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# Gay-Lussac's Law

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Gay-Lussac's Law

How much propane is in the tank?

Propane tanks are widely used with barbeque grills. But it’s not fun to find out half-way through your grilling that you’ve 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.

Joseph Louis Gay-Lussac.

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

$\frac{P}{T}=k \quad \text{and} \quad \frac{P_1}{T_1}=\frac{P_2}{T_2}$

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.

#### Sample Problem: Gay-Lussac’s Law

The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25°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°C?

Step 1: List the known quantities and plan the problem.

Known

• $P_1=3.00 \ \text{atm}$
• $T_1=25^\circ \text{C}=298 \text{ K}$
• $T_2=845^\circ \text{C}=1118 \text{ K}$

Unknown

• $P_2=? \ \text{atm}$

Use Gay-Lussac’s law to solve for the unknown pressure $(P_2)$ . The temperatures have first been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for $V_2$ .

$P_2=\frac{P_1 \times T_2}{T_1}$

Now substitute the known quantities into the equation and solve.

$P_2=\frac{3.00 \text{ atm} \times 1118 \text{ K}}{298 \text{ K}}=11.3 \text{ atm}$

The pressure increases dramatically due to large increase in temperature.

#### Summary

• Pressure and temperature at constant volume are directly proportional.

#### Practice

Work the problems found at the web site below:

#### Review

1. Explain Gay-Lussac’s Law in terms of the kinetic-molecular theory.
2. What would a graph of pressure vs. temperature show us?
3. What is the difference in containers in Charles’ Law and Gay-Lussac’s Law?