Deriving the Kinetic Theory Equation

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 = mu1 – (-mu1 ) = 2 mu1
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 2lx
=
x component of the velocity
u1
So Newton’s 2nd Law gives us:2
change of momentum − 2mu1 − mu1
force on molecule =
=
=
time taken
( 2l x / u1 )
lx
And Newton’s 3rd Law gives us:+ mu1
force on wall = − force on molecule =
lx
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 lz
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 + ... + p N
Where p2, p3… refer to the pressures of all the
other molecules up to N molecules.
mu1 2 mu2 2 mu3 2
mu N 2
p=
+
+
+ ... +
V
V
V
V
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 + u 2 + 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:2
2
2
c1 = u1 + v1 + w1
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 = Nmc rms
3
Can also be re-written as:-
1
2
p = ρcrms
3
Because:-
Nm
ρ=
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
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
m(c1 + c 2 + c3 + ... + c N )
N
c1 + c2 + c3 + ...+ c N
=
N
2
crms
2
2
2
2
= 1 2 mcrms
2
2
Mean Kinetic Energy
pV = nRT
pV =
1
2
Nmcrms
3
1
2
nRT = Nmcrms
3
1
2
nRT = nN A mcrms
3
R
1
2
T = mcrms
NA
3
kT =
1
2
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.