The kinetic theory of gases and the
gas laws
Kinetic theory/ideal gas
We can understand the behaviour of
gases using a very simple model, that of
an “ideal” gas.
The model makes a
few simple assumptions;
Ideal gas assumptions
• The particles of gas (atoms or molecules)
obey Newton’s laws of motion.
You should know these
by now!
Ideal gas assumptions
• The particles in a gas move with a range
of speeds
Ideal gas assumptions
• The volume of the individual gas particles
is very small compared to the volume of
the gas
Ideal gas assumptions
• The collisions between the particles and
the walls of the container and between the
particles themselves are elastic (no kinetic
energy lost)
Ideal gas assumptions
• There are no forces between the particles
(except when colliding). This means that
the particles only have kinetic energy (no
potential)
Ideal gas assumptions
• The duration of a collision is small
compared to the time between collisions.
Pressure – A reminder
Pressure is defined as the normal
(perpendicular) force per unit area
P = F/A
It is measured in Pascals, Pa (N.m-2)
Pressure – A reminder
What is origin of the pressure of a gas?
Pressure – A reminder
Collisions of the gas particles with the side
of a container give rise to a force, which
averaged of billions of collisions per
second macroscopically is measured as
the pressure of the gas
Change of
momentum
The behaviour of gases
The behaviour of gases
http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp
When we heat a gas at constant volume,
what happens to the pressure? Why?
Let’s do it!
The behaviour of gases
http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp
When we heat a gas at constant volume,
what happens to the pressure? Why?
P α T (if T is in
Kelvin)
The behaviour of gases
When we compress
(reduce the volume) a
gas at constant
temperature, what
happens to the
pressure? Why?
Let’s do it!
The behaviour of gases
When we compress
(reduce the volume) a
gas at constant
temperature, what
happens to the
pressure? Why?
pV = constant
The behaviour of gases
When we heat a gas a constant pressure,
what happens to its volume? Why?
The behaviour of gases
When we heat a gas a constant pressure,
what happens to its volume? Why?
V α T (if T is in
Kelvin)
Explaining the behaviour of
gases
In this way we are explaining the
macroscopic behaviour of a gas (the
quantities that can be measured like
temperature, pressure and volume) by
looking at its microscopic behaviour (how
the individual particles move)
The gas laws
We have found experimentally that;
At constant temperature, the pressure of
a fixed mass of gas is inversely
proportional to its volume.
p α 1/V
or
pV = constant
This is known as Boyle’s law
The gas laws
At constant pressure, the volume of a
fixed mass of gas is proportional to its
temperature;
VαT
or
V/T = constant
This is known as Charle’s law
If T is in Kelvin
The gas laws
At constant volume, the pressure
of a fixed mass of gas is
proportional to its temperature;
pαT
or
p/T = constant
If T is in Kelvin
This is known as the Pressure law
The equation of state
By combining these three laws
pV = constant
V/T = constant
p/T = constant
We get pV/T = constant
Or
p1V1
T1
Remember, T
must be in
Kelvin
= p2V2
T2
An example
At the top of Mount Everest the temperature is
around 250K, with atmospheric pressure around
3.3 x 104 Pa. At seas level these values are
300K and 1.0 x 105 Pa respectively. If the
density of air at sea level is 1.2 kg.m-3, what is
the density of the air on Mount Everest?
“Physics”, Patrick Fullick, Heinemann
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa
respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the
air on Mount Everest?
Take 1kg of air at sea level
Volume = mass/density = 1/1.2 = 0.83 m3.
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa
respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the
air on Mount Everest?
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
At the top of Mount Everest
p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
An example
At the top of Mount Everest the temperature is around 250K, with atmospheric
pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa
respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the
air on Mount Everest?
Therefore at sea level
p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.
At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.
p1V1/T1 = p2V2/T2
(1.0 x 105 Pa x 0.83 m3)/300K = (3.3 x 104 Pa x V2)/250K
V2 = 2.1 m3,
This is the volume of 1kg of air on Everest
Density = mass/volume = 1/2.1 = 0.48 kg.m-3.
pV = constant
T
The equation of state
Experiment has shown us that
pV = nR
T
Where n = number of moles of gas and R = Gas constant
(8.31J.K-1.mol-1)
Remember, T
must be in
Kelvin
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x
1023 molecules. What is the pressure in
the container?
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x
1023 molecules. What is the pressure in
the container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Sample question
• A container of hydrogen of volume 0.1m3
and temperature 25°C contains 3.20 x
1023 molecules. What is the pressure in
the container?
# moles = 3.20 x 1023/6.02 x 1023 = 0.53
P = RnT/V = (8.31 x 0.53 x 298)/0.1 = 1.3 x 104 N.m-2
K.A.Tsokos “Physics for the IB Diploma” 5th Edition
Questions!
Page 84
Questions 15,16,17 and
18
Page 85
Question 19
Answers:Answers