Kinetic Theory
close particles
strong
attraction




(Essay)
distant
particles
weak
attraction
All matter is made up of very tiny particles, which are
constantly in motion.
The molecules repel other strongly when they are forced
too close together.
When they are slightly apart, they attract each other.
When they are far apart, they hardly attract or repel each
other at all.
Interpreting temperature by
kinetic theory



The motion of the particles in a body depends on its
temperature.
When the temperature rises, particles vibrate more rapidly
or move faster. Hence, the temperature of the body is a
measure of the average kinetic energy of the particles.
Two bodies have the same temperature if particles in each
body have the same average kinetic energy.
A gas model (Essay)
Particles move
around freely at
high speed



container
Gases have no fixed volume and shape.
They can quickly fill any space available.
In a gas, particles are very far apart. Therefore, the
attraction among them is weak.
A gas model
Particles move
around freely at
high speed



container
video
The particles move at random at very high speeds (500
ms-1 at room temperature!).
They collide with each other and bombard the walls of
the container.
When gas molecules hit the walls of container, they
produce forces on the surface, which give rise to gas
pressure.
Gas pressure
Pressure is defined as the perpendicular force exerted per unit
area.
Force perpendicular to the surface
Pressure
Area of surface
F
P
A
Pressure is measured in N m-2 or pascal (Pa).
1 Pa = 1 N m-2.
Normal atmospheric pressure or 1 standard pressure
= 1.01 x 105 Pa
Example 1
Find the pressure exerted on the 5 kg block in each of the following
cases.
(a)
10 cm
5 cm
Solution:
P= F/A
= 50 / (0.05 x 0.1)
= 10000 Pa
Example 1
Find the pressure exerted on the 5 kg block in each of the
following cases.
10 cm
(b)
5 cm
30o
Solution:
P= F/A
(F = normal reaction)
= 50 cos30o / (0.05 x 0.1)
= 8660 Pa
Example 2

The figure shows a cylinder with a
piston. If the cross section area of the
cylinder is 20 cm2, and find the
pressure of the gas. Given that the
atmospheric pressure is 1.01 x 105 Pa.

Answer: 2.05 x 105 Pa
0.2 kg
Measuring gas pressure
 A Bourdon
gauge is used to measure
gas pressures.
Measuring gas pressure


A bourdon gauge works in the same way as a paper tube whistle.
When a paper tube whistle is blown, it uncoils.
In a Bourdon gauge, the gas pressure makes the curved metal tube
stretch out slightly. This causes the pointer to move round the scale.
The higher the gas pressure, the more the pointer moves.
Pressure, volume and temperature
 When
a gas is heated, its pressure P,
volume V and temperature T will change.
We are going to investigate these relations.
Boyle’s law (P-V relationship)
 In
1660, Robert Boyle investigated the
relationship between the volume of a fixed
amount of gas and its pressure, keeping the
temperature of the gas constant.
Video
Experiment



Press the air pump gently to
change the pressure and the
volume of the trapped gas in
the glass tube without
changing its temperature.
Measure the pressure and the
volume of the gas.
Repeat the above step to get
more sets of pressure and
volume.
Results
P ∝ 1/V or PV = constant or P1V1 = P2V2
Boyle’s law: For a fixed mass of gas at a constant temperature,
the product of its pressure and volume is constant.
Example 2




The air pressure inside an inflated ball is 2 x 105 Pa.
The ball is now squeezed to one-third of its original
volume. Find the new pressure inside the ball.
Solution:
Let V be the original volume of the ball.
By P1V1 = P2V2
2 x 105 V = P2(V/3)
P2 = 6 x 105 Pa
Thus, the new pressure is 6 x 105 Pa
Charles’ law (V – T relationship)
 In
this experiment, we keep the pressure of
the gas constant and try to investigate the
relation between its volume and temperature.
Experiment



Gas column


Heat the water until its temperature is
increased by 10oC.
Record the volume and temperature of
the trapped gas.
Repeat the above step to get more sets
of volume and temperature.
The mercury thread should be able to
move up and down the capillary tube
freely.
The friction between the mercury thread
and the capillary tube should be small so
that the pressure of the trapped gas is
equal the atmospheric pressure.
Charles’ law (V – T relationship)



