IX.
UNPUBLISHED FRAGMENTS.
[Being portions of a supplement to the "Equilibrium of Heterogeneous Substances" in preparation at the time of the author's death, and intended to accompany a proposed reprint of his thermodynamic papers^{[1]}]
[A list of subjects found with the manuscript and printed below appears to indicate the scope of the supplementary chapters as planned by Professor Gibbs. As will be observed, however, the authors unfinished manuscript, except for a number of disconnected notes, relates to only two of these subjects, the first and fourth in the list]
On the values of potentials in liquids for small components. (Temperature coefficients.)
On the fundamental equations of molecules with latent differences.
On the fundamental equations for vanishing components.
On the equations of electric motion.
On the liquid state, $p=0$.
On entropy as mixedupness.
Geometrical illustrations.
On similarity in thermodynamics.
Cryohydrates.
On the Values of Potentials in Liquids for Substances which form but a Small Part of the whole Mass.^{[2]}
The value of a potential^{[3]} for a volatile substance in a liquid may be measured in a coexistent gaseous phase,^{[4]} and so far as the latter may be treated as an ideal gas or gasmixture,^{[5]} the value of the potential will be given by the equation (276), ["Equilib. Het. Subs."] which may be briefly written
$\mu ={\text{func}}(t)+at\log \gamma _{\text{gas}},$

[1]

where
$\mu$ is the potential of the volatile substance considered, either in the liquid or in the gas,
$t$ the absolute temperature,
$\gamma _{\text{gas}}$ the density of the volatile substance in the gas and a the constant of the law of Boyle and Charles. Since this last quantity is inversely proportional to the molecular weight we may set
$a={\frac {A}{M}},$


where
$M$ denotes the molecular weight, and
$A$ an absolute constant (the constant of the law of Boyle, Charles, and Avogadro),
^{[6]} and write the equation in the form
$\mu ={\text{func}}(t)+{\frac {At}{M}}\log \gamma _{\text{gas}},$

[2]

in which the value of the potential depends explicitly on the molecular weight.
The validity of this equation, it is to be observed, is only limited by the applicability of the laws of ideal gases to the gaseous phase; there is no limitation in regard to the proportion of the substance in question to the whole liquid mass. Thus at 20° Cent. the equation may be determined by the potential for water or for alcohol in a mixture of the two substances in any proportions, since the vapor of the mixture may be regarded as an ideal gasmixture. But at a temperature at which we approach the critical state, the same is not true without limitation, since the coexistent gaseous phase cannot be treated as an ideal gasmixture. At the same temperature however, if we limit ourselves to cases in which the proportion of water does not exceed ${\tfrac {1}{10}}$ of one per cent., and suppose the density of the watervapor, $\gamma _{\text{gas}}$, to be measured in a space containing only watervapor and separated from the liquid by a diaphragm permeable to water and not to alcohol, then the above equation would probably be applicable, since then the watervapor might probably be treated as an ideal gas. The same would be true (mutatis mutandis) of the potential for alcohol in a mixture of alcohol and water containing not more than ${\tfrac {1}{10}}$ of one per cent, of alcohol.^{[7]}
This law, however, which makes the potential in a liquid depend upon the density of the substance in some other phase is manifestly not convenient for use. We may get over this difficulty most simply by the law of Henry according to which the ratio of the densities of a substance in coexistent liquid and gaseous phases is (in cases to which the law applies) constant. If $\gamma$ be the density in the liquid phase and $\gamma _{\text{gas}}$ in the gas, we have
$\gamma _{\text{gas}}=c\gamma ,$

