Mole Fraction and Partial Pressure

Introduction

Dalton's law says ideal-gas partial pressures add to the total. Mole fraction tells how that total is divided among components. If oxygen makes up one-quarter of the molecules in an ideal mixture, it contributes one-quarter of the total pressure at the common state. The relation is simple, but it depends on counting moles, not comparing gas masses or initial separate-container pressures.

Core explanation

For component i, define mole fraction xᵢ = nᵢ/nₜₒₜ. All components' mole fractions add to one because Σnᵢ/nₜₒₜ = 1. In an ideal gas mixture, Pᵢ = nᵢRT/V and Pₜₒₜ = nₜₒₜRT/V. Dividing gives Pᵢ/Pₜₒₜ = nᵢ/nₜₒₜ = xᵢ, so Pᵢ = xᵢPₜₒₜ. This result follows from the shared temperature and volume; it is not an independent claim about molecule mass.

Suppose a mixture has 2 mol nitrogen and 1 mol oxygen. Total amount is 3 mol, so xN₂ = 2/3 and xO₂ = 1/3. If total pressure is 300 kPa absolute, partial pressures are 200 and 100 kPa. They sum to the total. A quick check is that no component mole fraction can be negative or exceed one, and no partial pressure can exceed total pressure for a nonreacting ideal mixture of positive amounts.

Mole fraction differs from mass fraction. A heavy gas can contribute a large share of mixture mass but a small share of molecule count and pressure. If one mole helium and one mole oxygen are mixed, each has mole fraction one-half and contributes half of ideal total pressure, yet oxygen supplies much more than half the total mass. Converting a mass ratio directly into a pressure ratio without molar masses is therefore wrong.

If composition is given as mole percentages, divide by 100. A mixture stated as 30 mol% gas A and 70 mol% gas B has xA = 0.30 and xB = 0.70. The same numbers can be called volume percentages for ideal gases at common P and T, because volume is proportional to mole amount. At different conditions, compare moles rather than raw volumes.

If total pressure and one partial pressure are known, xᵢ = Pᵢ/Pₜₒₜ. This can infer composition under the ideal model. For example, a component at 60 kPa in a 240-kPa mixture has x = 0.25. If total moles are 4.0, its amount is 1.0 mol. This chain of reasoning is often useful in gas collection or reaction mixture problems.

Adding a component at fixed volume and temperature changes both total pressure and mole fractions. Existing component partial pressures stay fixed if their nᵢ stays fixed, but their mole fractions decrease because nₜₒₜ grows. This distinction can seem surprising. For example, adding helium to a rigid vessel does not reduce the ideal oxygen partial pressure, but oxygen becomes a smaller fraction of the total pressure. If instead the mixture expands at fixed total pressure, component partial pressures can change.

The ideal relation may need modification for strongly nonideal mixtures, and measured gas pressure can include water vapour. In a wet gas sample, water is itself a component with a mole fraction and partial pressure. If a question seeks dry gas composition, remove or correct the water contribution rather than using the wet total as though all of it came from dry gases.

Step-by-step reasoning

1. Convert given masses to moles if needed; do not use mass percentages as mole fractions directly. 2. Add all component mole amounts to get nₜₒₜ. 3. Compute xᵢ = nᵢ/nₜₒₜ and check that the fractions sum to one. 4. Multiply each xᵢ by the common total absolute pressure. 5. Add resulting partial pressures to check the total and consider water vapour or reactions.

Visual explanation

Draw one mixture box with six blue dots and three red dots. A nearby circle chart shows blue 2/3 and red 1/3 by number of dots. Beside it draw a 300-kPa pressure bar divided into 200 and 100 kPa. Do not size dots by mass; the pressure shares reflect counts.

Real-world analogy

A class vote share equals votes for one choice divided by all votes. Multiplying that share by total votes recovers the choice's count. Mole fraction similarly partitions ideal total pressure through particle counts. Votes have no molecular motion, so the analogy is only arithmetic.

Real-world example

Atmospheric gas composition is often reported as mole or volume percent for dry air. At a known total pressure, a component's ideal partial pressure can be estimated by multiplying its mole fraction by total pressure. Humidity changes the mixture because water vapour adds its own partial pressure and alters dry-component fractions.

Why?

Why is a component's ideal pressure share equal to its mole share? At common T and V, every mole contributes the same RT/V to pressure. A component with a fraction xᵢ of total moles therefore contributes the same fraction of total ideal pressure.

Common misconception

“The heaviest gas in a mixture always has the largest partial pressure.” Partial pressure depends on mole count at common T and V, not molecular mass. One mole of helium and one mole of xenon contribute equal ideal partial pressures even though their masses differ greatly.

Worked example

A gas mixture contains 0.60 mol N₂, 0.30 mol O₂ and 0.10 mol Ar at 500 kPa absolute. Total amount is 1.00 mol. Mole fractions are 0.60, 0.30 and 0.10. Partial pressures are PN₂ = 300 kPa, PO₂ = 150 kPa and PAr = 50 kPa. Their sum is 500 kPa. If the listed mole amounts were instead masses in grams, this calculation would be invalid until each mass was divided by its molar mass.

Quick check

1. A gas has mole fraction 0.25 in an ideal mixture at 200 kPa total. What is its partial pressure? Answer: 0.25 × 200 = 50 kPa absolute.

Exam focus

Write xᵢ = nᵢ/nₜₒₜ and Pᵢ = xᵢPₜₒₜ. Convert masses to moles first, sum fractions to one and sum partial pressures to total. State the ideal-mixture and shared-state assumptions.

Advanced insight

For ideal gases, mole fraction equals volume fraction only when volumes are interpreted at the same P and T. In real mixture thermodynamics, fugacity replaces simple partial pressure for accurate equilibrium calculations at higher pressures; xᵢP remains the useful low-pressure limit.

Summary

Mole fraction is a component's share of total moles. For an ideal gas mixture at common V and T, that equals its share of total pressure: Pᵢ = xᵢPₜₒₜ. Mass fraction is different, and wet or reacting mixtures require careful definition of components.

Practice questions

1. A mixture has 2 mol A and 3 mol B at 250 kPa. Find PA. Answer: xA = 2/5 = 0.40, so PA = 0.40×250 = 100 kPa. 2. Why can one mole helium and one mole oxygen have equal partial pressures but unequal mass fractions? Answer: Their molecule counts are equal, so ideal pressure shares match, but oxygen molecules have greater molar mass. 3. If a component's partial pressure is 80 kPa in a 320-kPa mixture, what is its mole fraction? Answer: x = 80/320 = 0.25 under the ideal-mixture model.