# Charles's Law ```January 31, 2008
•
•
•
•
Go over homework
Charles’s Law
Homework -- Page 480 -- 11, 13, 14bc,
15bc, 17, 19, 20
Charles’s Law
Timberlake, Chemistry 7th Edition, page 259
Charles's Law
If n and P are constant, then
V = (nR/P) = kT
This means, for example, that
Temperature goes up as Pressure goes up.
V and T are directly related.
A hot air balloon is a good example of Charles's law.
Jacques Charles
(1746 - 1823)
Isolated boron and studied gases.
Balloonist.
Temperature
• Raising the temperature of a gas increases
the pressure if the volume is held constant.
• The molecules hit the walls harder.
• The only way to increase the temperature at
constant pressure is to increase the volume.
300 K
• If you start with 1 liter of gas at 1 atm
pressure and 300 K
• and heat it to 600 K one of 2 things
happens
600 K
300 K
• Either the volume will increase
to 2 liters at 1 atm
300 K
the pressure will increase to 2 atm.
600 K
Charles’s Law
Timberlake, Chemistry 7th Edition, page 259
Charles’s Law
Charles’ Law
Volume Temperature
V/T
The
volume
and
absolute
(mL)
(K)
(mL / K)
temperature
(K) of a gas
are
40.0
273.2
0.146
directly
related
44.0
298.2
0.148
–at constant
mass &amp; pressure
47.7
323.2
0.148
51.3
348.2
0.147
V
T
V
k
T
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Charles’ Law
The volume and absolute
temperature (K) of a gas are
directly related
–at constant mass &amp; pressure
V
T
V
k
T
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Practice Problem – Charles’s Law
A child blows a soap bubble that
contains air at 28 &deg;C and has a
volume of 23 cm3 at 1 atm. As
the bubble rises, it encounters
a pocket of cold air
(temperature 10 &deg;C). If there is
no change in pressure, will the
bubble get larger or smaller as
the air inside cools to 18 &deg;C?
Calculate the new volume of
the bubble.
V1
T1
23 cm3
28 C
23 cm3
301K
V2
T2


V2
18 C

V2
291K
V2 = 22 cm2
Practice Problem – Charles’s Law
A sample of gas at 15 &deg;C (at 1
atm) has a volume of 2.58 L.
The temperature is then raised
to 38 &deg;C. Does the volume of
the gas increase or decrease?
What is the new volume?
V1
T1

2.58 L
15 C
2.58 L
288 K

V2
38 C

V2
311K
V2
T2
V2 = 2.79 L
• For a gas at constant temperature
and pressure, the volume is
directly proportional to the number
of moles of gas
V1
n1

V2
n2
At constant temperature and pressure
Temperature must be converted to Kelvin
This means, for example, that
As the number of moles goes up volume goes up.
n and V are directly related.
Suppose we have a 12.2 L sample
V1
V2
containing 0.50 mol of oxygen gas, O2,

n1
n2
at a pressure of 1 atm and a
temperature of 25 &deg;C. If all of this
oxygen is converted to ozone, O3, at
3 O2  2 O3
the same temperature and pressure,
what will be the volume of the ozone
formed?
2 mol O3
0.50 mol O2 x
 0.33 mol O3
3 mol O2
12.2L
0.50 mol

V2
0.33 mol
V2 = 8.1 L
A 1.5 mol sample of helium at a certain temperature and pressure has
a volume of 31.4 L. A second sample of helium at the same
temperature and pressure has a volume of 42.4 L. How many
moles of helium are in the second sample?
V1
n1
31.4 L
1.5 mol

V2
n2

42.4 L
n2
n2 = 2.03 mol
```