Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/346

346 {| Voltmeter, 10.9.
 * colspan=2|Centim. per minute.
 * = .1195
 * = .1205
 * = .120
 * }
 * = .1205
 * = .120
 * }
 * = .120
 * }
 * = .120
 * }
 * }
 * }
 * }

Therefore specific ionic velocity

which is appreciably greater than for the stronger solutions.

The success of these preliminary determinations and the closeness of their agreement with theory led me to consider whether any improvement in the method was possible, by which further and more accurate results could be obtained.

In order to eliminate entirely the effect of the discontinuity of potential gradient it is necessary to use pairs of salts which have the same specific resistance for the same strength of solution. This very much limits one’s choice, but an example will be found below. In such cases as that of copper and ammonium chlorides where the velocities in opposite directions are sensibly the same, it is fair to conclude that this effect is negligible, but it was worth while to find one case at any rate where it could not in the least affect the result.

The next improvement was in devising a means of getting over the necessity of directly estimating the potential gradient.

Consider the kation of one salt AC at the junction. Let its specific ionic velocity be $$v_1$$, and its actual velocity $$v$$, under a slope of potential $$= d\mathrm{V}/dx$$. Let $$r$$ be the specific resistance of the solution. Let $$\gamma$$ be the total current passing, and $$\mathrm{R}$$ the total resistance of the column of liquid from the anode to the junction.

Then we have $$v = v_1\,d\mathrm{V}/dx$$.

Since $$\gamma$$ is constant along the tube we get by ’s law

Now since the tube is of uniform bore near the junction $$d\mathrm{R}/dx =$$ the increase of resistance which would occur if we supposed the column of liquid from the electrode to the junction, whose resistance we have called $$\mathrm{R}$$, to be increased in length by unity.

This increase of resistance $$= r/\mathrm{A}$$, where $$\mathrm{A}$$ represents the area of cross-section of the tube at the junction.

Therefore

and we get

or