Partial Pressures and Dalton’s Law

General Chemistry · Gas Laws and Ideal Gas Calculations

Dr. Ray’s Learning Center · covers: Partial Pressures and Dalton’s Law

20–30 min
6 figures
1 simulator
3 worked examples
1 practice type: partial pressures

Before this lesson:

The Ideal Gas Law

Related (optional):

Related: Molality and Mole Fraction

Air is not one gas. Every breath you take is a mixture of nitrogen, oxygen, argon, carbon dioxide and water vapor, and the pressure a barometer reads is the push of all of them together. A diver, a respiratory therapist and a chemist collecting a gas in a bottle over water all need the next question answered: how much of that pressure comes from one of the gases?

This lesson answers it with Dalton’s law of partial pressures. Each gas in a mixture pushes as if it were alone in the container, and the pushes simply add. You will split a total pressure with the mole fraction, $\chi$ , find a total pressure from the total moles with $PV = nRT$ , and correct a gas collected over water for the water vapor mixed into it.

Moles of each gasadd them for the total
divideby the total moles
Mole fractiona share, no unit
multiplyby the total pressure
Partial pressureof that gas

No total pressure given: use the total moles in the ideal gas law.Collected over water: subtract the vapor pressure of water.

Figure 1. From moles to a share of the pressure: total moles, mole fraction, then times the total pressure.

Learning Objectives

After this lesson, you should be able to:

Calculate partial pressures from mole fractions and total pressure.
Correct a gas collected over water for water vapor pressure.

Key Ideas

1. Dalton’s Law and the Mole Fraction

In a mixture of gases that do not react, the partial pressure of one gas is the pressure it would exert if it were alone in the same container at the same temperature. Dalton’s law says the total pressure is the sum of the partial pressures (add one term for each gas):

P_{\text{total}} = P_{\text{A}} + P_{\text{B}} + P_{\text{C}}

The mole fraction of gas A, written $\chi_{\text{A}}$ (the Greek letter chi), is its share of all the moles. It has no unit, it lies between 0 and 1, and the mole fractions of all the gases add up to 1:

\chi_{\text{A}} = \dfrac{n_{\text{A}}}{n_{\text{total}}}

Because every gas in the mixture has the same volume and temperature, the pressures split in the same proportion as the moles. So the partial pressure of a gas is its mole fraction times the total pressure (Figure 2):

P_{\text{A}} = \chi_{\text{A}}\,P_{\text{total}}

Read the same relation backwards when the pressures are given: $\chi_{\text{A}} = \dfrac{P_{\text{A}}}{P_{\text{total}}}$ .

Figure 2. Each gas gets the same share of the pressure as its share of the moles. In Worked Example 1, nitrogen has most of the moles, so it has most of the pressure.

Key idea: a partial pressure is a share of the total pressure, and the share is the mole fraction. Divide by the total moles, never by the moles of one other gas.

2. Each Gas Obeys the Ideal Gas Law

Each gas fills the whole container, so each one has the full volume V and the temperature T of the mixture. Its partial pressure follows from its own moles:

P_{\text{A}} = \dfrac{n_{\text{A}}RT}{V}

Add these for every gas and the total pressure follows from the total moles (Figure 3). You do not need the identity of the gases, only how many moles there are in all:

P_{\text{total}} = \dfrac{n_{\text{total}}RT}{V}

As in any ideal gas calculation, use $T$ in $\text{K}$ , and the value of R must match the pressure unit: 0.08206 for atm, 62.36 for torr, 8.314 for kPa (each in L, mol and K).

Figure 3. Each gas fills the whole flask and pushes as if it were alone. The total pressure follows from the total moles, and it is the sum of the partial pressures.

Key idea: the total pressure depends only on the total moles, the volume and the temperature.

3. Collecting a Gas over Water

A common way to collect a gas in the lab is to bubble it into an upside-down bottle filled with water. The gas pushes the water out, but it also picks up water vapor, so the bottle holds a mixture. When the water level inside the bottle is the same as the level outside, the pressure inside equals the atmospheric pressure (Figure 4). By Dalton’s law, the pressure of the dry gas is what is left after the water vapor is taken away:

P_{\text{gas}} = P_{\text{atm}}-P_{\text{H}_2\text{O}}

The vapor pressure of water depends only on the temperature. This lesson uses Whitten 9e Table 12-4, which lists it from 19 °C to 28 °C; for example 22.38 torr at 24 °C. Every problem states the vapor pressure it uses.

Figure 4. With the water levels equal, the gas in the bottle is at the atmospheric pressure. Part of that pressure is water vapor; the rest belongs to the collected gas.

Data note: Table 12-4 is systematically 0.04 % to 0.09 % below the NIST values; the difference is below 0.1 % and changes no 3 s.f. answer in this lesson.

Key idea: subtract the vapor pressure of water from the atmospheric pressure before you use PV = nRT for the collected gas.

Try it: mix gases and split the pressure

Add up to three gases to a flask, set the moles, the volume and the temperature, and watch each gas take its share of the pressure. Load the settings of Worked Example 1 or 2, then open the second tab to collect a gas over water and see how much of the bottle’s pressure is water vapor.

