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# 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:
start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript, equals, start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 2, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 3, end text, end subscript, point, point, point
• Dalton's law can also be expressed using the mole fraction of a gas, x:
start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, equals, x, start subscript, 1, end subscript, start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript

## 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, start text, K, end text), and the pressure is around 1, start text, a, t, m, end text.
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, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, and oxygen gas, start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis. The mixture contains 6, point, 7, start text, m, o, l, end text hydrogen gas and 3, point, 3, start text, m, o, l, end text oxygen gas. The mixture is in a 300, start text, L, end text container at 273, start text, K, end text, and the total pressure of the gas mixture is 0, point, 75, start text, a, t, m, end text.
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, start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, we can completely ignore the oxygen gas and use the ideal gas law:
start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, start text, V, end text, equals, start text, n, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, start text, R, T, end text
Rearranging the ideal gas equation to solve for start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, 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, point, 50, start text, a, t, m, end text. 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:
start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript, equals, start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 2, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 3, end text, end subscript, point, point, point
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, start subscript, 1, end subscript, equals, start text, m, o, l, e, space, f, r, a, c, t, i, o, n, space, o, f, space, g, a, s, space, 1, end text, equals, start fraction, start text, m, o, l, e, s, space, o, f, space, g, a, s, space, 1, end text, divided by, start text, t, o, t, a, l, space, m, o, l, e, s, space, o, f, space, g, a, s, end text, end fraction
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:
start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, equals, x, start subscript, 1, end subscript, start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript
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, point, 0, start text, L, end text of nitrogen gas at 2, point, 00, start text, a, t, m, end text, and another container with 12, point, 0, start text, L, end text of oxygen gas at 2, point, 00, start text, a, t, m, end text. The temperature of both gases is 273, start text, K, end text.
If both gases are mixed in a 10, point, 0, start text, L, end text 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 start text, P, end text, start text, V, end text,and start text, T, end text 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:
start text, n, end text, equals, start fraction, start text, P, V, end text, divided by, start text, R, T, end text, end fraction
Solving for nitrogen and oxygen, we get:
start text, n, end text, start subscript, start text, N, end text, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, start text, a, t, m, end text, right parenthesis, left parenthesis, 24, point, 0, start text, L, end text, right parenthesis, divided by, left parenthesis, 0, point, 08206, start fraction, start text, a, t, m, end text, dot, start text, L, end text, divided by, start text, m, o, l, end text, dot, start text, K, end text, end fraction, right parenthesis, left parenthesis, 273, start text, K, end text, right parenthesis, end fraction, equals, 2, point, 14, start text, m, o, l, space, n, i, t, r, o, g, e, n, end text
start text, n, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, start text, a, t, m, end text, right parenthesis, left parenthesis, 12, point, 0, start text, L, end text, right parenthesis, divided by, left parenthesis, 0, point, 08206, start fraction, start text, a, t, m, end text, dot, start text, L, end text, divided by, start text, m, o, l, end text, dot, start text, K, end text, end fraction, right parenthesis, left parenthesis, 273, start text, K, end text, right parenthesis, end fraction, equals, 1, point, 07, start text, m, o, l, space, o, x, y, g, e, n, end text

### Step 2 (method 1): Calculate partial pressures and use Dalton's law to get $\text P_\text{Total}$start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript

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, point, 0, start text, L, end text container:
start text, P, end text, equals, start fraction, start text, n, R, T, end text, divided by, start text, V, end text, end fraction
start text, P, end text, start subscript, start text, N, end text, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, point, 14, start text, m, o, l, end text, right parenthesis, left parenthesis, 0, point, 08206, start fraction, start text, a, t, m, end text, dot, start text, L, end text, divided by, start text, m, o, l, end text, dot, start text, K, end text, end fraction, right parenthesis, left parenthesis, 273, start text, K, end text, right parenthesis, divided by, 10, start text, L, end text, end fraction, equals, 4, point, 79, start text, a, t, m, end text
start text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 1, point, 07, start text, m, o, l, end text, right parenthesis, left parenthesis, 0, point, 08206, start fraction, start text, a, t, m, end text, dot, start text, L, end text, divided by, start text, m, o, l, end text, dot, start text, K, end text, end fraction, right parenthesis, left parenthesis, 273, start text, K, end text, right parenthesis, divided by, 10, start text, L, end text, end fraction, equals, 2, point, 40, start text, a, t, m, end text
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 $\text P_\text{Total}$start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript 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:
start text, P, end text, start subscript, start text, N, end text, start subscript, 2, end subscript, end subscript, equals, x, start subscript, start text, N, end text, start subscript, 2, end subscript, end subscript, start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript, equals, left parenthesis, start fraction, 2, point, 14, start text, m, o, l, end text, divided by, 3, point, 21, start text, m, o, l, end text, end fraction, right parenthesis, left parenthesis, 7, point, 19, start text, a, t, m, end text, right parenthesis, equals, 4, point, 79, start text, a, t, m, end text
start text, P, end text, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, equals, x, start subscript, start text, O, end text, start subscript, 2, end subscript, end subscript, start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript, equals, left parenthesis, start fraction, 1, point, 07, start text, m, o, l, end text, divided by, 3, point, 21, start text, m, o, l, end text, end fraction, right parenthesis, left parenthesis, 7, point, 19, start text, a, t, m, end text, right parenthesis, equals, 2, point, 40, start text, a, t, m, end text
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:
start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript, equals, start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 2, end text, end subscript, plus, start text, P, end text, start subscript, start text, g, a, s, space, 3, end text, end subscript, point, point, point
• Dalton's law can also be expressed using the mole fraction of a gas, x:
start text, P, end text, start subscript, start text, g, a, s, space, 1, end text, end subscript, equals, x, start subscript, 1, end subscript, start text, P, end text, start subscript, start text, T, o, t, a, l, end text, end subscript

