Dalton's Law of Partial Pressures
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You should be comfortable manipulating the ideal gas law \(PV=nRT\) to solve for any variable, and you should understand that the total pressure of a gas sample is an intensive property determined by the collective kinetic energy of its molecules.
Why this matters
Understanding how individual gas components contribute to total pressure is essential for predicting chemical behaviour in mixtures.
In the previous lecture, we treated gases as pure substances, calculating density or molar mass for a single species. However, most real-world gas systems—such as the atmosphere, the air in a scuba tank, or the headspace in a chemical reactor—contain mixtures of different gases. Dalton’s Law of Partial Pressures provides the framework for decomposing the total pressure of such a mixture into the contributions of each individual component. This concept is not merely an additive rule; it is a direct consequence of the kinetic molecular theory, which posits that gas particles do not interact significantly with one another. By mastering this law, you gain the ability to calculate the effective concentration of a specific gas in a mixture, which is a prerequisite for understanding gas-phase equilibrium, reaction rates, and the physiological effects of gas mixtures on biological systems.
The Concept of Partial Pressure
Imagine a sealed container holding a mixture of two different gases, say nitrogen and oxygen. If you were to remove all the oxygen molecules and leave only the nitrogen, the nitrogen molecules would continue to move with the same average speed and collide with the container walls with the same frequency and force as they did when the oxygen was present. The pressure exerted by the nitrogen alone in this hypothetical scenario is defined as its partial pressure. Dalton’s insight was that in a real mixture, the total pressure is simply the sum of these hypothetical individual pressures. This works because, in the ideal gas approximation, molecules of one gas do not "know" that molecules of another gas are present. They do not attract or repel each other in a way that alters their individual kinetic energy or collision frequency. Therefore, the total pressure is not a new, emergent property of the mixture, but rather a linear superposition of the pressures each gas would exert if it occupied the volume alone. This principle allows us to treat a complex mixture as a collection of independent gases, each contributing a specific fraction of the total force per unit area on the container walls.
Mathematical Formulation and Mole Fractions
Mathematically, Dalton’s Law states that the total pressure \(P_{\text{total}}\) of a mixture of non-reacting gases is the sum of the partial pressures \(P_i\) of the individual components. For a mixture containing gases A, B, and C, this is expressed as \(P_{\text{total}} = P_A + P_B + P_C\). To connect this to measurable quantities, we apply the ideal gas law to each component. The partial pressure of gas A is given by \(P_A = \frac{n_A RT}{V}\), where \(n_A\) is the number of moles of gas A, \(R\) is the universal gas constant, \(T\) is the absolute temperature, and \(V\) is the total volume of the container. Notice that \(R\), \(T\), and \(V\) are common to all components in the mixture. By factoring these out, we can relate the partial pressure to the mole fraction. The mole fraction \(x_A\) is defined as the ratio of the moles of component A to the total moles of all gases, \(x_A = \frac{n_A}{n_{\text{total}}}\). Substituting this into the pressure equation yields \(P_A = x_A \frac{n_{\text{total}} RT}{V}\). Since \(\frac{n_{\text{total}} RT}{V}\) is simply the total pressure \(P_{\text{total}}\), we arrive at the crucial relationship \(P_A = x_A P_{\text{total}}\). This shows that the partial pressure is directly proportional to the mole fraction. If a gas constitutes 20% of the molecules in the mixture, it exerts 20% of the total pressure. This linear relationship is a powerful tool for stoichiometry in gas-phase reactions, as it allows us to convert between pressure units and molar quantities without needing to know the volume or temperature explicitly, provided the total pressure is known.
Application to Gas Collection Over Water
A classic and practical application of Dalton’s Law is the collection of a gas by displacement of water. When a gas is generated in a laboratory reaction and collected in an inverted tube over water, the gas in the tube is not pure; it is saturated with water vapour. The total pressure measured by the barometer or manometer is the sum of the pressure of the dry gas and the vapour pressure of water at that specific temperature. The vapour pressure of water is a temperature-dependent constant; for example, at 25 °C, it is approximately 23.8 mmHg. If the total pressure in the collection tube is 760 mmHg, the partial pressure of the dry gas is not 760 mmHg, but rather \(760 - 23.8 = 736.2\) mmHg. Failing to account for this water vapour contribution leads to significant errors in calculating the moles of the target gas. This correction is essential in quantitative analysis, such as determining the molar mass of a volatile liquid or the yield of a gas-producing reaction. The principle extends to any situation where a gas is in contact with a volatile liquid, such as in humid air, where the partial pressure of water vapour affects the partial pressures of oxygen and nitrogen, thereby influencing their solubility in liquids and their biological availability.