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Strategy: if the total pressure is given, use mole fractions; if the volume and temperature are given, use the total moles with PV = nRT; if the gas was collected over water, subtract the vapor pressure of water first.

Worked Example 1: Partial Pressures from Mole Fractions

Problem: A cylinder contains 1.20 mol of N₂, 0.300 mol of O₂ and 0.0500 mol of CO₂. The total pressure is 1.85 atm. Find the partial pressure of each gas.

Step 1. Add the moles of all the gases.

n_{\text{total}} = 1.20\,\text{mol} + 0.300\,\text{mol} + 0.0500\,\text{mol} = 1.55\,\text{mol}

Step 2. Find the mole fraction of each gas. mol cancels, so there is no unit:

\chi_{\mathrm{N_2}} = \dfrac{1.20\,\text{mol}}{1.55\,\text{mol}} \approx 0.7742

\chi_{\mathrm{O_2}} = \dfrac{0.300\,\text{mol}}{1.55\,\text{mol}} \approx 0.1935

\chi_{\mathrm{CO_2}} = \dfrac{0.0500\,\text{mol}}{1.55\,\text{mol}} \approx 0.03226

Step 3. Multiply each mole fraction by the total pressure.

\begin{aligned} P_{\mathrm{N_2}} &= 0.7742 \times 1.85\,\text{atm} \\ &\approx 1.432\,\text{atm} \approx 1.43\,\text{atm} \end{aligned}

\begin{aligned} P_{\mathrm{O_2}} &= 0.1935 \times 1.85\,\text{atm} \\ &\approx 0.3580\,\text{atm} \approx 0.358\,\text{atm} \end{aligned}

\begin{aligned} P_{\mathrm{CO_2}} &= 0.03226 \times 1.85\,\text{atm} \\ &\approx 0.05968\,\text{atm} \approx 0.0597\,\text{atm} \end{aligned}

Final Answer: N₂ 1.43 atm, O₂ 0.358 atm, CO₂ 0.0597 atm

Check: the partial pressures add up to the total pressure, as Dalton’s law requires.

\begin{aligned} 1.43\,\text{atm} + 0.358\,\text{atm} + 0.0597\,\text{atm} &= 1.8477\,\text{atm} \\ &\approx 1.85\,\text{atm} \end{aligned}

Moles of N₂ $1.20\,\text{mol}\ \mathrm{N_2}$ given
$\times \frac{1}{1.55\,\text{mol}}$ over the total moles
Mole fraction $0.7742$ 4 significant figures
$\times 1.85\,\text{atm}$ times the total pressure
Partial pressure $1.43\,\text{atm}$ Final answer to 3 significant figures

Figure 5. Worked Example 1 for nitrogen, one step at a time: its mole fraction, then its share of the total pressure. Oxygen and carbon dioxide follow the same two steps.

Rounded operands: Step 3 uses mole fractions rounded to 4 significant figures. The unrounded mole fractions give the same final answers.

Your turn: partial pressure or mole fraction

Same steps as Worked Example 1: total moles, mole fraction, then times the total pressure. If the pressures are given, divide one by the total to get the mole fraction.

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Worked Example 2: Total Pressure from the Total Moles

Problem: A 12.0 L flask contains 0.250 mol of CH₄ and 0.640 mol of O₂ at 27.0 °C. What is the total pressure in the flask, in atm?

Step 1. Convert the temperature to kelvin and add the moles.

T = (27.0 + 273.15)\,\text{K} = 300.15\,\text{K}

n_{\text{total}} = 0.250\,\text{mol} + 0.640\,\text{mol} = 0.890\,\text{mol}

Step 2. Use the total moles in the ideal gas law. With the pressure in atm, use R in L·atm:

\begin{aligned} P_{\text{total}} &= \dfrac{(0.890\,\text{mol})(0.08206\,\dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}})(300.15\,\text{K})}{12.0\,\text{L}} \\ &\approx 1.827\,\text{atm} \approx 1.83\,\text{atm} \end{aligned}

Final Answer: 1.83 atm

Check gas by gas: $P_{\mathrm{CH_4}} \approx 0.5131\,\text{atm}$ and $P_{\mathrm{O_2}} \approx 1.314\,\text{atm}$ , and their sum is about 1.827 atm, the same total. The mole fraction of methane is $\chi_{\mathrm{CH_4}} = \dfrac{0.250}{0.890} \approx 0.281$ , and $0.2809 \times 1.827\,\text{atm} \approx 0.513\,\text{atm}$ again.

Your turn: total pressure from the total moles

Same two steps as Worked Example 2: kelvin and total moles, then $P = \dfrac{nRT}{V}$ with the total moles.

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Worked Example 3: A Gas Collected over Water

Problem: A student collects 248 mL of O₂ over water at 24 °C. The water level inside the bottle is the same as outside, and the atmospheric pressure is 751 torr. The vapor pressure of water at 24 °C is 22.38 torr (Whitten 9e Table 12-4). How many moles of O₂ were collected, and what is their mass?

