Three moles of an ideal gas at `27^(@)C` are compressed reversibly and isothermally from a volume of `10 dm^(3)` to `5 dm^(3)`. Calculate the work done on the gas.

Explanation

`T = 27^(@)C = 27 + 273 = 300 K, V_(1) = 10 dm^(3), V_(2) = 5 dm^(3)`
`R = 8.314 J//K//Mol, n = 3 "mole"`
`R = 8.314 w = -2.303 nRT log ((V_(2))/(V_(1))) = -2.303 xx 3 xx 8.314 xx 300 xx log[(5)/(1)]`
`= -2.303 xx 3 xx 8.314 xx 300 xx bar(1).6990`
`= -2.303 xx 3 xx 8.314 xx 300 xx (-0.3010) " " because -1 + 0.6990 = -3010`.
`= +5,186.96 "joule" = 5.187 kJ`