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Unformatted text preview: Chapter 6 Diffusion:
Random Walksin
Air and Water This chapter deals with the process of random motion and the way in which it can act
as a mechanism of transport. The focus here is on the process of molecular diffusion
and its biological consequences. For instance, we will see how the diffusion of
gases can limit the size of plants and animals and explain why an ostrich egg is
more porous than that of a hummingbird. We will explore the role of diffusive
transport in foraging strategy: Is it better to go looking for food or to wait for the
food to come to you? And we will see how the mean free path of molecules in air
sets the minimal size of insect tracheoles. The principles outlined in this chapter are quite general, however, and we will
make considerable use of them in later chapters. 6.1 The Physics 6.1.1 Molecular Velocity In chapter 3 we deﬁned temperature in terms of the average kinetic energy of
molecules using the relationship = , (6.1) where m is the mass of the molecule, u its speed, and k is Boltzmann’s constant,
1.38 x 10"23 J K‘ 1. The brackets, 0, denote that velocity is averaged over time for
a single molecule, or averaged over many molecules at any one time. This simple
expression has some intriguing consequences. Consider, for instance, a typical molecule of air (nitrogen) with a weight per
molecule of about 4.7 x 10“26 kg. At room temperature (290 K), we have seen
(chapter 3) that the average speed of a nitrogen molecule is . @2) = ﬂ .~. scams—1. (6.2) m This is a sizable speed.1 For example, an unimpeded nitrogen molecule would
ﬁnish the hundredyard dash in about a ﬁfth of a second. And this speed is not
peculiar just to nitrogen. At 290 K oxygen moves at an average speed of 475 m 5“,
carbon dioxide at 405 m s“, and water vapor at a brisk 634 m s‘ 1. The high velocity of molecules at room temperature raises visions of extremely
effective transport systems. If you want to move oxygen from point Ato point B, you simply take the gas one molecule at a time, point it in the right direction,2 and '0f the total kinetic energy of a molecule, 1/3 (on average) is associated with motion along each
particular axis. Thus, at 290 K the average velocity of a nitrogen molecule along the 9: axis is kT
1/<uZ) = g z 292ms". (63) The overall speed of 508 ms‘ is (uZ) + (v2) + (wz). let it go. Thermal kinetic energy does the rest! But these visions seem at odds with
everyday experience. If you put a drop of dye in a glass of still water, it may take
hours to spread throughout the container. Similarly, it can take several minutes
for the scent from a bottle of perfume opened in one corner of a room of still air
to reach the opposite corner. How can this be if molecules are moving at several
hundred m s”1 ? ' Even worse, these everyday experiences are at odds with careful experiments.
In an ordinary glass of water, convection currents created as the water evaporates
account for most of the movement of dye. If one makes sure that no convection is
present, a drop of dye may take months (rather than hours) to spread throughout
the container. Similarly, convection currents in air account for most of the transport
of perfume. Clear evidence of these currents are the motes of dust that dance in a
sunbeam. In the absence of convection, scent takes hours (rather than minutes) to
spread across a room. Our task in this chapter is to reconcile these experimental results regarding
the macroscopic transport of molecules in ﬂuids with the microscopic behavior
of individual molecules. This reconciliation is found inthe process of random
motion. Each molecule moves at considerable speed, but it cannot move very far
in a straight line without colliding with another ﬂuid molecule. Like a billiard ball
on a crowded table, at each collision the molecule ﬂies off in a new direction only
to collide again. The direction taken by a molecule after a collision is purely a
matter of chance. If it hits another molecule squarely, it may rebound back from
where it came. A glancing blow results in a less drastic change in direction. As a
result of its collisions, each molecule performs a random walk through space. It is
to the characteristics of such walks that we now turn our attention. 6.1.2 Random Walks The concept of a random walk is best understood through the use of a simple
example. Consider motion along the x axis. We start by placing a hypothetical
particle at the origin, a: = 0, and every T seconds we allow the particle to step
a distance 6 along the axis. Thus the average speed of the particle is 6/7'. The
direction of each step, however, is purely a matter of chance. Half the time the
particle steps to the right, half the time to the left. In practice, we can simulate this
process by tossing a coin before'each step. If the coin comes up heads, the particle
moves to the right; if tails, it moves to the left. After each step in this random walk we note the position of the particle. In this
manner we can track the particle along the axis as a function of the number of steps
taken, or, equivalently, as a function of time. Because each step is chosen at random,
the precise path taken by a particle is impossible to predict (ﬁg. 6.1). However,
if we repeat the experiment a number of times, we can perceive a pattern in the
average manner in which particles move. It is this predictability of the average
that renders the concept of a random walk so useful. To explore the properties of the average, we examine the statistics of parti
cle motion. To do so, we track the random walks of N particles. Let rim) be the 2You may object to the idea of “pointing” a thermally agitated molecule. Because thermal motion
is disordered, the act of aiming a molecule in essence cools it. It would be possible in theory, however,
to let each molecule rattle around in a box until by chance it is moving in the proper direction, and then
suddenly release it. 85 Diﬂusion:
Random Walks in
Air and Water 86 Chapter 6 Fig. 6.1 Six particles, all starting at the same
spot, take quite different paths in a random walk.
