# 11.6: Gay-Lussac's Law: Temperature and Pressure


##### Learning Objectives
• Explain Gay-Lussac's Law.

Propane tanks are widely used with barbeque grills. But it's not fun to find out halfway through grilling that you have run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.

## Gay-Lussac's Law

When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment.

Figure $$\PageIndex{1}$$: Joseph Gay-Lussac.

The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law:

$\dfrac{P}{T} \: \: \: \text{and} \: \: \: \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \nonumber$

A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid.

##### Example $$\PageIndex{1}$$

The gas in an aerosol can is under a pressure of $$3.00 \: \text{atm}$$ at a temperature of $$25^\text{o} \text{C}$$. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of $$845^\text{o} \text{C}$$?

###### Solution
Solutions to Example 11.10.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."

Given:

$$P_1 = 3.00 \: \text{atm}$$

$$T_1 = 25^\text{o} \text{C} = 298 \: \text{K}$$

$$T_2 = 845^\text{o} \text{C} = 1118 \: \text{K}$$

Find: $$P_2 = ? \: \text{atm}$$

List other known quantities. The temperatures have first been converted to Kelvin.
Plan the problem.

First, rearrange the equation algebraically to solve for $$P_2$$.

$P_2 = \dfrac{P_1 \times T_2}{T_1} \nonumber$

Calculate.

Now substitute the known quantities into the equation and solve.

$P_2 = \dfrac{3.00 \: \text{atm} \times 1118 \: \cancel{\text{K}}}{298 \: \cancel{\text{K}}}= 11.3 \: \text{atm} \nonumber$

Think about your result. The pressure increases dramatically due to a large increase in temperature.

## Summary

• Pressure and temperature at constant volume are directly proportional.

11.6: Gay-Lussac's Law: Temperature and Pressure is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.