Combined Gas Law

combined gas law component parts

Core Concepts

The combined gas law defines the relationship between pressure, temperature, and volume. It is derived from three other names gas laws, including Charles’ law, Boyle’s law, and Gay-Lussac’s law. Below we explain the equation for the law, how it is derived, and provide practice problems with solutions.

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The Combined Gas Law

The combined gas law relates pressure, temperature, and volume when everything else is held constant (mainly the moles of gas, n). The most common form of the equation for the combined gas law is as follows:

    \begin{gather*} {\frac{PV}{T} = k} \end{gather*}

P is the pressure of the gas. T is the temperature of the gas. V is the volume of the gas. And k is a constant. The exact value of k will depend on the moles of gas.  

The combined gas law is also often written as two different time points. That is:

    \begin{align*} {\frac{P_{1}V_{1}}{T_{1}} &= k} \\ {\frac{P_{2}V_{2}}{T_{2}} &= k} \end{align*}

Both k’s are the same value and therefore can be set equal to each other. Resulting in the below equation:

    \begin{align*} {\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}} \end{align*}

The relationship of the combined gas law works as long as the gasses act as ideal gasses. Generally, this will be true when the temperature is high and pressure is low. You can learn more about what makes a gas an ideal gas in the article ‘The Ideal Gas Law’.

Derivation of the Combined Gas Law

The combined gas law is derived from combining Charles’ Law, Boyle’s Law, and Gay-Lussac’s Law.

Charle’s law gives the relationship between volume and temperature. That is  {\frac{V}{T} = k} . Further, Boyle’s law tells us that  P*V =k . And finally, Gay-Lussac’s law tells us that  P*T =k .

Charles' law, Boyle's law and Gay-Lussac's law combined to make the combined gas law

When all these relationships are combined into one equation, we get the combined gas law.

When the combined gas law is expanded and the moles of gas (n) are not held constant, you get the ideal gas law. You can also work backwa from the ideal gas law to get the other gas laws by holding different variables constant. In the case of the combined gas law, that would happen by holding the moles of gas (n) constant.

Example Problem 1

Suppose you have a sample of gas at 303K in a container with a volume of 2L and pressure of 760mmHg. The sample shifts to a temperature of 340 K and the volume increases slightly to 2.1L. What is the pressure of the sample now?


Here we are looking at two different states. The original state with subscript 1, and the second state with subscript 2. First, write out the variables we know:

    \begin{align*} {V_1 &= 2\text{L}} \\ {T_1 &= 303\text{K}} \\ {P_1 &= 760\text{mmHg}} \\ {V_2 &= 2.1\text{L}} \\ {T_2 &= 304\text{K}} \\ {P_2 &= text{?}} \end{align*}

We know all the variables except P2. We can also tell we are looking at a before and after state, so we want to use the following equation.

    \begin{align*} {\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}} \end{align*}

Next, we rearrange the equation so it is solving for P2. First, multiply each side by T2.

    \begin{align*} {T_2 * \frac{P_{1}V_{1}}{T_{1}} = P_{2}V_{2}} \end{align*}

Then divide each side by V2.

    \begin{align*} {\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}} = P_{2}} \end{align*}

Now we plug in the variables we know and solve.

    \begin{align*} {\frac{\left(760\text{mmHg}\right)\left(2\text{L}\right)\left(340\text{K}\right)}{\left(303\text{K}\right)\left(2.1\text{L}\right)} = P_{2} = 812\text{mmHg}} \end{align*}

Our final pressure is 812 mmHg. Also notice that all the units cancel except the units for pressure.

Example Problem 2

 You collect a gas at 620 mmHg and 177 K. At the time of collection, it takes up a volume of 1.3 L. What will the volume of the gas be when it moves to standard temperature and pressure?


Here we are looking at two different states of the gas, state 1 and state 2. Therefore we will use the following form of the combined gas law.

    \begin{align*} {\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}} \end{align*}

The first step is to determine the variables we know. The problem gives us pressure, temperature, and volume for the original state 1. Further, we have pressure and temperature for state 2 as 760 mmHg and 273K. The only variable we don’t know is volume 2, which is what we then solve for.

    \begin{align*} {V_1 &= 1.3\text{L}} \\ {T_1 &= 177\text{K}} \\ {P_1 &= 620\text{mmHg}} \\ {V_2 &= \text{?}} \\ {T_2 &= 273\text{K}} \\ {P_2 &= 760 \text{mmHg}} \end{align*}

To make the math simpler, let us rearrange the equation to solve for V2 before plugging in values. To do this, we multiply both sides by T2 and then divide by P2.

    \begin{align*} {\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = V_{2}} \end{align*}

We then plug in the values we know and solve.

    \begin{align*} {\frac{\left(620\text{mmHg}\right)\left(1.3\text{L}\right)\left(273\text{K}\right)}{\left(177\text{K}\right)\left(760\text{mmHg}\right)} = V_{2} = 1.6\text{L}} \end{align*}

Thus, the new volume of the gas is 1.6L. So as the temperature and pressure of the gas increased, the volume of the gas also increased.