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DennyCh2 - Chapter 6 Diffusion Random Walksin Air and Water...

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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 hundred-yard 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 A-to 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". (6-3) 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.._— 10-0 - 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 root-mean—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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