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Question

Calculate pressure exerted by 1 mole of a van der Waal gas at a temperature of 80.0821K in a 12 L container, if the volume of the molecule is assumed to be negligible and van der Waal's constant a=2 atm L2 mol2

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Solution

Using van der Waal's equation for 1 mol:
(P+aV2m)(Vmb)=RT
Since the volume is negligible, neglecting b in the above equation we get:
(P+aV2m)Vm=RT
Given : T=80.0821K
PVm+aVm=R×80.0821
PV2m8Vm+a=0
According to the question, Vm=12 L=0.5 L so,
P(0.5)28(0.5)+2=0
P=8 atm

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