## 43Diffusion

### 43–1Collisions between molecules

We have considered so far only the molecular motions in a gas which is in thermal equilibrium. We want now to discuss what happens when things are near, but not exactly in, equilibrium. In a situation far from equilibrium, things are extremely complicated, but in a situation very close to equilibrium we can easily work out what happens. To see what happens, we must, however, return to the kinetic theory. Statistical mechanics and thermodynamics deal with the equilibrium situation, but away from equilibrium we can only analyze what occurs atom by atom, so to speak.

As a simple example of a nonequilibrium circumstance, we shall
consider the diffusion of ions in a gas. Suppose that in a gas there
is a relatively small concentration of ions—electrically charged
molecules. If we put an electric field on the gas, then each ion will
have a force on it which is different from the forces on the neutral
molecules of the gas. If there were no other molecules present, an ion
would have a constant acceleration until it reached the wall of the
container. But because of the presence of the other molecules, it
cannot do that; its velocity increases only until it collides with a
molecule and loses its momentum. It starts again to pick up more
speed, but then it loses its momentum again. The net effect is that an
ion works its way along an erratic path, but with a net motion in the
direction of the electric force. We shall see that the ion has an
average “drift” with a mean speed which is proportional to the
electric field—the stronger the field, the faster it goes. While the
field is on, and while the ion is moving along, it is, of course,
*not* in thermal equilibrium, it is trying to get to equilibrium,
which is to be sitting at the end of the container. By means of the
kinetic theory we can compute the drift velocity.

It turns out that with our present mathematical abilities we cannot
really compute *precisely* what will happen, but we can obtain
approximate results which exhibit all the essential features. We can
find out how things will vary with pressure, with temperature, and so
on, but it will not be possible to get precisely the correct numerical
factors in front of all the terms. We shall, therefore, in our
derivations, not worry about the precise value of numerical
factors. They can be obtained only by a very much more sophisticated
mathematical treatment.

Before we consider what happens in nonequilibrium situations, we shall need to look a little closer at what goes on in a gas in thermal equilibrium. We shall need to know, for example, what the average time between successive collisions of a molecule is.

Any molecule experiences a sequence of collisions with other
molecules—in a random way, of course. A particular molecule will, in
a long period of time $T$, have a certain number, $N$, of hits. If we
double the length of time, there will be twice as many hits. So the
number of collisions is proportional to the time $T$. We would like to
write it this way:
\begin{equation}
\label{Eq:I:43:1}
N = T/\tau.
\end{equation}
We have written the constant of proportionality as $1/\tau$, where
$\tau$ will have the dimensions of a time. The constant $\tau$ is the
average time between collisions. Suppose, for example, that in an hour
there are $60$ collisions; then $\tau$ is one minute. We would say
that $\tau$ (one minute) is the *average time* between the
collisions.

We may often wish to ask the following question: “What is the
*chance* that a molecule will experience a collision during the
next *small interval* of time $dt$?” The answer, we may
intuitively understand, is $dt/\tau$. But let us try to make a more
convincing argument. Suppose that there were a very large number $N$
of molecules. How many will have collisions in the next interval of
time $dt$? If there is equilibrium, nothing is changing *on the
average* with time. So $N$ molecules waiting the time $dt$ will have
the same number of collisions as *one* molecule waiting for the
time $N\,dt$. That number we know is $N\,dt/\tau$. So the number of
hits of $N$ molecules is $N\,dt/\tau$ in a time $dt$, and the chance,
or probability, of a hit for any one molecule is just $1/N$ as large,
or $(1/N)(N\,dt/\tau) = dt/\tau$, as we guessed above. That is to say,
the fraction of the molecules which will suffer a collision in the
time $dt$ is $dt/\tau$. To take an example, if $\tau$ is one minute,
then in one second the fraction of particles which will suffer
collisions is $1/60$. What this means, of course, is that $1/60$ of
the molecules happen to be close enough to what they are going to hit
next that *their* collisions will occur in the next second.

When we say that $\tau$, the mean time between collisions, is one
minute, we do not mean that all the collisions will occur at times
separated by exactly one minute. A particular particle does not have a
collision, wait one minute, and then have another collision. The times
between successive collisions are quite variable. We will not need it
for our later work here, but we may make a small diversion to answer
the question: “What *are* the times between collisions?” We
know that for the case above, the *average* time is one minute,
but we might like to know, for example, what is the chance that we get
no collision for *two* minutes?

