Charles’ Law
We now have two distinct laws for gases:
Boyle’s Law: the pressure and volume of a gas are inversely proportional at a constant temperature
P1V1 = P2V2 at constant T
Gay-Lussac’s Law: the pressure and Kelvin temperature of a gas are directly proportional
at a constant volume
P1 = P2
at constant V
T1
T2
What happens if we have constant Pressure and examine the remaining two variables, volume
and temperature? Or to put it another way, what happens to a balloon-like container when it
gets heated or cooled?
Experiment
In this very brief experiment, a balloon will be filled with air and then placed in liquid nitrogen.
Nitrogen is normally a gas but if you get it to –196 oC or 77 K, it turns into a liquid so liquid
nitrogen is VERY cold. Examine the results below:
http://www.dkimages.com/discover/
previews/836/20084202.JPG
Balloon filled with air at
room temperature, 22 oC.
Balloon placed in liquid
nitrogen gets smaller.
Balloon is taken out of liquid nitrogen and returns to
room temperature, 22 oC.
Results
When the balloon got colder, it got smaller. When the balloon got hot again, it got larger. Is
that a direct or an inverse relationship? Direct, of course. Which of the other relationships
we have seen was also a direct relationship? Gay-Lussac’s Law which was:
P1 = P2
at constant V
T1
T2
The only difference between Gay-Lussac’s Law and Charles’ Law is that we are at constant
Pressure with variable Volume instead of the reverse. Consequently:
Charles’ Law
If we graph the volume of a balloon’s volume versus its temperature, we end up with a familiar
line:
Notice what is similar to what we have seen before? There should be about zero volume when
there is zero temperature. Consequently, absolute zero is the temperature where the volume is
zero on the graph. If you notice, this temperature is still –273 oC or 0 K.
http://butane.chem.uiuc.edu/cyerkes/Chem102AEFa07/Lecture_Notes_102/Lecture%2016_files/image005.jpg
V1 = V2 at constant P
T1 T2
http://mike66546.tripod.com/Physciweb/
Chapter8/charlesim.gif
http://boomeria.org/chemtextbook/fig18-6.jpg
http://educ.queensu.ca/~science/main/concept/gen/g09/R.%20Tripp/charles%27s%20law2.jpg
Jacques Charles was a French scientist
who lived from 1746 to 1823. In addition
to many of his scientific contributions he
made the first recorded flight in a hydrogen-filled balloon in 1783.
Charles’ Law Questions:
A) A sample of gas in a 7.5 L balloon is at a temperature of 22 oC. What volume will
the balloon expand to at 50 oC?
V1 = 7.5 L
T1 = 22 oC = 295 K
V2 = X
T2 = 50 oC = 323 K
7.5 L =
295 K
X
323 K
Cross multiply and divide:
(7.5 L)(323 K) =
X(295 K)
(7.5 L)(323 K) =
(295 K)
X
X = 8.2 L
B) At what temperature will a sample of gas in a balloon reach a volume of 120 L if at
standard temperature (0 oC) the gas is at 200 L?
V1 = 120 L
T1 = X
V2 = 200 L
T2 = 0 oC = 273 K
120 L =
X
200 L
273 K
Cross multiply and divide:
(120 L)(273 K)
=
(120 L)(273 K)
(200 L)
=
X (200 L)
X
X = 164 K
If you want it in oC then you must subtract 273
X = 164—273 = -109 oC
Questions
1.
2.
3.
4.
Charles’ Law is the relationship between what two variables?
Which variable must be constant in Charles’ Law?
What temperature scale must be used in Charles’ Law problems?
Sketch the curve of the graph of Volume vs. Temperature:
V
T
5. A sample of gas in a balloon has a volume of 2400 mL at room temperature (22 oC). How
big will the balloon be at 75 oC?
6. A sample of gas in a balloon has a volume of 10 L at 200 K. What will the volume of the
balloon be if the temperature drops to 100 K?
7. A sample of gas in a balloon has a volume of 12 L at 25 oC. At what temperature (both Celcius and Kelvin) will the balloon expand to 18 L?
8. A gas in a balloon has a volume of 2500 mL at standard temperature. What will the volume
be at 300 K?