## Try it: Evaporation in a closed system

### Part 1

A closed system with a volume of 2, point, 0, start text, L, end text contains radon gas and liquid water, and the container is allowed to equilibrate at 27, degrees, start text, C, end text until the total pressure is constant.
What is the partial pressure of radon if the total pressure is 780, start text, t, o, r, r, end text and the water vapor partial pressure is 1, point, 0, start text, a, t, m, end text?
start text, a, t, m, end text

### Part 2

Some helium gas is added to the system, and the total pressure increases to 1, point, 20, start text, a, t, m, end text.
What is the new partial pressure of radon?
start text, a, t, m, end text

## Want to join the conversation?

• In the very first example, where they are solving for the pressure of H2, why does the equation say 273L, not 273K? • For example 1 above when we calculated for H2's Pressure, why did we use 300L as Volume? Isn't that the volume of "both" gases? Why didn't we use the volume that is due to H2 alone? • Under the heading "Ideal gases and partial pressure," it says the temperature should be close to 0 K at STP. Shouldn't it really be 273 K? • Oxygen and helium are taken in equal weights in a vessel . The pressure exerted by helium in the mixture is • Since oxygen is diatomic, one molecule of oxygen would weigh 32 amu, or eight times the mass of an atom of helium. If you have equal amounts, by mass, of these two elements, then you would have eight times as many helium particles as oxygen particles. Therefore, the pressure exerted by the helium would be eight times that exerted by the oxygen.
• In question 2 why didn't the addition of helium gas not affect the partial pressure of radon? please explain further. • In the first question, I tried solving for each of the gases' partial pressure using Boyle's law
P_1*V_1 = P_2*V_2

For Nitrogen :
P2 = P_N2 = P1*V1/V2 = 2*24/10 = 4.80atm
For Oxygen:
P2 = P_O2 = P1*V1/V2 = 2*12/10 = 2.40atm

P_total = 4.80+2.40 = 7.20atm which is pretty close to the 7.19atm calculated here.

Can anyone explain what is happening lol
EDIT: Is it because the temperature is not constant but changes a bit with volume, thus causing the error in my calculation? • How did he get 3.21 moles in the mole fraction section.
(1 vote) • In step 1, we were asked to find the total pressure. The pressure of nitrogen is 2atm and that of oxygen is 2atm. Why could we not say that total pressure is equal to pressure of the individual gases(nitrogen and oxygen). Ptotal= Pn + Po. Ptotal= 2+2 which is equivalent to 4atm. I understand the fact that we`ve the pressure of nitrogen as P1, P1 is equivalent to n1RT/V and P total is equivalent to P1 + P2. P2= n2RT/V. So we find P1 and P2 and add them together using n1RT/V + n2RT/V will give the total pressure.
(1 vote) • Using Dalton’s law as you’re suggesting only works if all the gases are already mixed together in the same container where the volume and temperature are the same. For the nitrogen and oxygen gas example, they begin in two separate containers with different volumes. And finally we mix them together in a third container with a different volume from the previous two containers.

The 2atm pressures represent the pressures of the gases in their original containers before mixing. Their sum, 4atm, isn’t automatically the pressure in the third container when they are mixed because of the different volumes.

Using the idea gas to create new equations to substitute for the partial pressures in Dalton’s law like you’ve done is how they solved the problem in method 1.  