Calculate the work done by `1.0 mol` of an idela gas when it expands from `10atm` to `2atm` at `27^(@)C`.

Explanation

Process is reversible, since the pressure is decreased from `10 atm` to `2atm` in one step.
`w_(("irr,isotherm)) =- P_(ex)(V_(2)-V_(1))`
`=- P_(ex) ((nRT)/(P_(2))-(nRT)/(P_(1)))`
`=- P_(ex) xx nRT ((1)/(P_(2))-(1)/(P_(1)))`
`= - 2xx1xx 0.082 xx 300 ((1)/(2)-(1)/(10))`
`=- 2 xx 0.082 xx 300 xx (4)/(10)`
`= 19.68 L-atm`
`= 119.68 xx 101.3J =- 1993.58J`
(since `1L-atm = 101.3J)`