Charles’s law states that if a given quantity of gas
is held at a constant pressure, its volume is directly
proportional to the absolute temperature. Think of it
this way. As the temperature of the gas increases, the
gas molecules will begin to move around more quickly
and hit the walls of their container with more
force—thus the volume will increase. Keep in mind
that you must use only the Kelvin
temperature scale when working with temperature in all gas law formulas!

Change in Volume When Temperature Increases

Change in Volume When the Temperature Decreases (Watch
the barrel first)

NOTE** Charles Law
problems must have the
PRESSURE CONSTANT. Try using Charles’s law to solve the following
problem.

Example

A sample of gas at 15ºC and 1 atm has a volume of
2.50 L. What volume will this gas occupy at 30ºC and
1 atm?

OK, Here is our formula, .
So, just plug in the numbers.

NO

As a Chemistry teacher I am
REQUIRED to give you the temperature in Celsius and
make you convert to Kelvin. It is the Law of all
Chemistry Teachers.

So convert the temperatures to Kelvin. T_{1}=
15C +273=288K and T_{2} =30C +273=303K

Now Plug in and Solve

2.50 L

=

V_{2}

V_{2}
= 2.63 L

288K

303K

This makes sense—the temperature is increasing
slightly, so the volume should increase slightly.
Again
be
careful of questions like this. It’s tempting to
just use the Celsius temperature, but you must first
convert to Kelvin temperature (by adding 273) to get
the correct relationships!

Why must the temperature be
absolute?

If temperature is measured on a Celsius (non
absolute) scale, T can be negative. If we plug
negative values of T into the equation, we get
back negative volumes, which cannot exist. In
order to ensure that only values of V= 0 occur,
we have to use an absolute temperature scale
where T= 0. The standard absolute scale is the
Kelvin (K) scale. The temperature in Kelvin can
be calculated via T_{k} = T_{C}
+ 273.15. A plot of the temperature in Kelvin.
Charles' law predicts that volume will be zero
at 0 K. 0 K is the absolutely lowest temperature
possible, and is called absolute zero.

from-www.sparknotes.com/.../
ideal/section2.rhtml

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Answers

1. Calculate the decrease in temperature when 2.00L
at 20.0^{o}C is cooled to 1.00L.

V_{1}=2.00L

V_{1}

=

V_{2}

T_{1}=20.0^{o}C
+ 273=293K

T_{1 }

T_{2}

V_{2}=1.00L

2.00L

=

1.00L

T_{2}=?

293K _{ }

T_{2 }

( 2.00L)(T_{2})
=(1.00L)(293K)

T_{2}=
146.5K

DT=146.5K-273K=-146.5K

2. 600.0mL of air is at 20.0^{o}C. What is the volume at
60.0^{o}C?

V_{1}=600.0mL

V_{1}

=

V_{2}

T_{1}=20.0^{o}C
+ 273=293K

T_{1 }

T_{2}

V_{2}=?

600.0mL

=

V_{2}

T_{2}=60.0^{o}C
+273= 333K

293K _{ }

333K

( 600.0mL)(333K )
=(V_{2})(298K)

V_{2}=
681.9.mL

AP Try These-Highlight
to reveal Answers

1. A ballon
filled to a volume 7.00 x 10^{2}mL at a temperature of
of 20.0^{o}C. The ballon then is cooledat constant
pressure to a temperature of 1.00 x 10^{2}K. What is the
final volume of this balloon?

V_{1}=7.00 x 10^{2}mL

V_{1}

=

V_{2}

T_{1}=20.0^{o}C+
273=293K

T_{1}

T_{2}

V_{2}=?

7.00 x 10^{2}mL

=

V_{2}

T_{2}=1.00 x 10^{2}K

293K

1.00 x 10^{2}K

(1.00 x 10^{2}K)(7.00 x 10^{2}mL)=(V_{2})(293K)

239mL = V_{2 }

2. A sample of gas at 15^{o}C
and 1.0 atm has a volume of 2.58L. What volume
will this gas occupy at 38^{o}C and 1 atm?