What is Partial Pressure?

Core Concepts

In this article, we learn about partial pressure and it’s importance with respect to Dalton’s Law, the Ideal Gas Law, mole fractions, and L’Chatelier’s Principle.

Topics Covered in Other Articles

Pressure and Partial Pressure

Physical chemists tend to have a lot of interest in the pressure of gases. In chemistry, a gas’s pressure (a state function) is the force it exerts on some surface, usually the walls of a container. The gas exerts this force from the collisions of gas particles against the container, which results in an outward force.

With a mix of gases in some container, the total pressure of the mix involves the collisions of each gas. Physical chemists use the term partial pressure to describe the pressure due to the collisions of one specific gas in the mix. The individual partial pressures relate to the total pressure of the mix according to Dalton’s Law.

gas mixture, each with a partial pressure

Partial Pressure and Dalton’s Law

Dalton’s Law says that the partial pressures of each gas in a mixture simply add up to the total pressure of the mixture.

$\begin{gather*} {P_{total}=\sum P_{i}=P_{1}+P_{2}+P_{3}+\cdots P_{n}} \end{gather*}$

As you can observe, Dalton’s Law makes the math easy for finding the total pressure from partial pressures.

For example, let’s say we have a mix of $0.25\text{atm}$ of oxygen, $0.50\text{atm}$ of nitrogen, and $1.25\text{atm}$ of methane. Using Dalton’s Law, we add each partial pressure to find a total pressure of $2.00\text{atm}$ against the walls of the container:

$\begin{gather*} {P_{total} = P_{\text{O}_{2}}+P_{\text{N}_{2}}+P_{\text{CH}_{4}}} \\ {P_{total} = 0.25\text{atm} + 0.50\text{atm} + 1.25\text{atm} = 2.00\text{atm}} \end{gather*}$

Dalton’s Law is certainly useful in finding total pressure, assuming we know already know the partials. But in most practical research settings, you can rather easily measure the total pressure of a mixture using barometers. In these circumstances, we might instead want to know the partial pressures of each component gas in the mixture.

So how do we find partial pressures from total pressures? The answer involves using a different physical law, specifically the Ideal Gas Law

Partial Pressure and the Ideal Gas Law

Pressure (including partial pressure) relates to the gas’s temperature ( $T$ ), volume ( $V$ ), and moles ( $n$ ) according to the Ideal Gas Law:

$\begin{gather*} {PV=nRT} \end{gather*}$

$R$ = Ideal Gas Constant

The relationship between pressure and each of these other variables makes sense from the perspective of molecular collisions. Increasing temperature raises the speeds of the gas particles, which increases the force of their collisions against the container, thus increasing pressure. Decreasing volume shortens the paths of each gas molecule in the container, increasing collisions, thus increasing pressure. Increasing moles of a gas similarly increases the collisions in the container, thus increasing pressure.

In gas mixes, all gases have the same temperature and volume, due to sharing the same container. This means that a gas’ partial pressure in a gas mix depends entirely on the moles of that gas.

Thus, if we know the total pressure of a mixture, we can calculate the partial pressure of a gas if we know the gas’s mole fraction.

Partial Pressures and Mole Fractions

In chemistry, a mole fraction ( $\chi$ ) is the (unitless) ratio of the moles of a component in a mixture to the moles of every component in the mixture. Put differently, a mole fraction is the proportion of one component in a mixture. If a mixture of $2\text{mol}$ of gas particles includes $1\text{mol}$ of nitrogen, then nitrogen’s mole fraction of 0.5.

$\begin{align*} {{\chi}_{A}&= \frac{n_{A}}{n_{total}}} \\ {{\chi}_{\text{N}_{2}}&= \frac{n_{\text{N}_{2}}}{n_{total}}=\frac{1\text{mol}}{2\text{mol}}=0.5} \end{align*}$

Importantly, in a gas mixture, a gas’ mole fraction equals the ratio between its partial pressure and the total pressure.

$\begin{align*} {{\chi}_{A}&= \frac{n_{A}}{n_{total}}= \frac{P_{A}}{P_{total}}} \end{align*}$

Thus, you can multiply a gas’ mole fraction by the total pressure to get its partial pressure. If we know this same mixture has $2.50\text{atm}$ of total pressure, nitrogen must have a partial pressure of $1.25\text{atm}$ .

$\begin{align*} {{\chi}_{\text{N}_{2}}&= \frac{P_{\text{N}_{2}}}{P_{total}}} \\ {P_{\text{N}_{2}}={\chi}_{\text{N}_{2}}\left( P_{total}\right) &= 0.5\left( 0.250\text{atm}\right) =1.25\text{atm}\vphantom{\frac12}} \end{align*}$

Partial Pressure and L’Chatlier’s Principle

As a result of partial pressure’s direct relationship with mole proportions, chemists generally consider it the gaseous form of “concentration”. This means that partial pressure serves as an important concept when studying the equilibrium dynamics of a gas mixture.

Let’s say we have a gas mixture that also serves as a reaction mixture. Specifically, we have $1\text{atm}$ each of nitrogen, hydrogen, and ammonia. Through the Haber Process, nitrogen and hydrogen famously react to form ammonia according to this reversible reaction equation:

$\begin{align*} {\text{N}_{2} + 3\text{H}_{2} \rightleftarrows 2\text{NH}_{3}} \end{align*}$

Importantly, let’s say our gas mixture currently exists at equilibrium. This means that the partial pressures of each gas remain constant over long periods.

However, according to L’Chatlier’s Principle, if we place stress on this equilibrium, the reaction will shift in response. Specifically, if we raise the partial pressure of ammonia, we place a stress of pressure on the system. In response, the reaction “shifts left”, consuming some of that excess ammonia and producing nitrogen and hydrogen, until equilibrium reestablishes.

Additionally, if we raise the pressure of the entire system, we also see an equilibrium shift. You can do this by adding an inert gas, like krypton or xenon, for instance. Importantly, this doesn’t change the partial pressures of the reaction gases.

To lessen the stress on the system, our reaction would shift to the side of the reaction equation with fewer moles of gas. This, in turn, decreases the total partial pressures between the gases involved in the reaction.

In our example, there are $4 \text{mol}$ of gas on the left and $2 \text{mol}$ on the right. Thus, the addition of an inert gas shifts our reaction “right”, producing more ammonia.

Partial Pressure Practice Problems

Problem 1

At a given temperature you have $1\text{atm}$ of hydrogen gas. First, you add enough chlorine gas to reduce the mole fraction of hydrogen gas to $\chi _{H_{2}} = \frac{1}{3}$ . What are the partial pressures of hydrogen and chlorine?