Dalton’s Law and Partial Pressures
Learn and apply Dalton’s Law and Partial Pressures in the published Chemistry course sequence.
Continue where you stopped
The core idea
On this page
Dalton’s Law and Partial Pressures: Orientation
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
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.
B. Workflow: partial pressure from composition
- Convert any % to a fraction (e.g. 40% → 0.40).
- Compute xᵢ from moles (or volume fraction if ideal).
- 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
- Find p_(dry gas) = pₜₒₜₐₗ - p_H2O (only if p_H2O is given).
- 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
Problem
Study the worked solution
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.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
Problem
Try this before viewing the solution
Hints
Hint 1: pressure-ratio
Hint 2: ideal-mixture
View solution step by step
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.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
Learner attempt
Try this before viewing the solution
View solution step by step
State Dalton's law
Method
Sum every component partial pressure.Reason
Each gas contributes independently to wall collisions.Working
pₜₒₜₐₗ = ∑ pᵢ.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
Problem
Try this before viewing the solution
View solution step by step
Correct pressure
2 marksMethod
Subtract water’s partial pressure.Reason
The measured total contains hydrogen plus water vapour.Working
p_H2 = 101-3.17 = 97.8 kPa.Calculate amount
2 marksMethod
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.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit the pressure correction separately from the gas calculation.
Challenge 5
Use percentage composition
Problem
Try this before viewing the solution
Hints
Hint 1: fraction
Hint 2: ideal
View solution step by step
Translate representation
Method
Use x_CO2 = 0.120.Reason
Ideal-gas volume percentage gives mole fraction.Working
12.0% = 0.120.Find contribution
Method
Multiply by total pressure.Reason
pᵢ = xᵢpₜₒₜₐₗ.Working
p_CO2 = (0.120)(240) = 28.8 kPa.
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.).