Page:Scientific Memoirs, Vol. 2 (1841).djvu/505

the same proportion. But when two constituents combine, by both reciprocally penetrating one another, each extends beyond the entire space of the compound, on which account the intensity of the force proper to each constituent decreases by combination, in the same proportion as the space of the compound is greater than the space which each constituent occupied before the combination. Consequently if $z$ denote the relation of the space which the constituent $\mathrm {A}$, in the disc $\mathrm {M}$, occupied previous to combination to that space which the compound in the disc $\mathrm {M}$ occupies; and also, since we admit that both constituents, before and after the combination, occupy the same extent of space, $1-z$ will denote the same relation relatively to the constituent $\mathrm {B}$; then, since $m$ and $n$ designate the electroscopic forces of the constituents $\mathrm {A}$ and $\mathrm {B}$ previous to combination, $mz$ and $n(1-z)$ will represent the latent electroscopic forces of the constituents $\mathrm {A}$ and $\mathrm {B}$, which correspond to each chemical equivalent of the disc $\mathrm {M}$; and, at the same time, it follows from the above, that the variable values $z$ and $1-z$ cannot exceed the limits $0$ and $1$.

In order to ascertain the portion of the free electricity $u$ pertaining to each constituent, we will assume that it is distributed over the single constituents in proportion to their masses. If, therefore, we represent respectively by $\alpha$ and $\beta$ the masses of the constituents $\mathrm {A}$ and $\mathrm {B}$, on the supposition that one alone, to the exclusion of the other, occupies the entire disc, then $\alpha z$ and $\beta (1-z)$ will represent the masses of the constituents $\mathrm {A}$ and $\mathrm {B}$ united in the disc $\mathrm {M}$; consequently the portions

${\frac {\alpha uz}{\alpha z+\beta (1-z)}},\quad {\text{and}}\quad {\frac {\beta u(1-z)}{\alpha z+\beta (1-z)}}$

of the free electricity $u$ appertain to the constituents $\mathrm {A}$ and $\mathrm {B}$; instead of which, for the sake of conciseness, we will write

$\alpha \mathrm {U} z,\quad {\text{and}}\quad \beta \mathrm {U} (1-z)$.

If we now take into consideration what was stated in §30, respecting the motive force of the galvanic circuit, it is immediately evident that the tendency of the constituent $\mathrm {A}$ to move along the circuit, is expressed by

$2i(m+\alpha \mathrm {U} )z\mathrm {S}$,

or that of the constituent $\mathrm {B}$ by

$2i(n+\beta \mathrm {U} )(1-z)\mathrm {S}$.