Calculate the maximum work done in expanding `16g` of oxygen at `300K` occupying volume of `5dm^(3)` and isothermally untill the volume become `25dm^(3)`?

Explanation

Since, it is the maximum work, thus, the process is reversible.
`nO_(2) = (w)/(m) = (16)/(32),R = 8.314J`,
`T = 300K, V_(1) = 5dm^(3), V_(2) = 25dm^(3)`
`w =- 2.303nRT log_(10). (V_(2))/(V_(1))`
`=- 2.303 xx (16)/(32) xx 8.314 xx 300 log_(10). (25)/(5)`
`= - 2.01 xx 10^(3)J`