Two flasks joined by a stopcock.

Two flasks at different pressures joined by a stopcock; opening it gives a common pressure from total moles, with partial pressures stacked (Dalton).

Choose V₂ so the final pressure is 2 atm when 4 atm × 2 L is joined to 1 atm × V₂.

Averaging the two pressures.

• A mixture of ideal gases that do not react, μ₁ moles of gas 1, μ₂ of gas 2 and so on, in volume V at temperature T, obeys PV = (μ₁ + μ₂ + …)RT.
• So P = μ₁RT/V + μ₂RT/V + … = P₁ + P₂ + …. Here P₁ = μ₁RT/V is the partial pressure of gas 1: what it would exert alone in the same volume at the same temperature.
• Dalton's law of partial pressures: in a mixture of ideal gases, adding up the partial pressures gives the total pressure.
• Example 12.1: water has density 1000 kg m⁻³; its vapour at 100 °C and 1 atm has 0.6 kg m⁻³. The same mass spreads over 1000/0.6 times the volume, so molecules fill only a fraction 6 × 10⁻⁴ of the vapour's volume.
• Example 12.2: a mole of water is 18 g = 0.018 kg, so one molecule has mass 0.018/(6 × 10²³) = 3 × 10⁻²⁶ kg. Taking the molecule's density as that of liquid water, its volume is 3 × 10⁻²⁶/1000 = 3 × 10⁻²⁹ m³. Setting that equal to (4/3)πr³ gives r ≈ 2 × 10⁻¹⁰ m = 2 Å.
• Example 12.3: vapour gives each molecule about 1.67 × 10³ times the room it had in the liquid. A thousandfold volume means a tenfold length (cube root), so the radius of each molecule's share goes from 2 Å to 20 Å, and the average spacing is 2 × 20 = 40 Å.
• Example 12.4: neon and oxygen share a vessel with partial pressures 3 : 2. With V and T common, P ∝ μ, so the ratio of moles, and hence of molecules, is 3/2.
• Mass densities: ρ₁/ρ₂ = (μ₁M₁)/(μ₂M₂) = (3/2) × (20.2/32.0) = 0.947 for neon (20.2 u) to oxygen (32.0 u).

Teaching this idea? Put the simulation on your own page or LMS with this code.