http://www.ck12.org Chapter 14. The Behavior of Gases
Known
- V 1 = 2.20 L
- T 1 = 22°C = 295 K
- T 2 = 71°C = 344 K
Unknown
- V 2 =? L
Use Charles’s law to solve for the unknown volume (V 2 ). The temperatures have first been converted to Kelvin.
Step 2: Solve.
First, rearrange the equation algebraically to solve for V 2.
V 2 =
V 1 ×T 2
T 1
Now substitute the known quantities into the equation and solve.
V 2 =
2 .20 L×344 K
295 K
= 2 .57 L
Step 3: Think about your result.
The volume increases as the temperature increases. The result has three significant figures.
Practice Problem- A 465 mL sample of gas at 55°C is cooled to standard temperature (0°C). What is its new volume?
- To what Celsius temperature does 750. mL of gas originally at−12°C need to be heated to bring the volume
to 2.10 L?
Watch an animation of Charles’s Law at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Animations/
CharlesLaw.html.
A balloon cooled in liquid nitrogen illustrates Charles’s Law at http://www.youtube.com/watch?v=Gi5wPnkBEYI.
You can watch an experiment of Charles’s Law at http://www.youtube.com/watch?v=5M8GR6_zIps. The accom-
panying lab document is found at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Labs/Charles_La
w_Lab_web_01-02.doc.
Watch popping film canisters at http://education.jlab.org/frost/canister.html.
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. French chemist, Joseph Gay-Lussac (1778-1850), discovered the relationship
between the pressure of a gas and its absolute temperature.Gay-Lussac’s Lawstates that the pressure of a given