Partial Pressures in an N2/O2 Mixture

Medium

Problem

A gas mixture contains 0.60 mol of nitrogen (N2) and 0.40 mol of oxygen (O2) at a total pressure of 2.00 atm. Calculate the partial pressure of each gas.

Given

Moles of nitrogen: n(N2) = 0.60 mol
Moles of oxygen: n(O2) = 0.40 mol
Total pressure of the mixture: P_total = 2.00 atm

Find

The partial pressures of N2 and O2 in atm.

Principle

Dalton's law of partial pressures: in a mixture of non-reacting gases, each gas exerts the pressure it would exert if it alone occupied the container. That partial pressure equals the gas's mole fraction times the total pressure. Since mole fractions always sum to one, the partial pressures always sum to the total pressure.

Formula

P_i = X_i P_{\mathrm{tot}},\quad X_i = \frac{n_i}{\sum n}

Pi = Xi x P_total ; Xi = ni / n_total

Solution

Total moles: n_total = 0.60 + 0.40 = 1.00 mol.
Mole fraction of N2: X(N2) = 0.60 / 1.00 = 0.60.
Mole fraction of O2: X(O2) = 0.40 / 1.00 = 0.40. Check: 0.60 + 0.40 = 1.00.
Partial pressure of N2: P(N2) = 0.60 x 2.00 atm = 1.20 atm.
Partial pressure of O2: P(O2) = 0.40 x 2.00 atm = 0.80 atm.
Check: 1.20 + 0.80 = 2.00 atm, exactly the total pressure.

PN2 = 1.20 atm and PO2 = 0.80 atm; they sum to 2.00 atm.

Unit check

Mole fractions are dimensionless ratios of moles, so mole fraction x atm gives atm — each partial pressure correctly carries the pressure unit, and their sum reproduces the given total.

What this means physically

Each gas contributes pressure in proportion to its share of the molecules, as if it alone occupied the container. The sum of the parts always rebuilds the total — the same idea behind expressing air's composition as 78% nitrogen and 21% oxygen by volume.

Common mistake

Students add the partial pressures from nowhere or assign pressure by mass instead of by moles. Dalton's law runs on mole fractions, not mass fractions — 0.60 mol vs 0.40 mol is what matters, not the grams.