The Elementary Theory of the Brownian Motion

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Review by Bryan Hassell: AP 255 Fall 11

From: The Elementary Theory of the Brownian Motion by A. Einstein, Zeit. fur Elektrochemie , 14, 1908, pp. 235-239

Keywords: brownian motion, osmotic pressure, diffusion and diffusion coefficient

(*) First he notes that "we mean by Brownian motion that irregular movement which small particles of microscopic size carry out when suspended in liquid"

Introduction

The train of thought is as follows: First investigate how the process of diffusion in an undissociated dilute solution depends on the distribution of osmotic pressure in the solution and on the mobility of the dissolved substance in the solvent. This will yield an equation for the diffusion coefficient in the case when a molecule of the dissolved substance is great compared with a molecule of the solvent and will only depend on viscosity of the solvent and diameter of solute molecules. Next, relate the diffusion process to that of irregular motions of the solute molecules and look for how we can find the average magnitude of these motions from the diffusion coefficient or viscosity of the solvent and diameter of solute molecules.

<math>\S </math> 1. Diffusion and Osmotic Pressure

Figure 1
Figure 2

Say the vessel in figure 1 (Z) is filled with a dilute solution and the interior is divided into two sections A and B by a movable piston, k, which forms a semi-permeable partition. If the concentration in A is greater than B, an external force in the left direction must be applied to the piston in order to keep equilibrium. This force will be equal to the difference of the two osmotic pressures. If the force cannot act on the piston, it will move to the right until the concentrations in A and B are equal. So from this it is seen that it is the osmotic pressure which brings upon the equalization of concentrations in diffusion. *Nernst made this the foundation of his investigations into the connection between ionic mobility, diffusion coefficient and EMF in concentration cells. Looking at figure 2 in which diffusion is taking place within cylinder Z of unit area cross section, the osmotic pressure force p acts on the surface E from left to right and the force p' acts on the surface E' from right to left. The resultant force is thus p-p'. If x is the distance from the left to the surface E and x+dx is the distance to E', then dx is the volume of liquid we are analyzing. The osmotic pressure which acts on the dissolved substance in the unit volume is then

<math>K={p-p' \over dx} = -{p'-p\over dx} = -{dp\over dx}</math>

and substituting <math>p=RT\nu</math> from the ideal gas law we get

<math>K=-RT{d\nu \over dx}.</math>

To investigate the motions due to diffusion, to which the forces give rise, we need to know how great a resistance the solvent offers to a movement of the dissolved substance. If a force k acts on a molecule, it will impart a proportional velocity v of the form

<math>v={k\over \mathfrak{R}}</math>

where <math>\mathfrak{R}</math> is a constant and is the frictional resistance of the molecule. Assuming a spherical particle, from hydrodynamics, for a sphere moving in a liquid the resistance is <math>\mathfrak{R}=6\pi\eta\rho</math> where <math>\eta</math> is the viscosity and <math>\rho</math> the radius of the sphere. It can also be assumed that the molecules of the solute are large compared with that of the solvent so this resistance may be applied to single solute molecules, therefore we can estimate the mass of solute diffusing across a cross section of the cylinder per unit time. So there will be <math>\nu N</math> actual molecules in the volume and the force <math>k</math> is distributed over all of these so we can see that the velocity which the force is able to impart to the <math>\nu N</math> molecules we get

<math>v={1\over \nu N}\cdot {k\over \mathfrak{R}}</math>

If we say that this force k is equal to the osmotic force K then we get the following equation

<math>v\nu=-{RT\over N}\cdot {1\over \mathfrak{R}}\cdot {d\nu\over dx}</math>

The left hand side represents the mass of the dissolved substance which is carried per second by diffusion though the unit area. You can see then that the pre factor to <math>d\nu/dx</math> on the right hand side is the diffusion coefficient or that

<math>D={RT\over N}\cdot {1\over \mathfrak{R}}</math> and if assuming again spherical molecules, larger than that of the solute <math>D={RT\over N} {1\over 6\pi\eta\rho}</math>

So diffusion may be seen to depend on constants characteristic of the substance and the viscosity of the solvent and radius of the molecule.

<math>\S </math> 2. Diffusion and Irregular Motion of the Molecules

Figure 3

We can look at diffusion from another point of view from the molecular theory of heat. Again looking at the diffusion process in one direction, we imagine that we know the coordinates of all solute molecules at time <math>t</math> and at time <math>t+\tau</math> where <math>\tau</math> is an interval so short that the concentration varies slightly over this time. During <math>\tau</math>, through thermal motion, the first solute molecule moves <math>\Delta_1</math>, the second <math>\Delta_2</math>, etc where these displacements are both positive and negative. They'll also be of different magnitude but assuming a dilute solution, the magnitude is controlled by the surrounding solvent not the solute molecules so in portions of the solution of different concentrations these displacements will be on average of equal magnitude. To investigate the mass of the substance which diffuses in time <math>\tau</math> through the unit area of cross section we first eliminate the individual displacements <math>\Delta_1</math>,<math>\Delta_2</math> etc with just <math>\Delta</math>, the mean value. Looking at figure 3, we can see that during time <math>\tau</math>, only the molecules which are less than a distance <math>\Delta</math> from E will be able to cross plane E. Half the molecules between plane <math>Q_1</math> and E experience <math>+\Delta</math> or in gram-molecules, <math> {1 \over 2} v_1\Delta</math>. This is the mass of solute passing from left to right and so similarly that which is passing from right to left in time <math>\tau</math> is <math> {1 \over 2} v_2\Delta</math>, <math> v_1 </math> and <math> v_2</math> being the concentration in the middle layers <math>M_1</math> and <math>M_2</math> respectively. Then the quantity which diffuses across will be the difference between the two, <math> {1 \over 2} \Delta(v_1-v_2)</math>. If x is the distance from the left of the cylinder than in differential form,

<math> {v_2-v_1\over \Delta} = {dv\over dx}</math>

or the quantity which diffuses across E in time <math>\tau</math> is

<math> -{1\over 2}\Delta^2{dv\over dx}</math>

So in a unit of time, that which diffuses across E is

<math> -{1\over 2}{\Delta^2\over \tau}{dv\over dx}</math> or another value of diffusion coefficient <math> D = {1\over 2}{\Delta^2\over \tau}</math>. Here we see that the value <math>\Delta</math> is the length of the path of a solute molecule on average during time <math>\tau</math> or that <math> \Delta = \sqrt{2D\tau}</math>

<math>\S </math> 3. Movement of the Single Molecules: Brownian Motion

Equating the values for D form the previous sections we can solve for <math>\Delta</math> as,

<math>\Delta = \sqrt{{2RT\over N\mathfrak{R}}}\sqrt{\tau}</math>

So on average the path of a molecule is not proportional to the time but to the square root of time. If the solute molecule is large compared to that of the solvent and is spherical we can use the drag force used before to obtain

<math>\Delta = \sqrt{{RT\over N}{1\over 3\pi\eta\rho}}\cdot\sqrt{\tau}</math>

noting that we can calculate the mean displacement from temperature, viscosity and radius of the molecule. He then goes on to use this for some useful calculations: i.e. for an undissociated dissolved substance whose diffusion coefficient is known you may calculate <math>\Delta</math> for example sugar at room temperature if <math>\tau</math> =1 then <math>\Delta=27.6\mu m</math>. From this you can deduce from the number N and the molecular volume of solid sugar that the diameter of a molecule of sugar is about 1000 times smaller than the diameter of the particle he initially investigates (not shown here, where <math>\Delta=0.8\mu m</math>).