268 Mr. O. W. Richardson on the Solubility and 
where A and a are the inverses of the solubilities of the 
undissociated gas and of the products of dissociation 
respectively. 
By eliminating the concentration from these equations we 
obtain an interesting relation between the constants, viz. : 
qr oy ky 
AS 
In other words, the solubility of the products of dissociation 
is determined absolutely by the solubility of the undissociated 
substance, together with the two dissociation constants. In 
the simplest case, where the two dissociation constants are 
equal, the solubility of the dissociation products is the nth 
root of that of the original substance. 
These results may be confirmed and extended by treating 
the subject thermodynamically. We can obtain a reversible 
cycle, at constant temperature, as follows :—Suppose we have 
a cylinder whose walls are perfect conductors of heat and 
supplied with a piston at each end. Across the middle of the 
cylinder is a slice of the solution we are considering. The 
two sides of the slice are bounded by semipermeable mem- 
branes, one end allowing only undissociated, and the other 
only dissociated molecules to pass. Initially the external gas 
is in equilibrium with that inside the solution at both ends. 
Since the diaphragms are only permeable to one of the two 
gases present, this does not necessarily imply equilibrium 
between the internal and external gas at any one end as regards 
both constituents, but only as regards one constituent. 
According to the result we obtained before, this would involve 
equality in the total pressures as well; since, as we have already 
seen, there is one total pressure for which the constituent 
gases are in equilibrium with the internal ones. We shall 
show that this equality follows thermodynamically ; although 
any further proof cannot be regarded as strictly necessary, 
since the result is merely a particular case of Gibbs’s general 
theorem regarding the equilibrium of mixed systems. 
Let us suppose the partial pressures of the undissociated 
gas (X,,) and of the dissociated gas (X) on the side permeable . 
to X,, are P, and p, respectively, the corresponding quantities 
on the other side being P, and p,;. Then a volume V, of gas 
is forced through the X, diaphragm, into the solution, a 
corresponding quantity being withdrawn through the X side 
so as to maintain the total internal pressure constant. Owing 
to the supposed difference of pressure on the two sides, the 
volume V, withdrawn will not be the same as Vg, but is given 
by the modified law of Boyle and Charles for a dissociating 
As ieee 
