Dalton's Law of Partial Pressures

A mixture of gases is really several independent gases sharing the same container. Each gas behaves as though the others were not there. It pushes on the walls with its own share of the total force. Dalton's Law of Partial Pressures formalizes this idea. It lets you break a total pressure into contributions from individual gases. Crucially, it also lets you correct for water vapor. This correction applies whenever a gas is collected by bubbling it through water. It builds directly on the Ideal Gas Law and on Gas Stoichiometry. It also sets up the molecular picture developed in the Kinetic-Molecular Theory of Gases.

Key takeaways

In a mixture, each gas exerts a partial pressure $P_i$ as if it alone occupied the container.
Dalton's Law: $P_{\text{total}} = P_1 + P_2 + \cdots + P_n$ .
Partial pressure relates to composition through the mole fraction: $P_i = \chi_i P_{\text{total}}$ .
Gases collected "over water" are wet — subtract the water's vapor pressure to get the dry gas pressure.
The moles of a component follow $n_i = \chi_i n_{\text{total}}$ , feeding straight back into gas stoichiometry.
Key idea: total pressure is just the sum of independent contributions, each proportional to how many moles of that gas are present.

The core insight

Every gas in a mixture acts independently: its partial pressure depends only on its own moles, the volume, and the temperature — never on what else is in the container.

The physical picture

Imagine three flasks of identical volume and temperature, one holding pure nitrogen, one pure oxygen, one pure carbon dioxide, each exerting some pressure on its own walls. Now imagine pouring all three gases into a single flask of that same volume, at that same temperature. The molecules do not fuse or interfere with each other's pressure contributions — they simply collide with the walls, all at once, and the total force per unit area is the sum of what each gas would have contributed alone. This is the content of Dalton's Law: total pressure is an additive quantity because, in an ideal gas, molecules of different identity do not experience special extra interactions with each other beyond ordinary collisions. This additivity is a direct consequence of the Ideal Gas Law applied separately to each component. Since $PV = nRT$ holds for each gas independently (ideal gases don't "feel" each other beyond elastic collisions), the pressures simply stack up. The mole fraction $\chi_i$ tells you what fraction of the total moles — and therefore the total pressure — belongs to each gas. A common and important application is collecting a gas by water displacement: gas bubbles up through water into an inverted, water-filled container. The gas that arrives is never pure — it carries water vapor with it, because the space above any liquid water is always partly filled with vapor at that liquid's equilibrium vapor pressure. The measured total pressure inside the collection vessel is the sum of the "dry" gas pressure and the vapor pressure of water at that temperature.

Figure

Three separate gases combined into one mixture; total pressure is the sum of the parts

Figure

Gas collected over water: total pressure equals dry gas plus water vapor pressure

The core relations

P_{\text{total}} = P_1 + P_2 + \cdots + P_n = \sum_i P_i

where:

$P_{\text{total}}$ is the total pressure measured on the container,
$P_i$ is the partial pressure of gas $i$ — the pressure it would exert alone in the same volume $V$ at the same temperature $T$ .

P_i = \chi_i \, P_{\text{total}}

where:

$\chi_i = n_i / n_{\text{total}}$ is the mole fraction of gas $i$ ,
$n_i$ is the moles of gas $i$ , $n_{\text{total}} = \sum_i n_i$ .

P_{\text{gas(dry)}} = P_{\text{total}} - P_{\text{H}_2\text{O}}

where:

$P_{\text{gas(dry)}}$ is the pressure due to the collected gas alone,
$P_{\text{H}_2\text{O}}$ is the equilibrium vapor pressure of water at the working temperature (read from a table), when gas is collected over water.

Derivation ADeriving Dalton's Law from the Ideal Gas Law

The goal is to show that the total pressure of an ideal-gas mixture is the sum of the partial pressures, and that each partial pressure is the mole fraction times the total pressure.

Step 1. Consider $n_1, n_2, \dots, n_n$ moles of $n$ different ideal gases, all placed in the same container of volume $V$ at temperature $T$ . Because ideal gas molecules occupy negligible volume and do not interact except through elastic collisions, each species obeys the Ideal Gas Law independently, as if the others were absent:

P_i = \frac{n_i R T}{V}

Step 2. The total pressure measured by a gauge on the container is generated by every collision from every molecule, regardless of species. Summing the individual contributions:

P_{\text{total}} = \sum_i P_i = \sum_i \frac{n_i R T}{V} = \frac{RT}{V}\sum_i n_i

Step 3. Let $n_{\text{total}} = \sum_i n_i$ be the total moles of gas present. Then

P_{\text{total}} = \frac{n_{\text{total}} R T}{V}

which is exactly the ideal gas law applied to the mixture as a whole — confirming that a gas mixture obeys $PV = nRT$ with $n$ taken as total moles.