[3]

and by substitution in equation [2] we have
$\mu ={\text{func}}(t)+{\frac {At}{M}}\log c\gamma ,$


or$\mu ={\text{func}}(t)+{\frac {At}{M}}\log \gamma ,$

[4]

where the function of the temperature has been increased by
${\frac {At}{M}}\log c$.
With this value of the potential, which is manifestly demonstrated only to be used so far as the law of Henry applies, in connection with the general equation (98), ["Equilib. Het. Subs."] viz.,
$dp={\frac {\eta }{v}}dt+{\frac {m_{1}}{v}}d\mu _{1}+{\frac {m_{2}}{v}}d\mu _{2}...+{\frac {m_{n}}{v}}d\mu _{n},$


we may calculate the osmotic pressure, etc., etc., as we shall see more particularly hereafter.
I. Osmotic pressure.
II. Lowering freezing point.
III. Diminishing pressure of other gas.
IIIa. Effect on total pressure.
IIII. Raising boiling point with one pressure.
IIIIa. Raising boiling point with two pressures.
V. Interpolation formula for mixtures of liquids.
In fact, when $\gamma _{\text{D}}$^{[8]} is small, we have approximately
$\gamma _{\text{D}}d\mu _{\text{D}}={\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}=Atd{\frac {\eta _{\text{D}}}{v}},$

[5]

where
$n_{\text{D}}$ denotes the number of molecules of the form (
$D$). Hence we have for the solution
$dp={\frac {\eta }{v}}dt+\gamma _{\text{S}}d\mu _{\text{S}}+At\,d{\frac {n_{\text{D}}}{v}}\cdot$

[6]

If $t$ is constant, and also $\mu _{\text{S}}$,—a condition realized in equilibrium, when the solution is separated from the pure solvent by a diaphragm permeable to the solvent but not to the solutum, the equation reduces to
$dp=At\,d{\frac {n{\text{D}}}{v}}\cdot$


Whence$pp'={\frac {At}{M_{\text{D}}}}\gamma _{\text{D}}=At{\frac {n_{\text{D}}}{v}},$

[7]

$p'$ being the pressure where
$\gamma _{\text{D}}=0$, i.e., in the pure solvent. Here
$pp'$ is the socalled osmotic pressure, and
${\frac {At}{M_{\text{D}}}}\gamma _{\text{D}}$ is the pressure as calculated
^{[9]} by the laws of Boyle, Charles, and Avogadro for the solutum in the space occupied by the solution. The equation manifestly expresses van't HofF's law.
For a coexistent solid phase of the solvent, with constant pressure, the general equation gives
$0=\eta dt+m_{\text{S}}d\mu _{\text{S}}+v\,At\,d\gamma _{\text{D}}$


for the solution, and
$0=\eta 'dt+m_{1}d\mu _{\text{S}}$


for the solid coexistent phase. Here
$t$ and
$\mu _{\text{S}}$ have necessarily the same values in the two equations, and we may suppose the quantity of one of the phases to be so chosen as to make the values of
$m_{\text{S}}$ equal in the two equations. This gives
$(\eta '\eta )dt=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$

[8]

In integrating from $\gamma _{\text{D}}=0$ to any small value of $\gamma _{\text{D}}$, we may treat the coefficients of $dt$ and $d\gamma _{\text{D}}$ as having the same constant values as when $\gamma _{\text{D}}=0$. This gives
$(\eta '\eta )\Delta t=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}={\frac {At}{M_{\text{D}}}}m_{\text{D}}.$


If we write
$Q_{\text{S}}$ for
${\frac {t(\eta '\eta )}{m_{\text{S}}}}$ (the latent heat of melting for the unit of weight of the solvent), we get
$\Delta t={\frac {At}{M_{\text{D}}}}m_{\text{D}}{\frac {t}{Q_{\text{S}}m_{\text{S}}}},$


or$\Delta t={\frac {\frac {m_{\text{D}}}{M_{\text{D}}}}{\frac {m_{\text{S}}}{M_{\text{S}}}}}{\frac {At^{2}}{Q_{\text{S}}M_{\text{S}}}}={\frac {m_{\text{D}}}{M_{\text{D}}}}{\frac {At^{2}}{Q_{\text{S}}m_{\text{S}}}}\cdot$

[9]

