OSMOTIC PRESSURE 41 



and pressure in A would fall. This demonstrates the conversion 

 of unordered molecular motion into ordered hydraulic motion 

 capable of doing work. 



The relative velocities of diffusion of any two gases even 

 through a thin porous septum are inversely proportional to the 

 square roots of their densities. 



This does not hold if the septum is not very thin, for then the 

 velocity is decreased by the impact of the molecules on the inside 

 surface of the pores. 



Graham further found that it was possible to procure semi- 

 permeable membranes that is, a membrane which would allow 

 free passage to one gas but not to another. What will happen 

 in such a case ? Suppose B can pass freely through the septum 

 while A cannot. Both gases are at 1 atmos. pressure. Then 

 B will diffuse through the membrane, and fill up the next space 

 as if A were not there, i.e. there will finally be | atmos. of B on 

 both sides. The total pressure on A will be 1| atmos. The 

 excess of pressure is due to the gas A, which cannot pass through 

 the septum. So that by taking the difference of pressure on the 

 two sides of a semi-permeable membrane, we obtain the partial 

 pressure of the gas to which the membrane is impermeable. (See 

 Respiration, Chap. XXIII.) 



An attempt may now be made to apply these laws (which are 

 only absolutely true for perfect gases) to solutions in the first 

 instance to simple solutions only. 



All these symbols seem applicable to a substance in solution 

 except P. What is the pressure of a solute ? This may be 

 determined in a way similar to the determination of gaseous 

 pressure. If an osmometer be fitted up (Part II. p. 398) with 

 a solution of sugar inside and water outside, in a short time 

 the fluid inside will increase in volume and will rise in the osmo- 

 meter tube developing a hydrostatic pressure. To what is this 

 pressure due ? Obviously water (and pressure) will be transferred 

 from a point where its pressure is high to a point where its pres- 

 sure is low. In some way or other the presence of sugar (or other 

 solute) has lowered the pressure of the water. Can this be 

 explained by reference to the kinetic energy ? Reasoning back- 

 wards, it may be argued that the sugar acts as a drag upon the 

 water molecules that is, the bombardment of the membrane 

 becomes unbalanced. The pure solvent is able to exert a greater 