On average, however, the particles go nowhere at all. Distance From Origin 0 20 40 60 80 100 120 Steps location on the 3: axis of the ith particle after the particle has taken n steps. We
know from our assumptions that ' trim) = 33,(n — 1) :i: 6. (6.4) In other words, the position of the particle after n steps is either one step length to
the right or one step length to left of the position of the particle after 71. — 1 steps.
After we have repeated our experiment N times, we can compute the average position of particles after n steps: ($01)): 51:1(n) + $202]. . . + zN(n) 1 N
= ~ Zoe(n), (66)
N i=1 (6.5) where the symbol )3 means that we add the position after 11 steps of all particles numbered 1 to N. I 
We can expand this expression by replacing rim) With 55,(n — 1) :1: 6, which we know to be equivalent (Eq. 6.4). Thus, N ($01)) = %Ztcc.<n — 1) i 61 (6.7)
i=1
1 N 1 N
= NZxZ(n— 1)+N216. (6.8) But, since half the time 6 is to the right and half the time to the left, the average of
i6 must be 0. As a consequence, ($01)) = (x01 — 1)). (6.9) In other words, the average position of a particle after n steps is the same as the
position after n — 1 steps. Taken to its logical conclusion, this means that, on
average, particles starting at the origin remain at the origin. This is just what you might expect given equal probability of stepping right or left. The fact that particles go nowhere on average does not imply that all particles
remain at the origin, however. All that the average tells us is that for every particle
that ends up at x = +3, for instance, there is likely to be a particle that ends up at
a: = —3. As long as an equal number of particles have net movements left and right,
the average can still be 0, although there is considerable spread of particles.3 To
see how far away from the origin particles are likely to travel, we need to calculate
our average in a different way. It would be possible to avoid the problem of averaging positive with negative
values by taking the absolute value of each particle’s location after n steps and
averaging these values as before. Instead, it has become traditional to circumvent
the problem by taking the square of the location on the :3 axis. Because the squares
of both positive and negative numbers are always positive, we can, by averaging
the squares of locations, gain information as to the average distance traveled from
the origin. Again we begin by expressing the location after n steps in terms of the
location after 71 —_ 1 steps. Expanding the expression, we see that4 em) = [scan — 1) i 612 (6.10)
=m§(n—1)i26m,(n— 1)+62. (6.11) When we average this value for N particles, we see that N
(35201)) = % 2 [953(1) » 1) i 26mm — 1) + 62] (6.12)
2 l l 1:] 1 N 1 N
= — _1 __ 2 , ~ __ 2 .
N 23:47: )+ N I; d: 613,02 1)+ N g 6 (613) l N 1 N
= NZ}:£3(""1)+N;52 (6.14)
= ($201  1)) +52 (6.15) In this computation we have again made use of the fact that the average of i6 = O. This result tells us that the average (or mean) square of position after 71 steps is
greater by 62 than the average square of position after n — 1 steps. Now, we know
from our initial assumptions that the mean square position after 0 steps is O [i.e.,
(m(0)) = 0], so (222(1)) = 62. Aftertwo steps, the average square of position is 262.
and so forth. Thus, (93201)) = “52 (6.16) In other words, the square of the displacement from the starting point increases
directly with the number of steps taken. We can convert from the square of dis
placement to displacement itself simply by taking the square root of eq. 6.16: V = 5rrms = (6.17) 3The distinction between average and individual movement reminds me of the story of two statis
ticians who went duck hunting. A bird ﬂew past. and both hunters ﬁred. One shot passed 2 m in front
of the duck, the other 2 m behind. The statisticians proceeded to congratulate themselves because on
average they had hit the bird dead center. 4Recall that :rZ(n — l) is the z position of the ith particle after (71 ~ 1) steps. The square of this
value is $301 — 1), not x (n — 1)2. 87 Diffusion:
Random Walks in
Air and Water 88 . Chapter 6 The value \/ (x2(n)) is called the root mean square or rms displacement, and it
is a measure of how far, on average, particles have moved from their starting point after n steps. Readers familiar with statistics may note that the mean square
displacement is the same as the variance of displacement, and the root mean square
displacement is the standard deviation. We now return to our original assumptions regar
that the particle takes a step every 7' seconds. Thus, the number ‘ .
equal to 15/7, where t is the time since the particle started its walk. Inserting this expression for n in eq. 6.16, we see that ding the random walk and recall
of steps taken is 2
(32%)) = 6715. (6.18) particle (in other words, the probable ’" The mean square distance traveled by a '
time at a rate governed by the ratio of square of distance) increases directly with
62 to T. 6.2 The Diffusion Coefﬁcient We now are in a position to relate this theoretical consideration of a random
walk to the process of diffusion. To do so, we deﬁne a diﬂusion coeﬁ‘iczent, ’D: 'D (6.19) ill 62
5;. The reason for the factor of l / 2 will become clear later in this chapter; it eventually . . . . 2 _
makes the math simpler. The diffusmn coefﬁCient has the units m s .
