Kinetic Molecular Theory of Gases
Kinetic molecular theory, KMT for short, is very different from
thermodynamics although it deals with some of the same variables, such as
pressure, temperature, volume, and density.
Thermodynamics does not care whether molecules exist or not.
Essentially all the results of thermodynamics can be worked out without any
assumptions concerning the particle nature of matter. KMT, however, is a
molecular theory (or a molecular model) of matter. (Molecular theories of
matter are sometimes called "microscopic" models as opposed to the
"macroscopic" model of bulk material. KMT is a microscopic model of a
gas.)
1. The Kinetic-Molecular Gas 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)
 The molecules have mass, m.
 The molecules are contained in a volume, V.
 The gas is at some temperature, T (which is the Kelvin temperature).
 The dimensions of the molecules are small compared to the average
distance between molecules and compared to the size of the container.
 There is no potential energy of interaction between the molecules.
 The molecular motion is random, or chaotic, so that the gas is
isotropic. "Isotropic" means that the properties of the gas are the same
in every direction.
2. The Pressure of a Gas
Pressure is defined as the force per unit area:
In the kinetic-molecular theory of gases, pressure is the force exerted
against the wall of a container by the continual collision of molecules
against it. To calculate the pressure we need to determine the force exerted
by gas molecules colliding with wall A.
The force exerted by a molecule of mass m colliding with wall A can
be calculated from
.
The last quantity in the above equation can be determined if we know the
change in velocity per collision with wall A and the time between collisions
with wall A.
A collision with wall A will reverse only the x-component of the
velocity. If we assign the average initial x-component of the velocity before
collision as -ux and the final x-component of the velocity after collision to
ux then the change in velocity with each collision with wall A is
∆u = ux – (-ux) = 2 ux
The time between collisions with wall A will again depend upon the
x-component of the velocity and the distance travelled by the gas molecule
(along x) between collisions. In our box a gas molecule, after colliding with
wall A, would have to travel along x to the opposite wall, a distance of a,
and back again to wall A, for a total distance travelled along x of 2a. Thus
the time between collisions with wall A would be
and the force exerted by one gas molecule of mass m colliding with wall A
becomes
Rearranging we find:
pV = mux2
per gas molecule.
For N molecules we have:
N
P
i
(mux2 )i m N 2
m
  (ux )i  N ux2
V
V i
V
Recognizing that the velocity is related to its components by the
Pythagorean Theorem and that, on average, each of the components are
equal ( ux  u y  uz ) we find:
u 2  ux  u y  uz  3ux
2
Then we obtain:
Rearranging we find:
P
2
2
2
m u2
N
V
3
PV 
This equation can also be written as:
1
N mu 2
3
PV 
1
n M u2
3
(1)
(1bis)
3. Kinetic Energies and Temperature:
The average kinetic energy of one molecule is:
ke 
1
mu 2
2
Equation (1) can then be changed to:
PV 
PV 
2 1
N ( mu 2 )
3 2
2
2
N ke  n( N A ke)
3
3
PV 
2
n KE
3
(2)
where KE is the kinetic energy of Avogadro’s number of molecules.
Comparing this last equation (2) with:
we can find that:
KE 
PV = nRT
3
RT
2
(3)
and:
ke 
with: k =
KE
3 R
3

T  kT
NA
2 NA
2
(4)
R
called Boltzmann’s constant.
NA
4. Molecular Speeds:
The kinetic energy of Avogadro’s number of molecules (that’s one
mole of gas) can be written:
KE  N A (
1
1
mu 2 )  M u 2
2
2
(5)
where M = m NA , is the molar mass.
From equations (3) and (5) we can obtain:
u2 
3RT

M
3kT
m
(6)
The term u 2 is known as the root-mean square speed (rms)
Equation 6 is the first of three quantities that we will define and use to
describe the velocities of molecules in a gas. We will see that all three of
the velocity measures will have the same order of magnitude but will
differ slightly in their exact value.
Example:
As an example, let's calculate the rms velocity of a nitrogen molecule
at 25oC. Sometimes it is easier to work with a mole of molecules
instead of individual molecules. It is easy to see that.
rms 
u2 
3RT

M
3x8.314 x 298
 515 m s 1
0.028
5. Distribution of Molecular Speeds:
-
Average speed ( u ):
u
-
8RT
8kT

M
m
(7)
Most probable speed (α):

