On the Motion of Particles Suspended in Liquids at Rest, as Required by the Molecular-Kinetic Theory of Heat

§ 3. Theory of the diffusion of small suspended spheres.Show the German source

Let suspended particles be distributed irregularly in a liquid.Show the German source We shall investigate their state of dynamic equilibrium under the assumption that on the individual particles there acts a force $K$ that depends on position but not on time.Show the German source For the sake of simplicity let it be assumed that the force everywhere has the direction of the $X$-axis.Show the German source

Let $\nu$ be the number of suspended particles per unit volume; then in the case of thermodynamic equilibrium $\nu$ is a function of $x$ such that, for an arbitrary virtual displacement $\delta x$ of the suspended substance, the variation of the free energy vanishes.Show the German source We therefore have:Show the German source $$\delta F = \delta E - T \delta S = 0.$$$F$Free energy$E$System energy$T$Absolute temperature$S$EntropyExplain this equationShow the German source Let it be assumed that the liquid has cross-section 1 perpendicular to the $X$-axis and is bounded by the planes $x = 0$ and $x = l$.Show the German source We then have:Show the German source $$\delta E = -\int_0^l K \nu \delta x d x$$$E$System energy$l$Column length$K$External force per particle$\nu$Number densityExplain this equationShow the German source andShow the German source $$\delta S = \int_0^l R \frac{\nu}{N} \frac{\partial \delta x}{\partial x} d x = -\frac{R}{N} \int_0^l \frac{\partial \nu}{\partial x} \delta x d x.$$$S$Entropy$l$Column length$R$Gas constant$\nu$Number density$N$Avogadro's numberExplain this equationShow the German source The equilibrium condition sought is therefore:Show the German source $$-K \nu + \frac{R T}{N} \frac{\partial \nu}{\partial x} = 0$$$K$External force per particle$\nu$Number density$R$Gas constant$T$Absolute temperature$N$Avogadro's numberExplain this equationShow the German source orShow the German source $$K \nu - \frac{\partial p}{\partial x} = 0.$$$K$External force per particle$\nu$Number density$p$Osmotic pressureExplain this equationShow the German source The last equation states that the force $K$ is held in equilibrium by forces of osmotic pressure.Show the German source

We use equation (1) to determine the diffusion coefficient of the suspended substance.Show the German source We can regard the state of dynamic equilibrium just considered as the superposition of two processes running in opposite directions, namely
1. a motion of the suspended substance under the action of the force $K$ acting on each individual suspended particle,
2. a process of diffusion, which is to be regarded as a consequence of the irregular motions of the particles due to the molecular motion of heat.Show the German source

If the suspended particles are spherical (sphere radius $P$) and the liquid has the coefficient of friction $k$, the force $K$ imparts to the individual particle the velocity^[1)]Show the German source $$\frac{K}{6 \pi k P},$$$K$External force per particle$k$Dynamic viscosity$P$Particle radiusExplain this equationShow the German source and through the unit of cross-section there pass per unit timeShow the German source $$\frac{\nu K}{6 \pi k P}$$$\nu$Number density$K$External force per particle$k$Dynamic viscosity$P$Particle radiusExplain this equationShow the German source particles.Show the German source

If, further, $D$ denotes the diffusion coefficient of the suspended substance and $\mu$ the mass of a particle, then as a result of diffusion there pass through the unit of cross-section per unit timeShow the German source $$-D \frac{\partial (\mu \nu)}{\partial x} \quad \text{Gramm}$$$D$Diffusion coefficient$\mu$Particle mass$\nu$Number densityExplain this equationShow the German source orShow the German source $$-D \frac{\partial \nu}{\partial x}$$$D$Diffusion coefficient$\nu$Number densityExplain this equationShow the German source particles.Show the German source Since dynamic equilibrium is to prevail, we must have:Show the German source $$\frac{\nu K}{6 \pi k P} - D \frac{\partial \nu}{\partial x} = 0.$$$\nu$Number density$K$External force per particle$k$Dynamic viscosity$P$Particle radius$D$Diffusion coefficientExplain this equationShow the German source

From the two conditions (1) and (2) found for dynamic equilibrium, the diffusion coefficient can be calculated.Show the German source One obtains:Show the German source $$D = \frac{R T}{N} \frac{1}{6 \pi k P}.$$$D$Diffusion coefficient$R$Gas constant$T$Absolute temperature$N$Avogadro's number$k$Dynamic viscosity$P$Particle radiusExplain this equationShow the German source The diffusion coefficient of the suspended substance therefore depends, apart from universal constants and the absolute temperature, only on the coefficient of friction of the liquid and on the size of the suspended particles.Show the German source

Bern, May 1905.Show the German source

(Received 11 May 1905.)Show the German source