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The container shown in Fig. 13.6 has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2= 3.0 litre. The chambers contain μ1 = 4.0 and μ2 = 5.0 moles of a gas at pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

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V1 = 2.0 litre V2 = 3.0 litre

μ= 4.0 moles μ2 = 5.0 moles

P1 = 1.00 atm P2 = 2.00 atm

P1V1 = μ1RT1       P2V2 = (2/3)μ2RT2

μ = μ1 + μ2         V = V1V2 

For 1 mole PV = (2/3)E

For μ1 moles       P1V1 = (2/3)μ1E1

For μ2 moles     P2V2 = (2/3)μ2E2

Total energy is (μ1E1 + μ2E2) = 3/2(P1V1 + P2V2)

PV = (2/3)Etotal = (2/3)μEper mole

P(V1 + V2) = (2/3) x (3/2)(P1V1 + P2V2)

P = (P1V2 + P2V2/(V1 + V2)

= (1.00 x 2.0 + 2.00 x 3.0/2.0 + 3.0)atm

= (8.0/5.0) = 1.60atm

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