A volume of 100 ml of a liquid contained in an isolated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar by which the volume of liquid is decreased by 1 ml. The change in enthalpy, ∆H, of the liquid (in bar ml)
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answer is 09900.00.
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Detailed Solution
ΔU=qtwFor adiabatic process q = 0 ΔU=wΔU=PΔV=−PV2−V1ΔU=−100(99−100)=100 bar mL Now ΔH=ΔU+Δ(PV)ΔH=100+[100×99−1×100]ΔH=9900 bar mL