We shall find the answer to the general question: “What is the
probability that a molecule will go for a time $t$ without having a
collision?” At some arbitrary instant—that we call $t = 0$—we
begin to watch a particular molecule. What is the chance that it gets
by until $t$ without colliding with another molecule? To compute the
probability, we observe what is happening to all $N_0$ molecules in a
container. After we have waited a time $t$, some of them will have had
collisions. We let $N(t)$ be the number that have *not* had
collisions up to the time $t$. $N(t)$ is, of course, less than
$N_0$. We can find $N(t)$ because we know how it changes with time. If
we know that $N(t)$ molecules have got by until $t$, then $N(t + dt)$,
the number which get by until $t + dt$, is *less* than $N(t)$ by
the number that have collisions in $dt$. The number that collide
in $dt$ we have written above in terms of the mean time $\tau$ as $dN =
N(t)\,dt/\tau$. We have the equation
\begin{equation}
\label{Eq:I:43:2}
N(t + dt) = N(t) - N(t)\,\frac{dt}{\tau}.
\end{equation}
The quantity on the left-hand side, $N(t + dt)$, can be written,
according to the definitions of calculus, as $N(t) +
(dN/dt)\,dt$. Making this substitution, Eq. (43.2) yields
\begin{equation}
\label{Eq:I:43:3}
\ddt{N(t)}{t} = -\frac{N(t)}{\tau}.
\end{equation}
The number that are being lost in the interval $dt$ is proportional to
the number that are present, and inversely proportional to the mean
life $\tau$. Equation (43.3) is easily integrated if we
rewrite it as
\begin{equation}
\label{Eq:I:43:4}
\frac{dN(t)}{N(t)} = -\frac{dt}{\tau}.
\end{equation}
Each side is a perfect differential, so the integral is
\begin{equation}
\label{Eq:I:43:5}
\ln N(t) = -t/\tau + (\text{a constant}),
\end{equation}
which says the same thing as
\begin{equation}
\label{Eq:I:43:6}
N(t) = (\text{constant})e^{-t/\tau}.
\end{equation}
We know that the constant must be just $N_0$, the total number of
molecules present, since all of them start at $t = 0$ to wait for
their “next” collision. We can write our result as
\begin{equation}
\label{Eq:I:43:7}
N(t) = N_0e^{-t/\tau}.
\end{equation}
If we wish the *probability* of no collision, $P(t)$, we can get
it by dividing $N(t)$ by $N_0$, so
\begin{equation}
\label{Eq:I:43:8}
P(t) = e^{-t/\tau}.
\end{equation}
Our result is: the probability that a particular molecule survives a
time $t$ without a collision is $e^{-t/\tau}$, where $\tau$ is the
mean time between collisions. The probability starts out at $1$ (or
certainty) for $t = 0$, and gets less as $t$ gets bigger and
bigger. The probability that the molecule avoids a collision for a
time equal to $\tau$ is $e^{-1} \approx 0.37$. The chance is less than
one-half that it will have a greater than average time between
collisions. That is all right, because there are enough molecules
which go collision-free for times much *longer* than the mean
time before colliding, so that the average time can still be $\tau$.

We originally defined $\tau$ as the average time *between*
collisions. The result we have obtained in Eq. (43.7)
also says that the mean time from an *arbitrary* starting instant
to the *next* collision is *also* $\tau$. We can demonstrate
this somewhat surprising fact in the following way. The number of
molecules which experience their *next* collision in the
interval $dt$ at the time $t$ after an arbitrarily chosen starting time
is $N(t)\,dt/\tau$. Their “time until the next collision” is, of
course, just $t$. The “average time until the next collision” is
obtained in the usual way:
\begin{equation*}
\text{Average time until the next collision} =
\frac{1}{N_0}\int_0^\infty t\,\frac{N(t)\,dt}{\tau}.
\end{equation*}
Using $N(t)$ obtained in (43.7) and evaluating the
integral, we find indeed that $\tau$ is the average time from
*any* instant until the next collision.

