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

The work done on the gas in taking it from D to A is :

Options

A. `-414 R`
B. `+414R`
C. `-690 R`
D. `+690R`

Answer

B

Explanation

`D` to `A` is isothermal process
Work done by the gas in `SD` to `A` is
`W_(DA) = nRT In (V_(2))/(V_(1)) = nRT In (P_(1))/(P_(2))`
`= (2)(R )(300)In (10^(5))/(2xx10^(5)) = (600R)[-In2]`
`=- (600R) (0.693) =- 414 R`
`W_(DA) = - 414R`, it is work done by the gas
So work done on the gas is `+414 R`