5.2
The kinetic theory
Smelling durian
Explanation of the gas laws in terms of
molecular motion
Check-point 6
Assumptions of the kinetic theory of ideal
gases
p-V relationship due to molecular motion
Temperature and molecular motion
Check-point 7
Book 1 Section 5.2 The kinetic theory
Smelling durian
Do you know why the smell can spread
without the help of wind?
Related to the movement of gas particles.
P.2
Book 1 Section 5.2 The kinetic theory
1 Explanation of the gas laws in
terms of molecular motion
a Random motion of gas particles
Gas laws tell you what will happen to a
gas under certain conditions, but do not
tell you why.
Kinetic theory explains the reasons behind.
Expt 5d
Observing random motion of molecules
P.3
Book 1 Section 5.2 The kinetic theory
a Random motion of gas particles
The smoke particles move about
constantly along zigzag paths.
Einstein proposed that:
• bombarded by a large
number of air molecules
• from all directions but
not in equal number
 Random motion
Simulation
5.4 Random motion of molecules
P.5
Book 1 Section 5.2 The kinetic theory
1 Explanation of the gas laws in
terms of molecular motion
b Kinetic theory model
The three dimensional kinetic
theory model consists of a
transparent tube containing a
large number of ball-bearings.
Tube: closed by a movable
polystyrene piston
Motor-driven vibrator: sets
the ball-bearings in motion
P.6
Book 1 Section 5.2 The kinetic theory
b Kinetic theory model
The ball bearings represent gas molecules.
Physical quantity
Kinetic theory model
Pressure
Weight of the piston
(and cardboard discs)
Temperature
Voltage applied to
the motor
Volume
Height of the piston
Expt 5e
The three-dimensional kinetic
theory model
Book 1 Section 5.2 The kinetic theory
P.7
b Kinetic theory model
Step 1 of the expt illustrates the following:
Condition
voltage –
weight of
piston 
height of
piston –
voltage 
Result
height of
piston 
weight of
piston 
Interpretation
temp constant,
pressure   volume 
Simulating Boyle’s law
volume constant,
temp   pressure 
Simulating pressure law
P.9
Book 1 Section 5.2 The kinetic theory
b Kinetic theory model
Condition
Result
weight of
piston –
height of
piston 
voltage 
weight of piston –
voltage –
number of balls 
height of
piston 
Interpretation
pressure constant,
temp   volume 
Simulating Charles’ law
pressure and temp
constant,
number of molecules 
 volume 
In step 2, the big ball moves in a jerky manner.
 illustrates the random motion of smoke particles
P.10
Book 1 Section 5.2 The kinetic theory
1 Explanation of the gas laws in
terms of molecular motion
c Qualitative explanation using kinetic theory
According to the kinetic theory,
• a gas is made up of a huge
number of tiny particles,
each of which has a mass.
• All the particles are in
constant, random motion.
• They collide with each other and the walls of
the container continually.
Book 1 Section 5.2 The kinetic theory
P.11
c Qualitative explanation using kinetic
theory
The theory can be used to explain p andT:
• p: force exerted by the molecules colliding on the
container walls.
p  if the molecules hit the walls more often
or with a greater ∆ momentum.
• T : ave KE of the molecules.
T   molecules gain energy  move faster
Simulation
5.5 Kinetic theory and gas laws
P.12
Book 1 Section 5.2 The kinetic theory
c Qualitative explanation using kinetic
theory
Interpretation of Boyle’s law:
When a gas is compressed, the molecules have
less volume to move in.
They hit the walls more often  greater pressure
P.13
Book 1 Section 5.2 The kinetic theory
c Qualitative explanation using kinetic
theory
Interpretation of Pressure law:
When temp , the molecules move faster.
Volume fixed  molecules hit the walls more often.
 speed  ∆ momentum in each collision 
∴ greater pressure
P.14
Book 1 Section 5.2 The kinetic theory
c Qualitative explanation using kinetic
theory
Interpretation of Charles’ law:
Temp   molecules move faster
 Frequency of collisions and ∆ momentum in
each collision 
To keep pressure constant  volume must 
(so that the frequency of collisions )
P.15
Book 1 Section 5.2 The kinetic theory
c Qualitative explanation using kinetic
theory
Relationship between volume and number of
molecules:
Number of molecules   frequency of collisions 
To keep pressure constant, volume must 
P.16
Book 1 Section 5.2 The kinetic theory
Check-point 6 – Q1
In the kinetic theory model, state what each
of the following represents:
(a) ball bearings
Gas molecules.
(b) weight of the piston
Pressure.
(c) voltage applied
Temperature.
(d) height of the piston
Volume.
P.17
Book 1 Section 5.2 The kinetic theory
Check-point 6 – Q2
When pushing an inverted empty glass into water,
why does water not rise into it?
A The air molecules inside
the glass bombard the
water surface.
B There is no space in the
air inside the glass.
C The weight of the air
inside the glass is acting
on the water surface.
P.18
Book 1 Section 5.2 The kinetic theory
2 Assumptions of the kinetic theory
of ideal gases
1. All the molecules are identical with the same mass.
2. All the molecules are in constant, random motion.
3. The size of each molecule is negligible.
4. The duration of a collision is negligible.
5. All collisions are perfectly elastic.
6. Intermolecular forces are negligible.
A real gas does not satisfy these assumptions.
 Good approximation for high T and low p
Book 1 Section 5.2 The kinetic theory
P.19
3 p -V relationship due to molecular
motion
Consider a cubic container of length l containing
N identical molecules in random motion.
Suppose a molecule of mass m
travels with a velocity v
perpendicular to wall A.
It collides elastically with the
wall and bounces backwards
with the same speed.
P.20
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
∆ momentum of the molecule = –mv – mv
= –2mv
By conservation of momentum, total momentum
of the molecule and the wall should be the same
before and after collision,
i.e. total ∆ momentum = 0
 ∆ momentum of the wall due to this collision
= +2mv
P.21
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
This molecule will be
back and collide with
wall A again after
travelling through a
round-trip of length 2l.
for every time interval of
2l
v
, the wall will
experience ∆ momentum of +2mv.
P.22
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
Average force exerted on the wall by this molecule:
F=
change in momentum
time interval for the change
=
2mv
2l
=
mv 2
l
v
Pressure exerted on the wall by this molecule:
p=
force
area
=
mv 2
l
l2
=
mv 2
l3
=
mv 2
V
P.23
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
Most molecules collide
with the wall at an angle
 need to resolve the
velocity vectors into
components.
No change in the y-direction
 ∆ momentum in the x-direction only
P.24
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
For an elastic collision,
∆ momentum = –mvx – mvx
= –2mvx
Round-trip time =
2l
vx
 pressure exerted on wall A by the molecule:
p =
mvx 2
V
Book 1 Section 5.2 The kinetic theory
P.25
3 p -V relationship due to molecular
motion
There are N molecules in the container.
 total pressure exerted on the wall:
p =
=
m
V
m
V
(vx1 2 + vx2 2 + … + vxN 2)
N v̅ x 2
where v̅ x 2 =
......................... (1)
vx1 2 + vx2 2 + … + vxN 2
N
P.26
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
By symmetry, v̅ x 2 = v̅ y 2 = v̅ z 2 ..................... (2)
(due to random motion)
By Pythagoras’ theorem, c 2 = vx 2 + vy 2 + vz 2
Mean square value of velocity:
c̅ 2 = v̅ x 2 + v̅ y 2 + v̅ z 2 ..................... (3)
P.27
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
Combining (2) and (3): c̅ 2 = 3v̅ x 2
Substituting this into (1): p =
pV =
1
3
̅ c 2 

