The slow diffusion of a dissolved substance represents a difference in degree, not in kind, between gases and dissolved substances. Even in gases, we have such frequent collisions that the mean free path of an oxygen molecule at 0° and atmospheric pressure is only 0.00001 cm., whereas the velocity, the total path covered in one second, is 42,500 cm.
Exp. If a bulb containing a few drops of bromine is broken at the bottom of a tall cylinder, the bromine vapor is seen to diffuse rather slowly into the upper part of the cylinder, the bromine molecules, in their passage upward, rebounding from the air molecules, with which the cylinder is filled. If a second cylinder is first evacuated, and the bromine bulb is broken in vacuo, the vapor is seen to fill the cylinder instantly, the high velocity of the bromine molecules being thus revealed.
But a third question, of fundamental importance in the comparison of the condition of a substance existing as a gas and its condition in a solution of the same concentration and temperature, results from a consideration of the frequency of the impacts of the solute molecules against the solvent, growing out of the reduction of the lengths of the mean free paths of the solute molecules.[41] In order to be able to take this fact properly into account, it will be necessary to consider somewhat more precisely the manner in which, according to the kinetic theory, gas pressure is produced.
We may consider that we have in a cube of unit volume (1 c.c.) n molecules of a gas, each of mass m and average velocity u cm. per second. We may assume that one-third of the total number [p030] of molecules moves in each of the three dimensional directions.[42] A single molecule of mass m, striking the surface with a velocity u and rebounding with the same velocity in the opposite direction, will exert on the surface a force of 2 m u units. But, with a velocity of u cm. per second, it will reach the opposite wall and return to the surface we are considering, u / 2 times in one second. A single molecule will consequently exert a force 2 m u × u / 2 or m u2 on the surface, and the n / 3 molecules moving in the same direction will exert a force n / 3 × m u2 on the unit surface. This represents, therefore, the pressure of such a gas, as calculated on the basis of the assumptions of the kinetic theory. Now, when a gas is so strongly compressed, that the bulk of the molecules is not negligible in comparison with the total volume of the gas, the number of impacts on unit surface in unit time becomes sensibly greater than n / 3 × u / 2, since the distance to be covered between successive blows on the surface will be sensibly less than 2 cm., in a cube of unit volume. If we imagine, for the sake of a rough illustration, that one-third of the molecules in 1 c.c. are united into one spherical mass (indicated by A in Fig. 7), moving upwards and downwards, it is obvious that the distance covered between two successive blows on a surface is not 2 cm., but that distance diminished by twice the diameter of the sphere. For strongly compressed gases, the total number of impacts on unit surface is therefore sensibly greater than n / 3 × u / 2, and the pressure is proportionately greater. According to van der Waal's correction for this effect, P = n / 3 × m u2 / (1 − b), where b represents the volume actually occupied by the molecules in 1 c.c. of the gas.[43]
Fig. 7.
Now, for solute molecules, the "free space" of movement, as we may call it, is, similarly, very considerably reduced by the presence [p031] of the solvent, and the reduction of this free space, as Nernst has shown, will have the same effect on the pressure produced against unit surface of the solvent by the bombardment of the solvent by the solute, as the reduction of the free space has on the gas pressure when a gas is strongly compressed. The resulting pressure on unit surface of the solution must thus be increased, from the pressure Pgas, which would be exerted by the solute against the walls of a vessel, if it were present as a gas of the same concentration, at the same temperature, to Pgas / (1 − v), where v represents the real volume occupied by the solvent and (1 − v) the free space for the solute molecules in unit volume of solution.[44] If osmotic pressure is the result of such a bombardment of the solvent by the molecules of the solute, one might, therefore, expect to find the osmotic pressure very much greater than the gas pressure of the same substance in the same volume at the same temperature. However, in all the experimental determinations (by means of semipermeable membrane, vapor pressure, boiling-point and freezing-point measurements) of the osmotic pressure as defined on p. [10], this corrective factor cancels out again.[45] According to the kinetic theory, the osmotic pressure of a substance in dilute solution should, consequently, be found by experiment to be equal to the gas pressure which a gas, of the same molecular concentration, would exert at the same temperature.[46]
We find thus that the significant coincidence between the osmotic pressure of a substance in dilute solution, as defined and measured according to van 't Hoff, and the gas pressure which the substance would exert, if it were present as a gas in the same volume and at the same temperature, is in agreement with the fundamental assumptions of the kinetic theory. This theory, consequently, gives us an adequate theoretical explanation of [p032] osmotic pressure, as it does of gas pressure. As van 't Hoff says,[47] "if the osmotic pressure follows Gay-Lussac's law and is proportional to the absolute temperature, then, like gas pressure, it will become zero at 0° absolute temperature and will vanish when molecular movements come to rest. It is therefore natural to look for the cause of osmotic pressure in kinetic phenomena and not in attractions."[48]
Chapter III Footnotes
[29] L'Hermite, Compt. rend., 39, 1177 (1854); van 't Hoff, Lectures on Physical Chemistry, Part II, p. [37].