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9476 / 2027
The Gaseous State

Lesson 3 of 3

Partial pressures

Use mole fractions and correct gas collected over water.

A-Level 9476 (2026-2027)

  • I can use Dalton's law in gas mixtures.Syllabus

    Syllabus 9476, 3(d). Use Dalton's law in gas mixtures.

Each gas contributes its own partial pressure

For an ideal mixture, partial pressure equals mole fraction times total pressure.

Dalton's law states that the total pressure of a mixture of non-reacting ideal gases is the sum of their partial pressures. The partial pressure of one component is the pressure it would exert alone at the same temperature and total volume. Since pi = niRT/V, pi = xiptotal, where xi = ni/ntotal.

Worked exampleSplit the total pressure

A mixture contains 0.300 mol N2 and 0.200 mol O2 at a total pressure of 250 kPa. Find the partial pressures.

Try it first. Then show the worked stepsHide the worked steps
  1. Total amount = 0.500 mol; mole fractions are 0.600 and 0.400.
  2. p(N2) = 0.600 × 250 = 150 kPa.
  3. p(O2) = 0.400 × 250 = 100 kPa; their sum checks against 250 kPa.
Answer

150 kPa nitrogen and 100 kPa oxygen. Use mole fractions, not mass fractions.

Worked exampleGas collected over water is not dry gas

185 cm3 of gas is collected over water at 298 K. Total pressure is 105 kPa and water-vapour pressure is 3.17 kPa. Find the amount of dry gas, assuming negligible dissolution.

Try it first. Then show the worked stepsHide the worked steps
  1. p(dry gas) = 105 - 3.17 = 101.83 kPa.
  2. Use p = 101830 Pa and V = 185 × 10-6 m3.
  3. n = 101830 × 185 × 10-6/(8.31 × 298).
Answer

n = 7.61 × 10-3 mol. Using the full pressure would incorrectly count water vapour as the collected gas.

In a practical determination, check for leaks, allow the sample to reach the measured temperature, and correct any pressure difference from unequal liquid levels. Water collection is unsuitable when appreciable gas dissolves or reacts with water. For a volatile-liquid molar-mass experiment, incomplete vaporisation, air remaining in the vessel or loss of vapour can invalidate the assumed gas amount.

Partial pressures also enter Kp expressions in equilibrium calculations. Adding an inert gas at fixed volume raises the total pressure but does not change the partial pressures of the existing ideal-gas components: each still has the same ni, T and V.

Check your understanding

A mixture of 0.100 mol helium and 0.300 mol argon has a total pressure of 200 kPa. What is the partial pressure of helium?

  1. 50 kPa, from its mole fraction of 0.25
  2. 20 kPa, from 0.100 x 200 kPa
  3. 100 kPa, half the total for each gas
Show the answer

50 kPa, from its mole fraction of 0.25

x(He) = 0.100 / 0.400 = 0.250. p(He) = x(He) x p(total) = 0.250 x 200 = 50 kPa.