The kinetic molecular theory
Chapter 10 of the textbook
Elementary Physics 1
This is an article from my home-page: www.olewitthansen.dk
Ole Witt-Hansen
1977 (2017)
Contents
1. A model of an ideal gas..........................................................................................................46
2. Experimental demonstration of the motion of the molecules ................................................46
2.1 Diffusion in a glass tube.......................................................................................................46
2.2 Experiments with porous containers ....................................................................................47
2.3 Brownian movements ..........................................................................................................48
2.3.1 The light mill.....................................................................................................................48
3. Collision of molecules against the wall of a container ..........................................................49
4. The equation of state for ideal gases ......................................................................................50
The kinetic molecular theory
46
1. A model of an ideal gas
Hitherto we have described the behaviour of gases from a series of simple experiments from which
we have derived the laws of ideal gases summarized in the equation of state for ideal gasses.
In this section we shall derive the equation of state from a mechanical model, and as a
consequence derive the fundamental relation between the absolute temperature and the mean
kinetic energy of a molecule at that temperature.
We know that all substances consist of molecules, and it seems therefore natural to perceive the
different phases of substances (solid, liquid, gas) from the motion of the molecules. A point of
view that we have already have brought forward in the introduction to thermodynamics.
In the gas phase the molecules move freely having numerous collisions against the walls and
against each other.
In our model for an ideal gas, we shall thus perceive the molecules as small elastic bullets (having
the sizes of 10-10 m) which move with velocities that we shall later calculate to be some hundred
meters per second at room temperature.
If the molecules are confined in a container they collide against the walls more than a hundred
times per second. The force on the walls per unit area, which is the result of these collisions gives
the explanation of what appears as the pressure of the gas.
Before we exploit the model, we shall consider some classical experiments which demonstrate the
dynamic atomic structure of a gas.
2. Experimental demonstration of the motion of the molecules
To observe the motion of the molecules directly one must apply waves, where the wavelength is
comparable to the size of the molecules. But this is not possible, not even in an electron
microscope.
On the other hand there exist numerous simple experimental observations which clearly
demonstrate that the molecules in a liquid or in a gas are in constant motion, even if the gas or
liquid is at rest as a whole.
If you open a bottle of ammonia in a room, some of the ammonia will evaporate, and you may
register the characteristic smell after a short while several meters away from the open bottle.
And this happens even if there has been no convection of the air in the room.
The phenomenon is called disorderly heat motion or diffusion. It is explained by the constant
disorderly motion of the ammonia molecules. Since the diffusion of the ammonia molecules is
accompanied by numerous collisions with the molecules in the air, the diffusion speed is vastly
lower than the velocity of the molecules.
Diffusion should not be confused with air streaming, where the gas moves in a collective motion.
Air streaming is also called convection.
2.1 Diffusion in a glass tube
If you take a glass tube, which is about one meter, and simultaneously put two tufts of absorbent
cotton, where one have been soaked with ammonia (NH3) and the other with hydroichloric acid
(HCl) in each end, you will after a short while register a ring of mist in the tube.
The kinetic molecular theory
47
The ring comes about because the two gases diffuse into the tube, and when the meet they react
chemically, since
NH3 + HCl -> NH4Cl
One should notice, however, that the ring does not manifests itself in the middle of the tube, but it
is displaced to the end with the lighter gas NH3 with atomic weight: 17 = 14 + 3∙1. So it appears
that gasses having a lower atomic weight diffuse faster than heavier gases like HCl with atomic
weight 36.5.
If we measure the time it takes before the ring appears, then we may find an estimate of the
diffusion velocities for the two substances by dividing the two distances by the time.
These velocities are several times less the velocities of the molecules in their free movement, but
nevertheless it seems reasonable to assume that the diffusion velocities are proportional to the
velocities of the free molecules. In any case the experiment shows that the lighter molecules
diffuse faster than the heavier ones.
Later we shall show that molecules of different gasses have the same average kinetic energy.
If we therefore assume that HCl and NH3 have the same (mean) kinetic energy, we may conclude
that the ration between the diffusion velocities is the square root of the inverse ratio between the
m NH 3
v HCl
17
2
2
masses. (2.2) 12 mHCl v HCl  12 m NH 3 v NH 3



