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

Answer 1

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