`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

There are two process involved and both are under adiabatic condition.
Process first:
1bar, `100mL rarr 100 "bar", 100 mL`
Since change in volume `= 0`
`:. W = 0`
`dU = q +W`
`dU = W` (Addiabatic condition `q = 0`)
`:. qU = 0`
Process second:
`100 "bar", 100 mL rarr 100 "bar", 90 mL`
`W =- P (V_(2) - V_(1))`
`W =- 100 (99 - 100)`
`:. W = + 100 mL-atm`
Since container is still under adiabatic condition
`dU = 100 mL-atm`
Therefore, overall
`dU = 0 + 100 = 100 mL-atm = 0.1 L-atm`
For `DeltaH`:
`DeltaH_(1) = DeltaU + PDeltaV + DeltaPV`
`= DeltaU + 0 + 100 (99) = 9900 mL-atm`
`DeltaH_(2) = DeltaH + P DeltaV + DeltaPV`
`= 100 + P(V_(2) - V_(1)) +0 = 100 +100 (-1) = 0 mL-atm`
`DeltaH = DeltaH_(1) + DeltaH_(2) = 9900 mL-atm = 9.9 L-atm`