Using this deﬁnition, we can restate our conclusion regarding the random walk: ($20)) = 21% (6.20)
wrms = ($26)) = V (6.21) Eq. 6.21 is of special importance, and is therefore worth dwelling upon for a
moment. It says that the distance a particle is likely to travel from its point of
origin while performing a random walk (the rms distance) increases not With time, as one might guess, but rather with the square root of time. In other words, if a particle travels an average of 1cm in 1 s, it takes an average of 100s to travel 10 cm, and 10, 000 s to travel 100 cm. . ' I ”
As a result, if we try to express the rate of travel as a “diffusmn velocrty we come to a curious conclusion. Setting velocity equal to rms distance per time, /2D
“diffusion velocity” = (6.22) The shorter the period over which we measure velocity, the greater the velocity
we measure. Conversely, the longer the period of measurement, the slower the
velocity measured. The fact that the rate of transport in diffusion is a function of
time is one of the principal characteristics of diffusive processes. ‘
Note that eq. 6.22 can be applied only when t > T. It was calculated assuming
that the particle changes its direction at random. For periods shorter than 7", we have
assumed that the particle is in the midst of a step and not subject to directional change. Therefore, at t < T, our assumption is violated and eq. 6.22 leads to
spurious results. We now examine the diffusion coefﬁcients for three molecules with biological
importance: oxygen, carbon dioxide, and water vapor. 6.2.1 Diffusion Coeﬂicients in Air The diffusion coefﬁcients for oxygen and water vapor in air can be described by the equation, D(T,p> = Mk! ’2. (6.23) P where ’D(T, p) is the diffusion coefﬁcient at absolute temperature T and pressure
p and p0 is one normal atmosphere. The coefﬁcients k; and k2 vary from one type
of molecule to the next. For oxygen, [01 = 1.13 x 10‘9 m2 s“1 and k2 = 1.724
(Marrero and Mason 1972). For water vapor, k1 = 0.187 x 10’9 m2 s‘1 and
k2 = 2.072. The form of eq. 6.23 tells us that the diffusion coefﬁcient rises with
any increase in temperature, but decreases with any increase in pressure. The diffusion coefﬁcient for carbon dioxide in air is described by a slightly
more complex equation (Marrero and Mason 1972), D(T, p) = lek1 exp (~k3/T) 1%, (6.24) where k3 is an additional parameter. In this case k1 = 2.70 x 10‘9m2 5“, k2 =
1.590, and k3 = 102.1 K. . These results are summarized in table 6.1 and ﬁgure 6.2. The diffusivity of
water vapor is about 50% greater, and the diffusivity of 02 about 30% greater,
than that of carbon dioxide. For all three gases, the diffusion coefﬁcient is about
30% higher at 40° C than at 0° C. At the top of Mount Everest, where the pressure
is only a third of that at sea level, the diffusion coefﬁcients of these gases would
be three times those shown here. The distance (arms) that an oxygen molecule is expected to travel in air is shown
as a function of time in ﬁgure 6.3. In one second a molecule will, on average, move
about a centimeter from its starting point. Several hours are required, however,
before the molecule reliably travels one meter. 6.2.2 Diffusion Caeﬁicients in Water The diffusion coefﬁcients of oxygen and carbon dioxide are about 10,000 times
smaller in water than they are in air (table 6.2). This tremendous difference in
diffusivity will dominate our discussion of the biological consequences of diffu
sion, but there are two further contrasts to note. First, the variation in diffusion
coefﬁcient with temperature is much greater in water than in air. For example, at
40° C the diffusion coefﬁcient of 02 in water is 3.2 times that at 0° C, whereas in
air it is increased by a factor of only about 1.3. The variation is somewhat less for
C02; 7) in water is about 2.1 times higher at 40° C than at 0° C. Second, the dependence on temperature affects the rank order of the diffusivities
in water (table 6.2; ﬁg. 6.4). At low temperature, the diffusion coefﬁcient of C02
is larger than that of 02, but above about 6° C, the diffusivity of oxygen is greater
than that of carbon dioxide. ‘, The distance (arrms) that an oxygen molecule is expected to travel in water is
shown as a function of time in ﬁgure 6.3. These distances are surprisingly small. 89 M Diﬂ’usion:
Random Walks in
Air and Water Table 6.1 The diffusion coefﬁcients of various
gases in air at one atmosphere. Diffusion Coefﬁcient, TC) C) Dm (m2 s" x 10*) 02 C02 H20 0 17.9 13.9 20.9
5 18.5 14.4 21.7
10 19.1 14.9 22.5
15 19.7 15.4 23.3
20 20.3 16.0 24.2
25 20.9 16.5 25.1
30 21.5 17.0 26.0
35 22.1 17.6 26.8
40 22.7 18.1 27] Source: Calculated from data presented by
Marrero and Mason (1972). Notes: Note that these values represent the best
ﬁt to empirical measurements, but that actual
measured values may vary by d: 5% from
those shown here. The symbol D771 is used
here to distinguish molecular diffusion from
other types of diffusion (e.g., the diffusion of
heat). 05.: L,” ‘E . i E 90
Chapter 6 For example, relying on diffusion alone, a molecule moves only about 5 cm in a
million seconds.
