t = no of steps of length s

Transcription

1 s t = no of steps of length s Figure : Schematic of the path of a diffusing molecule, for example, one in a gas or a liquid. The particle is moving in steps of length s. For a molecule in a liquid the steps are of length approximately equal to the size of a molecule, i.e., s 0.nm. The time taken for one step is this length divided by the speed of a molecule. Dissipation: Making entropy In thermodynamics you learnt that entropy always increases, it never decreases. Now, we will look at how the entropy increases in a specific system: a single diffusing molecule. We will see how and why diffusion increases entropy. So, this section looks at dynamics, in particular at what are called dissipative dynamics. Dissipative dynamics are, by definition, dynamics that increase the entropy, as opposed to reversible dynamics which keep the total entropy constant. To keep things simple let us consider a single particle that starts off at the origin, x = 0, at time t = 0. It will then diffuse away from the origin and this diffusion will increase the entropy associated with the particle. To keep the notation simple we will set the length of a single hop of the molecule s =, i.e., we will work in units of length equal to that of a single hop. For a molecule in a liquid a single hop of a molecule is around 0. nm. Also, our units of time will be the time for a single hop, around a picosecond in a liquid. Thus t = is the time elapsed for a single hop, t = the time for two hops etc. We will also work in one dimension. So, after a time t, the particle has hopped t times. Let l of these hops be to the left and r = t l be to the right. The position of the particle x is just the number of hops it has made to the right minus the number it has made to the left x = r l = t l. So, the particle has made a total of t hops, each of which has equal probability of being to the left or to the right, and we want to know what is the probability that l of these hops were to the left. This probability is just the number of different possible sets (arrangements) of l hops to the left and r to the right, divided by the total number of different sets of t hops. The total number is just t. For hop there are possibilities, left and right, for hops there are = 4 possibilities, left left, left right, right left and right right, etc. The number of sets of t hops that contain l hops of one type, to the left, and r = t l of the other type of hop, is t!/l!r!. Thus the probability of the particle taking l hops to the left and hence being at position x is p(x) = t t! l!r!. () You recognise this as being just the same (except with t substituted for n) as the expression for the probability of l particles out of a total of n (equivalent to t here) being in the left half of a box. Thus the maths is just the same as for the box divided into two. As before, we take the log of the probability, find that it has a maximum at l = r = t/, which corresponds to x = 0, and then Taylor expand at that point. The Taylor-series

2 expansion around x = 0 gives us p(x) exp ( t ) x t. () Before the Gaussian held for large n, here it holds for large t. This is because it is the result of a Taylor series in which they have dropped all terms beyond the quadratic. This is OK for t but not for t that is not much larger than. Now, we know that p(x) must be normalised, it must add up to x= p(x) =. (3) For t there are many terms in this sum and we can approximate it by an integral p(x)dx =. (4) Introducing an, as yet unknown, normalisation constant c, p(x) = c exp ( t ) x, (5) we find c by substituting this expression for p into the integral c exp ( t ) x dx =. (6) The integral is the standard Gaussian integral that is in your formula booklets and so we have that and c = /(πt) /. Thus, we have that p(x, t) = exp ( αx ) ( π ) / dx = (7) α c (πt) / =, (8) exp ( t ) (πt) / x. (9) Note that the probability, p(x, t) of finding the particle at position x is a function of time. In Fig. we have plotted the probability as a function of x for different times. Now, p(x, t) is symmetric around x = 0, and so the mean value of x, x = 0. This is because the particle is as likely to go left as right. However, the mean of the square of the distance from the origin, x, is non-zero. Using standard integrals over Gaussian functions (given for example in the little booklet you get in exams) it can be calculated and is x = x p(x, t)dx = and so the root-mean-square distance moved is (πt) / x exp ( t ) x dx = ( πt 3 ) / = (πt) / t (0) x = t/ / ()

