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Question

Using van der Waals equation, find the constant a (in atm L2mol2) when two moles of a gas confined in 4L flask exerts a pressure of 11.0 atmospheres at a temperature of 300K. The value of b is 0.05Lmol1. (R=0.082atmL/Kmol) (as nearest integer)

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Solution

By Vanderwaal's Gas Equation (P+an2v2)(vnb)=nRT

Given P=11,n=2,V=4,b=0.05,R=0.082T=300

(11+a4)×(40.1)=2×0.082×30011+a4=12.61a=6.46
Answer is 6.

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