Dalton's Law and Partial Pressures

Because ideal particles do not interact, each gas in a mixture contributes pressure independently. The total pressure is P_total = P₁ + P₂ + P₃ and so on. Nothing about the identity of the gases matters, only how many moles of each are present.

For component i in an ideal mixture, P_i = n_iRT/V. Summing gives P_total = (Σn_i)RT/V, so P_i/P_total = n_i/n_total = χ_i. Each partial pressure uses the whole container volume, not a separate fraction of the container assigned to that gas. This independence assumes no chemical reaction between components and conditions close enough to ideal behavior.

The mole fraction of a component, written as chi, is its moles divided by the total moles. Multiplying the mole fraction by the total pressure gives the partial pressure of that gas. Mole fractions have no units and always sum to one, which is a fast way to check your work.

If masses are given, convert them before finding fractions. A mixture of 4.00 g He and 32.0 g O₂ contains about 1.00 mol of each, so each contributes half the total pressure despite the eightfold mass difference. At P_total = 3.00 atm, each partial pressure is 1.50 atm. The larger mass does not imply larger pressure because the faster motion of light particles compensates for their smaller mass in the ideal model.

When a gas is bubbled into an inverted container of water, the trapped sample is a mixture of the gas and water vapour. The measured total pressure therefore includes the vapour pressure of water at that temperature, which you look up in a table. Subtract it to obtain the pressure of the dry gas.

Subtract the water vapor pressure only when the gas is saturated with water vapor at the measurement temperature and liquid water is available to establish equilibrium. Use the same pressure units in the subtraction. If liquid levels inside and outside an inverted collection tube differ, the trapped total pressure may differ from atmospheric pressure because of the water column; equalize levels or use the supplied pressure correction before subtracting P_water. Then use the dry gas's partial pressure in n_gas = P_gasV/(RT). Do not include water-vapor moles as product gas.

For a sealed rigid flask, a noncondensing component with fixed moles follows P₂ = P₁T₂/T₁. A vapor that condenses does not retain fixed gas-phase moles, so apply that scaling only until it reaches saturation. With liquid water present, the final equilibrium water partial pressure is its vapor pressure at the final temperature. If the available water is insufficient for saturation, its pressure is lower.

Suppose Ar initially has P = 0.800 atm and water vapor 0.400 atm at 80 °C in a rigid flask. At 20 °C, Ar gives 0.800(293.15/353.15) = 0.664 atm. Scaling the original water vapor without condensation would predict 0.332 atm, above the supplied 20 °C saturation pressure of 0.0231 atm. Water therefore condenses, leaving P_water = 0.0231 atm and P_total ≈ 0.687 atm. Total pressure does not simply scale with T because some gas-phase water molecules leave the vapor.