2 RT
2kT

M
m
(8)
The three speeds (root-mean square, average speed and the most
probable speed) are in the ratios:
rms : u : α = 1.00 : 0.92 : 0.82
(9)
Example 1:
The root-mean square speed, rms, of N2 at 298 K is 515 m s-1.
What are the average speed and the most probable speed of this gas at
the same temperature?
Answer: by using equation (9) we can obtain
u
0.92

 u  0.92 rms  0.92 x 515  473.8 m s 1
rms
1

rms

0.82
 u  0.82 rms  0.82 x 515  422.3 m s 1
1
Example 2: Calculate the ratio
u
for the following 7 gas molecules:
rms
Molecule
1
2
3
4
5
6
7
Speed / m s-1
370
400
380
420
300
440
480
Answer:
u = (370 + 400 + 380 + 420 + 300 + 440 + 480) / 7 = 398 m s-1
rms = (3702 + 4002 + 3802 + 4202 + 3002 + 4402 + 4802) / 7 = 402 m s-1
u / rms = 398 / 402 = 0.99
6. The free mean path and collision number:
Let us consider a particular molecule A with diameter d and moving
with a speed u . This molecule will collide in 1 s all molecules that have
their centers in the cylinder of the figure below:
The volume of the cylinder whose radius is equal to diameter of the
molecule d is:
V  d2u
The number of molecules in the cylinder is:
N   d2 u N*
where N* is the number of molecules per cubic meter:
N* 
N PN A

V
RT
The free mean path, L, is known as the distance traveled between collisions;
that's the length of the cylinder, u , divided by the number of collisions
occurring in the cylinder:
L
u
1

*
 d u N  d 2 N*
2
Taking in account that collisions goes from glancing collisions (a) to headon collisions (b), the average speed is 2 u for collisions in right angles (c)
as we can see it on the figure below:
The free mean path becomes:
u
1
L

(10)
 d 2 2 u N*
2 d 2 N *
The number of collisions that a molecule A makes per second is denoted Z1:
Z1 
u
 2  d 2 u N*
L
(11)
The molecule A, in relation to the other molecules, travels with an effective
speed equal 2 u .
The number of collisions occurring in a unit volume per unit time is denoted
by Z11:
1
1
Z11  Z1 N * 
 d 2 u ( N * )2
(12)
2
2
The factor 1/2 ensures that each collision will not be counted twice.
7. Numerical values of collision properties:
Let us consider again the N2 (d = 3.74x10-10 m) gas at 25 0C and 1 atm. First
we have to calculate N* (number of molecules per cubic meter), starting
from:
PV = nRT
and using the relations:
N* 
N
V
and n 
N
NA
then we get:
N* 
P N A 101325 x6.022 x1023

 2.46 x1025 m 3
RT
8.314 x 298
We can calculate now the free mean path, L:
L
1
1

 6.50 x108 m
10 2
25
2
*
1.414 x3.14 x(3.74 x10 ) x 2.461x10
2 d N
and the number of collisions of molecule A per second ( u was already
calculated):
Z1  2  d 2 u N *  1.414 x3.14 x(3.74 x1010 ) 2 x515 x 2.461x1025  7.31x109 collisions s 1
and the number of collisions per m3 per second:
Z11 
1
 d 2 u ( N * )2  8.99 x1034 collisions m3 s 1
2
8. Real gases: Van Der Waals' equation.
The Dutch chemist van der Waals has attributed the failure of the PV = nRT
relation to duplicate the behavior of real gases to the neglect of:
a) The volume occupied by the gas molecules
b) the attractive force among the molecules
and has proposed these corrections:
 Correction of the volume
The presence of molecules of nonvanishing size means that a certain
volume, called the excluded volume, is not available for molecules to move
in. The excluded volume for one mole is represented by b, then the more
appropriate equation would be:
P( V – nb) = nRT
b is characteristic of each gas and must be determined empirically.
The relation between b and the size of the molecule can be seen on the
figure below:
Molecular volume
Excluded volume
(per pair of molecules)
1
excl
V
 4  d 3 
14
3
   d   4    
2 3

 3  2  
and b is equal to:
 4  d 3 
1
b  N AVexcl
 4N A     
 3  2  
(13)
Example:
What is the value of b for a molecule of N2 gas (d = 2.88x10-10 m) ?
Answer:
3

 4  d 3 
 2.88 x1010  
23 4
   3.01x105 m3mol 1
b  4 N A       4 x6.023x10  3.14
3
2
3
2

   

 
 Correction of pressure:
The attraction acts with the confining pressure to hold the molecules
together. the gas is confined, therefore, not only by the external pressure ,
but also by the intermolecular attractions which contribute with a term
proportional to (n / V)2. The proportionality factor is denoted by a . Finally
the equation becomes:
2

n 
 P  a  V  nb   nRT
 V  

(13)
Example:
What is the real pressure of 3 mol of CO2 in a 10 L container at STP ?
Van der Waals constants are : a = 2.01 atm L2 mol-2, b = 0.0319 L mol-1
Answer:
2
2
nRT
3x0.082 x 273
n
3
P
 a  
 2.01   6.6 atm
V  nb   V  10  3x0.0319
 10 