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Deriving the Kinetic Theory
Equation
Thermal Physics Lesson 6
Learning Objectives

Derive the kinetic theory equation.
Here we go…

We consider a molecule of mass m in a box of
dimensions lx, ly and lz with a velocity
components u1, v1 and w1 in the x, y and z
directions respectively.
A derivation of two halves…

The strategy is to derive the pressure for one
molecule on one face (1st half) and then sum all
the pressures from all the molecules (2nd half).

The first part involves Newton’s second law
which states that the force on a body is equal to
the rate of change of momentum.
Also note…

The speed c of the molecule is given by:-
c1  u1  v1  w1
2


2
2
2
This comes from doing Pythagoras’ theorem in 3
dimensions.
We will use this later in the 2nd half
Find the force on one molecule…

When the molecule impacts with Face A the xcomponent of its momentum is changed from mu1 to
-mu1.
Change in momentum  mu 1 – (-mu1 )  2 mu 1

But we want the rate of change of momentum, so
need to divide by the time taken between collisions
with Face A.
Time, t, between collisions…
t
total distance to opposite face and back 2l x

x component of the velocity
u1
So Newton’s 2nd Law gives us:change of momentum
 2mu1  mu1
force on molecule 


time taken
(2l x / u1 )
lx
And Newton’s 3rd Law gives us:force on wall   force on molecule 
 mu1
lx
2
2
Pressure

Recall that to work out the pressure:-
force
pressure 
area

So the pressure p1 on face A:2
force
mu1
mu1
p1 


area of face A(  l y l z ) l x l y l z
V

2
So that’s the first bit done! – the pressure from one molecule
in one direction.
Summing the pressures
The total pressure on face A can be calculated
by summing the pressures of all of the
molecules:-
p  p1  p2  p3  ...  pN
Where p2, p3… refer to the pressures of all the
other molecules up to N molecules.
2
2
2
mu3
muN
mu1 mu2
p


 ... 
V
V
V
V
2
Summing the pressures

The last line can be rewritten as:-
Nmu
p
V

2
Where the mean square x-component velocity is given
by:2
2
2
2
u1  u2  u3  ...  u N
u 
N
2

And similar equations can be derived for the y and z
2
2
2
2
components
v  v  v  ...  v
2
v 
1
2
3
N
N
Almost there…

But we want our equation to include the root mean
square speed in all directions. Using Pythagoras’s
theorem, the speed for one molecule is given by:-
c1  u1  v1  w1
2

2
2
2
You can show that the root mean square speed for all
the molecules is:-
crms  u  v  w
2
2
2
2
Almost there…

Because the motions are random we can write:-
u v w
2

2
2
Otherwise there would be a drift of particles in one
direction. So using the above two equations:-
1
2
u  crms
3
2

So now we can write:-
2
Nmu
1 Nmcrms
p

V
3 V
2
Final Rearrangement
The Kinetic Theory Equation:-
1
2
pV  Nmcrms
3
Can also be re-written as:-
1
2
p  crms
3
Because:-
Nm

V
Recap





Use Newton’s Second Law
Use Newton’s Third Law
Calculate pressure due to one molecule
(force/area)
Sum the pressures of all the molecules.
Rewrite the speed in terms of root mean square
speed.
Deriving Ideal Gas Equation

From Boyle’s Law:
1
V
p

From Pressure Law:
V T

From Avogadro’s Law:
V n


Combining these three:
nT
V
p
Rewriting using the gas
constant R:
 nT 

V  R
 p 
Therefore:-
pV  nRT
Mean kinetic energy
The mean kinetic energy of a
molecule is the total kinetic energy of
all the molecules/total number of
molecules.

1
mc1  12 mc2  12 mc3  ...  12 mcN
N
2
2

1
2
2
The root mean square speed is defined
as:
…and so the mean kinetic energy of
one molecule is gvien by…

2
2
2
c  c2  c3  ...  c N
 1
N
2
crms
2
m(c1  c2  c3  ...  c N )
N
2
2
2
1
2
2
2
mcrms
2
2
Mean Kinetic Energy
pV  nRT
pV 
1
2
Nmcrms
3
nRT 
1
2
Nmcrms
3
1
2
nRT  nN A mcrms
3
R
1
2
T  mcrms
NA
3
1
2
kT  mcrms
3
Note we have two
equations with pV
on the left hand side.
(N=nNA)
The Boltzmann
constant, k, is
defined as R/NA.
Mean Kinetic Energy
3
31
2
kT 
mcrms
2
23
1
3
2
mcrms  kT
2
2
1
3
2
mcrms  RT
2
2
1
3
2
mcrms  nRT
2
2
Multiply both sides by 3/2.
Mean kinetic energy of a
molecule of an ideal gas.
Kinetic energy of one mole of
gas
(because R=NAk)
Kinetic energy of n moles of gas.