Step 1. Subtract the vapor pressure of water. The levels are equal, so the bottle is at 751 torr:

\begin{aligned} P_{\mathrm{O_2}} &= 751\,\text{torr}-22.38\,\text{torr} \\ &= 728.62\,\text{torr} \approx 728.6\,\text{torr} \end{aligned}

Step 2. Convert the temperature to kelvin and the volume to liters.

T = (24 + 273.15)\,\text{K} = 297.15\,\text{K}

V = 248\,\text{mL} \times \dfrac{1\,\text{L}}{1000\,\text{mL}} = 0.248\,\text{L}

Step 3. Solve PV = nRT for the moles. With the pressure in torr, use R in L·torr:

\begin{aligned} n_{\mathrm{O_2}} &= \dfrac{(728.6\,\text{torr})(0.248\,\text{L})}{(62.36\,\dfrac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}})(297.15\,\text{K})} \\ &\approx 0.009751\,\text{mol} \approx 0.00975\,\text{mol} \end{aligned}

Step 4. Convert the moles to grams.

\begin{aligned} m_{\mathrm{O_2}} &= 0.009751\,\text{mol} \times 32.00\,\text{g/mol} \\ &\approx 0.3120\,\text{g} \approx 0.312\,\text{g} \end{aligned}

Final Answer: 0.00975 mol of O₂, 0.312 g

Atmosphere $751\,\text{torr}$ levels equal
$-22.38\,\text{torr}$ water vapor at 24 °C
Dry O₂ $728.6\,\text{torr}$ 4 significant figures
$\times \frac{0.248\,\text{L}}{(62.36)(297.15\,\text{K})}$ n from PV over RT
Moles of O₂ $0.009751\,\text{mol}$ 4 significant figures
$\times \frac{32.00\,\text{g}}{1\,\text{mol}}$ mol to g
Mass of O₂ $0.312\,\text{g}$ Final answer to 3 significant figures

Figure 6. Worked Example 3 one step at a time: take away the water vapor first, then use the ideal gas law for the dry oxygen alone.

Water vapor is a small but real part of the gas in the bottle: $\chi_{\text{H}_2\text{O}} = \dfrac{22.38}{751} \approx 0.02980$ , about 3 % of the moles.

Rounded operands: Steps 3 and 4 carry 728.6 torr and 0.009751 mol, each rounded to 4 significant figures. The unrounded values give the same final answers.

Your turn: a gas collected over water

Same steps as Worked Example 3: subtract the stated vapor pressure, convert to kelvin and liters, then $n = \dfrac{PV}{RT}$ .

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Common Mistakes

Common mistake: Giving the total pressure as the partial pressure. In Worked Example 1, 1.85 atm is the push of all three gases together. The N₂ alone exerts only its share, about 1.43 atm.

Common mistake: Dividing by the moles of one other gas instead of the total. 1.20 mol of N₂ divided by 0.300 mol of O₂, times 1.85 atm, gives 7.40 atm, more than the total pressure, which is impossible. The mole fraction divides by the total, 1.55 mol.

Common mistake: Leaving a gas out of the total. In Worked Example 2, using the methane alone gives 0.513 atm, which is only its partial pressure. The total pressure needs the moles of every gas.

Common mistake: Forgetting the water vapor. In Worked Example 3, using 751 torr for the O₂ gives 0.0101 mol, too high. The bottle also holds water vapor: subtract 22.38 torr first.

Common mistake: Using the temperature in °C. Putting 24 instead of 297.15 into Worked Example 3 gives 0.121 mol, more than ten times too much. Gas laws need kelvin.

Exam tip: a partial pressure is never larger than the total pressure, and the partial pressures must add up to the total. A mole fraction lies between 0 and 1.

Key Takeaway

In a mixture of gases, each gas exerts its own partial pressure, and the partial pressures add up to the total. The share of each gas is its mole fraction: moles of that gas over the total moles. With the volume and temperature, the total pressure follows from the total moles. A gas collected over water is mixed with water vapor, so subtract the vapor pressure of water from the atmospheric pressure.

The relations to keep at hand:

P_{\text{total}} = P_{\text{A}} + P_{\text{B}} + P_{\text{C}}

P_{\text{A}} = \chi_{\text{A}}\,P_{\text{total}}

P_{\text{gas}} = P_{\text{atm}}-P_{\text{H}_2\text{O}}

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC Commission on Isotopic Abundances and Atomic Weights, standard atomic weights 2021 (abridged values), the same table as the other stoichiometry lessons. Gas constant: CODATA 2022 recommended values (exact R in SI units, rounded to 0.08206, 62.36 and 8.314 in the lesson). Vapor pressure of water: K. Whitten, R. Davis, M. Peck and G. Stanley, General Chemistry, 9th ed., Table 12-4, checked against the NIST Chemistry WebBook (water, saturation properties). Dalton’s law, mole fractions and collecting a gas over water: K. Whitten, R. Davis, M. Peck and G. Stanley, General Chemistry, 9th ed., section 12-11, and J. Moore, C. Stanitski and P. Jurs, Chemistry: The Molecular Science, 5th ed., section 8-6. The simulator and the practice problems are audited; see the lesson audit log.