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Dalton’s Law and Partial Pressures

Separate a total pressure into component pressures before calculating.

  • GCE A-Level H2 Chemistry 9476-2027

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Dalton’s law questions are “mixture” questions: break the total pressure into components (or remove water vapour), then only then do any pV = nRT calculation.

Definitions (Must Know)

A. Dalton’s law of partial pressures

The total pressure of a mixture of gases equals the sum of the partial pressures: pₜₒₜₐₗ = p₁ + p₂ + …

B. Partial pressure, pᵢ

The partial pressure pᵢ is the pressure gas i would exert if it alone occupied the container at the same temperature and volume.

For an ideal gas mixture: pᵢ = xᵢ pₜₒₜₐₗ

C. Mole fraction, xᵢ

The mole fraction is: xᵢ = nᵢ/nₜₒₜₐₗ

D. Dry gas pressure (gas collected over water)

If a gas is collected over water, the measured pressure includes water vapour.

If the water vapour pressure is provided, the dry-gas pressure is: p_(dry gas) = pₜₒₜₐₗ - p_H2O

Gas collection over water setupA delivery tube enters beneath the open end of an inverted measuring cylinder in a water trough. Bubbles rise into a gas space containing dry gas and water vapour.Inverted measuring cylinderWater troughCollected gasdry gas +water vapourwater levelDelivery tube enters below the open end;gas bubbles rise and displace water.Pressure bookkeepingp(total) = p(dry gas) + p(H₂O)Use p_dry gas in pV = nRT.Subtract water vapour pressure first.
The measured pressure contains both dry gas and water vapour. Subtract the water-vapour pressure before applying the ideal gas equation.

Key Ideas (What Earns Marks)

  • xᵢ is between 0 and 1, and ∑ xᵢ = 1.
  • For ideal gases at the same T and p, mole fraction = volume fraction (so % by volume works like % by moles).
  • Partial pressure problems are usually one workflow:
    1. find xᵢ
    2. use pᵢ = xᵢ pₜₒₜₐₗ
  • For “over water” questions, subtract p_H2O first, then use pV = nRT for the dry gas.
Quick Recall
  • Total pressure is shared: every gas contributes part of pₜₒₜₐₗ.
  • In an ideal mixture, the fraction of moles equals the fraction of pressure: xᵢ = pᵢ/pₜₒₜₐₗ.
  • If “collected over water” is mentioned, the gas is mixed with water vapour.

Detailed Explanations

A. Why pᵢ = xᵢ pₜₒₜₐₗ is true (one-line derivation)

For an ideal mixture at the same T and V: pᵢV = nᵢRT and pₜₒₜₐₗV = nₜₒₜₐₗRT

Divide the first by the second: pᵢ/pₜₒₜₐₗ = nᵢ/nₜₒₜₐₗ = xᵢ

So pᵢ = xᵢ pₜₒₜₐₗ.

The gauge below measures the pressure each gas exerts on the walls separately, so you can test pₜₒₜₐₗ = p_N₂ + p_He and pᵢ = xᵢ pₜₒₜₐₗ.

0.8 mol of nitrogen and 0.2 mol of helium in 25 cubic decimetres at 300 K. The gauge is taking its first reading.

p(N₂)
— kPa
p(He)
— kPa
Total pressure
— kPa
Mole fraction of N₂
0.80
Gas
mol
mol
dm³

Drag the piston or use this slider.

K

Try this

0 of 5 done

  1. Halve the volume of the gas at the same temperature and let the pressure settle. (not done yet)

  2. Double the absolute temperature at the same volume and let the pressure settle. (not done yet)

  3. Mix helium with the nitrogen, then change the amount of helium. Compare the partial pressures with the total. (not done yet)

  4. Measure an unknown gas at two volumes or temperatures and work out its Mr from each reading. (not done yet)

  5. Compare the real-gas curves at 330 K or below and at 800 K or above. (not done yet)

B. Workflow: partial pressure from composition

  1. Convert any % to a fraction (e.g. 40% → 0.40).
  2. Compute xᵢ from moles (or volume fraction if ideal).
  3. Multiply by total pressure: pᵢ = xᵢ pₜₒₜₐₗ.

Mini example: 20% O₂ in 100 kPa total pressure → p_O₂ = 0.20 × 100 = 20 kPa.

Check the sum after calculating each component: in this example, 20 + 80 = 100 kPa = pₜₒₜₐₗ.

C. Workflow: composition from pressures

If partial pressures are given: xᵢ = pᵢ/pₜₒₜₐₗ

Then convert xᵢ to a percentage if asked.

D. Gas collected over water

When a gas is collected over water, the gas becomes mixed with water vapour. Dalton’s law applies:

pₜₒₜₐₗ = p_(dry gas) + p_H2O

  1. Find p_(dry gas) = pₜₒₜₐₗ - p_H2O (only if p_H2O is given).
  2. Use p_(dry gas)V = nRT to find n.

Mini example: pₜₒₜₐₗ = 101 kPa and p_H2O = 3.17 kPa → p_dry = 97.8 kPa.

The pressure check is 97.8 + 3.17 = 100.97 kPa, which rounds to the measured 101 kPa.

Worked Examples

Modelled example 1

Find partial pressure from moles

Core

Problem

A mixture contains 0.20 mol O₂ and 0.80 mol N₂ at 100 kPa. Find p_O₂.
Study the worked solution

  1. Find mole fraction

    Method

    Divide oxygen moles by total moles.

    Reason

    Mole fraction is the component’s share of all gas particles.

    Working

    x_O₂ = 0.20/(0.20 + 0.80) = 0.20.

