InterviewSolution
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InterviewSolution
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`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`. |
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Answer» 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` |
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