Gas Mixtures and Dalton's Law

Introduction

A gas vessel can contain several species at once. A pressure gauge measures their combined pressure, while a reaction or measurement may depend on the contribution of just one gas. Dalton's law provides that contribution for an ideal mixture: calculate each gas's share by its mole fraction. This works because each ideal-gas component uses the same vessel volume and temperature.

Core explanation

Let a vessel contain gas amounts n₁, n₂ and so on, with total n total = Σnᵢ. For an ideal mixture at common temperature T and volume V, each component has partial pressure Pᵢ = nᵢRT/V. Summing yields P total = ΣPᵢ = n total RT/V. Divide the component expression by the total expression: Pᵢ/P total = nᵢ/n total = xᵢ, where xᵢ is gas i's mole fraction. Therefore Pᵢ = xᵢP total. The mole fractions sum to one, and partial pressures sum to total pressure. These two closure checks catch many arithmetic errors.

The gas amounts must be the final amounts after any reaction. If 0.20 mol H₂ and 0.10 mol O₂ react completely to water and the water condenses, the gas mixture is not the original two gases. With exactly the 2:1 stoichiometric ratio, both reactants can be consumed. The final gas amount depends on reaction extent and whether water remains vapour or liquid. Dalton's law describes a specified mixture state; it does not itself calculate chemical change.

Suppose 0.200 mol N₂ and 0.100 mol O₂ are placed in a 5.00 L vessel at 300 K and remain unreacted. The total amount is 0.300 mol. With R = 0.08206 L atm mol⁻¹ K⁻¹, P total = (0.300)(0.08206)(300)/5.00 = 1.48 atm. Mole fractions are xN2 = 2/3 and xO2 = 1/3, giving partial pressures about 0.985 atm and 0.492 atm. Their sum is 1.477 atm before rounding. One cannot divide total pressure equally between the two species just because there are two names on the vessel label; their mole amounts differ.

If total pressure and gas composition are measured, the relation can be reversed. For example, xO2 = 0.21 and total pressure 1.00 atm imply an oxygen partial pressure of 0.21 atm under the ideal-mixture approximation. In a wet gas, water vapour is another component and contributes its own partial pressure. If dry-gas fractions are reported, their denominator excludes water; multiplying them by the wet total pressure would be wrong without accounting for water.

Real mixtures can show nonideal interactions, especially at high pressure. Dalton's law is exact within the ideal-gas model and an approximation for real gases. It is also crucial to identify whether the problem asks for a gas-phase mole fraction, a dissolved-solution mole fraction, or a mass fraction; these are different ratios.

Step-by-step reasoning

1. Determine which gases are actually present at the state of interest, after any reaction or condensation. 2. Convert each gas quantity to moles and sum only gas-phase components for n total. 3. Compute each xᵢ = nᵢ/n total and check Σxᵢ = 1. 4. Find P total from a measurement or n totalRT/V; then calculate Pᵢ = xᵢP total. 5. Check ΣPᵢ = P total to the precision of the reported values.

Visual explanation

Draw a single box containing twice as many blue dots as red dots. The blue and red dots share the same walls and temperature. A horizontal pressure bar is split into blue two-thirds and red one-third; the two segments together equal the gauge pressure. The picture represents mole fractions, not different regions of volume occupied by each gas.

Real-world analogy

Imagine a shared monthly bill divided among roommates according to their recorded usage. The total bill is the sum of each person's contribution, and a roommate's fraction of use determines the same fraction of the bill. For ideal gases, mole fraction plays the role of usage fraction in dividing total pressure into partial pressures.

Real-world example

A gas analyser reports mole fractions of several components in a sealed container. If a process requires the partial pressure of a reactant gas, multiply its measured fraction by the vessel's absolute total pressure. The calculation is meaningful only for the gas-phase sample as measured; water vapour and any unmeasured carrier gas must be included in the total composition.

Why?

Why is Pᵢ proportional to nᵢ? In the ideal-gas model every component in the common volume V at the common T obeys PᵢV = nᵢRT. The factors RT/V are the same for all components, so the ratio of their pressure contributions equals the ratio of their mole amounts.

Common misconception

“Each gas fills only its own fraction of the vessel.” Ideal gases mix throughout the available volume, so each partial pressure is defined for the whole vessel volume. Another mistake is to let mole fractions sum to more than one by mixing dry-basis and wet-basis component amounts.

Worked example

A 10.0 L container at 298 K holds 0.300 mol He, 0.200 mol Ne and 0.100 mol Ar. The total is 0.600 mol, giving xHe = 0.500, xNe = 0.333 and xAr = 0.167. Using R = 0.08206 L atm mol⁻¹ K⁻¹, total pressure is (0.600)(0.08206)(298)/10.0 = 1.47 atm. Partial pressures are approximately 0.734 atm He, 0.489 atm Ne and 0.245 atm Ar. They add to about 1.47 atm within rounding. The gas identities affect molar masses but not these ideal partial-pressure shares when their mole amounts are given.

Quick check

1. A nonreacting ideal mixture contains 0.20 mol A and 0.80 mol B at 5.0 atm total. What is P A? Answer: xA = 0.20/(0.20 + 0.80) = 0.20, so P A = 0.20 × 5.0 = 1.0 atm.

Exam focus

Write the denominator n total explicitly and include only components in the requested gas phase. Use final mole amounts after reaction. On a wet-gas problem, state whether composition and pressure are on wet or dry bases before multiplying.

Advanced insight

For real mixtures, a component's chemical potential is often expressed through fugacity rather than partial pressure alone. Fugacity approaches partial pressure in the ideal-gas limit. This refinement explains why Dalton's simple sum remains an excellent introductory model but may need correction in compressed industrial mixtures.

Summary

For a nonreacting ideal mixture in one vessel, P total = ΣPᵢ and Pᵢ = xᵢP total, with xᵢ = nᵢ/n total. Find the actual final gas composition first, maintain a consistent wet or dry basis, and verify that both mole fractions and partial pressures close to their totals.

Practice questions

1. A mixture has 0.30 mol N₂, 0.10 mol O₂ and 0.10 mol Ar. What is xO2? Answer: n total = 0.50 mol and xO2 = 0.10/0.50 = 0.20. 2. If that mixture has total pressure 2.50 atm, find P N2 and P Ar. Answer: xN2 = 0.30/0.50 = 0.60, so P N2 = 1.50 atm; xAr = 0.20, so P Ar = 0.500 atm. 3. Why might multiplying a dry oxygen fraction by wet total pressure overestimate its partial pressure? Answer: The dry fraction excludes water from its denominator, while wet total pressure includes water vapour. The fractions must first be converted to the same composition basis.