DIFFUSION IN GAS PHASE
Mental Experiment
Consider a container of gas in thermal equilibrium and introduce a very small amount of
different gas (“special” gas) in some place of the container.
Consider an imaginary plane of unit area
(A diagram on your page 1)
perpendicular to the X direction and compute the net flow of the molecules of the special
gas across that plane during time t . In order to count the number of special molecules
that cross the plane in a time t we first count the number of molecules in a volume t
from the plane. Here  is the average actual molecular velocity in the X direction.



The number of special molecules that passes from left to right is nt and from right to

left n t . Here n-, n+ are the number density (number in unit volume) of special
molecules to the left and to the right of our dividing plane.


The net molecular current (molecule/cm2s) from left to right, i.e. in the direction of the xaxis then is
jx 
nt  n t
 (n  n  )
t
(1)
The question arises where do we count our densities, n-, n+? We must count at positions

where molecules started their flight, i.e at a distance equal to the mean free path
from plane.

Now we would like to use a continuous function c to represent the concentration of
molecules. Then
1
away
n   n 
dc
dc
x  2
dx
dx
(2)
Ignore the factor 2 and substitute eq. (2) into eq. (1).
jx   

dc
dc
 D
dx
dx
(3)
Hence, molecular diffusivity D is proportional to the mean free path and mean molecular

velocity
D 
(4)
 The above development uses many approximations,  instead of  x , n+ and n- measured
at distance , while for molecules that do not travel perpendicularly to our plane it is a
slant distance that counts. A more careful analysis reveals


1 dc
jx  
3 dx
(3a)
1

3
(4a)
so that

D
The mean free path

expresses how far the molecules travel between collisions (on the
average).

We can also talk about the average time between collisions  . Then
 
(5)
By collision cross section we mean the area within which the center of our

particle
(molecule)
must
be
located
if
it
is
to
collide
with a particular other molecule.

 c2   (r12  r22 ) . Then
 c2 no 1
(6)
There is one collision on the average when the particle goes a distance


gathering molecules could just cover the total area.
Some additional comments:

2
in which the
If we have special molecules (S) and background molecules, then there is a specific force
F acting on S imparting an average velocity on S.
 drift 
F
m
(7)
Drift velocity is proportional to this force! (no general name for constant of

proportionality). In electrical problems where F  qE (charge x field)
 drift  qE where  is mobility).
Let us adopt that name

 drift  F

Then  


(7a)

(8).
m
Ionic conductivity
 drift  qE  q
V
b
(  ddrift x t ) ions arrive at a plate. There are ci ions per unit volume. The number reaching
 the plate is (c i A  drift t) . Each ion carries charge q. The charge collected =

qc i A  drift t. Current I 

V V
I  q 2c i 
b R
b
R 2
q c i

ch arge collcted
 qc i A drift
unit time
Relationship between resistances and molecular properties



m
____________________________________________________________

So referring to the diffusion flux of (3a) we can represent it as:
j x  i kT
dc
dx
(3b)
D  i kT


(4b)
The diffusion coefficient is kT times the mobility coefficient.
The consequence of Ficks first law is the expression for the flux repeated below.
3
jx  D
dc
dx
Ficks 2nd law then states:

( Acc)  (in )  (out)
c
x  j x x  j x x x
t
Upon dividing with x and taking the limit one gets:

c j x
 2c

D 2
t x
x

The response to an impulse at x = 0 at t= 0 is felt instantaneously at all x (although
exponentially small).

Infinite speed of signal propagation by diffusion equation is a paradox due to the
theoretical nature of the diffusion equation. However, the molecules do not cover the
free path  instantaneously. So the time when concentrations are counted at the left and
right side of the plane differs from the time when the molecular current is measured by
approximately the mean time between collisions or   /  . Mathematically, the flux is

evaluated at t   based on concentration difference that existed at t.
j x (t   )  D
c(t)
x

(9)

This yields (by expansion in Taylor series)

jx (t   )  jx (t) 
jx
c
  D
t
x
(10)
This formulation avoids the infinite speed of propagation.

It has been derived by
Maxwelll (1867)
Fock (1926)
Davidov (1935)
Goldstein (1951)
Davies (1954)
Cattineo (1948)
Vermotte (1958)
4
Derivation of equation (10) is valid if c varies slowly so that one term Taylor expansion
is permissible.
When that is not the case and concentration changes noticeably during  or on the length
a more general approach is needed.
-
Consider 1D general case when c can change considerably during a time period of

 or on the length of mean path  .
-
No convective flow


Consider particles in random walk along x axis, v+, v- are velocities in each direction.