$m_{\text{S}}Q_{\text{S}}$ is the latent heat of so much of the solvent as occurs in the solution. (Or make
$m_{\text{S}}=1$.)
Raoult makes $\Delta t\propto {\frac {m_{\text{D}}}{M_{\text{D}}}}$, with exceptions.
With a coexistent gaseous phase of the solvent (the solutum being not volatile), we have for the solution
$dp=\gamma _{\text{S}}d\mu _{\text{S}}+{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},$


and for the gaseous phase
$dp=\gamma _{\text{S}}'d\mu _{\text{S}}.$


Here, on account of the coexistence of the phases,
$p$ and
$\mu _{\text{S}}$ and
$dp$ and
$d\mu _{\text{S}}$ have the same values. Hence
$dp={\frac {\gamma _{\text{S}}}{\gamma _{\text{S}}'}}dp+{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$ $dp{\frac {(\gamma _{\text{S}}\gamma _{\text{S}})}{\gamma _{\text{S}}}}={\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$


Say$dp={\frac {\gamma _{\text{S}}'}{\gamma _{\text{S}}}}{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},$


$pP={\frac {\gamma _{\text{S}}'}{\gamma _{\text{S}}}}{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$^{[10]} 
[10]

or^{[11]}${\frac {dp}{p}}={\frac {M_{\text{S}}}{\gamma _{\text{S}}}}{\frac {d\gamma _{\text{D}}}{M_{\text{D}}}},$


${\frac {pP}{p}}={\frac {M_{\text{S}}}{\gamma _{\text{S}}}}{\frac {\gamma _{\text{D}}}{M_{\text{D}}}}\cdot$

[11]

$M_{\text{D}}$ is the molecular weight [of solutum] in solution;
$M_{\text{S}}$ is the molecular weight [of solvent] in vapor.
But the foregoing equation suggests a generalization which is not confined to cases in which the law of Henry has been proved. The letter $M$ in the equation has been defined as the molecular weight of the substance in the form of gas. Now the molecular weight which figures in the relation between the potential and the density of a substance in a liquid would naturally be the molecular weight of the substance as it exists in the liquid. It is therefore a natural supposition suggested by the equation that, in the case where Henry's law holds good, and consequently eq. [4], the molecular weight of the solutum is the same in the liquid and in the gaseous phase; that in case the law of Henry and eq. [4] do not hold, it may be on account of a difference in the molecular weight in the gas and the liquid, and that the eq. [4] may still hold if we give the proper value to $M$ in that equation, viz., the molecular weight in the liquid.
But as these considerations, although natural, fall somewhat short of a rigorous demonstration, let us scrutinize the case more carefully. It is easy to give an a priori demonstration of Henry's law and equation [4] in cases in which there is only one molecular formula for the solutum in liquid and in gas, so long as the density both in liquid and in gas is so small that we may neglect the mutual action of the molecules of the solutum. In such a case the molecules of the solutum will be divided between the liquid and the gas in a (sensibly) constant ratio (the volume of the liquid and gas being kept constant), simply because every molecule, moving as if there were no others, would spend the same part of its time in the vapor and in the liquid as if the others were absent, and the number of the molecules being large, this would make the division sensibly constant. This proof will apply in cases in which the law of Henry can hardly be experimentally demonstrated, because the density of the solutum as gas is so small as to escape our power of measurement. Also in cases in which a semipermeable diaphragm is necessary, an arrangement very convenient for theoretical demonstrations, but imperfectly realizable in practice. (Also in cases in [which a] difference of level is necessary, with or without diaphragm.) But in every case when the law of Henry is demonstrably untrue for dilute solutions, we may be sure that there is more than one value of the molecular weight of the solutum in the phases considered.
This theoretical proof will apply to cases in which experimental proof is impossible:
(1) When the density in gas is too small to measure.
(2) When the density in gas is too great, either the total density or the partial. (Diaphragm or vertical column.)
(3) When the liquid (or other phase) is sensitive to pressure and not in equilibrium with the gas.
Will the various theorems exist in these cases?
If one or both appear in a larger molecular form, the densities of $\gamma _{\text{M}}$ and $\gamma _{\text{M}}'$^{[12]} are proportional and
$\gamma _{2{\text{M}}}\propto \gamma _{\text{M}}^{2},$$\gamma _{\text{2M}}'\propto \gamma _{\text{M}}^{'2},$ 
$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$ [12]

