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Gas laws – Gay Lussac’s Law

The physical behaviour of a gas is described using four main variables :

  1. Temperature
  2. Pressure
  3. Volume
  4. Amount (mol number)

Gay Lussac’s Law

Joseph Louis Gay-Lussac was a French chemist and physicist who was investigating the relation between the pressure and the temperature of a fixed amount of gas at constant volume. Gay-Lussac’s law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant. So as the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.

Gas laws - Gay Lussac's Law

Imagine you have a balloon inside a container that ensures it has a fixed volume. You heat the balloon.

What is happening to the temperature of the gas inside the balloon ? The temperature increases

What will happen to the pressure inside the balloon ? The pressure increases

Gas laws - Gay Lussac's Law

Gay-Lussac’s law can be applied to reduce the time to cook food. In a pressure cooker, food cooks faster than in an ordinary pot because under a high pressure the temperature should increases and becomes hotter than it would under normal atmospheric pressure. A pressure cooker has a valve that allows some vapor to escape when the pressure exceeds a certain value to prevent the risk of explosion.

Gas laws - Gay Lussac's Law

Let’s consider a gas under two different conditions of temperature (T1 and T2) and pressure (P1 and P2) with a constant volume. Gay Lussac’s Law leads to the mathematical expression :

Gas laws - Gay Lussac's Law

The temperature should be measured in Kelvin (K). (0 oC = 0 + 273 = 273 K)

Practice problem – Gay Lussac’s Law

Aerosol cans carry labels: “warning not to burn the cans or store them above a certain temperature”.

The gas in a used aerosol can is at a pressure of P1 = 103 kPa at T1 = 25oC. If the can is thrown onto a fire, what will the pressure be when the temperature reaches T2 = 928oC ?

Gas laws - Gay Lussac's Law

Solution

Because you will use a gas law, start by expressing the temperatures in kelvin.

T1 = 25oC + 273 = 298 K

T2 = 928oC + 273 = 1201 K

Gay Lussac’s law : P1/T1 = P2/T2

P2 = (P1 x T2) / T1 = (103 x 1201) / 298 = 415.1 KPa

P2 = 415.1 KPa which makes sense because when the temperature increases, the pressure should also increase. T2 > T1 so P2 > P1 .