 0,68  23
v NH 3
mHCl
38,5
So the distances from the end of the tube to the white ring, should be in the proportion 2:3, in
fairly good accordance with results from the experiments.
2.2 Experiments with porous containers
The freely moving molecules in liquids and gases may also be demonstrated using porous
cylinders. Porous means that they do not leak liquids, but can be penetrated by molecules and ions.
In figure (2.2) a porous cylinder of clay is provided with a cork holding a glass tube that is
connected to a manometer with liquid. The porous clay cylinder is first place in an upside down
cup, and a light gas e.g. Helium is led into the cup. You will observe the pressure rises in the glass
container. This is because helium is lighter, and therefore moves faster than atmospheric air, and it
therefore diffuses into the clay cylinder than the air diffuses out, resulting in a slight increase of
pressure as measured on the manometer.
The kinetic molecular theory
48
If the glass cup is turned, and now a heavier gas e.g. propane is lead down in the cup, we will
observe the opposite picture that the pressure in the clay cylinder is lowered for the same reason as
in the first case, since the heavier molecules diffuse slower.
If the glass cup is removed the pressure will fall below from what it was before the experiment
started. The reason for that is the lighter gas will diffuse out faster the air diffuses in.
In the experiment with the heavier gas, exactly the opposite will happen, but for the same reason.
2.3 Brownian movements
The botanist Robert Brown (1773 – 1858) discovered that pollen from flowers being dispersed in
water and observed through a microscope performed microscopic irregular movements.
His first thought was that it could be living organisms, but the movements continued, even if the
dispersed liquid was heated to the boiling point, so he rejected that hypothesis.
Not until the beginning of the 1900s, it became clear, that the the Brownian movements were
caused by collisions by the fast water molecules and the much heavier pollen particles.
The easiest way to observe Brownian movements is to mix some smoke particles in a gas.
Strongly lit, and observed in a microscope, it is possible to observe very small irregular
movements of he smoke particles. Even if the smoke particles are a million times heavier than the
molecules, their large velocities are sufficient to give the pollen particles enough momentum, so
that it can be observed.
A. Einstein performed some theoretical calculations (random walk) of the Brownian movements.
Thereby he obtained a theoretical formula of how far in average a pollen particle would move
away from a position in a given amount of time. The density of molecules was one of the variables
in the formula, and so he could compute this density of molecules from the observation of pollen.
This brought him in a position to obtain a value for Avogadro’s number.
Since this value was in accordance with the already known value, Einstein’s experiment became a
confirmation of the kinetic molecular theory.
2.3.1 The light mill
A light mill is a an evacuated glass bulb, where a small ”mill” can rotate without friction on the
pin of a needle, as shown in figure (2.3).
On the one side the wings are inked black, while the other side is blank. When the light mill is
placed in the sunlight the mill will start to rotate, as if there was a force acting on the black side.
The phenomenon can easily be explained from the
kinetic molecular theory. The black sides of the wings
will be heated more, because light is absorbed better by
black surfaces, but reflected by blank surfaces.
The heating means the gas molecules near the black
sides of the wings acquire more kinetic energy, (greater
velocity), and therefore they can transfer a larger
amount of momentum when colliding with the black
sides of the wings, than when colliding with the blank
sides.
The kinetic molecular theory
49
3. Collision of molecules against the wall of a container
To be able to calculate the gas pressure from the collisions of the molecules against the walls of a
container, we must first find the momentum that is transferred, when a single molecule hits the
wall. The situation is illustrated in the two figures (3.2) (a) and (b).
In the following, we shall make use of Newton’s 2. law, when written with change of momentum
instead of change of velocity.



