Chemistry Reference
In-Depth Information
The Formula for Charles's Law
V
1
T
1
V
2
T
2
=
It is very important that you work with Kelvin temperature, whenever
you do a Charles's Law problem. If you work in Celsius you won't get the
correct answer. You may recall that we discussed the problem with the
Celsius scale in Chapter 2. You can have negative values for temperatures
in the Celsius scale, which, when solving a Charles's Law problem for vol-
ume, would result in a negative value for volume. Can you imagine what a
gas with a negative volume would be like? Neither can I, and that is why we
must always convert our temperatures to Kelvin when solving Charles's
Law problems.
Although you need to work in Kelvin, the temperature that you are
given in a particular problem may be in Celsius degrees. This upsets some
students, and they consider this an example of a “trick” question. There is,
however, a logical reason for your teacher to give you some Charles's Law
problems in Celsius degrees. Most students in chemistry laboratories work
with thermometers marked with the Celsius scale. In these real-life situa-
tions, you will need to convert between temperature units before doing
calculations. Some problems are designed to get you ready for these real-
life experiences.
Now, let's look at an example of a problem that requires Charles's Law
to solve.
Example 1
A student collects a 250. cm
3
sample of a gas at 21.0
o
C. Assuming
the pressure of the gas remains constant, at what temperature will
the volume of this gas be 500. cm
3
?
The first thing that we want to do is change the Celsius degrees to Kelvin,
using the formula K =
o
C + 273 that we learned in Chapter 2. I like to do
that right away, directly on the written problem, to avoid making mistakes.
Because 21 + 273 = 294, I will change the temperature to 294 K, as shown
below.
A student collects a 250. cm
3
sample of a gas at 21.0
o
C 294 K.
Assuming the pressure of the gas remains constant, at what
temperature will the volume of this gas be 500. cm
3
?
