One mole of a liquid (`1` bar, `100 mL)` is taken in an adiabatic container and the pressure increase steeply to `100bar`. Then at a constant pressure of `100` bar, volume decreases by `1mL` . Find `DeltaU` and `DeltaH`.

Explanation

Volume of 1mol liquid `= 100mL` at pressure 1 bar
Volume of 1mol liquid `= 99 mL` at pressure `100` bar
The process being irreversibles as it is steeply changed ftom 1bar to 100 bar.
`:. W =- P (V_(2) - V_(1)) =- 100 xx (99 - 100) = 100 "bar" mL`
Also `DeltaU = q +w`
`DeltaU = w( :. q = 0`, adiabatic neture due to insulation)
`DeltaU = 100 "bar" mL`
Also `DeltaH = DeltaU + P DeltaV = DeltaU +(P_(2)V_(2) - P_(1)V_(1))`
`= 100 +(100 xx 99 - 1xx100)`
`= 9900` bar mL