### 43–2The mean free path

Another way of describing the molecular collisions is to talk not
about the *time* between collisions, but about *how far* the
particle moves between collisions. If we say that the average time
between collisions is $\tau$, and that the molecules have a mean
velocity $v$, we can expect that the average *distance* between
collisions, which we shall call $l$, is just the product of $\tau$
and $v$. This distance between collisions is usually called the *mean
free path:*
\begin{equation}
\label{Eq:I:43:9}
\text{Mean free path $l$} = \tau v.
\end{equation}

In this chapter we shall be a little careless about *what kind of
average* we mean in any particular case. The various possible
averages—the mean, the root-mean-square, etc.—are all nearly equal
and differ by factors which are near to one. Since a detailed analysis
is required to obtain the correct numerical factors anyway, we need
not worry about which average is required at any particular point. We
may also warn the reader that the algebraic symbols we are using for
some of the physical quantities (e.g., $l$ for the mean free path) do
not follow a generally accepted convention, mainly because there is no
general agreement.

Just as the chance that a molecule will have a collision in a short time $dt$ is equal to $dt/\tau$, the chance that it will have a collision in going a distance $dx$ is $dx/l$. Following the same line of argument used above, the reader can show that the probability that a molecule will go at least the distance $x$ before having its next collision is $e^{-x/l}$.

The average distance a molecule goes before colliding with another molecule—the mean free path $l$—will depend on how many molecules there are around and on the “size” of the molecules, i.e., how big a target they represent. The effective “size” of a target in a collision we usually describe by a “collision cross section,” the same idea that is used in nuclear physics, or in light-scattering problems.

Consider a moving particle which travels a distance $dx$ through a gas which has $n_0$ scatterers (molecules) per unit volume (Fig. 43–1). If we look at each unit of area perpendicular to the direction of motion of our selected particle, we will find there $n_0\,dx$ molecules. If each one presents an effective collision area or, as it is usually called, “collision cross section,” $\sigma_c$, then the total area covered by the scatterers is $\sigma_cn_0\,dx$.

By “collision cross section” we mean the area within which the center of our particle must be located if it is to collide with a particular molecule. If molecules were little spheres (a classical picture) we would expect that $\sigma_c = \pi(r_1 + r_2)^2$, where $r_1$ and $r_2$ are the radii of the two colliding objects. The chance that our particle will have a collision is the ratio of the area covered by scattering molecules to the total area, which we have taken to be one. So the probability of a collision in going a distance $dx$ is just $\sigma_cn_0\,dx$: \begin{equation} \label{Eq:I:43:10} \text{Chance of a collision in $dx$} = \sigma_cn_0\,dx. \end{equation}

We have seen above that the chance of a collision in $dx$ can also be written in terms of the mean free path $l$ as $dx/l$. Comparing this with (43.10), we can relate the mean free path to the collision cross section: \begin{equation} \label{Eq:I:43:11} \frac{1}{l} = \sigma_cn_0, \end{equation} which is easier to remember if we write it as \begin{equation} \label{Eq:I:43:12} \sigma_cn_0l = 1. \end{equation}

This formula can be thought of as saying that there should be one
collision, on the average, when the particle goes through a
distance $l$ in which the scattering molecules *could* just cover the
total area. In a cylindrical volume of length $l$ and a base of unit
area, there are $n_0l$ scatterers; if each one has an area $\sigma_c$
the total area covered is $n_0l\sigma_c$, which is just one unit of
area. The whole area is *not* covered, of course, because some
molecules are partly hidden behind others. That is why some molecules
go farther than $l$ before having a collision. It is only *on the
average* that the molecules have a collision by the time they go the
distance $l$. From measurements of the mean free path $l$ we can
determine the scattering cross section $\sigma_c$, and compare the
result with calculations based on a detailed theory of atomic
structure. But that is a different subject! So we return to the
problem of nonequilibrium states.

### 43–3The drift speed

We want to describe what happens to a molecule, or several molecules, which are different in some way from the large majority of the molecules in a gas. We shall refer to the “majority” molecules as the “background” molecules, and we shall call the molecules which are different from the background molecules “special” molecules or, for short, the $S$-molecules. A molecule could be special for any number of reasons: It might be heavier than the background molecules. It might be a different chemical. It might have an electric charge—i.e., be an ion in a background of uncharged molecules. Because of their different masses or charges the $S$-molecules may have forces on them which are different from the forces on the background molecules. By considering what happens to these $S$-molecules we can understand the basic effects which come into play in a similar way in many different phenomena. To list a few: the diffusion of gases, electric currents in batteries, sedimentation, centrifugal separation, etc.

