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

Approved

thnx...............