Dalton's law of partial pressure

Definition of partial pressure and using Dalton's law of partial pressures.

Key points

• The pressure exerted by an individual gas in a mixture is known as its partial pressure.
• Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
• Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:
$\text {P}_{\text{Total}} = \text P_{\text {gas 1}} + \text P_{\text {gas 2}} + \text P_{\text {gas 3}} ...$
• Dalton's law can also be expressed using the mole fraction of a gas, $x$:
$\text P_{\text {gas 1}} = x_1 \text {P}_{\text{Total}}$

Introduction

Picture of the pressure gauge on a bicycle pump.
The pressure gauge on this bicycle pump measures the pressure of the air inside the tire in units of pounds per square inch. Photo by Andreas Kambanis from flickr, CC BY 2.0
In day-to-day life, we measure gas pressure when we use a barometer to check the atmospheric pressure outside or a tire gauge to measure the pressure in a bike tube. When we do this, we are measuring a macroscopic physical property of a large number of gas molecules that are invisible to the naked eye. On the molecular level, the pressure we are measuring comes from the force of individual gas molecules colliding with other objects, such as the walls of their container.
Let's take a closer look at pressure from a molecular perspective and learn how Dalton's Law helps us calculate total and partial pressures for mixtures of gases.

Ideal gases and partial pressure

In this article, we will be assuming the gases in our mixtures can be approximated as ideal gases. This assumption is generally reasonable as long as the temperature of the gas is not super low (close to $0\,\text K$), and the pressure is around $1\,\text {atm}$.
The exact temperature and pressure range under which a gas behaves ideally depends on the gas. However, the guidelines of ~$1\,\text{atm}$ and not temperatures close to $0\,\text K$ are true for most gases. That rule also covers most situations you might encounter in your chemistry class.
For a more detailed discussion, you can check out this article on non-ideal gas behavior.
This means we are making some assumptions about our gas molecules:
• We assume that the gas molecules take up no volume.
• We assume that the molecules have no intermolecular attractions, which means they act independently of other gas molecules.
Based on these assumptions, we can calculate the contribution of different gases in a mixture to the total pressure. We refer to the pressure exerted by a specific gas in a mixture as its partial pressure. The partial pressure of a gas can be calculated using the ideal gas law, which we will cover in the next section, as well as using Dalton's law of partial pressures.

Example 1: Calculating the partial pressure of a gas

Let's say we have a mixture of hydrogen gas, $\text H_2(g)$, and oxygen gas, $\text O_2(g)$. The mixture contains $6.7\,\text{mol}$ hydrogen gas and $3.3\,\text{mol}$ oxygen gas. The mixture is in a $300\,\text L$ container at $273\,\text K$, and the total pressure of the gas mixture is $0.75\,\text{atm}$.
The contribution of hydrogen gas to the total pressure is its partial pressure. Since the gas molecules in an ideal gas behave independently of other gases in the mixture, the partial pressure of hydrogen is the same pressure as if there were no other gases in the container. Therefore, if we want to know the partial pressure of hydrogen gas in the mixture, $\text P_{\text H_2}$, we can completely ignore the oxygen gas and use the ideal gas law:
$\text P_{\text H_2}\text V = \text {n}_{\text H_2}\text{RT}$
Rearranging the ideal gas equation to solve for $\text P_{\text H_2}$, we get:
\begin{aligned}\text P_{\text H_2} &= \dfrac{\text{n}_{\text H_2}\text{RT}}{\text V}\\ \\ &=\dfrac{(6.7\,\text {mol})(0.08206\,\dfrac{\text {atm} \cdot \text L} {\text {mol} \cdot \text K})(273\,\text K)}{300\,\text L}=0.50\,\text {atm}\end{aligned}
Thus, the ideal gas law tells us that the partial pressure of hydrogen in the mixture is $0.50\,\text {atm}$. We can also calculate the partial pressure of hydrogen in this problem using Dalton's law of partial pressures, which will be discussed in the next section.