We begin by concentrating on the basic process: an $S$-molecule in a
background gas is acted on by some specific force $\FLPF$ (which might
be, e.g., gravitational or electrical) and *in addition* by the
not-so-specific forces due to collisions with the background
molecules. We would like to describe the *general* behavior of
the $S$-molecule. What happens to it, *in detail*, is that it
darts around hither and yon as it collides over and over again with
other molecules. But if we watch it carefully we see that it does make
some net progress in the direction of the force $\FLPF$. We say that
there is a *drift*, superposed on its random motion. We would
like to know what the speed of its drift is—its *drift
velocity*—due to the force $\FLPF$.

If we start to observe an $S$-molecule at some instant we may expect that it is somewhere between two collisions. In addition to the velocity it was left with after its last collision it is picking up some velocity component due to the force $\FLPF$. In a short time (on the average, in a time $\tau$) it will experience a collision and start out on a new piece of its trajectory. It will have a new starting velocity, but the same acceleration from $\FLPF$.

To keep things simple for the moment, we shall suppose that after each collision our $S$-molecule gets a completely “fresh” start. That is, that it keeps no remembrance of its past acceleration by $\FLPF$. This might be a reasonable assumption if our $S$-molecule were much lighter than the background molecules, but it is certainly not valid in general. We shall discuss later an improved assumption.

For the moment, then, our assumption is that the $S$-molecule leaves
each collision with a velocity which may be in any direction with
equal likelihood. The starting velocity will take it equally in all
directions and will not contribute to any net motion, so we shall not
worry further about its initial velocity after a collision. In
addition to its random motion, each $S$-molecule will have, at any
moment, an additional velocity in the direction of the force $\FLPF$,
which it has picked up *since* its last collision. What is the
*average* value of *this* part of the velocity? It is just
the acceleration $\FLPF/m$ (where $m$ is the mass of the $S$-molecule)
times the *average* time *since* the last collision. Now the
average time *since* the *last* collision must be the same
as the average time *until* the *next* collision, which we
have called $\tau$, above. The *average* velocity from $\FLPF$,
of course, is just what is called the drift velocity, so we have the
relation
\begin{equation}
\label{Eq:I:43:13}
v_{\text{drift}} = \frac{F\tau}{m}.
\end{equation}
This basic relation is the heart of our subject. There may be some
complication in determining what $\tau$ is, but the basic process is
defined by Eq. (43.13).

You will notice that the drift velocity is *proportional* to the
force. There is, unfortunately, no generally used name for the
constant of proportionality. Different names have been used for each
different kind of force. If in an electrical problem the force is
written as the charge times the electric field, $\FLPF = q\FLPE$, then
the constant of proportionality between the velocity and the electric
field $\FLPE$ is usually called the “mobility.” In spite of the
possibility of some confusion, *we* shall use the term
*mobility* for the ratio of the drift velocity to the force for
*any* force. We write
\begin{equation}
\label{Eq:I:43:14}
v_{\text{drift}} = \mu F
\end{equation}
in general, and we shall call $\mu$ the mobility. We have from
Eq. (43.13) that
\begin{equation}
\label{Eq:I:43:15}
\mu = \tau/m.
\end{equation}
The mobility is proportional to the mean time between collisions
(there are fewer collisions to slow it down) and inversely
proportional to the mass (more inertia means less speed picked up
between collisions).

To get the correct numerical coefficient in Eq. (43.13),
which is correct as given, takes some care. Without intending to
confuse, we should still point out that the arguments have a subtlety
which can be appreciated only by a careful and detailed study. To
illustrate that there are difficulties, in spite of appearances, we
shall make over again the argument which led to Eq. (43.13)
in a reasonable *but erroneous* way (and the way one will find in
many textbooks!).

