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He/Ar mixtures in two piston cylinders: the mass ratio m1/m2 that makes P1 = 5 P2 at equal volume and temperature

Two piston cylinders hold He and Ar with the same two masses swapped between the gases. Press start and the lab does the rest: every mass becomes molecules (a gram of He is ten times as many as a gram of Ar), PV = nRT turns equal volume and temperature into P1/P2 = n1/n2, a live bisection scans the mass ratio until the first piston needs five times the pressure, the pistons lock level, the result is checked by putting it back and by doubling the temperature, and two common wrong answers are audited before the answer is revealed. No prediction stage - input, run, watch, result.

Sub-topics covered

• Moles from mass and molar mass
• Ideal gas equation PV = nRT
• Total moles of a gas mixture (Dalton's law)
• Pressure proportional to moles at fixed V and T
• Setting up and solving a ratio equation

What it teaches

• n = m/M: with M(Ar) = 10 M(He), a mass of He contains ten times as many moles as the same mass of Ar
• Cylinder 1: n1 = m1/M + m2/10M = (10m1 + m2)/10M; cylinder 2: n2 = (10m2 + m1)/10M
• At equal V and T, PV = nRT gives P1/P2 = n1/n2 - R, T and V cancel
• P1 = 5P2: 10m1 + m2 = 5(10m2 + m1), so 5m1 = 49m2 and m1/m2 = 49/5 = 9.80
• General form: m1/m2 = (pk - 1)/(k - p), which exists only when 1/k < p < k

Simulation ID

• ADV-2026-P1-CHE-Q09 · revision 1

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