Gas column


simulation
Heat the water until its temperature
is increased by 10oC.
Record the volume and temperature
of the trapped gas.
Repeat the above step to get more
sets of volume and temperature.
The mercury thread should be able to
move up and down the capillary tube
freely.
The friction between the mercury
thread and the capillary tube should
be small so that the pressure of the
trapped gas is equal the atmospheric
pressure.
Absolute zero



The straight line cuts the axis at 273oC which is known as the
absolute zero of temperature.
Absolute zero is the theoretical
lowest attainable temperature.
Theoretically, the gas at absolute
zero occupies zero volume and
exerts no pressure. However, in
practice, it will liquefy before it
can reach this temperature.
Kelvin scale

When the temperature is measured in Kelvin (K), the volume of
the gas is directly proportional to its temperature shown in the
figure below.
volume V

0
Temperature T (K)
Note: Absolute temperature (K) = Celsius temperature + 273
i.e. melting point of ice: 0oC = 273 K
room temperature:
27oC = (27 + 273) K = 300 K
boiling point of water: 100oC = (100 + 273) K = 373 K
volume V
Temperature T
(K)
0
V  T or
V
 constant
T
V1 V2
or

T1 T2
Charles’s law: For a fixed mass of gas at a constant pressure, its
volume is directly proportional to its absolute temperature.
Example 2
An inflated balloon contains 4 x 10-3 m3 of air at 27oC. It is
put into a large tank of liquid nitrogen at -173 oC. What is the
new volume of the balloon?


Solution:
By Charles’ law,
V1/T1 = V2/T2
4 x 10-3 / (27 + 273) = V2 / (-173 + 273)
V2 = 1.33 x 10-3 m3
Thus, the new volume of the balloon is 1.33 x 10-3 m3
Pressure law (P – T relationship)




Experiment
Heat the water unit its
temperature increases by about
10oC. Stir the water thoroughly
and wait for several minutes to
allow the air in the flask to
reach the temperature of the
water.
Measure the temperature of the
air by the thermometer and its
pressure by the Bourdon gauge.
Repeat the above steps to get
more sets of pressure and
temperature.
Results

The pressure of the gas increases linearly with its temperature.
When the temperature is measured in Kelvin (K), the pressure is
directly proportional to its temperature shown in the figure below.
pressure P
0
Temperature T (K)
P1 P2
P

 constant or
P  T or
T1 T2
T
Pressure law: For a fixed mass of gas at a constant volume,
its pressure is directly proportional to its absolute temperature.
Example 3
When the flask used in the experiment is put into a bath of
melting ice, the pressure is 8.7 x 104 Pa. What is the pressure
when the flask is put into a bath of boiling water?
Solution:



Temperature of melting ice = 0 oC = 273 K
Temperature of boiling water = 100 oC = 373 K
By pressure law, P1/ T1 = P2 / T2
8.7 x 104 / 273 = P2 / 373
P2 = 1.19 x 105 Pa
Thus, the new pressure is 1.19 x 105 Pa
General gas law

Boyle’s law:
Charles’ law:
Pressure law:

All the gas laws can be summarized by a general equation:


PV = constant if T is constant
V/T = constant if P is constant
P/T = constant if V is constant
PV
 constant or
T
P1V1 P2V2

T1
T2
Example 4
A weather balloon contains 5 m3 of helium at the normal
atmospheric pressure of 100 k Pa and at temperature of 27 C.
What will be its volume when it rises to an altitude where the
pressure is 80 k Pa and the temperature is 7 C?
 Solution:
P1V1 P2V2

By
T1
T2
(100 k)(5)/(27 + 273) = (80 k)(V2) / (7 + 273)
V2 = 5.83 m3
Thus, its new volume is 5.83 m3
Equation of state and molar gas constant
For 1 mole of gas (i.e. 6.02 x 1023 gas particles),
PV
R
T
(same for all gases) where R is called the universal molar
gas constant. R = 8.31 J mole-1 K-1
For n moles of gas,
PV
 nR  PV  nRT
T
Equation of state and molar gas constant
If m is the mass of a gas and M is the mass of 1 mole of
m
gas, the number of mole n 
M
m
RT
From PV  nRT , we have PV 
M
If N is the number of molecules in a gas and NA = Avogadro
N
number = 6.02 x 1023, the number n 
of mole
NA
N
From PV  nRT , we have PV 
RT
NA
.
R = 8.31 J K-1 mol-1
Example 5
Show that the volume of 1 mole of gas occupied at s.t.p. is 22.4
litres.
Note: s.t.p. stands for standard temperature and pressure.
Standard temperature = 0oC or 273 K
Standard pressure = 1 atmospheric pressure
= 1.01 x 105 Pa