3 p(x,t) x Figure : Plots of Gaussian p(x, t), Eq. (9), as a function of x, for two different times t. The solid curve is for t = 0 and the dashed curve for t = 0. At the later time the probability distribution function is broader, the particle can be found over a longer part of the x axis. It is this increase in the uncertainty of the position (which defines its state here) of the particle that leads to the increase in its entropy. i.e., x increases linearly in time the root-mean-square displacement increases as the square root of time. This is characteristic of diffusion, if the time elapsed is doubled the typical distance travelled increases only by a factor of /.4. Now, Shannon s expression for the entropy is S = i p i lnp i () For the times t that we are considering we can approximate this sum by an integral, i.e., p i lnp i = p(x) ln p(x)dx (3) i and so the entropy at time t is given by S(t) = p(x, t)lnp(x, t)dx (4) where we have indicated that the probability p is a function both of position x and time t, i.e., the probability that the particle is at position x depends on time t. As the entropy is an integral over x it does not depend on x, only on t. Now, to do this integral we need to substitute the probability of Eq. (9) into this equation. This can be done but we can get an approximate answer and simplify the maths a bit if we realise that at time t the particle is mostly between x = t / and x = t /. If we neglect p(x, t) outside of this range then we can approximate the Gaussian function probability by a step function { /(t p(x, t) = / ) t / x t / 0 outside this range (5) 3

4 Figure 3: Ink diffusing in water. N.B. The swirls indicate that there is also flow (which is different from diffusion) going on. This makes the picture look pretty but is not taken into account in the problem and is not covered by the course. The probability is /(t / ) due to normalisation. The in the denominator is needed to ensure that the integral with respect to x over p(x, t) equals one. With this approximation for p(x, t) S = = ln( t /) t / t / p(x, t)lnp(x, t)dx = ln t / t/ ( ) t / dx = ( t / ln t /) t / dx t / ( t / = ln t /) = ln + ln(t). (6) So, we are interested in how the entropy S varies with time. We have found that it increases, as we should expect nd Law of Thermodynamics says it should. This is clear if we take the time derivative of the entropy ds/dt = /(t) > 0 it is always greater than 0. The entropy increases because as time passes the length (volume for 3-dimensional diffusion) over which the particle may be found increases. Thus the uncertainty in the position of the particle increases, and so its entropy is larger. Another way of saying this starts by noting that at t = 0, before diffusion occurs, the particle is definitely at the origin. So it is definitely in one state and so has an entropy of S = 0. After a time t it can be any one of order t / positions, i.e., t / states. If we assume that these states are all roughly equally likely then we can use the fact that the entropy S = ln Ω for a system with Ω equally likely states and again get that S = lnω = ln(t / ) = (/)lnt..0. Equipartition Before we consider diffusion in a gas we need to use the Equipartition Theorem. This tells us the average kinetic energy of, amongst other things, molecules in a classical gas. We will need the average kinetic energy in order to estimate how rapidly molecules diffuse. Also, once we know the average kinetic energy we can calculate the average velocity and momentum. Now, we know that the probability that an atom has a velocity component v along one direction, say the 4

5 x axis, is given by ( p(v) exp K.E. ) ) = exp ( mv kt kt Thus the energy is a quadratic function of velocity. Equipartition then tells us that the mean K.E. equals (/)kt. As the mass m is constant we can take it out of the average (7) K.E. = mv = kt (8) m v = kt ( v ) / = ( kt m ) / (9) the typical, root-mean-square, velocity is proportional to the square root of kt and inversely proportional to the square root of the mass m. Let us consider the Earth s atmosphere. Here N is the most abundant molecule. For N, m = kg and kt = 4 0 J at room temperature. Thus ( v ) / = 300 m s which is about the speed of sound. In fact the speed of sound in a gas is approximately set by the speed of molecules in it.. Diffusion in a gas We have seen that diffusion is essentially a series of hops with random directions, and that the typical distance travelled increases only as the square root of time. We found this to be true in dimension but it is also true in 3 dimensions. In 3 dimensions we have a vector displacement r(t), which has a magnitude r(t). The mean of the square of r increases linearly with time. Thus for a particle that was at the origin at time t = 0, at some later time t r t. (0) Now, dimensional analysis tells us that the proportionality constant between r and t has dimensions of length squared divided by time. Essentially, the length in the diffusion constant is the size of the hops we considered in the previous section, and the time is the time a single hop takes. The proportionality constant defines the diffusion constant D. In fact it is 6 times D, so we have r = 6Dt r = (6Dt) /. () The factor of 6 appears because the diffusion constant is defined so that the flux of molecules per unit area j = D c(r), where c(r) is concentration of molecules per unit volume at position r. This equation is called Fick s Law. It leads to the diffusion equation you saw in level two c t = D c x. () Note that D appears both when we consider a microscopic phenomenon, and a macroscopic phenomenon. The microscopic phenomenon is the motion of a single molecule that is diffusing. The macroscopic phenomenon is the flux of particles per second through some large area, which is due to the motion of many many molecules not one. The same constant, D, controls both. This means that gases of molecules that diffuse rapidly also have large fluxes j, when there is a gradient of the concentration c(r)... Calculation of an estimate for D In this section we will estimate the diffusion constant D in gases like the Earth s atmosphere as it is this that allows us to work out how far molecules can diffuse in a given time, which molecules diffuse rapidly and which diffuse slowly etc. 5

