Gr11 Chem Unit 2: Gases and the Atmosphere, Lesson 6: Volume, Temperature
Charles' Law
Jaques Charles was a disturbing and scary guy. Though he came up with a really
handy law for determining what the relationships between the volume and temperature
of a gas are, his private life was far more bizarre. Some say that if you go by the old
Charles mansion at the edge of town, you can still hear the moaning and wailing of his
ghost, forever roaming the night.*
Anyhow, what Charles determined through his studies was that when you change the
temperature of a gas, the volume changes. Not surprising - you probably know already
that if you heat something, it tends to get bigger. What he found, though, was that if you
divide the volume by the temperature of a gas at one temperature, you get a
constant. Just like Boyle found, if you change the volume or temperature of this gas,
you get the same constant. From this, Charles came up with this statement:
V1/T1 = V 2/T2
Where the subscript "1" indicates the initial volume and temperature and the subscript
"2" indicated the volume and temperature after the change. Temperature, incidentially,
needs to be given in Kelvins and not in Celsius - this is because if you have a
temperature below zero degrees Celsius, the calculation works out so the volume of the
gas is negative, and you can't have a negative volume.
Let's see an example of this equation in action:
Question: If we have 2 L of methane gas at a temperature of 40 degrees Celsius, what
will the volume be if we heat the gas to 80 degrees Celsius?
Answer: The first thing we have to do is convert the temperatures to Kelvins (by adding
273), because Celsius can't be used in this equation. To do this, we get that the initial
temperature is 40 + 273 = 313 K and the final temperature is 80 + 273 = 353 K. We're
now ready to start sticking these numbers into the equation:
2 L / 313 K = x L / 353 K
x = 2.26 L
*This story isn't true.
(Source: http://misterguch.brinkster.net/gaslaws.html)
Gr11 Chem Unit 2: Gases and the Atmosphere, Lesson 6: Volume, Temperature
Charles’s Law
1)
If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to
550 C, what will the new volume of the balloon be?
2)
Calcium carbonate decomposes at 12000 C to form carbon dioxide and calcium oxide. If 25 liters
of carbon dioxide are collected at 12000 C, what will the volume of this gas be after it cools to
250 C?
3)
I have 130 liters of gas in a piston at a temperature of 2500 C. If I cool the gas until the volume
decreases to 85 liters, what will temperature of the gas be?
Gr11 Chem Unit 2: Gases and the Atmosphere, Lesson 6: Volume, Temperature
Charles’s Law – Solutions
1)
If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to
550 C, what will the new volume of the balloon be?
45L
x

298K 328K
x  50 L
2)
Calcium carbonate decomposes at 12000 C to form carbon dioxide and calcium oxide. If 25 liters
of carbon dioxide are collected at 12000 C, what will the volume of this gas be after it cools to
250 C?
25 L
x

1473K 298 K
x  5.1L
3)
I have 130 liters of gas in a piston at a temperature of 2500 C. If I cool the gas until the volume
decreases to 85 liters, what will temperature of the gas be?
130 L 85 L

523K
x
x  340 K
x  69 0 C
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