`2.8g` of `N_(2)` gas at `300K` and `20atm` was allowed to expand isothermally against a constant external pressure of `1atm`. Calculate `DeltaU, q,` and `W` for the gas.

Explanation

`DeltaU = 0, DetaU = q +w or w =- q`
`w =- p DeltaV =- p (V_(2)-V_(1))`
`V_(2)` and `V_(1)` can be calculated from `V = (nRT)/(p)`
`V_(1) = (0.1 xx 0.082 xx 300)/(20) = 0.123L`
`V_(2) = (0.1 xx 0.082 xx 300)/(1) = 2.46 L`
`w =- (1atm) xx (2.46 - 0.123) L =- 2.337 L atm`
or `w =- 2.337 xx 101.3 = 236.7 J`