`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

`DeltaU = q +w`
For adiabatic process, `q = 0`, hence `DeltaU = w`.
and `w =- p (DeltaV) =- (V_(2)-V_(1))`
`DeltaU =- 100 (99-100) =- 100 (-1) = 100 "bar" mL`
Now `DeltaH = DeltaU +Delta(pV)`
`DeltaU +(p_(2)V_(2)-p_(1)V_(1))`
`= 100 +(100 xx 99 - 1 xx 100)`
So, `DeltaH = 9900 bar ml`