MCAT content library / Chemistry & physics / Gas mixtures: calculate partial pressure from composition
Track gas amount, mole fraction and total pressure separately so compression, added gas and diffusion do not become the same problem.
About 13 minutes, including the worked examples and checks.
Useful foundations
On this page
- Concept: separate a gas’s amount from its share of the mixture
- Pattern: count moles before assigning pressure
- Rule: identify what stays fixed during a change
- Application: distinguish wet and dry composition
- Application: compare the partial pressure of the same gas
- Check your reasoning
- Continue learning
- Sources
Concept: separate a gas’s amount from its share of the mixture
A mixture can have the same oxygen percentage at two different oxygen partial pressures. It can also have a lower oxygen percentage without losing any oxygen or lowering its partial pressure. To decide which has occurred, track the amount of each gas and the conditions that set pressure. This is the composition bookkeeping within AAMC Category 4B.1
For an ideal gas mixture, pi = niRT/V is the partial pressure of species i: the pressure that its amount would exert alone at the mixture’s temperature and full volume. Every species has access to that same V. Dalton’s law gives Ptotal = Σpi. These are absolute pressures, not gauge readings relative to the room.2
The mole fraction xi = ni/ntotal has no unit. Dividing the individual gas equation by the total gas equation gives pi = xiPtotal. Mole fractions sum to 1, and partial pressures sum to the measured total. A percentage must be converted to a fraction:30% means 0.30.
Pattern: count moles before assigning pressure
The original illustrative vessel below contains a dry, nonreacting mixture with total pressure 200 kPa at a uniform temperature. Assume ideal behavior and no condensation. Its amounts sum to 1.00 mol.
| Species | Amount (mol) | Mole fraction | Partial pressure (kPa) |
|---|---|---|---|
| N₂ | 0.60 | 0.60 | 120 |
| O₂ | 0.30 | 0.30 | 60 |
| He | 0.10 | 0.10 | 20 |
| Total | 1.00 | 1.00 | 200 |
The oxygen fraction is 0.30/1.00 =0.30, so its partial pressure is 0.30×200 =60 kPa. Nitrogen and helium contribute 120 and 20 kPa. They do not compress the oxygen into 30% of the container. Using 0.30V as oxygen’s volume would count its composition effect a second time and give the wrong pressure.
If composition is supplied by mass, convert each mass using its own molar mass first. For example, a separate mixture containing 8.0 g He and 28 g N₂ contains 2.0 mol He and 1.0 mol N₂, using molar masses 4.0 and 28 g/mol. Helium supplies two thirds of the moles, despite supplying only 8/36, or about 22%, of the mass. A mass fraction is not a pressure fraction.2
Rule: identify what stays fixed during a change
Use the original 1.00 mol mixture as the starting point for each independent change below. Temperature remains constant, and none of the gases reacts. V₀ denotes its original volume. The same oxygen amount can give different pressures if the boundary permits a different volume.3
| Change | Final total P (kPa) | Final O₂ fraction | Final pO₂ (kPa) |
|---|---|---|---|
| Compress the closed mixture to V₀/2 | 400 | 0.30 | 120 |
| Add 0.50 mol He; keep V₀ fixed | 300 | 0.20 | 60 |
| Add 0.50 mol He; keep total P at 200 kPa | 200 | 0.20 | 40 |
Compression changes pressure without changing composition
Halving the volume of the closed mixture doubles every ni/V. Total pressure doubles to 400 kPa, and oxygen partial pressure doubles to 120 kPa. No gas was added or removed, so each mole fraction stays the same. This requires constant temperature; an unqualified rapid compression need not preserve it.
Adding another gas does not always lower an existing partial pressure
With the vessel rigid, adding 0.50 mol He raises the total amount to 1.50 mol and total pressure to 300 kPa. Oxygen’s mole fraction falls to 0.30/1.50 =0.20. Yet its partial pressure remains 0.20×300 =60 kPa, because oxygen amount, temperature and volume are unchanged. Its share of the total has changed; its amount per volume has not.
With a movable boundary holding total pressure at 200 kPa, the same added helium makes the vessel expand to 1.50V₀. Oxygen’s fraction is again 0.20, but its partial pressure is now 40 kPa. In this case, oxygen amount per volume decreases. A statement such as “adding an inert gas dilutes oxygen” needs to specify whether dilution means a lower mole fraction, a lower concentration, or both.
Application: distinguish wet and dry composition
A pressure measurement includes gaseous water when water vapor is present. Suppose a different final mixture has measured total pressure 100 kPa and a supplied water-vapor partial pressure of 6.0 kPa. Its nonwater gases therefore contribute 94 kPa. If oxygen constitutes 25% of the dry gas moles, then pO₂ =0.25×94 =23.5 kPa, not 25 kPa.2
The dry fraction has a denominator that excludes water; the total pressure has not excluded it. Align those denominators before multiplying. If a gas is saturated over liquid water, the water-vapor pressure depends on temperature. Do not reuse one supplied vapor pressure after a temperature change or treat condensation as conserving the number of gaseous water molecules.