Deviations from Ideal Behaviour
While Dalton’s Law is an excellent approximation for many gases at moderate pressures and temperatures, it is strictly an ideal law. Real gases exhibit intermolecular forces, such as London dispersion forces or dipole-dipole interactions. In a mixture, these forces can be different for like-like, like-unlike, and unlike-unlike interactions. If the attractive forces between molecules of different types are stronger than the average of the like-like interactions, the total pressure may be slightly less than the sum of the partial pressures predicted by the ideal law. Conversely, if repulsive forces dominate, the pressure may be higher. These deviations become significant at high pressures where the volume of the molecules themselves is no longer negligible compared to the container volume, or at low temperatures where kinetic energy is low and intermolecular attractions become more influential. For most introductory calculations involving gases at standard temperature and pressure (STP) or slightly above, these deviations are small enough to be ignored. However, in high-precision engineering or when dealing with highly polar gases like ammonia or water vapour at high densities, more complex equations of state, such as the van der Waals equation or the virial equation, must be used to accurately predict the behaviour of gas mixtures.
The result
The ideal gas law
Three separately discovered empirical proportionalities — Boyle's law (\(V\propto1/P\) at fixed \(n,T\)), Charles's law (\(V\propto T\) at fixed \(n,P\)), and Avogadro's law (\(V\propto n\) at fixed \(T,P\), gas-stoichiometry-avogadro) — combine into a single joint proportionality \(V\propto nT/P\), which becomes an equation once a proportionality constant \(R\) (the universal gas constant, empirically found to be the same for every gas in the ideal limit) is introduced: \(PV=nRT\).
A common misconception is that partial pressure is an intrinsic property of the gas molecule itself, independent of the mixture. In reality, partial pressure is a state function dependent on the total pressure and the mole fraction. If you compress a mixture, the total pressure increases, and consequently, the partial pressure of every component increases proportionally, even though the chemical identity of the molecules has not changed. This distinction is crucial when comparing the reactivity of a gas in a pure state versus in a mixture, as the effective concentration (partial pressure) determines the collision frequency with other reactants. C
Worked example
Problem. A mixture of methane (\(\mathrm{CH_4}\)) and ethane (\(\mathrm{C_2H_6}\)) is contained in a 10.0 L vessel at 300 K. The total pressure of the mixture is 2.50 atm. If the mixture contains 0.500 mol of methane, calculate the partial pressure of ethane and the number of moles of ethane present.
- Calculate the total number of moles in the mixture using the ideal gas law \(PV=nRT\). Using \(R = 0.08206\ \mathrm{L\cdot atm\cdot mol^{-1}\cdot K^{-1}}\), \(n_{\text{total}} = \frac{PV}{RT} = \frac{2.50 \times 10.0}{0.08206 \times 300} = \frac{25.0}{24.618} \approx 1.0155\ \mathrm{mol}\).
- Determine the number of moles of ethane by subtracting the moles of methane from the total moles: \(n_{\text{ethane}} = n_{\text{total}} - n_{\text{methane}} = 1.0155 - 0.500 = 0.5155\ \mathrm{mol}\).
- Calculate the mole fraction of ethane: \(x_{\text{ethane}} = \frac{n_{\text{ethane}}}{n_{\text{total}}} = \frac{0.5155}{1.0155} \approx 0.5076\).
- Calculate the partial pressure of ethane using Dalton’s Law: \(P_{\text{ethane}} = x_{\text{ethane}} \times P_{\text{total}} = 0.5076 \times 2.50\ \mathrm{atm} \approx 1.27\ \mathrm{atm}\).
- Verify the result by calculating the partial pressure of methane: \(P_{\text{methane}} = 2.50 - 1.27 = 1.23\ \mathrm{atm}\). Check if \(P_{\text{methane}} = \frac{n_{\text{methane}}RT}{V} = \frac{0.500 \times 0.08206 \times 300}{10.0} = 1.23\ \mathrm{atm}\). The values are consistent.
Units check. The pressure is in atmospheres, consistent with the input data, and the moles are dimensionless counts, consistent with the ideal gas law usage.
Three quick tests
- If the mole fraction of oxygen in air is 0.21 and the total atmospheric pressure is 1.00 atm, what is the partial pressure of oxygen?
Answer
0.21 atm - A gas mixture has a total pressure of 5.0 atm. If the partial pressure of component A is 1.5 atm, what is the mole fraction of component A?
Answer
0.30 - Why does the partial pressure of a gas in a mixture increase when the temperature is increased at constant volume?
Answer
Because the average kinetic energy of the molecules increases, leading to more frequent and forceful collisions with the container walls.
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Lecture notes checked line by line by the editor; every number in the worked example was recomputed.