Calculate q,w, `DeltaU` and `DeltaH` for the reversible isothermal expansion of one mole of an ideal gas at `127^(@)C` from a volume of `10dm^(3)` to `20dm^(3)`.

Explanation

Since, the process is isothermal,
`DeltaU=DeltaH=0`
From first law of thermodynamics,
`DeltaU=q+w=0`
`q=-w`
`w=-2.30nRTlog((V_(2))/(V_(1)))`
`=-2.303xx1xx8.314xx400log((20)/(10))`
`=-2.303xx1xx8.314xx400xx0.3010`
`=-2305.3J` (Work is one by the system)
`q=-w=2305.3J` (Heat is absorbed by the system).