A thermally insulated partition divides a container into volumes V₁ = 2 L and V₂ = 3 L containing n₁ = 5 mol and n₂ = 4 mol of the same ideal gas, at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. After removing the partition and reaching equilibrium, find the final pressure.
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Question

A thermally insulated partition divides a container into volumes V₁ = 2 L and V₂ = 3 L containing n₁ = 5 mol and n₂ = 4 mol of the same ideal gas, at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. After removing the partition and reaching equilibrium, find the final pressure.

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G1 Physics · Accepted Answer

Step 1: The rigid insulated container does no external work and exchanges no heat, so total internal energy is conserved. Step 2: For the same ideal gas (constant degrees of freedom), internal energy is U = (f/2)nRT = (f/2)PV. Step 3: Hence p₁V₁ + p₂V₂ = p𝒇(V₁ + V₂). Step 4: p𝒇 = [(1 atm)(2 L) + (2 atm)(3 L)]/(2 + 3)L = (2 + 6)/5 atm. Final answer: 1.6 atm, Option B.