# Kinetic Theory of Gases Boyle's and Charles' Gas Laws

Dispatch
Draw a picture of a gas in a container. Then
1. What sort of path do the molecules travel in?
2. Do they all move at the same speed?
3. When temperature increases, what happens to
the motion of the particles?
4. What happens when they collide with the
walls of the container?
5. Why is it possible to compress a gas?
6. How does a gas exert pressure?
1
Kinetic Theory of Gases
Boyle’s Law, Charles’s Law
2
Pressure of a Gas Depends on:
• Number of molecules
• Volume in which they are
contained
• Average kinetic energy of
molecules
3
Boyle’s Law
If the amount and temperature
of a gas remain constant, the
pressure exerted by the gas
varies inversely with the
volume.
P  1/V or PV = k
4
Applying Boyle’s Law
P1V1 = P2V2
If some oxygen gas collected at
760 mm Hg is allowed to expand
from 5.0 L to 10.0 L without
changing the temperature, what
pressure will the oxygen gas
exert?
380 mm Hg
5
Charles’s Law
The volume of a quantity of gas,
held at a constant pressure,
varies directly with the Kelvin
temperature.
VT
or
V = kT
6
Applying Charles’ Law
V1/T1 = V2/T2
A 225 mL volume of gas is
collected at 58.0 oC. What volume
would this sample of gas occupy at
standard temperature, assuming
pressure remains constant.
186 mL
7
Combining Boyle’s and Charles’ Laws
P1V1 = P2V2
T1
417 mL
T2
The volume of a
gas measured at
755 mm Hg and
o
23.0 C is 455
mL. What is the
volume of the gas
at STP?
8
Sample Problem
At STP, the volume of a gas is
325 mL. What volume does it
occupy at 20.0 oC and 756 mm
Hg?
351 mL
9
Sample Problem
A gas occupies a volume of 245
mL at 30.0 oC and 746 mm Hg.
Assuming no change in
pressure, at what temperature
will the gas occupy a volume of
490. mL?
606 K
10
Sample problem
A chemist collects 8.00 mL of
gas at STP. If the temperature
remains constant, at what
pressure will the gas volume be
reduced to 4.00 mL?
1520 mm Hg
11
Gay-Lussac’s Law
The pressure of a quantity of
gas, held at a constant volume,
varies directly with the Kelvin
temperature.
PT
or
P = kT
12