`100mL` of a liquid is contained in an insulated container at a pressure of 1bar. The pressure is steeply increased to `100`bar. The volume of the liquid is decreased by `1mL` at this constant pressure. Find `DeltaH` and `DeltaU`.

Explanation

`P_(1) = 1 "bar", P_(2) = 100 "bar"`
`V_(1) = 100mL V_(2) = 99mL`
For adiabatic process, `q = 0`
`DeltaU = q+w`
`DeltaU = w`
`w =- P DeltaV =- 100(99-100) = 100 bar mL`
`DeltaH = DeltaU +DeltaPV`
`= 100 +P_(2)V_(2) - P_(1)V_(1)`
`= 100 +(100xx99) -(1xx100)`
`= 100 +9900 - 100 = 9900 bar mL`