138 Dr. Thomas Ewan on 



The osmotic pressure of the solution being P, and the pres- 

 sure of the gas/. 



Then in both cases, in order to keep the temperature 

 constant, a certain amount of heat must be added, which is 

 the equivalent of the external and internal work done by 

 the solution (or gas) in expanding. For the solution the 



work done is PK— - - J^Q where dQ is the heat evolved on 



adding dw gr. solvent to the solution without doing external 

 work. 



For the gas or vapour the work done is according to 



Van der Waals pdV + ^dY 



The heat of dilution evidently represents the internal 

 work done in the expansion. 



According to Van der Waals we have 



pdV + ^ 2 dV = ^jdV 



Assuming that a similar equation is true for the sum of 

 external and internal work done when a solution expands, 

 we get 



ptt^ T^n RT Kdw 



or 



P(V-6)-jg|(V-6) = RT . . . (10) 



As R is the gas constant for 1 gr. mol. of substance, V 

 must be taken also as the volume in which a gram 

 mol. is contained. 



Compare equation (10) with (9), viz. : — 



P^ -J^M ^ = ET^ .... (9) 

 ctw 



nv is the volume by which the solution diminishes when 

 n gr. mols. of solvent are withdrawn from it without changing 

 its concentration, it may, therefore, be regarded as the 