$\gamma _{\text{2M}}=\gamma _{\text{M}}=\gamma _{\text{M}}'=\gamma _{\text{2M}}',$

hence one equation of form,
$\mu _{\text{M}}={\frac {At}{M}}\log \gamma _{\text{M}}$ proves all.
Let us next consider the case in which the solutum appears with more than one molecular formula in the liquid or gas or both. Now there are two cases, that in which the quantities of the substance with the different molecular formulæ are independently variable, and that in which they are not. In the [first] case there is no question. If, for example, hydrogen appears with the molecular formula H_{2}O and also in molecules with the molecular formula H_{2}O, these are to be treated as separate substances, and we have the two equations
$\mu _{{\text{H}}_{2}}={\text{func}}(t)+{\frac {At}{M_{{\text{H}}_{2}}}}\log \gamma _{{\text{H}}_{2}},$


and
$\mu _{{\text{H}}_{2}{\text{O}}}={\text{func}}(t)+{\frac {At}{M_{{\text{H}}_{2}{\text{O}}}}}\log \gamma _{{\text{H}}_{2}{\text{O}}},$


and also if free oxygen is present
$\mu _{{\text{O}}_{2}}={\text{func}}(t)+{\frac {At}{M_{{\text{O}}_{2}}}}\log \gamma _{{\text{O}}_{2}}.$


But when the quantities of the substance associated in the different molecular combinations are not independently variable, then we have the equation
$M_{1}\mu _{1}+M_{2}\mu _{2}=M_{12}\mu _{12},$

[13]

which is exact and certain, and the considerations adduced on
$p$. (
^{[13]}), which are not limited to gases, seem to show that in this case the equations of the form (
^{[14]}) all continue to subsist, but we have also the equation of form (
^{[15]}).
It would therefore appear that we may regard the equation
$\mu ={\text{func}}(t)+{\frac {At}{M}}\log \gamma$


as expressing a general law of nature, where the letter
$M$ is the molecular weight corresponding to any molecular combination in the liquid and
$\gamma$ is the density of the matter which has that molecular formula, provided that the density
$\gamma$ is so small that of the molecules which it represents only a negligible fraction at any time are within the spheres of each other's attraction. It goes without saying that the law is approximative, as the last condition can only be satisfied approximately for any finite value of
$\gamma$. (Need of verification on account of the unknown
$M$.)
[
The author's manuscript for the proposed supplement ends, so far at least as a connected treatment is concerned, at this point. The following notes are appended.]
In case of one molecular formula in liquid and none in gas, we may give the molecules repelling forces which will make the gas possible. (?) [See p. 417.]
Deduce Ostwald's law in more general form.
Deduce interpolation formula.
What use can we make of Latent Differences? $\mu _{\text{A}},\mu _{\text{AA}},\mu _{\text{B}},\mu _{\text{BB}},\mu _{\text{AB}}$ all conform to law, I think.
[On the Equations of Electric Motion.]
[A somewhat abbreviated copy of a letter written four years earlier (in May 1899) to Professor W. D. Bancroft of Cornell University had been placed by Professor Gibbs between the pages of the manuscript, and was evidently intended to serve as a basis for the chapter "On the equations of electric motion" mentioned in the list on page 418.
Through the courtesy of Professor Bancroft the original letter has been placed at the disposal of the editors and is here given in full. The major portion of this letter was incorporated by Professor Bancroft in an article entitled "Chemical Potential and Electromotive Force" published after the death of Professor Gibbs, in the Journal of Physical Chemistry, vol. vii.,p. 416, June 1903.]
My dear Prof. Bancroft:
A working theory of galvanic cells requires (as you suggest) that we should be able to evaluate the (intrinsic or chemical) potentials involved, and your formula
$d\mu =Rt\,d\log p,$


is all right as you interpret it. I should perhaps prefer to write
$\mu _{\text{D}}=B+{\frac {At}{M_{\text{D}}}}\log \gamma _{\text{D}},$

