Vapor pressure and mixtures

In a closed jar some molecules at a liquid's surface always have enough energy to escape, and some vapor molecules land back. The pressure of the vapor when the two balance is the vapor pressure. It rises steeply with temperature, and a liquid boils when its vapor pressure equals the pressure above it. Water's vapor pressure reaches 1 atm at 100 °C; up a mountain the air pressure is lower, so water boils below 100 °C, and in a pressure cooker above it.

A phase diagram maps which phase is stable at each temperature and pressure. Lines mark where two phases coexist: the boiling line, the melting line and the sublimation line. They meet at the triple point, where solid, liquid and gas all coexist (for water 0.01 °C and 0.6 kPa). The boiling line ends at the critical point (water: 374 °C, 221 bar), beyond which liquid and gas can't be told apart.

In a mixture of two liquids each one's contribution to the vapor drops in proportion to how much of it is in the liquid. Raoult's law: pᵢ = xᵢ × Pᵢ*, where xᵢ is its mole fraction in the liquid (its share of the moles) and Pᵢ* the vapor pressure it would have on its own. The total pressure is the sum of the parts. It works best for similar molecules, like benzene and toluene.

The vapor's makeup follows from the partial pressures: yᵢ = pᵢ ÷ total. Take a liquid of 40% A (P* = 100 kPa) and 60% B (P* = 40 kPa). Then p_A = 0.4 × 100 = 40 kPa, p_B = 0.6 × 40 = 24 kPa, the total is 64 kPa, and the vapor is 40 ÷ 64 = 62.5% A. The vapor is richer in the more volatile liquid (the one with the higher vapor pressure).

That's the whole idea behind distillation: boil a mixture, condense the vapor, and it's richer in the light component. Repeat it many times up a column and you can separate them almost completely. Some mixtures break Raoult's law enough to form an azeotrope, where vapor and liquid have the same makeup; ethanol and water do at about 96% ethanol, so plain distillation can't get past it.