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Question

Using van der Waals' equation, calculate the constant ′a′ when two moles of a gas confined in a four litre flask exerts a pressure of 11 atm at a temperature of 300 K. The value of ′b′ is 0.05 litre mol−1

A
6.46 atm litre2mol2
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B
5.46 atm litre2mol2
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C
6.64 atm litre2mol2
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D
5.64 atm litre2mol2
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Solution

The correct option is B 6.46 atm litre2mol2
Vander waal's equation for n moles of gas is, [P+an2v2][vnb]=nRT.
Given, v=4L,P=11atm,T=300K,b=0.05litre/mol
n=2
Thus ,[11+a2242][42(0.05)]=2×0.082×300
a=6.46 atm litre2 mol2

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