`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

Volume of `1` mole liquid`=100 mL` at pressure `1` bar volume of `1` mole liquid `=99 mL` at pressure `100` bar. The process being irreversible as it is steepy changed from `1` bar to `100` bar.
`:. W=-P(V_(2)-V_(1))=-100xx(99-100)=100` bar `mL`
Also `DeltaU=q+w`
`DeltaU=w`(`:. q=0`, adiabtic nature due to insultation)
`DeltaU=100` bar `mL`
Also `DeltaH=DeltaU+PDeltaV=DeltaU+(P_(2)V_(2)-P_(1)V_(1))`
`=100+(100xx99-1xx100)`
`=9900` bar `mL`