Page:Scientific Papers of Josiah Willard Gibbs.djvu/469

Let⁠$P=e^{{\frac {X}{T}}x''},$⁠$\gamma _{0}''-\gamma _{0}'=\pm H_{0}(P-P^{-1}).$ Let⁠$\Delta _{0}=\gamma _{0}''-\gamma _{0}',$⁠$\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}=\pm \Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x}}{P-P^{-1}}},$

$c_{0}k_{0}\gamma _{0}\mp {\frac {c_{0}\gamma _{0}}{X}}=\pm c_{0}k_{0}\Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x}}{P-P^{-1}}}\cdot$

The condition of no electrical current gives

$\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x}}{P-P^{-1}}}\cdot$

Apply to both ends and add,

${\begin{aligned}\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}''+\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}'&=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x''}+e^{\pm {\frac {X}{T}}x'}}{P-P^{-1}}},\\&=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {P+P^{-1}}{P-P^{-1}}}\cdot \end{aligned}}$

If we set, to abridge,

${\begin{aligned}K_{1}'=\textstyle \sum _{1}\displaystyle c_{1}k_{1}\gamma _{1}',&K_{1}''=\textstyle \sum _{1}\displaystyle c_{1}k_{1}\gamma _{1}'',\\K_{2}'=\textstyle \sum _{2}\displaystyle c_{2}k_{2}\gamma _{2}',&K_{2}''=\textstyle \sum _{2}\displaystyle c_{2}k_{2}\gamma _{2}''.\end{aligned}}$

When the summations are for cations or anions separately, the last equation may be written

$K_{1}''+K_{2}''+K_{1}'+K_{2}'={\frac {P+P^{-1}}{P-P^{-1}}}(K_{1}''-K_{1}'-K_{2}''+K_{2}'),$

which gives⁠$P^{2}={\frac {K_{1}''+K_{2}'}{K_{2}''+K_{1}'}}\cdot$

Now ⁠	$\log P$	$={\frac {X}{T}}x''={\frac {-X}{T}}x',$	⁠	$2\log P$	$={\frac {X(x''-x')}{T}}={\frac {V'-V''}{T}}$
	${\frac {V'-V''}{T}}$	$=\log {\frac {K_{1}''+K_{2}'}{K_{1}'+K_{2}''}},$	⁠	$V''-V'$	$={\frac {Aat}{v}}\log {\frac {K_{1}'+K_{2}''}{K_{1}''+K_{2}'}}\cdot$

$K_{1}'$ is the part of the conductivity of the first electrolyte which is due to the cations.

If the first electrolyte contains only one cation (₁) and one anion (₂), and the second only one cation (₃) and one anion (₄), we have

	$V''-V'$	$={\frac {Aat}{v}}\log {\frac {c_{1}k_{1}\gamma _{1}'+c_{4}k_{4}\gamma _{4}''}{c_{3}k_{3}\gamma _{3}''+c_{2}k_{2}\gamma _{2}'}},$
or, since⁠	$c_{1}\gamma _{1}'$	$=c_{2}\gamma _{2}'=c_{3}\gamma _{3}''=c_{4}\gamma _{4}'',$
	$V''-V'$	$={\frac {Aat}{v}}\log {\frac {k_{1}+k_{4}}{k_{3}+k_{2}}},$

like the formula which you quote.