 v
We write Newton’s 2. law: F  ma , where the acceleration a 
is the change of velocity per
t


unit time. We have defined the momentum as p  mv , from which it follows that the change in


momentum is: p  mv : We then get:
(3.1)




v mv
F  ma  m

t
t

 p


F
. ( Ft  p is called the momentum of the force).
t

In the figure 3.2 (a) a molecule with momentum p1 approaches perpendicularly to the wall.
The collision is central and elastic. From the theory of central collisions, we know that the


molecule after the collision has the momentum p2 , which is equal to, but opposite directed to p1 ,


so that p2   p1 . The change in momentum is therefore.
(3.3)




pmolecule  p2  p1  2 p1
By any collision the momentum is conserved, which is the same as saying that the sum of changes
of momentum for the two bodies is zero. Therefore the wall gets a change in momentum given by.
(3.4)


pwall  pmolecule  0



pwall  2 p1
If the duration of the collision is Δt, the wall is affected by a force:
(3.5)



p
2p
Fwall  wall  1
t
t
The force from a single collision is of course unnoticeable, but here we must remember that the
number of molecules have the magnitude of Avogadro’s number NA =6.023 1023, while the mass
of the atoms is of magnitude 1 u =1.660 10-27 kg.
The kinetic molecular theory
50
In the figure 3.2 (b) is shown a molecule which approaches the wall at an oblique angle.
We assume that the wall is completely smooth, so there is no frictional force parallel to the wall.

The momentum of the molecule can be dissolved in a component p parallel to the wall, and a

component p perpendicular to the wall.

Since there can be no force component parallel to the wall, then p will remain unaltered.

The momentum p will on the other hand change to the opposite direction, and after the collision

be equal to  p . Thus we have:
(3.6)





pmolecule  p  p  0  ( p  p )
(3.7)


pwall  pmolecule  0

=>


pmolecule  2 p



pwall  2 p  2mv
The transfer of momentum to the wall, at an oblique angle is therefore given by (3.7), where m is

the mass of the molecule, and v is the perpendicular velocity component of the molecule to the
wall. Finally we notice that the speed of the molecule is unaltered, which follows from above, but
it also follows from the fact that the kinetic energy is conserved in an elastic collision.
4. The equation of state for ideal gases
We consider a cylinder shaped container with length L and cross section A, as shown in the figure
(4.1). In the figure is also sketched the trajectory for a molecule having mass m and velocity v.
We showed in the last paragraph, that even if the molecule changes direction in the collisions with
the walls, the speed v is unaltered.
If we submit a three right angled coordinate system, with the x-axis along the longitude of the
   
cylinder, we can dissolve the velocity vector in three components: v  v x  v y  v z
The speed of the molecule is constant in the x-direction. The velocity in the x-direction does not
change by a collision with the side walls, since the parallel velocity does not change in a collision

with the walls, and v x is parallel to the side walls.

The component v x changes to the opposite direction, when colliding with the end walls, but the
speed is unaltered. So without collisions with other molecules the speed of a molecule along the
cylinder is constant.
The kinetic molecular theory
51
The time for a molecule to pass the length L of the cylinder is therefore constant, and it is equal to
t ' L / vx . The time Δt before it hits the same end is twice as much: t  2 L / vx .
The momentum transferred to the end face is therefore, according to (3.7): p  2 p  2mv x .
When dividing by Δt , which is the time within the transfer of momentum takes place, we find an
expression for the average force on the end face coming from a single molecule.
2
p  2mv x mv x


2L
t
L
vx
Since every molecule in the container causes the same average force, the overall force on the end
face may simply be found by multiplying by the number of molecules in the container.
F 
(4.2)
In the considerations above, we have presupposed that a molecule moves without interference
from one end to another. This can not be upheld in reality, since any molecule has numerous
collisions with other molecules. However, in a collision between molecules, the momentum is
conserved in any direction and therefore also in the longitudinal direction, the x-direction.
Since the collisions are considered elastic, the kinetic energies are also conserved. The kinetic
energy that goes into the expression (4.2) is also independent of whether the molecules collide
with other molecules on their way back and forth between the ends.
Since the molecules do not have the same velocity, we can not just add (4.2) for all the molecules.
If there are N molecules in the container we may, however, write the sum as N times the average
kinetic energy of a molecule in the x-direction
The symbol < > is used to designate the average or the mean of a quantity. Thus we find:
(4.3)
N
F   F k
k 1
2
 mv x 
m
2
 N  F  N
 N  vx 
L
L
2
 v x  is the mean of the square of the velocity in the x-direction.
We shall rewrite the expression (4.3) a little, so that the force is expressed by the mean value of the
kinetic energy of a molecule.
The square of the velocity is v2 = vx2 + vy2 + vz2, and for the mean, we have:
< v2 >=< vx2 >+< vy2 >+< vz2 >
(4.4)
Since the directions x, y, z are arbitrarily chosen direction in a coordinate system, we must have
that the three mean values are the same:
(4.5)
< vx2 > = < vy2 > = < vz2 >
=> < v2 >=3< vx2 >  < vx2 > = 1/3< v2 >
When (4.5) is inserted in (4.3), the expression for the force, we have:
(4.6)
FN
m
2
 vx 
L

