Options

A. Zero
B. `276R`
C. `1076R`
D. `1904R`

Answer

B

Explanation

`W_(ABCDA) = W_(AB) +W_(BC) +W_(CD) +W_(DA)`
`= nR (DeltaT)_(AB) +nR(T_(B)) In (P_(B))/(P_(C))`
`+nR(DeltaT)_(CD) +nR(T_(D)) In (P_(D))/(P_(A))`
`= nR (200) +500 nR In 2 +nR (-200)`
`+300 nR In (1)/(2)`
`= 2 In 2[500R - 300R]`
`= (400R)(In2) - (400R) (0.693) = 276R`
`W_(ABCDA) = 276R`