A gas occupies `2L` at `STP`. It is provided `300J` heat so that it volume becomes `2.5L` at `1atm`. Caluclate the change in its internal enegry.

Explanation

Work done `=- P xx DeltaV = 1 xx (2.5 - 2.0)`
`=- 0.5 L atm`
Therefore, work is carried out at constant `P` and thus irreversible.
`=- (0.5 xx 1.987xx4.184)/(0.0821)J =0.5 xx 101.328J`
`=- 50.631J`
From the first law of thermodynamics,
`:. q = DeltaU - w`
`300 = DeltaU +50.63:. DeltaU = 249.37J`