598 Mr. L. Vegard : Contributions to 



We have still to find the work done by the force during 

 the variation. On the average the masses have been moved 

 a space dx s in the direction of the force, and accordingly 



ha = K (M^Wi + M 2 e»o) dx ; 



or by the aid of equation (86) : 



§a=K(M 1 -^M 2 )en 1 ^. "... (106) 



If the values found for 8co and 8a (equations (10a) and 

 (106)) are introduced into equation (3), we get as the con- 

 dition for equilibrium : 



/Mi B/ifa P, T) _ M 2 a f 2 (c, p, T) \5c =K /Mi _ M 2 \ 



\ Vi O'c r. 2 "dc )~dx \Vj v., J' 



If we now eliminate ^~ by the aid of equation (5) and 

 write — in view of equation (2) — n instead of x 3 



MteT)(M 1+ lMA|< :=K /M l _M ? Y 



OC \ t\ c v 2 J on \ t\ r 2 / 



From this equation it appears that when the solution is 

 in equilibrium, the concentration gradient is at every point 

 proportional to the force. 



In equation (11a) we will try to introduce quantities that 

 are more spontaneously accessible, and begin by finding 



other values for the ratios — - and — -. These quantities 



v l v 2 



mean the molecular masses divided by the molecular volumes. 

 To make clearer the meaning of these quantities, we may 

 suppose that we have a very big quantity of the solution at 

 the concentration c. If to the fluid is added a mass M t of 

 the first or M 2 of the second component, the volume will be 

 increased by the quantities i\ and r 2 respectively. 



Instead of an infinite, we consider a finite quantity and 

 add infinitely small masses "M. x dni and M. 2 dn 2 to the solution. 

 By doing that the volumes will be increased by di\ and dv 2 ; 

 and then 



M 1 M^w, , M, IsUln, 

 — = 7 ~ and - = — t — -. 

 Vi dv } v 2 dv 2 



If the whole volume of the solution is V, 



pV = M 1 n 1 + M 2 n 2 . 