F
2
3
N
 ½ mv 2 
L

F
2
3
N
 E kin 
L
The kinetic molecular theory
52
Where  Ekin  is the mean kinetic energy of a molecule.
The expression (4.6) then gives us the force on which the gas affect the end faces.
If we want to know the pressure we shall divide with the cross section area A.
F
P  . The volume of the container is V = A∙L, so we get:
A
F
N
N
(4.7)
P   23
 ½ mv 2   P  23  ½ mv 2 
A
A L
V
Finally we invent N = nMNA , where nM is the number of moles and NA is Avogadro’s number,
which is the number of molecules in one mole.
When the equation (4.7) is multiplied by the volume V, we arrive at an equation which directly can
be compared with the equation of state of ideal gasses: PV  n M RT
(4.8)
PV  23 nM N A  ½ mv 2 

PV  nM RT

2
3
N A  ½ mv 2  RT
The latter equation in (4.8) is the fundamental equation which delivers the connection between the
absolute temperature and the mean kinetic energy of a molecule. Solving for  ½mv 2  gives:
(4.9)
 ½ mv 2 
3
2
R
T
NA

In (4.9) we have defined a universal constant: k 
k = 1.38 10-23 J/K
 ½ mv 2  32 kT
R
, which is called Boltzmann’s constant.
NA
Boltzmann’s constant is expressed by two other universal constants, namely the gas constant
R =8.31 J/mol K and Avogadro’s number NA = 6,023 1023 molecules/mol.
Boltzmann was the first to clarify the connection between temperature and the kinetic energy of
the molecules and Boltzmann’s constant is simply the conversion factor between the Kelvin
temperature and the mean kinetic energy of a molecule, at that temperature.
As mentioned previously, the equation (4.9) implies the existence of an absolute zero point for
temperature, namely where the molecules are at rest, so that the kinetic energy is zero.
The derivation we have made is done under the assumption of one-atomic molecules. We say that
they have 3 degrees of freedom, since they can move freely in 3 directions. Poly-atomic molecules
can, however, also acquire energy in he form of vibration or rotation energy, so more generally:
(4.10)
<Ekin> = γkT
(The equipartition principle)
The constant γ depends on the nature of the gas only, and γ = 3/2 for an one-atomic gas.
More generally we write γ as n/2, where n is the number of degrees of freedom. For one-atomic
molecules n =3, corresponding to one degree of freedom along each of the three axes. For two-
The kinetic molecular theory
53
atomic molecules n =5. (Two extra degrees of freedom corresponding to rotational and vibration
energies)
The internal (thermal) energy of a gas, can be found as the sum of the individual kinetic energies
of the molecules.
(4.11)
E kin  N  ½ mv 2   nM N A  ½ mv 2   32 nM RT
(4.11) applies for a one-atomic gas. More generally Ekin  nM RT , where γ is the constant
mentioned above.
4.12 Example
Estimate the speed of ammonia molecules (NH3 , atomic weight 17) and hydrochloric acid (HCl, atomic weight 36.5)
at the temperature 20 0C.
Solution;
We apply:
 ½ mv 2   32 kT

 v2  
3kT
m
We allow ourselves to identify the mean velocity with the square root of the mean of the square, which is only
approximately correct.
The mass of an atom can be found by dividing the mass of one mole with Avogadro’s number: NA: m =M/NA.
mNH 3 
17 g / mol
 2.8 10 23 g  2.8 10 26 kg
23
6.023 10
 v NH3 
3 1.38 10 23 J / K 293 K
 6.6 10 2 m / s
2.8 10 26 kg
It follows from (2.1), which is justified by (4.9) that the ratio between the mean velocities is the square root of the
inverse ration between the mole masses.
 vHCl  
17
v NH3   4.5 10 2 m / s
36.5