Page:EB1911 - Volume 08.djvu/274

account the following statement is correct.) In a state of equilibrium, the density of each fluid at any point thus depends only on the partial pressure of that fluid alone, and is the same as if the other fluids were absent. It does not depend on the partial pressures of the other fluids. If this were not the case, the resistance to diffusion would be analogous to friction, and would contain terms which were independent of the relative velocity u₂ − u₁. (2) For slow motions the resistance to diffusion is (approximately at any rate) proportional to the relative velocity. (3) The coefficient of resistance C is not necessarily always constant; it may, for example, and, in general, does, depend on the temperature.

If we form the equations of hydrodynamics for the different fluids occurring in any mixture, taking account of diffusion, but neglecting viscosity, and using suffixes 1, 2 to denote the separate fluids, these assume the form given by James Clerk Maxwell (“Diffusion,” in Ency. Brit., 9th ed.):—

$\textstyle \rho _{1}{\frac {\mathrm {D} u_{1}}{\mathrm {D} t}}+{\frac {dp_{1}}{dx}}-\mathrm {X} _{1}\rho _{1}+\mathrm {C} _{12}\rho _{1}\rho _{2}(u_{1}-u_{2})+\mathrm {\And \!\!c.} =0,$

where

$\textstyle {\frac {\mathrm {D} u_{1}}{\mathrm {D} t}}={\frac {du_{1}}{dt}}+u_{1}{\frac {du_{1}}{dx}}+v_{1}{\frac {du_{1}}{dy}}+w_{1}{\frac {du_{1}}{dz}},$

and these equations imply that when diffusion and other motions cease, the fluids satisfy the separate conditions of equilibrium dp₁/dx − X₁ρ₁＝0. The assumption made in the following account is that terms such as Du₁/Dt may be neglected in the cases considered.

A further property based on experience is that the motions set up in a mixture by diffusion are very slow compared with those set up by mechanical actions, such as differences of pressure. Thus, if two gases at equal temperature and pressure be allowed to mix by diffusion, the heavier gas being below the lighter, the process will take a long time; on the other hand, if two gases, or parts of the same gas, at different pressures be connected, equalization of pressure will take place almost immediately. It follows from this property that the forces required to overcome the “inertia” of the fluids in the motions due to diffusion are quite imperceptible. At any stage of the process, therefore, any one of the diffusing fluids may be regarded as in equilibrium under the action of its own partial pressure, the external forces to which it is subjected and the resistance to diffusion of the other fluids.

5. Slow Diffusion of two Gases. Relation between the Coefficients of Resistance and of Diffusion.—We now suppose the diffusing substances to be two gases which obey Boyle’s law, and that diffusion takes place in a closed cylinder or tube of unit sectional area at constant temperature, the surfaces of equal density being perpendicular to the axis of the cylinder, so that the direction of diffusion is along the length of the cylinder, and we suppose no external forces, such as gravity, to act on the system.

The densities of the gases are denoted by ρ₁, ρ₂, their velocities of diffusion by u₁, u₂, and if their partial pressures are p₁, p₂, we have by Boyle’s law p₁＝k₁ρ₁, p₂＝k₂ρ₂, where k₁,k₂ are constants for the two gases, the temperature being constant. The axis of the cylinder is taken as the axis of x.

From the considerations of the preceding section, the effects of inertia of the diffusing gases may be neglected, and at any instant of the process either of the gases is to be treated as kept in equilibrium by its partial pressure and the resistance to diffusion produced by the other gas. Calling this resistance per unit volume R, and putting R＝Cρ₁ρ₂(u₁ − u₂), where C is the coefficient of resistance, the equations of equilibrium give

$\textstyle {\frac {dp_{1}}{dx}}+\mathrm {C} \rho _{1}\rho _{2}(u_{1}-u_{2})=0,$ and $\textstyle {\frac {dp_{2}}{dx}}+\mathrm {C} \rho _{1}\rho _{2}(u_{2}-u_{1})=0$

(1).

These involve

$\textstyle {\frac {dp_{1}}{dx}}+{\frac {dp_{2}}{dx}}=0$ or  p1 + p2＝P

(2).

where P is the total pressure of the mixture, and is everywhere constant, consistently with the conditions of mechanical equilibrium.

Now dp₁/dx is the pressure-gradient of the first gas, and is, by Boyle’s law, equal to k₁ times the corresponding density-gradient. Again ρ₁u₁ is the mass of gas flowing across any section per unit time, and k₁ρ₁u₁ or p₁u₁ can be regarded as representing the flux of partial pressure produced by the motion of the gas. Since the total pressure is everywhere constant, and the ends of the cylinder are supposed fixed, the fluxes of partial pressure due to the two gases are equal and opposite, so that

p1u1 + p2u2＝0 or k1ρ1u1 + k2ρ2u2＝ 0

(3).

