| dρ1 | + | d(ρ1u1) | = 0, and | dρ2 | + | d(ρ2u2) | = 0, |
| dt | dx | dt | dx |
leading to the equations of diffusion
| dρ1 | = | d | (K | dρ1 | ), and | dρ2 | = | d | (K | dρ2 | ), |
| dt | dx | dx | dt | dx | dx |
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
| −K × | I | dρ | . |
| ρ | 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.