Solution
By PV = nRT
(1.01 x 105)V = (1)(8.31)(273)
V = 0.0224 m-3 or 22.4 litres
Ideal gases


An ideal gas is a gas that obeys Boyle’s law (PV is
constant) for all pressures and temperatures.
However, real gases such as hydrogen, oxygen, nitrogen
or carbon dioxide are not ideal gases but they behave like
an ideal gas at high temperature and low pressure.
Example 6


Two insulated gas cylinders A and B are
connected by a tube of negligible volume
as shown.
Each cylinder has an internal volume of 2
x 10-2 m3. Initially, the tap is closed and
cylinder A contains 1.2 mol of an ideal gas
at a temperature of 37oC. Cylinder B
contains the same ideal gas at pressure
1.2 x 105 Pa and temperature 37 oC.
(a)
Calculate the amount, in mol, of the gas in cylinder B.
Solution:
Consider the gas in cylinder B
By PV = nRT
(1.2 x 105 )(2 x 10-2) = n(8.31)(37 + 273)
n = 0.932
The amount of gas in cylinder B is 0.932 mol
Example 6


Two insulated gas cylinders A and B are
connected by a tube of negligible volume
as shown.
Each cylinder has an internal volume of 2
x 10-2 m3. Initially, the tap is closed and
cylinder A contains 1.2 mol of an ideal gas
at a temperature of 37oC. Cylinder B
contains the same ideal gas at pressure
1.2 x 105 Pa and temperature 37 oC.
(a)
Calculate the pressure of the gas in cylinder A.
Solution:
Consider the gas in cylinder A
By PV = nRT
P(2 x 10-2) = (1.2)(8.31)(37 + 273)
P = 1.55 x 105 Pa
The pressure of gas in cylinder A is 1.55 x 105 Pa
Example 6


Two insulated gas cylinders A and B are
connected by a tube of negligible volume
as shown.
Each cylinder has an internal volume of 2
x 10-2 m3. Initially, the tap is closed and
cylinder A contains 1.2 mol of an ideal gas
at a temperature of 37oC. Cylinder B
contains the same ideal gas at pressure
1.2 x 105 Pa and temperature 37 oC.
(b) determine the final pressure of the
gas in the cylinder.
Solution
By PV = nRT
P(4 x 10-2) = (1.2 + 0.962)(8.31)(37 + 273)
P = 1.37 x 105 Pa
1.2 mol
0.962 mol
Example 7
There are two containers X and Y filled with the same type of ideal
gas as shown. They are connected by a tube. A steady state is
obtained with X held at 100 K and Y at 400 K. If the volume of X is
half that of Y and the mass of gas in X is m, what is the mass of gas
in Y, in terms of m?


Solution:
Let P be the pressure of the gas and V be the volume of
container X
For the gas in container Y, P(2V) = nYR(400) --- (1)
For the gas in container X, P(V) = nXR(100) --- (2)
(1)/(2):
2 = 4nY/nX
nY = ½ nX
Mass of gas in Y = ½ mass of gas in X = ½ m
Example 7 root mean square
Find (a) the mean (b) mean square, and (c) the root mean
square of 1, 3, 5, 7 and 9.
Solution:
 Mean = (1 + 3 + 5 + 7 + 9) / 5 = 5
 Mean square
= (12 + 32 + 52 + 72 + 92)/5
= (1 + 9 + 25 + 49 + 81) / 5 = 33
 Root mean square = [(12 + 32 + 52 + 72 + 92)/5] ½ = 5.74
Example 8
It is given that the density of hydrogen gas at s.t.p. is 0.09 kg
m-3. Find the r.m.s. speed of hydrogen molecules.
Solution:
 By P = ⅓(rcr.m.s.2)
1.01 x 105 = ⅓(0.09cr.m.s.2)
cr.m.s. = 1835 ms-1
Distribution of molecular speeds



Particles move around container
freely at high speed
Simulation 1
A gas contains a large number of
molecules in rapid motions.
In each collision between
molecules, some molecules gain
energy while others lose energy.
As a result, each molecule has a
different speed.
Simulation 2
Maxwell distribution



