Two moles of a monoatomic gas at `300K` are kept in a non conducting container enclosed by a piston . Gas is now compressed to increase the temperature from `300K` to `400K`. Find work done by the gaS `(R = (25)/(3) J//mol-K)`

Explanation

`DeltaQ = DeltaU +W`
Since container is non conduting therefore
`DeltaQ = 0 =DeltaU +W`
`rArr W =- DeltaU =- n(f)/(2)R DeltaT =- 2 xx (3)/(2) (400-300)`
`=-3 xx (25)/(3) xx 100J =- 2500J`