`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

Initial for `N_(2)`, using gas equation `PV = nRT`
`20 xx V_(1) = (2.8)/(28) xx 0.0821 xx 300 rArr V_(1) = 0.123L`
Finally for `N_(2)`
`1 xx V_(2) = (2.8)/(28) xx 0.0821 xx 300 rArr V_(2) = 2.463L`
`W =- P xx DeltaV` [`:'` work is doen against constant `P`,
`=- 1 xx (2.463 - 0.123)` therefore, irreversible]
`=- 2.340L atm`
`=- (2.340 xx 1.987)/(0.0821)cal`
`=- (2.340 xx 1.987 xx 4.184)/(0.0821)J`
`W =- 236.95J`
Now `q = DeltaU - W`
`q = 0 +236.95 = 236.95J`
(`:' DeltaU = 0`