Note:
Maxwell distribution is
not symmetric.
When temperature
increases, the
distribution curve
flattens out but the area
under the curve remains
unchanged.
Most probable speed c0: speed possessed by the most number of
molecules
Number of molecules with speed c
Mean speed cm: mean speed of molecules
Maxwell distribution
c c   c
cm 
1
2
N
N
Root mean square speed cr.m.s.:
0
co cm cr.m.s.
It is found that c0 < cm < cr.m.s.
c/m s-1
c1  c 2   c N
N
2
c r .m. s . 
2
2
Temperature, molar mass and root
mean square speed
1
PV  Nmc r2.m.s and
3
1
2
Nmc
we have
r .m . s  nRT
3
From
PV  nRT
For 1 mole of gas, n = 1, N = NA and the mass of the gas =
molar mass Mm = mNA
1
Nmc r2.m.s  nRT
3
3RT
1
2
N A mc r .m.s  RT ⇒ c r .m.s 
⇒
Mm
3
By cr .m.s
3RT

Mm
Note:
cr.m.s  T
 cr.m.s. increases with temperature.
 cr.m.s. decreases with the mass of the molecule or molar
mass of the gas.
1
c r . m. s 
Mm
Example 8
Find the root mean square speed of hydrogen molecules at
27oC.
It is given that R = 8.31 J K-1 mol-1.


Solution:
By cr .m.s

3RT

Mm
cr.m.s. = (3 x 8.31 x 300 / 0.002)½ = 1934 ms-1
Example 9
Find the ratio of the root mean square speed of hydrogen to
that of oxygen.
3RT

Mm

By c r .m.s

For hydrogen: ch = (3RT/0.002) ½
For oxygen: co = (3RT/0.032) ½
ch / co = (0.032/0.002)½ = 4
the ratio of the root mean square speed of hydrogen
to that of oxygen is 4.



Translational K.E. and Rotational K.E.

Translational motion

Rotational motion

Translational + rotational motion
Monoatomic and diatomic gases
 Monoatomic
means having only one atom
in the molecule. All of the noble gases are
monoatomic.
 Diatomic means having two atoms in the
molecule. Examples are oxygen, nitrogen
kinetic energy of a monoatomic gas

Translational K.E. = ½ mc2

Rotational motion = 0 (neglected)
Translational K.E. of a monoatomic gas
1
From PV  Nm c 2 and PV  nRT
3
1
Nmc 2  nRT
we have
3
For 1 mole of gas, n = 1, N = NA and
the mass of the gas = molar mass Mm = mNA
1
2
1
2
N
m
c
 RT
Nmc  nRT ⇒
A
3
3
Average translational kinetic energy of a molecule
R
1 2 3 R
3
 23
1
k


1
.
38

10
JK
where
T  kT
= mc 
NA
2
2 NA
2
which is known as Boltzmann constant.
Note:

All types of gas molecule have the same
average translational kinetic energy at the same
temperature. (  1 mc 2  3 kT
)
2

2
For diatomic gases such as hydrogen, oxygen
and nitrogen, the average kinetic energy
5
(translational + rotational) of a molecule = kT .
2
(Out of syllabus)
Avogadro’s law

Avogadro’s law states that under the same temperature
and pressure, equal volumes of different gases contain an
equal number of molecules.
Proof
Consider 2 gases under the same temperature, pressure and volume.
Let N1 and N2 be the number of molecules in gas 1 and gas 2
respectively.
1
For gas 1: P1V1  N 1 m1 c1 2
3
1
2
For gas 2: P2V2  N 2 m2 c 2
3
Same volume and pressure ⇒ V1 = V2 and P1 = P2
1
1
2
2
N
m
c

N
m
c
∴
1 1 1
2 2 2
3
3
Same temperature ⇒ same average kinetic energy of a molecule
∴
1
2
mc  m c
2
1 1
1
2
2
2 2
1
2
1
2

N
m
c
N 1 m1 c1
2 2 2
⇒ N1 = N2
3
3
Dalton’s law of partial pressures

Dalton's law of partial pressures states that the total
pressure exerted by a mixture of gases is the sum of the
pressures exerted by each gas if it were allowed to occupy
the whole volume alone.
p1
p
p2
p = p1 + p2
Dalton’s law of partial pressures
PV = nRT
p1
p2
n1 RT
P1 
V
n2 RT
P2 
V
P = P1 + P2
p
P

n1  n2 RT

V