Application: compare the partial pressure of the same gas
For passive diffusion between regions at the same temperature, compare each gas’s own partial pressure. Consider a membrane that permits oxygen diffusion but prevents bulk gas flow. On its left, total pressure is 200 kPa and oxygen fraction 0.20, giving pO₂ =40 kPa. On its right, total pressure is 150 kPa and oxygen fraction 0.40, giving pO₂ =60 kPa. The initial net oxygen movement is therefore right to left, despite the lower total pressure on the right.4
This predicts a direction under the stated model, not a numerical transport rate. Membrane properties and the size of the gradient also matter for rate. If several gases can diffuse, each can have its own direction; following total pressure alone misses that possibility. The physiology of ventilation and blood transport adds further mechanisms beyond this gas-mixture calculation.
For a new mixture problem, identify the composition basis, obtain moles or mole fractions, and use an absolute total pressure. For a change, list ni, T and V before deciding what happens to pi. If reaction, leakage, condensation or appreciable nonideal behavior occurs, revise the simple model instead of carrying the initial fractions forward unchanged.
ACTIVE PRACTICE · GAS LAW REASONING
Track Oxygen Partial Pressure, Not Just Mole Fraction
Start with the original ideal dry mixture at fixed temperature: 0.60 mol N₂, 0.30 mol O₂ and 0.10 mol He in volume V₀ at total pressure 200 kPa. Each scenario starts independently from that initial state.
At V₀/2 and fixed T: total pressure 400 kPa; oxygen mole fraction 0.30; oxygen partial pressure 120 kPa.
Use absolute pressures. Mole fraction describes composition; partial pressure also depends on the mixture’s total pressure. The wet-gas transfer has a different denominator and is not part of the initial dry vessel.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
An ideal dry mixture contains 6.0 g He and 14 g N₂ at total pressure 160 kPa. Use molar masses 4.0 g/mol for He and 28 g/mol for N₂. What is the helium partial pressure?
A. 40 kPa
Review this choice. This is the nitrogen partial pressure: nitrogen has 0.50 of the 2.00 total moles.
B. 80 kPa
Review this choice. There are two species, but they do not have equal amounts. Count their moles instead of assigning half the pressure to each.
C. 120 kPa
Correct. Helium contributes 6/4=1.50 mol and nitrogen 14/28=0.50 mol. Helium’s fraction is 1.50/2.00=0.75, so pHe=0.75×160=120 kPa.
D. 160 kPa
Review this choice. The total pressure includes both gases. Helium supplies only three quarters of the gas moles.
The principle to carry forward
For an ideal mixture, pressure fractions follow mole fractions, not species counts or mass fractions.
Check 2
A rigid vessel initially contains 0.40 mol O₂ and 0.60 mol N₂ at total pressure 100 kPa. An additional 1.00 mol Ar is admitted without losing any original gas. Temperature and volume remain fixed, and all gases behave ideally. What are the final total pressure and oxygen partial pressure?
A. Total 100 kPa; oxygen 20 kPa.
Review this choice. Holding volume fixed while doubling total moles doubles total pressure. Constant total pressure would require a different boundary condition.
B. Total 100 kPa; oxygen 40 kPa.
Review this choice. The oxygen value is correct, but the added gas raises total pressure to 200 kPa.
C. Total 200 kPa; oxygen 20 kPa.
Review this choice. The oxygen mole fraction halves to 0.20, but total pressure doubles. Multiplying 0.20×200 gives 40 kPa.
D. Total 200 kPa; oxygen 40 kPa.
Correct. Total moles double from 1.00 to 2.00, so total pressure doubles. Oxygen amount, temperature and volume do not change, so its partial pressure remains 40 kPa.
The principle to carry forward
A decreasing fraction does not imply a decreasing partial pressure when the total pressure changes at the same time.
Check 3
Two chambers at the same temperature contain only O₂ and N₂. The left has total pressure 100 kPa and oxygen mole fraction 0.30. The right has total pressure 160 kPa and oxygen mole fraction 0.125. Their separating membrane permits passive diffusion of both species but prevents bulk flow. What are the initial net diffusion directions?
A. O₂ left to right; N₂ left to right.
Review this choice. Oxygen does move left to right, but nitrogen’s partial pressure is 70 kPa on the left and 140 kPa on the right.
B. O₂ right to left; N₂ right to left.
Review this choice. Nitrogen follows that direction, but oxygen’s partial pressures are 30 kPa left and 20 kPa right. Total pressure does not set both directions.
C. O₂ left to right; N₂ right to left.
Correct. For O₂,0.30×100=30 kPa and 0.125×160=20 kPa. For N₂, the remaining partial pressures are 70 and 140 kPa. Each species initially diffuses down its own gradient.
D. O₂ right to left; N₂ left to right.
Review this choice. Both directions are reversed relative to their species-specific partial-pressure gradients.
The principle to carry forward
Calculate the pressure of each species on both sides; different gases can have opposing diffusion directions.
Continue learning
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Sources
1. AAMC: C/P Content Category 4B
2. OpenStax Chemistry 2e: Gaseous Mixtures and Reactions