  2. Scale total pressure

    Method

    Multiply by total pressure.

    Reason

    For an ideal mixture, pᵢ = xᵢpₜₒₜₐₗ.

    Working

    p_O₂ = (0.20)(100) = 20.0 kPa.

Guided practice 2

Recover mole and volume fraction

About 5 min

Problem

A helium–argon mixture is at 150 kPa; helium contributes 45 kPa. Find helium mole fraction and percentage by volume.

Try this before viewing the solution

Hints

Hint 1: pressure-ratio
Use xᵢ = pᵢ/pₜₒₜₐₗ.
Hint 2: ideal-mixture
For ideal gases, volume fraction equals mole fraction.
View solution step by step

  1. Find mole fraction

    Method

    Take the partial-to-total pressure ratio.

    Reason

    Dalton’s law links the ratio directly to mole fraction.

    Working

    x_He = 45/150 = 0.300.

  2. Convert to percentage

    Method

    Multiply the ideal-gas volume fraction by 100.

    Reason

    Volume fraction equals mole fraction at common conditions.

    Working

    30.0% by volume.

Common misconception 3

Add, do not average, partial pressures

Find and correct the mistake

Learner attempt

A mixture has component partial pressures 30 kPa and 70 kPa. A learner averages them to obtain 50 kPa total pressure. Correct the method.

Try this before viewing the solution

Required operation

View solution step by step

  1. State Dalton's law

    Method

    Sum every component partial pressure.

    Reason

    Each gas contributes independently to wall collisions.

    Working

    pₜₒₜₐₗ = ∑ pᵢ.

  2. Correct the result

    Method

    Add 30 and 70.

    Reason

    No division by the number of gases is part of the law.

    Working

    pₜₒₜₐₗ = 100 kPa.

Examiner practice 4

Correct gas collected over water

4 marks

Problem

H₂ is collected over water at 298 K. Total pressure is 101 kPa, water vapour pressure is 3.17 kPa and volume is 0.480 dm³. Calculate dry H₂ amount. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Correct pressure

    2 marks

    Method

    Subtract water’s partial pressure.

    Reason

    The measured total contains hydrogen plus water vapour.

    Working

    p_H2 = 101-3.17 = 97.8 kPa.

  2. Calculate amount

    2 marks

    Method

    Use the dry-gas pressure in pV = nRT.

    Reason

    Only hydrogen’s partial pressure belongs in its mole calculation.

    Working

    n = (97.8)(0.480)/(8.31)(298) = 1.90 × 10⁻² mol.

Challenge 5

Use percentage composition

Minimal support

Problem

An ideal gas mixture is 12.0% carbon dioxide by volume at total pressure 240 kPa. Calculate the carbon-dioxide partial pressure.

Try this before viewing the solution

Hints

Hint 1: fraction
Convert 12.0% to 0.120.
Hint 2: ideal
Volume fraction equals mole fraction.
View solution step by step

  1. Translate representation

    Method

    Use x_CO2 = 0.120.

    Reason

    Ideal-gas volume percentage gives mole fraction.

    Working

    12.0% = 0.120.

  2. Find contribution

    Method

    Multiply by total pressure.

    Reason

    pᵢ = xᵢpₜₒₜₐₗ.

    Working

    p_CO2 = (0.120)(240) = 28.8 kPa.

Common Mistakes

  • Using % as-is instead of converting to a fraction.
  • Using the total pressure instead of the dry-gas pressure for “over water” questions.
  • Forgetting that all the xᵢ values must add to 1.
  • Mixing pressure/volume units and using an inconsistent R.

Exam Tips

  • Keep units consistent and choose R to match (see Ideal Gas Model and pV = nRT).
  • When pressure is high enough for non-ideal behaviour to matter, compare with Real Gases and Deviations before final interpretation.
  • When asked for a partial pressure, write the final value with units (kPa/Pa).
  • If a question says “gas collected over water” but gives no water vapour pressure, don’t invent one (it should be provided).

Mind Stretchers

Mind stretcher 1Extension

A cylinder at 300 K contains O₂ and N₂ in a total volume of 2.00 dm³. The total pressure is 250 kPa. If the cylinder contains 0.0500 mol of O₂, calculate the moles of N₂ present (assume ideal behaviour).

Show Hint

Use the gas equation for total moles in the cylinder, then subtract the known oxygen amount.

Show Answer

Mark scheme:

  • Total moles: nₜₒₜₐₗ = pV/RT = (250 × 2.00)/(8.31 × 300) = 0.201 mol.
  • n(N₂) = nₜₒₜₐₗ - n(O₂) = 0.201 - 0.0500 = 0.151 mol.

Mind stretcher 2Extension

A sample of calcium carbonate reacts with excess hydrochloric acid and the CO₂ produced is collected over water at 298 K. The total pressure is 101 kPa and the water vapour pressure is 3.17 kPa (given). The volume of gas collected is 0.360 dm³. Calculate the mass of CaCO₃ that reacted.

Show Hint

Remove water vapour from the measured pressure before finding carbon dioxide moles and applying the reaction ratio.

Show Answer

Mark scheme:

  • Equation: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
  • p_dry = 101 - 3.17 = 97.8 kPa.
  • n(CO₂) = pV/RT = (97.8 × 0.360)/(8.31 × 298) = 1.42 × 10⁻² mol.
  • From the equation, n(CaCO₃) = n(CO₂).
  • Mass of CaCO₃ = nMᵣ = (1.42 × 10⁻²)(100.1) = 1.42 g (3 s.f.).
Syllabus and review details

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