the Osmotic Tlceory of Solutions. 613 



equation (27) gives the downward force acting on the solvent 

 molecules per unit volume of solution, in other words the 

 driving force. The driving force F 2 on the solute molecules 

 is equal and opposite to F 1 as already pointed out. 



When we require the driving forces in a uniformly rotating 

 mass of solution, we have only to replace dh by dr, and g by 

 dC/dr, w r here r represents radial distances from the axis of 

 rotation, and C the potential of the centrifugal force. To 

 pass to equilibrum conditions, we have to put F : = = F 2 ; 

 and (27) then becomes identical with (15) above, and (16) 

 and (17) follow as before. 



Thus all these equations can be obtained without taking 

 vapour-pressures into consideration and without assuming it 

 possible for the pure liquid solvent and solute to exist at such 

 a pressure as to be in osmotic equilibrium with the solution, 

 or even to be capable of separate existence at all. 



Note on the Application of the Preceding General Discussion 

 to a Diffusing Column of Solution*. 



As before, consider a volume element dr at any level in 

 the column. The solvent molecules in this element collec- 

 tively experience a force F^Zt, and the solute molecules an 

 equal and opposite force F 2 dr ; these forces tend to produce 

 opposite displacements of the solvent and solute relatively to 

 the mass centre of the volume dr. without causing any bodily 

 acceleration of the solution contained in dr. Draw through 

 the mass centre of the solution in dr an element d& of hori- 

 zontal surface ; let this element of surface he moving with the 

 mass centre in question, and let the rates of displacement of 

 solute and solvent be measured with respect to it. These 

 rates we shall denote by rn^ and m 2 , and define as the 

 algebraic total of grams ot solvent (or solute) passing down- 

 wards through d$ per unit of area per unit of time. 



Let Vj be the mean downward velocity, relative to <ZS, of 

 the solvent molecules in dr, and V 2 the corresponding velocity 

 (downward) of the solute molecules, then, p being the density 

 of the solution, Y^pc 1 = m l and Y. 2 pc 2 =m 2 . 



But V^pt'idr, \ 2 pc 2 dr are the respective momenta of the 

 solvent and solute, both reckoned relatively to the mass 

 •centre of the element ; their sum is therefore zero, and hence 



m-L + mci = 0. 

 A fuller discussion is reserved for a subsequent communication. 