(1)

or$\gamma _{\text{D}}d\mu _{\text{D}}={\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},$

(2)

for small values of
$\gamma _{\text{D}}$, where
$\gamma _{\text{D}}$ is the density of a component (say the mass of the solutum divided by the volume of the solution),
$M_{\text{D}}$ its molecular weight (viz., for the kind of molecule which actually exists in the solution),
$A$ the constant of Avogadro's Law
$\left({\frac {pv}{mt}}={\frac {A}{M}}\right),$ and
$B$ a quantity which depends upon the solvent and the solutum, as well as the temperature, but which may be regarded as independent of
$\gamma _{\text{D}}$ so long as this is small, and which is practically independent of the pressure in ordinary cases.
We may avoid 'hedging' in regard to $B$ by using the differential equation (2). We may simply say that this equation holds for changes produced by varying the quantity of ($D$), when $\gamma _{\text{D}}$ is small. It is not limited to changes in which t is constant, for the change in $\mu _{\text{D}}$ due to $t$ appearing in (1) (both explicitly, and implicitly in $B$) becomes negligible when multiplied by the small quantity $\gamma _{\text{D}}$.
The formula contains the molecular weight $M_{\text{D}}$, and if all the solutum has not the same molecular formula, the $\gamma _{\text{D}}$ must be understood as relating only to a single kind of molecule.
Thus if a salt (_{12}) is partly dissociated into the ions (_{1}) and (_{2}), we will have the three equations
$\mu _{1}=B_{1}+{\frac {At}{M_{1}}}\log \gamma _{1},$ $\mu _{2}=B_{2}+{\frac {At}{M_{2}}}\log \gamma _{2},$ $\mu _{12}=B_{12}+{\frac {At}{M_{12}}}\log \gamma _{12}.$


The three potentials are also connected by the relation
$M_{1}\mu _{1}+M_{2}\mu _{2}=M_{12}\mu _{12},$


which determines the amount of dissociation. We have, namely,
$M_{1}B_{1}+M_{2}B_{2}M_{12}B_{12}+At\log {\frac {\gamma _{1}\gamma _{2}}{\gamma _{12}}}=0,$


which makes constant, for constant temperature and solvent.
I may observe in passing that this relation, eq. (1) or (2), which is so fundamental in the modern theory of solutions, is somewhat vaguely indicated in my "Equilib. Het. Subs." (See [this volume] pp. 135–138, 156, and 164–165.) I say vaguely, because the coefficient of the logarithm is only given (in the general case) as constant for a given solvent and temperature. The generalization that this coefficient is in all cases of exactly the same form as for gases, even to the details which arise in cases of dissociation, is due to van't Hoff in connection with Arrhenius, who suggested that the "discords" are but "harmonies not understood," and that exceptions vanish when we use the true molecular weights. At all events, eq. (2) with (98) (E.H.S.) gives for a solvent ($S$) with one dissolved substance ($D$),
$dp={\frac {\eta }{v}}dt+{\frac {m_{\text{S}}}{v}}d\mu _{\text{S}}+{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$


If we integrate, keeping
$t$ constant and also
$\mu _{\text{S}}$ (by connection with the pure solvent through a semipermeable diaphragm), we have van't HofTs Law,
$pp'={\frac {At}{M_{\text{D}}}}\gamma _{\text{D}}.$


In the above case of dissociation the formula would be
$pp_{1}=At\left({\frac {\gamma _{1}}{M_{1}}}+{\frac {\gamma _{2}}{M_{2}}}+{\frac {\gamma _{3}}{M_{3}}}\right)\cdot$


For a coexistent solid phase of the solvent we have for constant pressure
$0=\eta dt+md\mu _{\text{S}}+v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},$ $0=\eta 'dt+md\mu _{\text{S}},$


