Kinetic Molecular Theory of Gases
Kinetic molecular theory, KMT for short, is very different from
thermodynamics.
Thermodynamics does not care whether molecules exist or not, KMT does,
it is a molecular model of matter (sometimes called a "microscopic" model
as opposed to the "macroscopic" model of bulk material).
Thermodynamics assumes equilibrium (infinite time), KMT assumes
molecules are in motion and must include time explicitly.
The Model
Assume that the gas consists of
N molecules (N is large, on the order of Avogadro's number, NA = 6.023 
1023 molecules/mol)
of mass, m,
contained in volume, V,
at temperature, T (which is the Kelvin temperature).
There is no potential energy of interaction between the molecules. (Potential
energy can be added in later if desired.)
The molecular motion is random so that the gas is isotropic. (This statement
also includes the notion that motion in any one of the three coordinate
directions is independent of the motion in the other two dimensions.)
We will use the "number density," N/V, the number of molecules per unit
volume, instead of the usual mass density.
Since we are dealing with individual molecules we will write the gas
constant on a molecular basis instead of a molar basis, and we will give it a
new name and symbol, k or kB. that is,
k  kB 
R
.
NA
We will assume that the ideal gas law holds. That is, pV = nRT = NkT.
If that's a problem, note that
(1)
pV  nRT 
NA
R
nRT  ( N A n)
T  NkT .
NA
NA
Consider one molecule.
The molecule moves in three spatial dimensions, x, y, z, so it has a velocity
which is a vector quantity. Write,
(2)
v  vx x  v y y  vx z .
(Note that we are using x , y , and z for unit vectors instead of i , j , and k .)
The molecule has a kinetic energy which we will write as E, defined as
usual,
(3)
1
1
1
E  mv 2  mv  v  m(vx2  v y2  vz2 ) .
2
2
2
The internal energy of the gas would be the sum of the kinetic energies of all
the molecules. That is,
(4)
internal energy =
 2mv .
1
2
i
i
It is impossible to track the motion of 1023 molecules so we must deal with
averages. We will write the average energy of one molecule as,
(5)
average energy =
1 2
mv ,
2
where the angle brackets indicate that we are taking the average of the
quantity enclosed within them. Since we can't add up all the individual
kinetic energies we will just multiply the average energy by the number of
molecules,
(6)
internal energy = N
1 2
1
mv  N m v 2 .
2
2
Since
(7)
v2  vx2  vy2  vz2 ,
then
(8)
v 2  vx2  v y2  vz2 .
Since the motion is random, hence isotropic, there is no preferred direction
for the motion. The average velocities (and the average of the velocities
squared) must be the same in every direction. That is,
(9)
vx2  v y2  vz2 ,
hence
(10)
v 2  3 vx2 ,
or
(11)
vx2 
1 2
v .
3
(We could have just as well used vy or vz here, it makes no difference.
The momentum of a molecule is defined as usual:
(12)
p  mv ,
or, in terms of components,
(13)
p x  mv x ,
etc.
Probably the simplest way to deal with random motion is to use probabilities
(or statistics). In order to use probabilities or statistics we need to define and
use the velocity probability distribution function, f (vx ) . (Sometimes we
will just call this the velocity distribution function, but it really is a
probability distribution function.) This function is defined so that (and this
is important)
f (vx )dvx = the probability that a molecule has x-component of velocity
between vx and vx + dvx.
(Probability distribution functions are very important and very useful in
chemistry and in many branches of science. We will define two more
velocity-related probability distribution functions in a similar manner in this
course and you will see two or three more probability distribution functions
in the introduction to quantum mechanics and in statistical mechanics.)
Since the molecule must have some velocity the sum of all these
probabilities must be equal to unity. The way we sum all these probabilities
is, of course, by integration. That is,

(14)

f (vx )dvx  1 .

(This is a nonrelativistic theory. Our molecules never get close to the speed
of light so we don’t have to worry about the fact that material particles can’t
go faster than the speed of light. This is reflected in kinetic molecular theory
in that the probability function for molecules being near the velocity of light
becomes vanishingly small.)
We don’t know the form of f yet, but we will figure out what it is later.
There are several properties of f that we can tell right away. We will
assume that our sample of gas is not going anywhere so the bulk velocity is
zero in every direction. This means that
(15)
f (v x )  f ( v x ) .
In words, this means that the properties of the velocity distribution function
must be the same in the negative x-direction as they are in the positive xdirection. Mathematically we would say that that f is an even function of vx.
From the fact that the gas is isotropic we conclude that
f (vx ), f (v y ), and f (vz ) all have the same functional form so that if we can
find the form of one of them we will know them all.
We can use these velocity probability distribution functions to find averages
of quantities that depend on velocity. (As already stated, we will denote
averages by angle brackets,
. We have already done this with average
kinetic energy and average of vx 2.)
Find averages as follows: Take the value of your molecular property that
you want to average and evaluate it at some velocity, vx. Multiply that value
by the probability that the molecule has that velocity and then sum over all
velocities. It’s easier to write this as an equation than to say it in words,
(16)
 


   f (v )dv
x
x
,

where you place the quantity you want averaged in the parentheses. For
example, if you wanted the average of the kinetic energy in the x-direction
you would write,

