Transport Properties
The phrase “transport properties” implies that something is moving or being
transported. We will mention three, but discuss only one in detail.
Mass transport is called “diffusion,”
energy transport is called “thermal conductivity,”
momentum transport is called “viscosity.”
We will discuss these three properties in one dimension. Three-dimensional
versions of our equations are available, but the fundamental principles are
the same as in one dimension and the math is easier so we will stay in one
dimension. All of these transport properties are described by the same form
of equation. In all cases there is the flux of some quantity, which we will
call, Jz, indicating that the flux is in the positive z direction,
J z  flux 
(mass or energy or momentum)
. (1)
s m2
Then there is a gradient of some other quantity, which drives the flux, which
we will indicate by d(quantity)/dz. The transport bulk equation says that the
flux is proportional to the gradient. That is, for diffusion,
N
  D V , (2)
dz
d
J z ,mass
where a gradient in the number density drives a mass flux and D is called the
diffusion coefficient. This is called “Fick’s law of diffusion” (sometimes
Fick’s first law of diffusion).
For heat conductivity a temperature gradient drives a heat (or energy) flux,
J z ,energy  
dT
, (3)
dz
and  is called the thermal conductivity (or the coefficient of thermal
conduction).
Viscosity is a little harder to visualize, but let’s try anyway. We have a large
(infinite) surface perpendicular to the z axis (call it the xy plane) and there is
a gas flowing along the surface in the x direction. The molecules right at the
surface of the plane are not flowing at all because they are in contact with
the stationary plane. But the bulk gas is flowing so as we move away from
the surface in the positive z-direction the velocity of the molecules increases
because they are being dragged along by the flowing gas. This increase in
velocity as we move away from the plane means that the gas is transporting
momentum in the z-direction. At any given plane (at some value of z) the
plane above it is passing down momentum to try to speed the plane up and
the plane below it is passing up momentum in the opposite direction trying
to slow it down. The net result is,
J z , x -momentum  
dvx
, (4)
dz
and  is called the coefficient of viscosity.
Our task is to see if we can find out how these various coefficients depend
on the parameters of our model of the gas, things like m, T, p, and etc.
Let’s try diffusion first:
Consider a plane at z = 0. There are molecules crossing this plane from the
 z direction and from the + z direction. The net flux of molecules is the
flux from the  x direction minus the flux from the + z direction. Let’s
assume that the molecules coming from the  z direction originate at a plane
located at z =  , and the molecules coming from the + z direction
originate from a plane located at z = + . Then, assuming that  is small,
Jz 
1 N
1 N
v    v  
4  V  4  V  
1  N 
N 
v      
4  V     V    
1
d N
  v 2  
4
dz  V 

(5a, b, c, d)
N
d 
1
V
  v   .
2
dz
Comparing Equation 5d with the bulk equation, Equation 2 we see that the
diffusion coefficient is,
D

2
v . (6)
The assumption that the molecules crossing the plane at z = 0 originate
from planes at z =  is a fairly crude approximation. In reality we should
integrate the flux from planes ranging from z = 0 to z = . A careful
treatment of the hard-sphere gas (which our model is) would give a
corrected result of
D
3 1
v  , (7)
8 2
which is not too far off from our approximate result since,
3
 1.18.
8
Kinetic molecular theory expressions for the other two transport properties
can be derived in a manner similar to the one used here. One considers the
net flux of either energy (as heat) or momentum across the plane, z = 0,
where the fluxes originate from planes at z = .
Just for the record, the elementary KMT expressions for the heat
conductivity and viscosity coefficients are,


1
CV N
v

2
NA V
1 8kT C V
1
N
(8a, b, c)
2  m N A N 2 d 2 V
V
kT C V 1
,
 m NA  d 2

and


m
N
v 
2
V
m 8kT
1
N
(9a, b, c)
2  m N 2 d 2 V
V
1
kTm
,
d2 
respectively. Our treatment is fairly crude. More sophisticated models will
change the factor 1/2 in Equations 8a and 9a to something more
complicated. However, the dependence on various parameters, such as
temperature, cross section, mass of the molecule, and so on, is correct.

WRS