We might have said: The mean time between collisions is $\tau$. After
a collision the particle starts out with a random velocity, but it
picks up an additional velocity between collisions, which is equal to
the acceleration times the time. Since it takes the time $\tau$ to
arrive at the *next* collision it gets there with the
velocity $(F/m)\tau$. At the beginning of the collision it had zero
velocity. So between the two collisions it has, on the average, a
velocity one-half of the final velocity, so the mean drift velocity
is $\tfrac{1}{2}F\tau/m$. (Wrong!) This result is wrong and the result
in Eq. (43.13) is right, although the arguments may sound
equally satisfactory. The reason the second result is wrong is
somewhat subtle, and has to do with the following: The argument is
made as though all collisions were separated by the mean
time $\tau$. The fact is that some times are shorter and others are longer
than the mean. Short times occur *more often* but make
*less* contribution to the drift velocity because they have less
chance “to really get going.” If one takes proper account of the
*distribution* of free times between collisions, one can show
that there should not be the factor $\tfrac{1}{2}$ that was obtained
from the second argument. The error was made in trying to relate by a
simple argument the *average final* velocity to the average
velocity itself. This relationship is not simple, so it is best to
concentrate on what is wanted: the average velocity itself. The first
argument we gave determines the average velocity directly—and
correctly! But we can perhaps see now why we shall not in general try
to get all of the correct numerical coefficients in our elementary
derivations!

We return now to our simplifying assumption that each collision knocks out all memory of the past motion—that a fresh start is made after each collision. Suppose our $S$-molecule is a heavy object in a background of lighter molecules. Then our $S$-molecule will not lose its “forward” momentum in each collision. It would take several collisions before its motion was “randomized” again. We should assume, instead, that at each collision—in each time $\tau$ on the average—it loses a certain fraction of its momentum. We shall not work out the details, but just state that the result is equivalent to replacing $\tau$, the average collision time, by a new—and longer—$\tau$ which corresponds to the average “forgetting time,” i.e., the average time to forget its forward momentum. With such an interpretation of $\tau$ we can use our formula (43.15) for situations which are not quite as simple as we first assumed.

### 43–4Ionic conductivity

We now apply our results to a special case. Suppose we have a gas in a vessel in which there are also some ions—atoms or molecules with a net electric charge. We show the situation schematically in Fig. 43–2. If two opposite walls of the container are metallic plates, we can connect them to the terminals of a battery and thereby produce an electric field in the gas. The electric field will result in a force on the ions, so they will begin to drift toward one or the other of the plates. An electric current will be induced, and the gas with its ions will behave like a resistor. By computing the ion flow from the drift velocity we can compute the resistance. We ask, specifically: How does the flow of electric current depend on the voltage difference $V$ that we apply across the two plates?

We consider the case that our container is a rectangular box of length $b$ and cross-sectional area $A$ (Fig. 43–2). If the potential difference, or voltage, from one plate to the other is $V$, the electric field $E$ between the plates is $V/b$. (The electric potential is the work done in carrying a unit charge from one plate to the other. The force on a unit charge is $\FLPE$. If $\FLPE$ is the same everywhere between the plates, which is a good enough approximation for now, the work done on a unit charge is just $Eb$, so $V = Eb$.) The special force on an ion of the gas is $q\FLPE$, where $q$ is the charge on the ion. The drift velocity of the ion is then $\mu$ times this force, or \begin{equation} \label{Eq:I:43:16} v_{\text{drift}} = \mu F = \mu qE = \mu q\,\frac{V}{b}. \end{equation} An electric current $I$ is the flow of charge in a unit time. The electric current to one of the plates is given by the total charge of the ions which arrive at the plate in a unit of time. If the ions drift toward the plate with the velocity $v_{\text{drift}}$, then those which are within a distance ($v_{\text{drift}}\cdot T$) will arrive at the plate in the time $T$. If there are $n_i$ ions per unit volume, the number which reach the plate in the time $T$ is ($n_i\cdot A\cdot v_{\text{drift}}\cdot T$). Each ion carries the charge $q$, so we have that \begin{equation} \label{Eq:I:43:17} \text{Charge collected in $T$} = qn_iAv_{\text{drift}}T. \end{equation} The current $I$ is the charge collected in $T$ divided by $T$, so \begin{equation} \label{Eq:I:43:18} I = qn_iAv_{\text{drift}}. \end{equation} Substituting $v_{\text{drift}}$ from (43.16), we have \begin{equation} \label{Eq:I:43:19} I = \mu q^2n_i\,\frac{A}{b}\,V. \end{equation} We find that the current is proportional to the voltage, which is just the form of Ohm’s law, and the resistance $R$ is the inverse of the proportionality constant: \begin{equation} \label{Eq:I:43:20} \frac{1}{R} = \mu q^2n_i\,\frac{A}{b}. \end{equation} We have a relation between the resistance and the molecular properties $n_i$, $q$, and $\mu$, which depends in turn on $m$ and $\tau$. If we know $n_i$ and $q$ from atomic measurements, a measurement of $R$ could be used to determine $\mu$, and from $\mu$ also $\tau$.

