`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) = 10mL V_(2) = 99` mL
For adiabatic process, q = 0
`DeltaU = q + w`
`DeltaU = w`
` w = -pDeltaV`
` = -100(99 - 100) = 100 bar mL`.
`Delta H = DeltaU + DeltapV`
` = 100 + p_(2)V_(2) -p_(1)V_(1)`
` = 100 + (100 xx 99) -(1 xx 100)`
` = 100 + 9900 - 100 = 9900 bar (mL)`.