1. Book 1 Section 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 1 2 3 4
Book 1 Section 5.2 The kinetic theory

2. 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.
Book 1 Section 5.2 The kinetic theory

3. 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
Book 1 Section 5.2 The kinetic theory

4. 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
Book 1 Section 5.2 The kinetic theory

5. 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
Book 1 Section 5.2 The kinetic theory

6. Book 1 Section 5.2 The kinetic theory
b Kinetic theory model
The ball bearings represent gas molecules.
Physical quantity
Kinetic theory model
Weight of the piston (and cardboard discs)
Pressure
Voltage applied to the motor
Temperature
Volume
Height of the piston
The three-dimensional kinetic theory model
Expt 5e
Book 1 Section 5.2 The kinetic theory

7. Book 1 Section 5.2 The kinetic theory
b Kinetic theory model
Step 1 of the expt illustrates the following:
Condition
Result
Interpretation
voltage –
weight of piston 
temp constant, pressure   volume 
Simulating Boyle’s law
height of piston 
height of piston –
voltage 
volume constant, temp   pressure 
Simulating pressure law
weight of piston 
Book 1 Section 5.2 The kinetic theory

8. Book 1 Section 5.2 The kinetic theory
b Kinetic theory model
Condition
Result
Interpretation
weight of piston –
voltage 
pressure constant, temp   volume 
Simulating Charles’ law
height of piston 
pressure and temp constant,
number of molecules   volume 
weight of piston –
voltage –
number of balls 
height of piston 
In step 2, the big ball moves in a jerky manner.
 illustrates the random motion of smoke particles
Book 1 Section 5.2 The kinetic theory

9. 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

10. 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.
T : ave KE of the molecules.
p  if the molecules hit the walls more often or with a greater ∆ momentum.
T   molecules gain energy  move faster
Simulation
5.5 Kinetic theory and gas laws
Book 1 Section 5.2 The kinetic theory

11. 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
Book 1 Section 5.2 The kinetic theory

12. 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
Book 1 Section 5.2 The kinetic theory

13. 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 )
Book 1 Section 5.2 The kinetic theory

14. 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 
Book 1 Section 5.2 The kinetic theory

15. 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.
Book 1 Section 5.2 The kinetic theory

16. 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.
Book 1 Section 5.2 The kinetic theory

17. 2 Assumptions of the kinetic theory of ideal gases
All the molecules are identical with the same mass.
All the molecules are in constant, random motion.
The size of each molecule is negligible.
The duration of a collision is negligible.
All collisions are perfectly elastic.
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

18. 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.
Book 1 Section 5.2 The kinetic theory

19. 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
Book 1 Section 5.2 The kinetic theory

20. 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. 2l v
for every time interval of , the wall will experience ∆ momentum of +2mv.
Book 1 Section 5.2 The kinetic theory

21. 3 p -V relationship due to molecular motion
Average force exerted on the wall by this molecule:
change in momentum
time interval for the change
2mv 2l v =
mv 2 l =
F =
Pressure exerted on the wall by this molecule:
mv 2 l
l 2 =
force
area
mv 2
l 3 =
mv 2 V =
p =
Book 1 Section 5.2 The kinetic theory

22. 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
Book 1 Section 5.2 The kinetic theory

23. 3 p -V relationship due to molecular motion
For an elastic collision,
∆ momentum = –mvx – mvx = –2mvx 2l vx
Round-trip time =
 pressure exerted on wall A by the molecule:
mvx 2 V = p
Book 1 Section 5.2 The kinetic theory

24. 3 p -V relationship due to molecular motion
There are N molecules in the container.
 total pressure exerted on the wall: m V =
(vx1 2 + vx2 2 + … + vxN 2) p m V =
N vx 2 ̅
(1) =
vx1 2 + vx2 2 + … + vxN 2 N
vx 2 ̅
where
Book 1 Section 5.2 The kinetic theory

25. 3 p -V relationship due to molecular motion
vx 2 ̅
vy 2
vz 2 =
By symmetry,
(2)
(due to random motion)
By Pythagoras’ theorem,
c 2 = vx 2 + vy 2 + vz 2
Mean square value of velocity:
c 2 = vx 2 + vy 2 + vz 2 ̅
(3)
Book 1 Section 5.2 The kinetic theory

26. 3 p -V relationship due to molecular motion
c 2 = 3vx 2 ̅
Combining (2) and (3): m V = p N
c 2 3 ̅
Substituting this into (1): pV 1 3 =
Nmc 2 ̅
The equation is true for containers of all shapes.
Book 1 Section 5.2 The kinetic theory

27. 3 p -V relationship due to molecular motion
Mean square speed and pressure
Example 6
Book 1 Section 5.2 The kinetic theory

28. 4 Temperature and molecular motion
a Molecular interpretation of temperature pV 1 3 =
Nmc 2 ̅
and
pV = nRT 1 3
Nmc 2 ̅
= nRT  1 2
mc 2 ̅
3RT 2 n N =
3RT
2NA = 
3RT
2NA =
KEaverage
Book 1 Section 5.2 The kinetic theory

29. a Molecular interpretation of temperature
3RT
2NA =
KEaverage 
T (in Kelvin)  KEaverage
3RT
2NA 3 2
nRT =
For n moles of gas,
KEtotal = nNA × 3 2
nRT =
∴KEtotal of a gas = total internal energy
Book 1 Section 5.2 The kinetic theory

30. 4 Temperature and molecular motion
b Root-mean-square speed of molecules
Distribution of speed of gas molecules depends on temp.
c 2 ̅
Typical speed = root of
 root-mean-square speed
3pV Nm =
3RT
mNA =
c 2 ̅
c rms =
Nm : total mass of the gas; mNA : molar mass
Book 1 Section 5.2 The kinetic theory

31. Book 1 Section 5.2 The kinetic theory
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 =
= 50
= 7.07 m s–1
Book 1 Section 5.2 The kinetic theory

32. Book 1 Section 5.2 The kinetic theory
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 = ? pV = 3
Nmc 2 ̅ By ,
3pV Nm =
3  120  103  40  10–6
6.5 × 10−5
c 2 = ̅
= 2.22  105 m s–1
Book 1 Section 5.2 The kinetic theory

33. 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  ( )
3.34  10−27  6.02  1023
= 2150 m s–1
Book 1 Section 5.2 The kinetic theory

34. 4 Temperature and molecular motion
Diffusion
Example 7
Book 1 Section 5.2 The kinetic theory

35. 4 Temperature and molecular motion
Air in the can
Example 8
Book 1 Section 5.2 The kinetic theory

36. Book 1 Section 5.2 The kinetic theory
The End
Book 1 Section 5.2 The kinetic theory