Gay Lussac’s Law

What is Gay Lussac’s Law?

Gay-Lussac’s law is a gas law that states the pressure of a gas varies directly with temperature when mass and volume are kept constant. As the temperature increases, the pressure will also increase. The concept is shown graphically below.

As temperature increases so does pressure, as described by Gay Lussac's law

This phenomenon occurs because as temperature increases, the kinetic energy of the gas molecules increases. The increased energy means the molecules collide with the walls of the container with more force, meaning higher pressure.

The Gay Lussac’s Law is also sometimes called Amonton’s Law. Amonton proved the same law by making a thermometer where the measured pressure was a readout for the current temperature. Gay-Lussac proved the law more precisely, so it is more often called by his name.

Gay Lussac’s Law Formula

Gay-Lussac’s law gives us a formula where pressure and temperature are related to a constant when volume and mass/moles are held constant. That is:

$\begin{gather*} {\frac{P}{T} = k} \end{gather*}$

We can also relate pressure and temperature at two different points then because they are both equal to the same constant value. That is:

$\begin{gather*} {\frac{P_1}{T_1} = k} \end{gather*}$

And

$\begin{gather*} {\frac{P_2}{T_2} = k} \end{gather*}$

Therefore,

$\begin{gather*} {\frac{P_1}{T_1} = \frac{P_2}{T_2}} \end{gather*}$

You may see this formula written in a variety of formats. For example, you can rearrange the variables to get

$\begin{gather*} {\frac{P_1}{P_2} = \frac{T_1}{T_2}} \end{gather*}$

The k in these equations can also be solved by rearranging the ideal gas law.

$\begin(align*} {PV&=nRT} \\ {\frac{P}{V}&=\frac{nR}{V}} \end{align*}$

We are holding the volume ( $V$ ) and moles ( $n$ ) constant. $R$ is always a constant. So, the whole right-hand side of the bottom equation is a constant.

Gay-Lussac’s Law Real Life Example

As the temperature drops in winter, you may notice that you get a low tire pressure alert in your car. This phenomenon happens because of the relationship we see in Gay Lussac’s law. The temperature drops, therefore, the amount of pressure in the tire drops because they are directly proportional. Both the total volume and the total mass of gas inside is remaining the same, however. (With large temperature swings the volume of the tire does change, but for small temperature changes it stays relatively the same)

The same is true with propane tanks. The tank may read lower or higher pressure depending on the temperature outside (as long as you don’t use the tank to grill!). As the temperature increases, the pressure gauge on the tank will read higher.

For a visual of pressure decreasing because temperature decreases, watch or try the collapsing can experiment! This experiment isn’t a perfect representation of Gay Lussac’s law but is a good example of pressure changes due to temperature. In Gay Lussac’s experiments, he had a rigid container with a set volume.

Gay-Lussac’s Law Example Problems

Problem: You are trying to dispose of an aerosol container that has a pressure of $2.00 \text{atm}$ at $20\degree \text{C}$ . When the container is disposed of, it may increase to a temperature of $110\degree \text{C}$ . What would the pressure be at this temperature?

Worked Solution:

First, we need to identify the variables we have. We know $P_{1} = 2.00 \text{atm}$ , $T_{1} = 20 \degree \text{C}$ or $293\text{K}$ , and that $T_{2} = 110\degree \text{C}$ or $383\text{K}$ . We are trying to solve for $P_2$ .

The equation for the Gay-Lussac’s law is:

$\begin{gather*} {\frac{P_1}{T_1} = \frac{P_2}{T_2}} \end{gather*}$

We can rearrange to solve for $P_2$ .

$\begin{gather*} {T_{2}\left( \frac{P_1}{T_1} \right) = P_2} \end{gather*}$

Now we plug in our values and solve.

$\begin{gather*} {(383\text{K})\left( \frac{2.00\text{atm}}{293\text{K}} \right) = P_2 = 2.61 \text{atm}} \end{gather*}$

Problem: Your car tire is reading $2.24 \text{atm}$ and the temperature outside is $80\degree \text{F}$ . What will the tire pressure be when the temperature outside is $0\degree \text{F}$ ? (Assume the volume of the tire does not change if it goes flat)

Worked Solution:

We are looking at a tire, so the number of moles and volume is constant which means we can use Gay-Luccass’s law.

Our $P_{1} = 2.24 \text{atm}$ and $T_{1} = 80\degree \text{F} (300\text{K})$ . We also know $T_{2} = 0\degree \text{F} (255\text{K})$ . We need to solve for $P_{2}$ .

From the problem above we have our rearranged equation. (If you don’t remember how to rearrange the equation, it is worked out in the previous problem)

$\begin{gather*} {T_{2}\left( \frac{P_1}{T_1} \right) = P_2} \end{gather*}$

Then we can plug in the numbers we know and calculate our answer.

$\begin{gather*} {(255\text{K})\left( \frac{2.24\text{atm}}{300 \text{K}} \right) = P_2 = 1.90 \text{atm}} \end{gather*}$

The new tire pressure is $1.90 \text{atm}$ . Significantly lower than the $2.24 \text{atm}$ of pressure in the tire before!

Who is Gay-Lussac?

Joseph Louis Gay-Lussac was a French chemist who lived from 1778 to 1850. He discovered and shared his famous Gay Lussac’s law in the early 1800s. Besides the law described above, he also developed many analytical chemistry techniques, discovered boron, and much more. He had a rivalry with Humphry Davy who discovered several other elements including calcium and potassium.

Fun fact! In an attempt to measure the magnetic field of the earth at high elevation, Gay-Lussac held the world record for the highest balloon flight for about fifty years. He reached a height of about 7,000 meters or 23,000 feet.

Gay Lussac’s Law Practice Problems

Problem 1

As the temperature of $1\text{atm}$ of nitrogen gas in a closed tank is cooled from $40\degree \text{C}$ to $10 \degree \text{C}$ , does the pressure exerted on the tank’s walls increase or decrease? Calculate the final pressure.