Calcualte `q, w, DeltaU`, and `DeltaH` for the reversible isothermal expansion of one mole of an ideal gas at `127^(@)C` from a volume of `20dm^(3)` to `40dm^(3)`.

Explanation

Since the process is isothermal,
`DeltaU = DeltaH = 0`
From first law of thermodynamics,
`DeltaU = q +w = 0`
`q =- w`
`w =- 2.303 nRT log ((V_(2))/(V_(1)))`
`=- 2.303 xx1xx 8.314 xx 400 "log" (40)/(20)`
`=- 2.303 xx1xx 8.314 xx 400 xx 0.3010`
`=- 2305.3J` (work is done by the system)
`q =- w = 2305.3J` (Heat is abosrbed by the system)