A container has two chambers of volumes V₁ = 2 litres and V₂ = 3 litres 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 (a) 1.4 atm (b) 1.8 atm (c) 1.3 atm (d) 1.6 atm

Answer: (d). For an ideal gas PV is proportional to internal energy, which is conserved when the partition is removed. p(V₁ + V₂) = p₁V₁ + p₂V₂. p = (1 × 2 + 2 × 3)/(2 + 3) = 8/5 = 1.6 atm. So the answer is (d) 1.6 atm.

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Gas Laws & Ideal Gas Equation

A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1\text{ atm}$ and $p_2 = 2\text{ atm}$ , respectively. When the partition is removed, the mixture attains an equilibrium pressure of

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Answer: (d)

For an ideal gas $PV$ is proportional to internal energy, which is conserved when the partition is removed.
$p(V_1 + V_2) = p_1V_1 + p_2V_2$ .
$p = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{8}{5} = 1.6$ atm.
So the answer is (d) 1.6 atm.

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