14.2. Gas Laws http://www.ck12.org
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.
The mathematical expressions for Gay-Lussac’s Law are likewise similar to those of Charles’s Law:
27.16 P
T
=k andP 1
T 1
=
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 14.3: 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 atm
- T 1 = 25°C = 298 K
- T 2 = 845°C = 1118 K
Unknown
- P 2 =? 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 P 2.
P 2 =
P 1 ×T 2
T 1
Now, substitute the known quantities into the equation and solve.
P 2 =
3 .00 atm×1118 K
298 K
= 11 .3 atmStep 3: Think about your result.
The pressure increases dramatically due to the large increase in temperature.
Practice Problem- The pressure in a car tire is 217 kPa at 24°C. After being driven on a hot summer day, the pressure in the tire
increases to 258 kPa. What is the Celsius temperature of the air in the tire? - A gas sample originally at−39°C is heated to standard temperature (0°C), at which point its pressure is
measured to be 808 mmHg. What was the original pressure of the gas?