﻿64:6 Prof. Porter and Mr. Hedges on the Law of 



plus the atmosphere (or ocean) of solvent (or second phase) 

 which surrounded it, and to which it was attached (so 

 that motion of one entailed similar motion of the other), 

 constituted the whole of the substance present. The radius 

 calculated by considering b as the effective volume is 

 28 X 10" 4 cm. Now it is hard to think that the whole of 

 the liquid in the domain of a particle should be so attached 

 to the particle as to form practically part of it, for, if this 

 were so, the whole suspension would be rigid. It might 

 be conceivable in the case of even a dilute concentration of 

 gelatine, for this can form a rigid gel ; no such rigidity is 

 observed in the case of gamboge. It is noteworthy, however, 

 that if b be regarded as an effective volume, then b and K 

 are not independent constants, for K is proportional to the- 

 effective volume of the particle. When the radius is calcu- 

 lated from K, it is found to be 1*6 X 10 -5 cm., which is very 

 much less than when calculated from b, but is certainly a 

 measure of the effective radius of the particles that were 

 examined. We return to this point later. 



It is safer, however, to regard b as merely an empirical 

 constant. It may be pointed out that Callendar's equation 

 for unsaturated steam is of the form 



p(v-b)='RT, 



where, however, b is even negative and represents the com- 

 bined effects arising from the finite size of the molecules 

 and the attractive forces which tend to " co-aggregate " the 

 molecules. It is noteworthy, however, that, if b is to become 

 large and remain positive, repulsive forces are required. 



Burton, in the second edition of his monograph on the 

 physical properties of colloidal solutions, page 87, attempts 

 to explain the present problem by considering repulsion 

 arising from the electrical charges in the particles. He 

 concludes that the forces on any layer due to the rest of the 

 solution will be of the form kne per unit charge on the 

 layer dh, where k, he considers, may be taken as a constant, 

 at any rate for regions near the surface , and e is the charge- 

 in electrostatic units on each particle. Consequently the 

 total force per unit area is kn 2 e 2 dh. A term of this kind is 

 taken to represent the excess electrical repulsion of all the 

 particles below dh over that due to those above dh. Perrin's 

 equation then becomes, for the forces balancing per unit 

 area, 



-~r- dn + kn 2 e 2 dh = nV(d — w)gdh. 