6 λ σ v volume = πσ λ Figure 4: Schematic diagram of a gas atom (left-hand side black circle) travelling a distance λ in a straight line at a speed u, before colliding with another gas atom (right-hand side black circle). After the collision direction of travel of the atom is very different from that before. It then moves off along this new direction, shown by the dotted arrow and circle. The diffusion constant D has dimensions of length squared over time. The length is the distance of a single hop and the time is the time taken during a single hop. The length of a hop being the distance over which a molecule moves in a straight line, which is the average distance between collisions. Collisions tend to randomise the direction of travel of molecules so a diffusing molecule undergoes a random walk in which the hops or steps are the distances between collisions. If the length of a typical hop is λ and the time per hop is τ, then we have D λ τ. (3) For example, if a particle is making hops of length nm and it makes 0 0 hops per second then the diffusion constant D (0 9 ) /0 0 = 0 8 m s. Note that this only gives an approximation to D. Calculating diffusion constants exactly can be done but is complicated. During this course we only consider simple approximate ways to calculate diffusion constants. So, to estimate the diffusion constant we need to estimate: ) the length of the hops, λ, and ) the time per hop, τ. For molecules in a gas (like the Earth s atmosphere) this is easy. A hop is the distance a molecule travels in a straight line, before colliding with another molecule and so heading off in another direction. The atoms behave just like billiard balls or ball bearings colliding with each other. Imagine a box full of tiny ball bearings (the atoms) each travelling at the speed of sound. The atoms are a few tenths of a nm across and the typical separation between the atoms is about ten times that, about nm. Atoms at room temperature are moving with a typical speed, u, of around the speed of sound, a few hundred m s. Each atom will collide with any other atom it gets within a few tenths of a nm of. We call this distance σ, it is the size of the atom, i.e., of the cloud of electrons that surrounds the nucleus. Thus, as it moves each atoms sweeps out a cylindrical volume of radius σ. See Fig. 4. If the typical distance between collisions with another atom is λ then it sweeps out a volume of πσ λ before it collides with another atom. Once we have λ we can find the time for a single hop, as it is just the hop length over the speed, τ = λ/u. Here u is a typical speed, which is the root mean square of the velocity u = ( v ) /. Now, by definition πσ λ is the volume an atom sweeps out before it encounters and collides with one other atom. But we know the average volume that contains one atom, it is just V/N, where V is the total volume and N is the total number of molecules. Thus we have that σ λ V/N or λ V σ N (4) 6

7 where we dropped the factor of π as we are only calculating a rough order of magnitude estimate. The V/N factor can be understood if we note that if we divided the box of volume V into N little cubes each of volume V/N then on average each cube would contain atom. These cubes have sides of length (V/N) /3. Then if we consider two adjacent cubes each with atom these two adjacent atoms will be of order (V/N) /3 apart. Thus the typical separation between atoms s = (V/N) /3. The hop length λ can then be written as λ s 3 /σ. Having determined λ we need the time per hop, which is the time the atom takes to move a distance λ. We call this time τ and it is given by τ = λ/u, for u the typical velocity of the atom. Now, from the equipartition theorem in the previous section u ( v ) / = ( kt m ) /. (5) Now that we know u we know τ = λ/u and we have everything. Thus the diffusion constant is approximately D λ /τ = λ /(λ/u) = λu s3 u σ. (6) In the Earth s atmosphere there are approximately N/V = 0 6 molecules per unit volume. Thus the typical separation s = (V/N) /3 = nm. The molecules are roughly a fifth this size, σ = 0.4 nm. Also, for O at room temperature v = (KT/m) / = (4 0 /5 0 6 ) / 300 m s. Thus the diffusion constant of the molecules in the atmosphere D ( 0 9 ) ( ) 0 5 m s. (7) This estimate is close to the measured value for the diffusion constant for O in air at 0 C, which is 0 5 m s. So, in s an oxygen molecule diffuses a distance of approximately (6D ) / (6 0 5 ) / 0 m= cm. A molecule diffuses rapidly over distances as short as a cm. However, as the distance increases only as the square root of time, the time required to move larger distances becomes quite long. For example, it takes 0, 000 s 3 hours to diffuse m, and a day to diffuse 0 m. In practice, movement over distances of a metre or more occurs not by diffusion but by convection and other mechanisms in which the air flows. 7