### 43–5Molecular diffusion

We turn now to a different kind of problem, and a different kind of
analysis: the theory of diffusion. Suppose that we have a container of
gas in thermal equilibrium, and that we introduce a small amount of a
different kind of gas at some place in the container. We shall call
the original gas the “background” gas and the new one the
“special” gas. The special gas will start to spread out through the
whole container, but it will spread slowly because of the presence of
the background gas. This slow spreading-out process is called
*diffusion*. The diffusion is controlled mainly by the molecules
of the special gas getting knocked about by the molecules of the
background gas. After a large number of collisions, the special
molecules end up spread out more or less evenly throughout the whole
volume. We must be careful *not* to confuse diffusion of a gas
with the gross transport that may occur due to convection
currents. Most commonly, the mixing of two gases occurs by a
combination of convection and diffusion. We are interested now only in
the case that there are *no “wind” currents*. The gas is
spreading only by molecular motions, by diffusion. We wish to compute
how fast diffusion takes place.

We now compute the *net flow* of molecules of the “special” gas
due to the molecular motions. There will be a net flow only when there
is some nonuniform distribution of the molecules, otherwise all of the
molecular motions would average to give no net flow. Let us consider
first the flow in the $x$-direction. To find the flow, we consider an
imaginary plane surface perpendicular to the $x$-axis and count the
number of special molecules that cross this plane. To obtain the net
flow, we must count as positive those molecules which cross in the
direction of positive $x$ and *subtract* from this number the
number which cross in the negative $x$-direction. As we have seen many
times, the number which cross a surface area in a time $\Delta T$ is
given by the number which start the interval $\Delta T$ in a volume
which extends the distance $v\,\Delta T$ from the plane. (Note that
$v$, here, is the actual molecular velocity, not the drift velocity.)

We shall simplify our algebra by giving our surface one unit of area. Then the number of special molecules which pass from left to right (taking the $+x$-direction to the right) is $n_-v\,\Delta T$, where $n_-$ is the number of special molecules per unit volume to the left (within a factor of $2$ or so, but we are ignoring such factors!). The number which cross from right to left is, similarly, $n_+v\,\Delta T$, where $n_+$ is the number density of special molecules on the right-hand side of the plane. If we call the molecular current $J$, by which we mean the net flow of molecules per unit area per unit time, we have \begin{equation} \label{Eq:I:43:21} J = \frac{n_-v\,\Delta T - n_+v\,\Delta T}{\Delta T}, \end{equation} or \begin{equation} \label{Eq:I:43:22} J = (n_- - n_+)v. \end{equation}

What shall we use for $n_-$ and $n_+$? When we say “the density on the
left,” how *far* to the left do we mean? We should choose the
density at the place from which the molecules started their “flight,”
because the number which *start* such trips is determined by the
number present at that place. So by $n_-$ we should mean the density a
distance to the left equal to the mean free path $l$, and by $n_+$, the
density at the distance $l$ to the right of our imaginary surface.

It is convenient to consider that the distribution of our special molecules in space is described by a continuous function of $x$, $y$, and $z$ which we shall call $n_a$. By $n_a(x,y,z)$ we mean the number density of special molecules in a small volume element centered on $(x,y,z)$. In terms of $n_a$ we can express the difference $(n_+ - n_-)$ as \begin{equation} \label{Eq:I:43:23} (n_+ - n_-) = \ddt{n_a}{x}\,\Delta x = \ddt{n_a}{x}\cdot 2l. \end{equation} Substituting this result in Eq. (43.22) and neglecting the factor of $2$, we get \begin{equation} \label{Eq:I:43:24} J_x = -lv\,\ddt{n_a}{x}. \end{equation} We have found that the flow of special molecules is proportional to the derivative of the density, or to what is sometimes called the “gradient” of the density.

