`2mol` of an ideal gas at `25^(@)C` is allowed to expand reversibly at constant temperature (isothermally) form a volume of `2L` to `10L` by reducing the pressure slowely. Calculate the work done by the gas `(w), DeltaU, q`, and `DeltaH`.

Explanation

`w =- 2.303 nRT log V_(2)//V_(1)`
`log V_(2)//V_(1) = log 10L//2 L = log 5 = 0.6990`
`T = 25^(@)C +273 = 298K`
`:.w =- 2.303 (2mol) (8.3143J mol^(-1) K^(-1)) (298K)(0.6990)`
`=- 7977.1J`
a. For an isothermal reversible change, `DeltaU = 0`
b. `DeltaU = q+w =-0` and `q +(-7977.1J) = 0, q = 7977.1 J`
c. `DeltaH = DeltaU +DeltaPV = DeltaU = Delta(nRT)`
Since `R` and `T` are constants
`:. DeltaH = 0+0 =0`