In a cyclic process shown in the figure an ideal gas is adial gas is adiabatically taken from `B`to`A`, the work done on the gas during the process `B`to`A` is `30,J` when the gas is taken from `A` to `B` the heat absorbed by the gas is `20 J` Then change in internat energy of the gas in the process `A` to `B` is :

Options

A. `20 J`
B. `-30 J`
C. `50 J`
D. `-10 J`

Answer

B

Explanation

`W_(BA)=-30 J, Q_(BA) =0`.
`:.DeltaU_(BA) =-W_(BA) =30 J`
Now, `DeltaU_(BA) =-U_(BA) =-30 J`.