It is clear that we have made several rough approximations. Besides
various factors of two we have left out, we have used $v$ where we
should have used $v_x$, and we have assumed that $n_+$ and $n_-$ refer
to places at the perpendicular distance $l$ from our surface, whereas
for those molecules which do not travel perpendicular to the surface
element, $l$ should correspond to the *slant* distance from the
surface. All of these refinements can be made; the result of a more
careful analysis shows that the right-hand side of
Eq. (43.24) should be multiplied by $1/3$. So a better answer
is
\begin{equation}
\label{Eq:I:43:25}
J_x = -\frac{lv}{3}\,\ddt{n_a}{x}.
\end{equation}
Similar equations can be written for the currents in the $y$-
and $z$-directions.

The current $J_x$ and the density gradient $dn_a/dx$ can be measured by macroscopic observations. Their experimentally determined ratio is called the “diffusion coefficient,” $D$. That is, \begin{equation} \label{Eq:I:43:26} J_x = -D\,\ddt{n_a}{x}. \end{equation} We have been able to show that for a gas we expect \begin{equation} \label{Eq:I:43:27} D = \tfrac{1}{3}lv. \end{equation}

So far in this chapter we have considered two distinct processes:
*mobility*, the drift of molecules due to “outside” forces; and
*diffusion*, the spreading determined only by the internal
forces, the random collisions. There is, however, a relation between
them, since they both depend basically on the thermal motions, and the
mean free path $l$ appears in both calculations.

If, in Eq. (43.25), we substitute $l = v\tau$ and $\tau =
\mu m$, we have
\begin{equation}
\label{Eq:I:43:28}
J_x = -\tfrac{1}{3}mv^2\mu\,\ddt{n_a}{x}.
\end{equation}
But $mv^2$ depends only on the temperature. We recall that
\begin{equation}
\label{Eq:I:43:29}
\tfrac{1}{2}mv^2 = \tfrac{3}{2}kT,
\end{equation}
so
\begin{equation}
\label{Eq:I:43:30}
J_x = -\mu kT\,\ddt{n_a}{x}.
\end{equation}
We find that $D$, the *diffusion* coefficient, is just $kT$
times $\mu$, the *mobility* coefficient:
\begin{equation}
\label{Eq:I:43:31}
D = \mu kT.
\end{equation}
And it turns out that the numerical coefficient in (43.31)
is exactly right—no extra factors have to be thrown in to adjust for
our rough assumptions. We can show, in fact, that (43.31)
must *always* be correct—even in complicated situations (for
example, the case of a suspension in a liquid) where the details of
our simple calculations would not apply at all.

To show that (43.31) must be correct in general, we shall
derive it in a different way, using only our basic principles of
statistical mechanics. Imagine a situation in which there is a
gradient of “special” molecules, and we have a diffusion current
proportional to the density gradient, according to
Eq. (43.26). We now apply a force field in the $x$-direction,
so that each special molecule feels the force $F$. According to the
*definition* of the mobility $\mu$ there will be a drift velocity
given by
\begin{equation}
\label{Eq:I:43:32}
v_{\text{drift}} = \mu F.
\end{equation}
By our usual arguments, the *drift current* (the *net*
number of molecules which pass a unit of area in a unit of time) will
be
\begin{equation}
\label{Eq:I:43:33}
J_{\text{drift}} = n_av_{\text{drift}},
\end{equation}
or
\begin{equation}
\label{Eq:I:43:34}
J_{\text{drift}} = n_a\mu F.
\end{equation}
We now *adjust* the force $F$ so that the drift current due
to $F$ just *balances* the diffusion, so that there is *no net
flow* of our special molecules. We have $J_x + J_{\text{drift}} = 0$,
or
\begin{equation}
\label{Eq:I:43:35}
D\,\ddt{n_a}{x} = n_a\mu F.
\end{equation}

