Calculate work done by `1` mole of ideal gas expand isothermally and irreversibly from pressure of `5` atm to `2` atm against a constant external pressure of `1` atm at `300 K` temperature.

Explanation

`int dW = -int P_("ext")dv`
`W_("irr") = -P_("ext")[V_(2)-V_(1)] = -P_("ext")((nRT)/(P_(2))-(nRT)/(P_(1))) = -P_("ext") xx nRT((1)/(P_(2))-(1)/(1P_(1)))`
`= -1 xx 1 xx .082 xx 300 ((1)/(2)-(1)/(5)) = -1 xx .082 xx 300 xx (3)/(10) = -7.38 L. atm = -747.8 J`
`W_("irr") = -0.7478 KJ`