Question

A container has two chambers of volumes V₁ = 2 L and V₂ = 3 L separated by a partition made of a thermal insulator. The chambers contain n₁ = 5 and n₂ = 4 moles of ideal gas at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

पर्याय

• 1.3 atm

• 1.6 atm

• 1.4 atm

• 1.8 atm

Answer 1

1.6 atm

Explanation:

Given: V₁ = 2 L

V₂ = 3 L

n₁ = 5

n₂ = 4

p₁ = 1 atm

p₂ = 2 atm

On removing the partition, the gases in the two chambers mix and acquire an equilibrium pressure.

Applying CОЕ,

E₁ + E₂ = E

`n_1[1/2 f_1RT_1] + n_2[1/2f_2RT_2] = (n_1 + n_2)(1/2fRT)`

Here, f₁ = f₂ = f

Since the same gas is present in each chamber. From the general gas equation, pV = nRT, we get

`1/2f_1(p_1V_1) + 1/2f_2(p_2V_2) = 1/2f(pV)`

= `(p_1V_1 + p_2V_2)/V`

= `(p_1V_1 + p_2V_2)/(V_1 + V_2)`

= `(1(2) + 2(3))/(2 + 3)`

= `(2 + 6)/(5)`

= `8/5`

= 1.6 atm