Dalton's law of partial pressures

Dalton's law of partial pressures states that the total pressure of a mixture of gases is the sum of the partial pressures of its components:
$\text {P}_{\text{Total}} = \text P_{\text {gas 1}} + \text P_{\text {gas 2}} + \text P_{\text {gas 3}} ...$
where the partial pressure of each gas is the pressure that the gas would exert if it was the only gas in the container. That is because we assume there are no attractive forces between the gases.
From left to right: A container with oxygen gas at 159 mm Hg, plus an identically sized container with nitrogen gas at 593 mm Hg combined will give the same container with a mixture of both gases and a total pressure of 752 mm Hg.
The partial pressure of a gas in a mixture is the same as the pressure of the gas in the container by itself. The sum of the partial pressures gives the total pressure of the gas mixture. Image adapted from OpenStax, CC BY 3.0
Dalton's law of partial pressure can also be expressed in terms of the mole fraction of a gas in the mixture. The mole fraction of a gas is the number of moles of that gas divided by the total moles of gas in the mixture, and it is often abbreviated as $x$:
$x_1=\text{mole fraction of gas 1}=\dfrac{\text{moles of gas 1}}{\text {total moles of gas}}$
Dalton's law can be rearranged to give the partial pressure of gas 1 in a mixture in terms of the mole fraction of gas 1:
$\text P_{\text {gas 1}} = x_1 \text {P}_{\text{Total}}$
Both forms of Dalton's law are extremely useful in solving different kinds of problems including:
• Calculating the partial pressure of a gas when you know the mole ratio and total pressure
• Calculating moles of an individual gas if you know the partial pressure and total pressure
• Calculating the total pressure if you know the partial pressures of the components

Example 2: Calculating partial pressures and total pressure

Let's say that we have one container with $24.0\,\text L$ of nitrogen gas at $2.00 \,\text {atm}$, and another container with $12.0\,\text L$ of oxygen gas at $2.00\,\text {atm}$. The temperature of both gases is $273\,\text K$.
If both gases are mixed in a $10.0\,\text L$ container, what are the partial pressures of nitrogen and oxygen in the resulting mixture? What is the total pressure?

Step 1: Calculate moles of oxygen and nitrogen gas

Since we know $\text P$, $\text V$,and $\text T$ for each of the gases before they're combined, we can find the number of moles of nitrogen gas and oxygen gas using the ideal gas law:
$\text n = \dfrac{\text{PV}}{\text{RT}}$
Solving for nitrogen and oxygen, we get:
$\text n_{\text{N}_2} = \dfrac{(2\,\text {atm}) (24.0\,\text {L})} {(0.08206\,\dfrac{\text {atm} \cdot \text L} {\text{mol} \cdot \text K})(273 \text K)} = 2.14\,\text {mol nitrogen}$
$\text n_{\text {O}_2} = \dfrac{(2\,\text {atm})(12.0\,\text {L})}{(0.08206 \,\dfrac{\text {atm} \cdot \text L}{\text {mol} \cdot \text K})(273 \,\text {K})} = 1.07\,\text{mol oxygen}$

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get $\text P_\text{Total}$

Once we know the number of moles for each gas in our mixture, we can now use the ideal gas law to find the partial pressure of each component in the $10.0\,\text L$ container:
$\text P = \dfrac{\text{nRT}}{\text V}$
$\text P_{\text{N}_2} =\dfrac{(2.14\,\text{mol})(0.08206\,\dfrac{\text{atm}\cdot \text L}{\text{mol} \cdot \text K})(273\,\text K)} {10\,\text L} = 4.79\,\text{atm}$
$\text P_{\text {O}_2}=\dfrac{(1.07\,\text{mol})(0.08206\,\dfrac{\text{atm}\cdot \text L}{\text{mol} \cdot \text K})(273\,\text K)} {10\,\text L} = 2.40\,\text{atm}$
Notice that the partial pressure for each of the gases increased compared to the pressure of the gas in the original container. This makes sense since the volume of both gases decreased, and pressure is inversely proportional to volume.
We can now get the total pressure of the mixture by adding the partial pressures together using Dalton's Law:
\begin{aligned}\text P_\text{Total}&=\text P_{\text{N}_2} + \text P_{\text {O}_2}\\ \\ &=4.79\,\text{atm} + 2.40\,\text{atm} = 7.19\,\text{atm}\end{aligned}

