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Question

A container is divided into two chambers by a partition. The volume of the first chamber is 4.5 L and the second chamber is 5.5 L. The first chamber contains 3.0 mol of gas at pressure 2.0 atm and the second chamber contain 4.0 mol of identical gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×101 atm. Value of x is _______. (up to two significant figures)

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Solution

By energy Conservation :

n1fRT12+n2fRT22=(n1+n2)fRT2

Using, PV=nRT, we get,

P1V1+P2V2=P(V1+V2)

P=P1V1+P2V2V1+V2

P=2×4.5+3×5.54.5+5.5

P=2.55 atm=25.5×101 atm26×101 atm

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