An inflatable pool float may
seem quite firm as it sits on a deck in the hot sun. However, minutes after you toss
to float into the cold pool, the same float may seem under-inflated. You may suspect
that the float has developed a slow leak, but that may not be the most likely explanation
for the apparent loss of air pressure. It may be that Charles's law is responsible.
Charles's law, discovered by Jacques Charles, states that the volume of a
quantity of gas, held at constant pressure, varies directly with the Kelvin temperature.
Gases expand as they are heated
and they contract when they are cooled. In other words, as the temperature of a
sample of gas at constant pressure increases, the volume increases. As the
temperature goes down, the volume decreases as well. The mathematical expression for
Charles's law is shown below:
V1/T1 = V2/T2
Remember that Charles's law calculations
must be done in the Kelvin scale.
Example 1. A 250 cm3 sample of neon is
collected at 44.0 oC. Assuming the pressure remains constant, what
would be the volume of the neon at standard temperature?
Solution:
First change the Celsius temperature to Kelvin.
K = oC + 273
K = 44.0 oC + 273
K = 317
Now list the given quantities and the unknown.
T1 = 317 K
V1 = 250 cm3
T2 = 273 K (standard temperature in Kelvin
V2 = ?
Now predict the results. The temperature is going down, so the volume
must go down as well.
Write the original formula and then isolate the unknown:
a) V1 V2
---- = ----
T1 T2
b) T2 x V1 V2
x T2
---- = ----
T1 T2
c) V2 = T2 x V1
-----------
T1
Now, substitute, solve and round to correct significant digits.
T1 = 317 K
V1 = 250 cm3
T2 = 273 K (standard temperature in Kelvin
V2 = ?
V2 = T2 x V1
-----------
T1
V2 = 273 K x 250 cm3
-----------------------
317 K
V2 = 215.2996845 cm3
V2 220 cm3
Finally, we check that our prediction was correct.
The volume did go down.
Example 2. A sample of oxygen gas has a volume of
2.73 dm3 at 21.0 oC. At what temperature would the gas have a
volume of 4.00 dm3?
Solution:
First change the Celsius temperature to Kelvin.
K = oC + 273
K = 21.0 oC + 273
K = 294 K
Now list the given quantities and the unknown.
T1 = 294 K
V1 = 2.73 dm3
T2 = ?
V2 = 4.00 dm3
Now predict the results. The temperature must go up in order for the
volume to go up.
Write the original formula and then isolate the unknown:
a) V1 V2
---- = ----
T1 T2
b) T2 = T1 x V2
-----------
V1
Now, substitute, solve and round to correct significant digits.
T1 = 294 K
V1 = 2.73 dm3
T2 = ?
V2 = 4.00 dm3
T2 = T1 x V2
-----------
V1
T2 = 294 K x 4.00 dm3
-----------------------
2.73 dm3
T2 = 430.7692308 K
T2 431 K
Finally, we check that our prediction was correct.
The temperature did go up.
Now, follow the links below and practice
what you have learned.
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
