Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/338

338 work consisted of a splendid series of determinations of the conductivities of salt solutions. Assuming that all the salt molecules dissolved are actively concerned in conveying the current, from simple notions of convection combined with ’s law he calculated the relative velocity of the two ions which would be necessary to give the observed conductivity.

Thus (following Dr. ’s abstract in the ‘B.A. Report,’ 1886), let $$n^3$$ be the number of active molecules in 1 cub. centim., $$q$$ the total charge of electricity of whichever kind possessed by the ions of each active molecule, and $$\mathrm{U}$$ the relative velocity with which the opposite ions are sheared past each other by a potential gradient, $$dv/dx$$; we shall then get from considerations of convection $$n^3q\mathrm{U}$$ for the intensity of current (or quantity of electricity conveyed per second through unit area normal to the flow), and from ’s law $$k(d\mathrm{V}/dx)$$ for the same quantity, where $$k$$ is the conductivity of the unit cube, therefore

If $$\mathrm{N}$$ be the number of monad gramme equivalents of the active substance present in the unit cube, and $$\eta$$ the electro-chemical equivalent of hydrogen

therefore

therefore

Whence the arithmetical sum of the opposite velocities of anion and kation ($$u$$ and $$v$$) when urged by a slope of potential of one volt per centimetre through a solution containing $$\mathrm{N}$$ monad gramme equivalents of active substance per cubic centimetre, and of specific conductivity $$k$$, is

In order to apply this to particular cases assumes that all the dissolved salt is active, so that $$\mathrm{N}$$ is known from the strength of the solutions. The ratio $$u/v$$ is taken from ’s migration experiments, and thus the absolute values of $$u$$ and $$v$$ can be obtained.

expresses his experimental results in terms of the specific molecular conductivity (i.e., $$k/\mathrm{N}$$), and finds that for increasing dilution this tends to a limiting value. Thus, by drawing curves with $$\mathrm{N}^\frac{1}{2}$$ (the reciprocal of the average distance