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100ml of a liquid contained in an insulated container at a pressure of 1 bar. The pressure is steeply increasedto 100 bar. The volume of the liquid is decreased by 1ml at this constant pressure. Find the ΔH & ΔU....
V1=100ml & V2=?, P1=1bar & P2=100bar ΔH=ΔU+nRT BY CHARLES LAW P1V1=P2V2 1*100=100*V2 100/100=V2 V2=1ml IT IS AN INSULATED CONTAINER SO NO TRANSFER OF HEAT TAKES PLACE THEREFORE ΔU=0 ΔH=0+nRT => ΔH=nRT PV=nRT [IDEAL GAS EQUATION] (P1-P2)*(V1-V2)=(1-100)*(100-1)=-99*99=-9801 IS WHAT I THINK SO I AM NOT SURE ENQUIRE OR ASK SOME ONE ELSE
V1=100ml & V2=?,
P1=1bar & P2=100bar
ΔH=ΔU+nRT
BY CHARLES LAW
P1V1=P2V2
1*100=100*V2
100/100=V2
V2=1ml
IT IS AN INSULATED CONTAINER SO NO TRANSFER OF HEAT TAKES PLACE
THEREFORE ΔU=0
ΔH=0+nRT => ΔH=nRT
PV=nRT [IDEAL GAS EQUATION]
(P1-P2)*(V1-V2)=(1-100)*(100-1)=-99*99=-9801
IS WHAT I THINK SO
I AM NOT SURE
ENQUIRE OR ASK SOME ONE ELSE
thnx...............
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