Under the “balance” conditions we find a steady (with time) gradient of density given by \begin{equation} \label{Eq:I:43:36} \ddt{n_a}{x} = \frac{n_a\mu F}{D}. \end{equation}

But notice! We are describing an *equilibrium* condition, so our
*equilibrium* laws of statistical mechanics apply. According to
these laws the probability of finding a molecule at the coordinate $x$
is proportional to $e^{-U/kT}$, where $U$ is the potential energy. In
terms of the number density $n_a$, this means that
\begin{equation}
\label{Eq:I:43:37}
n_a = n_0e^{-U/kT}.
\end{equation}
If we differentiate (43.37) with respect to $x$, we find
\begin{equation}
\label{Eq:I:43:38}
\ddt{n_a}{x} = -n_0e^{-U/kT}\cdot\frac{1}{kT}\,\ddt{U}{x},
\end{equation}
or
\begin{equation}
\label{Eq:I:43:39}
\ddt{n_a}{x} = -\frac{n_a}{kT}\,\ddt{U}{x}.
\end{equation}
In our situation, since the force $F$ is in the $x$-direction, the
potential energy $U$ is just $-Fx$, and $-dU/dx = F$.
Equation (43.39) then gives
\begin{equation}
\label{Eq:I:43:40}
\ddt{n_a}{x} = \frac{n_aF}{kT}.
\end{equation}
[This is just exactly Eq. (40.2), from which we
deduced $e^{-U/kT}$ in the first place, so we have come in a
circle]. Comparing (43.40) with (43.36), we get
exactly Eq. (43.31). We have shown that
Eq. (43.31), which gives the diffusion current in terms of
the mobility, has the correct coefficient and is very generally true.
Mobility and diffusion are intimately connected. This relation was first
deduced by Einstein.

### 43–6Thermal conductivity

The methods of the kinetic theory that we have been using above can be
used also to compute the *thermal conductivity* of a gas. If the
gas at the top of a container is hotter than the gas at the bottom,
heat will flow from the top to the bottom. (We think of the top being
hotter because otherwise convection currents would be set up and the
problem would no longer be one of heat *conduction*.) The
transfer of heat from the hotter gas to the colder gas is by the
diffusion of the “hot” molecules—those with more energy—downward
and the diffusion of the “cold” molecules upward. To compute the
flow of thermal energy we can ask about the energy carried downward
across an element of area by the downward-moving molecules, and about
the energy carried upward across the surface by the upward-moving
molecules. The difference will give us the net downward flow of
energy.

The thermal conductivity $\kappa$ is defined as the ratio of the rate at which thermal energy is carried across a unit surface area, to the temperature gradient: \begin{equation} \label{Eq:I:43:41} \frac{1}{A}\,\ddt{Q}{t} = -\kappa\,\ddt{T}{z}. \end{equation} Since the details of the calculations are quite similar to those we have done above in considering molecular diffusion, we shall leave it as an exercise for the reader to show that \begin{equation} \label{Eq:I:43:42} \kappa = \frac{knlv}{\gamma - 1}, \end{equation} where $kT/(\gamma - 1)$ is the average energy of a molecule at the temperature $T$.

If we use our relation $nl\sigma_c = 1$, the heat conductivity can be written as \begin{equation} \label{Eq:I:43:43} \kappa = \frac{1}{\gamma - 1}\,\frac{kv}{\sigma_c}. \end{equation}

We have a rather surprising result. We know that the average velocity
of gas molecules depends on the temperature but *not on the
density*. We expect $\sigma_c$ to depend only on the *size* of
the molecules. So our simple result says that the thermal
conductivity $\kappa$ (and therefore the *rate* of flow of heat in any
particular circumstance) is independent of the *density* of the
gas! The change in the number of “carriers” of energy with a change
in density is just compensated by the larger distance the “carriers”
can go between collisions.

One may ask: “Is the heat flow independent of the gas density in the limit as the density goes to zero? When there is no gas at all?” Certainly not! The formula (43.43) was derived, as were all the others in this chapter, under the assumption that the mean free path between collisions is much smaller than any of the dimensions of the container. Whenever the gas density is so low that a molecule has a fair chance of crossing from one wall of its container to the other without having a collision, none of the calculations of this chapter apply. We must in such cases go back to kinetic theory and calculate again the details of what will occur.