Particle interacts instantaneously with surroundings and as a consequence changes
directions. Each particle has the probability p+ that it will move with v+ and probability p
= 1 – p+ that it will move with v-. Time spent moving in a certain direction is a random
variable characterized by probability depending on this period.
We count again particles crossing from left to right and those from right to left, at a plane
at x.
The particles reaching x from the left at time t had their last collision at the time t   ,
and at the point x - v  (  0) and obtained the velocity v+ in the direction of plane at x,
moved during time  without collisions and did not disappear during time of travel due

to reaction.

The total number of collisions at point x  v  and at time t   , in terms of continuous

function c(x,t) describing the distribution of particles along the x-axis, is equal to
c( x  v  , t   ) /  where  is the mean period of time between collisions.

The flux of particles at x and at t from left to right is

pv 
j ( x, t)     G( )H( )c( x  v  , t   )d


(11)
o
where


G() = probability that a particle survives a time  without collisions
H()
= probability that a particle survives a time  without reacting.

Similarly


5
p.v p
j ( x, t) 

 G()H()c( x  v ;t  )d


(12)
o
If the probability that a particle moves without collision during time longer than a certain

value does not depend on the history of particle movement, then for arbitrary t1 and t2.
G(t1 )G(t 2 )  G(t1  t 2 )
(13)
so that
G(t)  e
t / 
If we consider a first order reaction , then
H(t)  ekt

(14)
Now

Mass balance requires
pv   pv
(15a)
The concentration is defined by
c

j

j
v

(15b)
The net flux in the positive x-direction is:
j  j   j

(15c)
Probabilities of all permissible events sum to one:

p+ +p- = 1
(15d)
Consider now j+ and for ease of writing drop the subscript + for now. Then, upon
substituting equations (13) and (14) into equation (12) we get:
j
pv


e
a
c( x  v, t   )d
(16)
o
with

1
a  (k  )
(17)

Integration by parts of equation (16) can be performed as indicated below:

dv  ea d
u  c( x  v, t   )


1
a
   ea
du  (v
c c
 )d
x t
to yield:
6
j

pv  1 a
1
 e c( x  v, t   )  
  a
a
o

 ea (v
o

c c
 )d
x t

(18)
Upon substituting the integration limit in the first term above and splitting the second

term into two, on multiplication by constant a one gets:
aj 
pv

c( x, t) 
pv


ve
o
a
c
pv
( x  v, t   )d
x


c
 e  t d
a
(19)
o
Differentiation of equation (16) with respect to position and time yields:

j pv

x 

j pv

t 

e
a
o
e
a
o
c
d
x
c
d
t
(20)
(21)
(1

Substitution of eqs (20) and (21) into eq (19) yields:
aj 
pv

c 
j j

x t
(22)
Now this has been derived and is valid for j+. A similar expression, only with the sign for

j
changed, because now one deals with c( x  v ;t   ), can be derived for j- . The
x
resulting equations are shown below:



1
p
j  j
(k  ) j     c      


t
x
(23)
1
pv
j
j
(k  ) j    c      


t
x
(24)
Divide equation (23) with   and (24) with v- and add them up.
1  j
j  p
p   1 j
1 j  j j 
(k  )         c            
       
   t  t  x x 



7
(25)
Recall and use in eq (25) eqs (15b) (15c) and (15d) to obtain:
1
1
c j
(k  )c  c  


t x
(26)
Upon cancellation of a common term this becomes:
c j
  kc  0
t x

(27)
Subtract now eq (24) from (23) to get:
p 
1
p v  
j
j
(k  )( j   j )        c  ( j   j )        

  t
x
x
 

(28)
Upon substitution of eqs (15a) (15c) we get:

1
j
j
j
(k  ) j        

t
x
x
(29)
Now we can add and subtract the following terms, given by expression (30), hence

effectively adding zero) to eq (29). Upon, use of (15c) and regrouping we get equation
(31):
 
j
j
j
j
     v   v 
x
x
x
x
(30)
 1
j
j
j
j
j
 v   v  
k  j       v
t
x
x
x
x
  

However from eq (15b) it follows that v v

c
j
j
 v   v  
x
x
x
(31)
(32)
Substitution into eq (31) yields:

j
j
c
(k  ) j    (v   v )  v v

t
x
x
1
(33)
Upon multiplication by  we get:
(k  1) j  

j
j
c
 (v   v )  v v
0
t
x
x

In summary hyperbolic equations with finite speed of propagation describe the

“diffusion” process more accurately. These equations are:
8
(34)
c j
  kc  0
(35)
t x
j
j
c
(1 k ) j     (v   v )  v v  0 (36)
t
x
x
These equations (35) and (36) need to be solved for concentration c(x,t) and flux j (x,t).


9