$m_{\text{S}}$ being for convenience taken the same in both phases.
Then
$(\eta '\eta )dt=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.$
In integrating for small values of
$\gamma _{\text{D}}$ we may treat the coefficients of
$dt$ and
$d\gamma _{\text{D}}$ as constant. This gives
$(\eta '\eta )\Delta t=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}=At{\frac {m_{\text{D}}}{M_{\text{D}}}},$


or if we write
$Q_{\text{S}}$ for (the latent heat of melting for the unit of weight of the solvent), we have
$\Delta t={\frac {At^{2}m_{\text{D}}}{Q_{\text{S}}M_{\text{D}}m_{\text{S}}}}\cdot$


This may be written
$\Delta t{\frac {m_{\text{S}}}{m_{\text{D}}}}{\frac {M_{\text{D}}}{M_{\text{S}}}}={\frac {At^{2}}{Q_{\text{S}}M_{\text{S}}}}\cdot$


According to Raoult, the first member of this equation has a value nearly identical for all solvents and solutes (supposed definite compounds). This would make the second member the same for all liquids of "definite" composition, when we give
$M_{\text{S}}$ the value for the molecule in the liquid state. I should think it more likely that these properties should hold for the two members of the equation
${\frac {\Delta t}{t}}{\frac {m_{\text{S}}}{m_{\text{D}}}}{\frac {M_{\text{D}}}{M_{\text{S}}}}={\frac {At}{Q_{\text{S}}M_{\text{S}}}},$


which are pure numbers (of no dimensions in physical units). In this form it has a certain analogy with van der Waals' law of "corresponding states."
With a coexistent vapor phase of the solvent, we have
${\begin{aligned}v\,dp&=m_{\text{S}}d\mu _{\text{S}}+v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},\\v'\,dp&=m_{\text{S}}d\mu _{\text{S}},\\(vv')dp&=v{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}},\\dp&={\frac {v}{v'v}}{\frac {At}{M_{\text{D}}}}d\gamma _{\text{D}}.\end{aligned}}$


We may regard
${\frac {v}{v'v}}$ as constant in integrating (for small
$\gamma _{\text{D}}$), which gives
$Pp={\frac {v}{v'v}}At{\frac {\gamma _{\text{D}}}{M_{\text{D}}}}\cdot$


Now
${\frac {At}{v'v}}={\frac {At}{v'}}={\frac {PM_{\text{S}}}{m_{\text{S}}}}$ nearly, which gives
${\frac {Pp}{P}}={\frac {m_{\text{D}}}{m_{\text{S}}}}{\frac {M_{\text{S}}}{M_{\text{D}}}},$ which is Raoult's Law.
Raoult found values about 5 per cent, larger than this, which agrees very well with the fact that
${\frac {At}{v'v}}$ is somewhat larger than
${\frac {PM_{\text{S}}}{m_{\text{S}}}}\cdot$ It is also to be observed that
$M_{\text{D}}$ relates to the molecules in the solution, but
$M_{\text{S}}$ to the molecules in the vapor. Or, with a coexistent vapor phase of the solutum (alone or mixed with other vapors or gases), we have
${\begin{aligned}\mu &=B+{\frac {At}{M_{\text{D}}}}\log \gamma _{\text{D}},\\\mu &=B'+{\frac {At}{M_{\text{D}}}}\log \gamma _{\text{D}}',\\{\frac {B'B}{At}}M_{\text{D}}&=\log {\frac {\gamma _{\text{D}}}{\gamma _{\text{D}}'}},\end{aligned}}$


which makes
${\frac {\gamma _{\text{D}}}{\gamma _{\text{D}}'}}$ constant for the same solvent, solutum, and temperature, according to Henry's Law.
So for the galvanic cell which you first consider, I should write
$V''V'=a_{a}(\mu '\mu '')=a_{a}{\frac {At}{M_{a}}}\log {\frac {\gamma _{a}'}{\gamma _{a}''}},$