N 
3
V
 
m
Nmc̅ 2
The equation is true for containers of all shapes.
P.28
Book 1 Section 5.2 The kinetic theory
3 p -V relationship due to molecular
motion
Example 6
Mean square speed and
pressure
P.29
Book 1 Section 5.2 The kinetic theory
4 Temperature and molecular motion
a Molecular interpretation of temperature
pV =
1
3


1
2
Nmc̅ 2
1
3
mc̅
pV = nRT
and
Nmc̅ 2 = nRT
2
=
3RT
n
2
N
KEaverage =
=
3RT
2NA
3RT
2NA
Book 1 Section 5.2 The kinetic theory
P.31
a Molecular interpretation of temperature
KEaverage =
3RT
2NA
 T (in Kelvin)  KEaverage
For n moles of gas, KEtotal = nNA ×
3RT
2NA
=
∴KEtotal of a gas = total internal energy =
3
2
3
2
nRT
nRT
P.32
Book 1 Section 5.2 The kinetic theory
4 Temperature and molecular motion
b Root-mean-square speed of molecules
Distribution of speed of gas
molecules depends on temp.
Typical speed = root of
̅c 2
 root-mean-square speed
c rms = ̅ c
2
=
3pV
Nm
=
3RT
mNA
Nm : total mass of the gas; mNA : molar mass
Book 1 Section 5.2 The kinetic theory
P.33
Check-point 7 – Q1
Find the velocity of a particle
if the velocity components
along x, y and z directions are
3 m s−1, 4 m s−1 and 5 m s−1
respectively.
v= v
2
= vx 2 + vy 2 + vz 2
= 32 + 42 + 52
= 50
= 7.07 m s–1
Book 1 Section 5.2 The kinetic theory
P.34
Check-point 7 – Q2
Air pressure inside a container of 40 cm3 = 120 kPa
Total mass of the air = 6.5 × 10−5 kg
Mean square speed of the air molecules = ?
By pV =
̅c 2
=
Nmc̅ 2
3
3pV
Nm
=
,
3  120  103  40  10–6
6.5 × 10−5
= 2.22  105 m s–1
P.35
Book 1 Section 5.2 The kinetic theory
Check-point 7 – Q3
Temp of gas X = 100 °C
Mass of a molecule of gas X = 3.34  10–27 kg
Root-mean-square speed of the molecules = ?
(R = 8.31 J mol–1 K–1, NA = 6.02  1023 mol–1)
Root-mean-square speed
=
3RT
mNA
=
3  8.31  (273 + 100)
3.34  10−27  6.02  1023
= 2150 m s–1
P.36
Book 1 Section 5.2 The kinetic theory
4 Temperature and molecular motion
Example 7
Diffusion
P.37
Book 1 Section 5.2 The kinetic theory
4 Temperature and molecular motion
Example 8
Air in the can
P.42
Book 1 Section 5.2 The kinetic theory
The End
P.49
Book 1 Section 5.2 The kinetic theory