Step 2 (method 2): Use ideal gas law to calculate $\text P_\text{Total}$ without partial pressures

Since the pressure of an ideal gas mixture only depends on the number of gas molecules in the container (and not the identity of the gas molecules), we can use the total moles of gas to calculate the total pressure using the ideal gas law:
\begin{aligned}\text P_{\text{Total}} &= \dfrac{(\text{n}_{\text N_2}+\text n_{\text{O}_2})\text{RT}}{\text V}\\ \\ &=\dfrac{(2.14\,\text{mol}+1.07\,\text{mol})(0.08206\,\dfrac{\text{atm}\cdot \text L}{\text{mol} \cdot \text K})(273\,\text K)}{10\,\text L}\\ \\ &=\dfrac{(3.21\,\text{mol})(0.08206\,\dfrac{\text{atm}\cdot \text L}{\text{mol} \cdot \text K})(273\,\text K)}{10\,\text L}\\ \\ &=7.19\,\text{atm}\end{aligned}
Once we know the total pressure, we can use the mole fraction version of Dalton's law to calculate the partial pressures:
$\text P_{\text {N}_2} = x_{\text N_2} \text {P}_{\text{Total}} =\left(\dfrac{2.14\,\text{mol}}{3.21\,\text{mol}}\right)(7.19\,\text{atm})=4.79\,\text {atm}$
$\text P_{\text {O}_2} = x_{\text O_2} \text {P}_{\text{Total}} =\left(\dfrac{1.07\,\text{mol}}{3.21\,\text{mol}}\right)(7.19\,\text{atm})=2.40\,\text {atm}$
Luckily, both methods give the same answers!
You might be wondering when you might want to use each method. It mostly depends on which one you prefer, and partly on what you are solving for. For instance, if all you need to know is the total pressure, it might be better to use the second method to save a couple calculation steps.

Summary

• The pressure exerted by an individual gas in a mixture is known as its partial pressure.
• Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
• Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:
$\text {P}_{\text{Total}} = \text P_{\text {gas 1}} + \text P_{\text {gas 2}} + \text P_{\text {gas 3}} ...$
• Dalton's law can also be expressed using the mole fraction of a gas, $x$:
$\text P_{\text {gas 1}} = x_1 \text {P}_{\text{Total}}$

Try it: Evaporation in a closed system

Part 1

A closed system with a volume of $2.0\,\text L$ contains radon gas and liquid water, and the container is allowed to equilibrate at $27\,^\circ \text C$ until the total pressure is constant.
What is the partial pressure of radon if the total pressure is $780\,\text {torr}$ and the water vapor partial pressure is $1.0\,\text{atm}$?
$\text{atm}$
We can start by making the units for pressure the same for both the radon and water vapor using the conversion factor, $1\,\text {atm}=760\,\text{torr}$:
$\text P_\text{total} = 780\,\cancel{\text {torr}} \times \dfrac{1\,\text{atm}}{760\,\cancel{\text{torr}}} = 1.03\,\text {atm}$
We can solve for the partial pressure of radon using Dalton's law:
$\text P_\text{total} = \text P_\text{radon} + \text P_\text{water}$
$\text P_\text{total} =1.03\,\text {atm} = \text P_\text{radon} + 1.00\,\text{atm}$
Rearranging to solve for the partial pressure of radon, we get:
$\text P_\text{radon} = 1.03\,\text {atm} - 1.00\,\text {atm} = 0.03\,\text {atm}$

Part 2

Some helium gas is added to the system, and the total pressure increases to $1.20\,\text{atm}$.
What is the new partial pressure of radon?
$\text{atm}$
The partial pressure of radon is independent of other gases in the container, so adding another gas to the mixture doesn't change its partial pressure (although it does increase the total pressure). The partial pressure of radon depends only on the number of moles of radon, the volume of the container, and the temperature, none of which are affected by adding the helium gas.
Therefore, the partial pressure of the radon gas is still $0.03\,\text{atm}$.