`5mol` of an ideal gas at `293K` is expanded isothermally from an initial pressure `0.4 kPa` to a final pressure of `0.1kPa` against a constant external pressure of `0.1kPa`.
a. Calcualte `q,w, Delta,U`, and `DeltaH`.
b. Calculate the corresponding value of `q,w, DeltaU`, and `DeltaH` if the above pressure is carried out reversibly.

Explanation

a. For an isothermal expansion against a constant pressure, we have:
`w =- P_(ex) (V_(2)-V_(1))`
`=- P_(ex) ((nRT)/(P_(2))-(nRT)/(P_(1)))=- P_(ex) ((1)/(P_(1))-(1)/(P_(2))) =- 9135J`
Since temperature is constant
`DeltaU = 0, DeltaH = 0,q =- W = 9135J`
b. For isothermal reversible expansion
`-w = 2.303nRT "log"(P_(1))/(P_(2))`
`= 2.303 xx5xx 8.324 xx 293 "log"(0.4)/(0.1)`
`rArr w =- 16888J`
`DeltaU = DeltaH = 0,q =- W = 16888J`