Page:A history of the theories of aether and electricity. Whittacker E.T. (1910).pdf/402

of salt, in passing from a high to a low concentration, are therefore capable of supplying energy, just as a compressed gas is capable of supplying energy when its degree of compression is reduced. To examine the matter quantitatively, let nf(n/V) denote the term in the available energy of a solution, which is due to the dissolution of n gramme-molecules of salt in a volume V of pure solvent; the function f will of course depend also on the temperature. Then when dn gramme-molecules of solvent are evaporated from the solution, the decrease in the available energy of the system is evidently equal to the available energy of dn gramme-molecules of liquid solvent, less the available energy of dn gramme-molecules of the vapour of the solvent, together with nf(n/V) less nf{n/(V-vdn)}, where v denotes the volume of one gramme-molecule of the liquid. But this decrease in available energy must be equal to the mechanical work supplied to the external world, which is dn.p₁(u′ – v), if p₁, denote the vapour-pressure of the solution at the temperature in question, and {{Wikimath|v′ denote the volume of one gramme-molecule of vapour. We have therefore

$dn.p_{1}(v^{\prime }-v)=$	-available energy of dn gramme-molecules of solvent vapour
	+available energy of dn gramme-molecules of liquid solvent
	+$nf(nV)-nf\{n/(V-vdn)\}$.

Subtracting from this the equation obtained by making n zero, we have

$dn.(p_{1}-p_{0})(v^{\prime }-v)=nf(n/V)-nf\{n/(V-vdn)\}$,

where p₀ denotes the vapour-pressure of the pure solvent at the temperature in question; so that

$(p_{1}-p_{0})(v^{\prime }-v)=-(n^{2}/V^{2})f^{\prime }(n/V)v$.

Now, it is known that when a salt is dissolved in water, the vapour-pressure is lowered in proportion to the concentration of the salt—at any rate when the concentration is small: in