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Question
The container shown in figure 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.
| V1 | V2 |
| µ1, p1 | µ2 |
| p2 |
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Solution
Consider the diagram,
| p1V1 | p2V2 |
| µ1 | µ2 |
Given, V1 = 2.0 L, V2 = 3.0 L
µ1 = 4.0 mol, µ2 = 5.0 mol
p1 = 1.00 atm, p2 = 2.00 atm
For chamber 1, p1, V1 = µ1RT1
For chamber 2, p2, V2 = µ2RT2
When the partition is removed the gases get mixed without any loss of energy. The mixture Now attains a common equilibrium pressure and the total volume of the system is the sum of the volume of individual chambers V1 and V2.
So, µ = µ1 + µ2, V = V1 + V2
From the kinetic theory of gases,
For `l` mole `pV = 2/3 E` ......`[(E = "Translational"),("Kinetic energy")]`
For `µ_1` moles, `p_1V_1 = 2/3 µ_1E_1`
For `µ_2` moles, `p_2V_2 = 2/3 µ_2E_2`
The total energy is `(µ_1E_1 + µ_2E_2) = 3/2 (p_1V_1 + p_2V_2)`
From the above relation, `pV = 2/3 E_"total" = 2/3 µE_"per mole"`
`p(V_1 + V_2) = 2/3 xx 3/2 (p_1V_1 + p_2V_2)`
`p = (p_1V_1 + p_2V_2)/(V_1 + V_2)`
= `((1.00 + 2.0 + 2.00 xx 3.0)/(2.0 + 3.0))` atm
= `8.0/5.0`
= 1.60 atm
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