Step 4. To connect partial pressure to composition, divide the expression for $P_i$ (Step 1) by the expression for $P_{\text{total}}$ (Step 3). The identical factors of $RT/V$ cancel:

\frac{P_i}{P_{\text{total}}} = \frac{n_i R T / V}{n_{\text{total}} R T / V} = \frac{n_i}{n_{\text{total}}} = \chi_i

Step 5. Rearranging gives the mole-fraction form used constantly in practice:

P_i = \chi_i P_{\text{total}}

This says a gas's share of the pressure equals its share of the moles — a direct bookkeeping consequence of ideality, with no fitting parameters.

Worked examples

Example 1 — Partial pressures in a mixed cylinder A 12.0 L cylinder at 305 K contains 0.420 mol N₂, 0.170 mol O₂, and 0.065 mol CO₂. Find each partial pressure and the total pressure.

Step 1. Total moles:

n_{\text{total}} = 0.420 + 0.170 + 0.065 = 0.655\ \text{mol}

Step 2. Total pressure via the mixture ideal gas law ( $R = 0.08206\ \text{L}\,\cdot\,\text{atm}\,\cdot\,\text{mol}^{-1}\text{K}^{-1}$ ):

P_{\text{total}} = \frac{n_{\text{total}} R T}{V} = \frac{(0.655)(0.08206)(305)}{12.0}

P_{\text{total}} = \frac{16.394}{12.0} = 1.366\ \text{atm}

Step 3. Mole fractions:

\chi_{\text{N}_2} = \frac{0.420}{0.655} = 0.6412, \qquad \chi_{\text{O}_2} = \frac{0.170}{0.655} = 0.2595, \qquad \chi_{\text{CO}_2} = \frac{0.065}{0.655} = 0.0992

Step 4. Partial pressures, $P_i = \chi_i P_{\text{total}}$ :

P_{\text{N}_2} = 0.6412 \times 1.366 = 0.876\ \text{atm}

P_{\text{O}_2} = 0.2595 \times 1.366 = 0.354\ \text{atm}

P_{\text{CO}_2} = 0.0992 \times 1.366 = 0.1355\ \text{atm}

Check. $0.876 + 0.354 + 0.1355 = 1.366$ atm, matching $P_{\text{total}}$ from Step 2 — the additivity check confirms internal consistency.

Limiting case. If CO₂ were removed entirely ( $n_{\text{CO}_2}\to 0$ ), $\chi_{\text{CO}_2}\to 0$ and $P_{\text{CO}_2}\to 0$ , while $P_{\text{total}}$ would drop to the sum of just the N₂ and O₂ terms — exactly what direct substitution predicts, since the formula contains no cross-terms between species.

Example 2 — Hydrogen gas collected over water Hydrogen gas is generated by reacting magnesium with hydrochloric acid and collected over water at 23.0 °C. The collection eudiometer reads a total pressure of 748.5 torr, and the volume of gas collected is 0.152 L. The vapor pressure of water at 23.0 °C is 21.1 torr. Find the moles of dry H₂ collected, and the mass of Mg consumed. Reaction: $\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$

Step 1. Subtract water vapor pressure to get the dry H₂ pressure:

P_{\text{H}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 748.5 - 21.1 = 727.4\ \text{torr}

Step 2. Convert to atm:

P_{\text{H}_2} = \frac{727.4}{760} = 0.9571\ \text{atm}

Step 3. Use the ideal gas law to find moles of H₂, with $T = 23.0 + 273.15 = 296.15$ K:

n_{\text{H}_2} = \frac{P V}{RT} = \frac{(0.9571)(0.152)}{(0.08206)(296.15)}

n_{\text{H}_2} = \frac{0.14548}{24.303} = 5.986\times10^{-3}\ \text{mol}

Step 4. Convert moles of H₂ to moles of Mg using the 1:1 stoichiometric ratio from the balanced equation (an application of Gas Stoichiometry):

n_{\text{Mg}} = n_{\text{H}_2} = 5.986\times10^{-3}\ \text{mol}

Step 5. Convert to mass using the molar mass of Mg (24.305 g/mol):

m_{\text{Mg}} = (5.986\times10^{-3}\ \text{mol})(24.305\ \text{g/mol}) = 0.1455\ \text{g}

Limiting case. If the gas had been collected dry (no water present, $P_{\text{H}_2\text{O}} = 0$ ), the correction step would vanish and $P_{\text{H}_2}$ would simply equal $P_{\text{total}}$ — showing the water correction is a strict subtraction, not a multiplicative factor, and disappears cleanly when there is no water vapor.