30 10a
10"
9A 20 A 10'1
3 5
’< c
Tm E 10 ’
E 7.
"E 1 '5.
Q 10 ’ 10.
L 104
La,“
0 .1. r ._l  4—1» 4—; I i. .I—L A _I. L _L.._— 100 
0 10 20 40 10"10'510“10"10"10" 10° 10' 102 10’ 10‘ 10’ 10° 101
Time (5) Temperature (“C) Fig. 6.2 The diffusion coefﬁcients of gases in air
increase slightly with an increase in temperature.
(Data from table 6.1) Fig. 6.3 The rootmean—square distance traveled
by an oxygen molecule is much greater in air
than in water. Note, however, that in either
medium it takes a very long time to travel a
meter by diffusion alone. Table 6.2 The diffusivity of 02 and C02 in
water at one atmosphere (from Armstrong 1979). Diffusion Coefﬁcient, T(° C) Dm (m2 s”l X 10‘9) 02 C02 Id—
0 0.99 1.15
5 1.27 1.30
10 1.54 1.46
15 1.82 1.63
20 2.10 1.77
25 2.38 1.92
30 2.67 2.08 _______.__._.___,_—.—’— Note: The symbol ’Dm is used here to distinguish
molecular diffusion from other types of
diffusion (e.g., the diffusion of heat). 6.2.3 Mean Free Path
Let us now examine the concept of a diffusion coefﬁcient in greater detail. We ﬁrst note that because the coefﬁcient has dimensions of length squared per time, it
can be treated as the product of velocity and a distance. Thus, the rate at which a
particle is diffusively transported depends both on how fast it moves while in free
ﬂight between collisions (u = 6/7) and on the average distance it moves before
again colliding (6 / 2), a distance called the mean free path, 1. We have already
calculated how fast ﬂuid molecules move at room temperature (6/7 x 500 m s” 1 [eq. 62]), and are now in a position to estimate their mean free path.
To do so, we rely on the diffusion coefﬁcients as presented in tables 6.1 and 6.2. We will discuss later in this chapter how these empirical measurements are
made, but for now accept them as given. The diffusion coefﬁcient for an oxygen
molecule in air is about 2 x 10"5 m2 s” 1. Noting that 6 = 219/ u, we calculate that
the mean free path in air is 8 x 10‘8 m. In other words, every 0.08 pm an oxygen
molecule collides with another air molecule and careens off in a new direction.
Traveling at its velocity of 500ms", the 02 molecule travels this distance in
about 0.00016 its. To put it another way, an air molecule experiences about 6.25
billion collisions every second, each one capable of changing the direction of its
ﬂight. With so many collisions going on, it is no wonder that air molecules behave
differently in bulk than might be expected from the behavior of a single molecule
on its own. In water, the diffusion coefﬁcient of an oxygen molecule is 10,000 times smaller
than that in air, 2 x 10—9 m2 s". We know, however, that at a given temperature,
an oxygen molecule dissolved in water has the same kinetic energy as one in air,
and therefore must have the same average velocity. As a consequence, the reduced
diffusion coefﬁcient in water must be due to a reduction in the mean free path.
Again equating 6 and 2D/u, we calculate that the mean free path of an oxygen
molecule in water is about 10“ '1 m, only a twentieth of the diameter of a hydrogen
atom! Because water molecules are packed so tightly together, there is virtually no
“free space” through which an oxygen molecule can ﬂy, and the randomly walking
molecule undergoes about 60,000 billion collisions per second. The net result is a...
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 Fall '07
 ANANDAKRISHNAN
 Velocity, Orders of magnitude, Molecular diffusion

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