From (2) (3) we find by elementary algebra

u₁/p₂＝−u₂/p₁＝(u₁ − u₂)/(p₁ + p₂)＝(u₁ − u₂)/P,

and therefore

p₂u₁＝−p₂u₂＝p₁p₂(u₁ − u₂)/P＝k₁k₂ρ₁ρ₂(u₁ − u₂)/P

Hence equations (1) (2) gives

$\textstyle {\frac {dp_{1}}{dx}}+{\frac {\mathrm {CP} }{k_{1}k_{2}}}(p_{1}u_{1})=0,$ and $\textstyle {\frac {dp_{2}}{dx}}+{\frac {\mathrm {CP} }{k_{1}k_{2}}}(p_{2}u_{2})=0;$

whence also substituting p₁＝k₁ρ₁, p₂＝k₂ρ₂, and by transposing

$\textstyle \rho _{1}u_{1}=-{\frac {k_{1}k_{2}}{\mathrm {CP} }}{\frac {d\rho _{1}}{dx}},$ and $\textstyle \rho _{2}u_{2}=-{\frac {k_{1}k_{2}}{\mathrm {CP} }}{\frac {d\rho _{2}}{dx}}.$

We may now define the “coefficient of diffusion” of either gas as the ratio of the rate of flow of that gas to its density-gradient. With this definition, the coefficients of diffusion of both the gases in a mixture are equal, each being equal to k₁k₂/CP. The ratios of the fluxes of partial pressure to the corresponding pressure-gradients are also equal to the same coefficient. Calling this coefficient K, we also observe that the equations of continuity for the two gases are

$\textstyle {\frac {d\rho _{1}}{dt}}+{\frac {d(\rho _{1}u_{1})}{dx}}=0,$ and $\textstyle {\frac {d\rho _{2}}{dt}}+{\frac {d(\rho _{2}u_{2})}{dx}}=0,$

leading to the equations of diffusion

$\textstyle {\frac {d\rho _{1}}{dt}}={\frac {d}{dx}}\left(\mathrm {K} {\frac {d\rho _{1}}{dx}}\right),$ and $\textstyle {\frac {d\rho _{2}}{dt}}={\frac {d}{dx}}\left(\mathrm {K} {\frac {d\rho _{2}}{dx}}\right),$

exactly as in the case of diffusion through a solid.

If we attempt to treat diffusion in liquids by a similar method, it is, in the first place, necessary to define the “partial pressure” of the components occurring in a liquid mixture. This leads to the conception of “osmotic pressure,” which is dealt with in the article Solution. For dilute solutions at constant temperature, the assumption that the osmotic pressure is proportional to the density, leads to results agreeing fairly closely with experience, and this fact may be represented by the statement that a substance occurring in a dilute solution behaves like a perfect gas.

6. Relation of the Coefficient of Diffusion to the Units of Length and Time.—We may write the equation defining K in the form

$\textstyle -\mathrm {K} \times {\frac {\mathrm {I} }{\rho }}{\frac {d\rho }{dx}}.$

Here −dρ/ρdx represents the “percentage rate” at which the density decreases with the distance x; and we thus see that the coefficient of diffusion represents the ratio of the velocity of flow to the percentage rate at which the density decreases with the distance measured in the direction of flow. This percentage rate being of the nature of a number divided by a length, and the velocity being of the nature of a length divided by a time, we may state that K is of two dimensions in length and −1 in time, i.e. dimensions L²/T.

Example 1. Taking K＝0·1423 for carbon dioxide and air (at temperature 0° C. and pressure 76 cm. of mercury) referred to a centimetre and a second as units, we may interpret the result as follows:—Supposing in a mixture of carbon dioxide and air, the density of the carbon dioxide decreases by, say, 1, 2 or 3% of itself in a distance of 1 cm., then the corresponding velocities of the diffusing carbon dioxide will be respectively 0·01, 0·02 and 0·03 times 0·1423, that is, 0·001423, 0·002846 and 0·004269 cm. per second in the three cases.

Example 2. If we wished to take a foot and a second as our units, we should have to divide the value of the coefficient of diffusion in Example 1 by the square of the number of centimetres in 1 ft., that is, roughly speaking, by 900, giving the new value of K＝0·00016 roughly.

7. Numerical Values of the Coefficient of Diffusion.—The table on p. 258 gives the values of the coefficient of diffusion of several of the principal pairs of gases at a pressure of 76 cm. of mercury, and also of a number of other substances. In the gases the centimetre and second are taken as fundamental units, in other cases the centimetre and day.

8. Irreversible Changes accompanying Diffusion.—The diffusion of two gases at constant pressure and temperature is a good example of an “irreversible process.” The gases always tend to mix, never to separate. In order to separate the gases a change must be effected in the external conditions to which the mixture is subjected, either by liquefying one of the gases, or by separating them by diffusion through a membrane, or by bringing other outside influences to bear on them. In the case of liquids, electrolysis affords a means of separating the constituents of a mixture. Every such method involves some change taking place outside the mixture, and this change may be regarded as a “compensating