Limiting cases & common traps

Situation	What happens	Common mistake
Only one gas present	$\chi_i = 1$ , $P_i = P_{\text{total}}$	Forgetting mole fraction reduces to 1 in the single-gas limit
Very humid collection (high $T$ )	$P_{\text{H}_2\text{O}}$ large, big correction needed	Ignoring the water-vapor subtraction entirely
Equal moles of two gases	$\chi_1 = \chi_2 = 0.5$ , pressures split evenly	Assuming equal masses also split pressure evenly (it's moles, not mass, that matters)
Mole fraction given as percent	$\chi_i$ must be a decimal fraction of 1, not "50"	Plugging in 50 instead of 0.50 into $P_i = \chi_i P_{\text{total}}$

The chemist behind it

John Dalton was an English scientist working in Manchester in the early 1800s, chiefly remembered for proposing that matter is made of discrete atoms with characteristic masses. His interest in the atmosphere — he kept meteorological diaries for over 50 years — led him to study how different gases behave when mixed. Measuring the total pressure of humid air and its components, he arrived at the empirical law that now bears his name: that each gas in a mixture contributes to the total pressure independently of the others, in proportion to its own quantity. This was a strikingly modern piece of quantitative reasoning for its time, since it implicitly assumed gas particles were small, numerous, and largely non-interacting — an assumption that would later be formalized in the Kinetic-Molecular Theory of Gases.

Origins · 1 minJohn Dalton and how this was discovered

John Dalton

1766–1844 · Manchester, England, early 1800s — atomic theory and atmospheric chemistry

Dalton combined careful measurement of atmospheric gases with a bold theoretical leap: that all matter consists of atoms of definite, characteristic mass. His partial-pressure law and his atomic theory together supplied much of the conceptual scaffolding for quantitative chemistry in the century that followed.

Ask the tutor: If I mix two gases at different temperatures before they equilibrate, does Dalton's Law still let me add their original pressures directly?

Does Dalton's Law work for real (non-ideal) gases?

Only approximately. Dalton's Law assumes gas particles don't interact except through elastic collisions, which is the same assumption behind the Ideal Gas Law. At high pressure or low temperature, real gas molecules attract or repel each other and take up finite volume, so measured total pressures deviate from the simple sum of ideal partial pressures. For typical introductory problems — moderate pressure, moderate temperature — the law is an excellent approximation.

Why do we subtract water vapor pressure rather than some other correction?

Whenever a gas is collected by bubbling it through liquid water into an inverted container, the space above the water inside that container reaches equilibrium: water molecules continuously evaporate and condense until the vapor exerts its equilibrium vapor pressure at that temperature. That water vapor mixes with the collected gas exactly as any two gases would mix, so Dalton's Law applies directly, and the dry gas pressure is the total pressure minus the tabulated vapor pressure of water at the working temperature.

Can mole fractions add up to something other than 1?

No. Since $\chi_i = n_i/n_{\text{total}}$ and $n_{\text{total}}$ is defined as the sum of all $n_i$ , the mole fractions of every component in a mixture must sum to exactly 1. If a computed set of mole fractions does not sum to 1, some component's moles were miscounted or omitted.

How is Dalton's Law used in gas stoichiometry problems?

Once you know a species' partial pressure, volume, and temperature, you can find its moles from the ideal gas law and feed that number into a balanced equation, exactly as in ordinary Gas Stoichiometry. This is especially common for gases collected over water, where the "wet" total pressure must first be corrected to the dry partial pressure before any mole calculation is valid.

Is partial pressure a real, measurable pressure, or just a bookkeeping number?

It is a real, physically meaningful pressure — it is exactly the pressure that gas would exert if all other gases were removed from the container while volume and temperature stayed fixed. It is not directly measured by a single gauge in a mixture (the gauge reads only the total), but it can be calculated from mole fraction and total pressure, or measured directly by physically isolating that gas.