(17)
1 2
1

mvx    mvx2  f (vx )dvx .
2
2

 
Other examples:

(18)
vx 
  v  f (v )dv ,
x
x
x


(19)
v
2
x

  v  f (v )dv ,
2
x
x
x


(20)
v
3
x

  v  f (v )dv ,
3
x
x
x

(21)
e
 bvx2




e  bvx f (vx )dvx ,
2

and so on.
Let’s take a look at the average of vx.

vx 
  v  f (v )dv
x
x

0
x


v
x
x
f (vx )dvx
0
 (v ) f (v )(dv )   v
x
x
x

x
f (vx )dvx
0
0
(22,a,b,c,d,e)
v

0

f (vx )dvx 

  vx f (vx )dvx 



v
f (vx )dvx
0
   vx f (vx )dvx 
0
x

v
x
f (vx ) dvx
0
0
In these equations we have made use of several things we know about
integrals and about the function f(vx). You can break an integral from – to
+ into the sum of two integrals, one from – to 0 and the other from 0 to
+. Also, when you interchange the upper and lower limits of integration
the integral changes sign. We have also used the fact that f is an even
function of vx.
However,

(23)
v
2
x

v
2
x
f (vx )dvx  0 .

We can calculate vx2 without knowing the detailed functional form of f by
considering the pressure of a gas against a wall. The pressure of a gas
against a wall is caused by collisions of gas molecules with the wall. Since
there are so many molecules colliding with the wall the pressure seems
smooth. We know that
(23)
pressure =
force
,
area
but according to Newton’s second law
(24)
force = rate of change of momentum ,
so that
(25)
pressure =
rate of change of momentum
.
area
Let’s consider a portion of the wall of area, A and a molecule with xcomponent of velocity, vx. Think about what happens in a small time
interval t. In the time t the molecule will travel a distance vxt in the xdirection. (Let vx be positive so that the molecule is traveling in the +x
direction.) The molecule will hit the wall in a time t if it is within a
distance vxt of the wall. This distance and the area, A, create a small
volume A vxt. Let the number density be N/V. Then this small volume
contains
(24)
N
Avx t
V
molecules. The number of these molecules which have velocity, vx, is
(25)
N
Avx t f (vx )dvx .
V
Consider, now, what happens when the molecule hits the wall. The
molecule has initial momentum, mvx directed toward the wall. Assuming
that the collision is elastic the molecule will bounce off the wall with
momentum, –mvx away from the wall. The change of momentum for the
molecule is
(26) final momentum – initial momentum = – mvx – (+mvx) = – 2mvx.
According to the laws of physics, momentum is conserved, so this
momentum had to be transferred to the wall. The momentum transferred to
the wall is +2mvx. Then the momentum transferred to the wall by all the
molecules with this velocity is
(27)
N
N
Avx t f (vx )dvx 2mvx  2 Amtvx2 f (vx )dvx
V
V
We obtain the total momentum transferred to the wall in time t by
integrating this expression from 0 to . (We integrate only over the positive
values of vx because the molecules with negative vx are going the other way
and will not hit the wall.) The total momentum transferred to the wall is
then,
At 2m
(28a,b)
N 2
N1  2
vx f (vx )dvx  At 2m

 vx f (vx )dvx
V 0
V 2 
N 2
 Atm
v .
V x
The rate of change of momentum is this quantity divided by t and the rate
of change of momentum per unit area is that quantity divided by At. Then
rate of change of momentum
A
N
At m
V v2
(29a,b,c)

x
At
N 2
m
vx
V
So, if we combine this with the ideal gas equation of state we get
p
(30)
pm
N 2
NkT
,
vx 
V
V
or
(31)
m vx2  kT ,
or
(32)
vx2 
kT
.
m
But we know that vx2 
1 2
kT
,
v 
3
m
so
(33)
v2 
3kT
m
from which we get
(34)
vrms 
v2 
3kT
.
m
We call this average vrms for "root-mean-square" average and it is defined by
Equation -----. ( Root-mean-square, or rms, averages are a common way to
describe the magnitudes of quantities which average to zero. For example,
common house current (electricity ) is usually 110 to 120 volts. However,
household electricity is "alternating current" so that the voltage alternates
between positive and negative such that the average voltage is zero. If you
place the probes of a direct current volt meter in the terminals of a wall
electrical outlet you will get a reading of zero - if it doesn't burn your volt
meter out. But you can easily determine that there is a voltage there by
shorting the terminals out with your fingers.)