8 5 4 u(x) / kt 3 / / (< x > ) = ( kt / φ ) x / nm Figure 5: Schematic of a one-dimensional quadratic potential well, u = (/)φx, for a force constant φ = 0.kT/nm. The approximate size of the root-mean square displacement from the minimum, ( x ) /, is shown. Fluctuations and Response There is a general relationship between how much a variable fluctuates due to thermal energy, and how much the mean value of a variable changes when an external field is applied. In its most general form, the theorem that relates the two won Lars Onsager the Nobel prize in chemistry in 968. We will consider how it works for a simple example. This is a (classical) particle moving along the x axis in a harmonic potential. So, the energy of the particle as a function of x, u(x), is given by u(x) = φx (8) Here the spring constant is φ (we can t call it k as we are using k for Boltzmann s constant). With this energy the Boltzmann weight for position x of the particle is exp[ φx /(kt)]. The probability of the particle being at position x, p(x), is proportional to this weight. So if we introduce the proportionality constant c, we have p(x) = c exp[ u(x)/kt] = c exp[ φx /(kt)] (9) As the problem is symmetric around the origin the particle is as likely to be left of the origin as right of the origin and the mean value of the position of the particle x = 0. This is of course useless as a measure of how far thermal motion of the particle can take the particle away from the origin. So we need the root of the mean of the square of x, ( x ) /. This can be estimated as we know that the particle will have about kt thermal energy at a temperature T. This is sufficient to get it a distance x from the origin given by φx = kt (30) If we approximate ( x ) / by the value of x in this equation we have φ x kt x kt φ ( x ) / ( kt φ ) / (3) where we dropped the factor / after the first equation as the value we get is just an estimate. So the typical distance from the origin varies with temperature as T /, and with the size of the force constant φ, as /φ /. 8

9 So, now we have characterised the size of the fluctuations of the position x of the particle, they are of order (kt/φ) /. We will now look at how the mean value, x, varies if we apply an external force. With an external force of strength f the energy as a function of position is now u(x) = φx fx (3) The contribution to the energy is fx because a force pushing the particle to the right (f > 0) lowers the energy to the right, at positive values of x. Thus the probability of the particle being at position x is now proportional to the new Boltzmann weight The mean value of x is then p(x) = c exp[ u(x)/kt] = c exp [ φx /(kt) + fx/kt ] (33) x = xp(x)dx = c xexp [ φx /(kt) + fx/kt ] (34) For non-zero f this will be non-zero. Now, to see the relationship between the response of the mean position to an external force, and the fluctuations we take the derivative of x with respect to f x f x f = ( c f = c kt = kt x xexp [ φx /(kt) + fx/kt ] ) dx x exp [ φx /(kt) + fx/kt ] dx = kt = c ( x ) xexp [ φx /(kt) + fx/kt ] dx kt x p(x)dx In words: the rate of change of the mean position, x, equals /kt times the square of the size of fluctuations in position, x. The larger are the fluctuations in x, the larger is the change in the mean of x when a force is applied. Another way of putting the same thing is to say that the size of the fluctuations, x, equals kt times the rate of change of the mean when an external force is applied. As we have estimated the size of the fluctuations we also know the rate of change of x with f (35) x f = kt x kt kt φ = φ (36) Finally, here we have only proved that there is a relationship between the fluctuations in the position of a particle, and the rate of change of the mean position of the particle with the strength of an external force. However, the relationship is in fact general. Consider any variable, call it a, that has a mean of zero without an external force, but fluctuates. Then if an external force f a is applied, such that it makes a contribution to the energy of f a a, then we have a = f a kt a (37) 9