$\gamma _{a}',\gamma _{a}''$ being the densities, supposed small, of the cation (a) in the two electrodes, which are supposed identical except for the dissolved (a). Here
$a_{a}$ has reference to the solution and
$M_{a}$ to the electrodes. It may be more convenient to divide a a into the factors
$E_{a},a_{\text{H}}$, where
$a_{\text{H}}$ is the weight of hydrogen which carries the unit of electricity, and
$E_{a}$ the weight of (a) which carries the same quantity of electricity as the unit of weight of hydrogen. In other words
$E_{a}$ is Faraday's "electrochemical equivalent" and
$a_{a}$ is Maxwell's "electrochemical equivalent." This gives
$V''V'=a_{\text{H}}At{\frac {E_{a}}{M_{a}}}\log {\frac {\gamma _{a}'}{\gamma _{a}''}},$


where
$a_{\text{H}}A$ is your
$R$ and
${\frac {M_{a}}{E_{a}}}$ your
$v,v'.$^{[16]}
The meagreness of the results obtained in my E.H.S. in the matter of electrolysis has a deeper reason than the difficulty of the evaluation of the potentials.
In the first place, cases of true equilibrium (even for open circuit) are quite exceptional. Thus the single case of unequal concentration of the electrolyte cannot be one of equilibrium since the process of diffusion cannot be stopped. Cases in which equilibrium does not subsist were formally excluded by my subject, and indeed could not be satisfactorily treated without the introduction of new ideas quite foreign to those necessary for the treatment of equilibrium.
Again, the consideration of the electrical potential in the electrolyte, and especially the consideration of the difference of potential in electrolyte and electrode, involves the consideration of quantities of which we have no apparent means of physical measurement, while the difference of potential in "pieces of metal of the same kind attached to the electrodes" is exactly one of the things which we can and do measure.
Nevertheless, with some hedging in regard to the definition of the electrical potential, we may apply
$V''V'=a_{a}(\mu _{a}'\mu _{a}'')$


to points in electrolyte (
$'$) and electrode (
$''$).
This gives
$V''V'=a_{a}\left(B+{\frac {At}{M_{a}}}\log \gamma _{a}\mu _{a}''\right),$


say,
$V''V'={\frac {a_{a}At}{M_{a}}}\log {\frac {\gamma _{a}}{G}}\cdot$
The
$G$ like the
$P$ of your formula seems to depend on the solvent, presumably varies with the temperature, but as Nernst remarks does not depend on the other ion associated with (
$a$), so long as the solution is dilute.
The case of unequal concentration, or, in general, cases in which the electrolyte is not homogeneous, I should treat as follows; Let us suppose for convenience that the cell is in form of a rectangular parallelepiped with edge parallel to axis of x and cross section of unit area. The electrolyte is supposed homogeneous in planes parallel to the ends, which are formed by the electrodes.
Of course we should have equilibrium if proper forces could be applied to prevent the migration of the ions and also of the part of the solutum which is not dissociated. What would these forces be? For the molecules (_{12}) which are not dissociated, the force per unit of mass would be ${\frac {d\mu _{12}}{dx}}\cdot$ (The problem is practically the same as that discussed in E.H.S. [this volume], pp. 144 ff.) If the unit of mass of the cation (_{1}) has the charge $c_{1}$, the force necessary to prevent its migration would be
${\frac {d\mu _{1}}{dx}}+c_{1}{\frac {dV}{dx}}\cdot$


For an anion (
_{2}) the force would be
${\frac {d\mu _{2}}{dx}}c_{2}{\frac {dV}{dx}}\cdot$^{[17]} 

Now we may suppose that the same ion in different parts of a dilute solution will have velocities proportional to the forces which would be required to prevent its motion. We may therefore write for the velocity of the cation (
_{1}),
${\frac {k_{1}}{c_{1}}}\left({\frac {d\mu _{1}}{dx}}+c_{1}{\frac {dV}{dx}}\right),$


and for the flux of the cation (
_{1}),
$\phi _{1}={\frac {k_{1}}{c_{1}}}\gamma _{1}\left({\frac {d\mu _{1}}{dx}}+c_{1}{\frac {dV}{dx}}\right)=At{\frac {k_{1}}{c_{1}M_{1}}}{\frac {d\gamma _{1}}{dx}}k_{1}{\frac {dV}{dx}}\gamma _{1};$

(3)

for the flux of the anion (
_{2}),
$\phi _{2}={\frac {k_{2}}{c_{2}}}\gamma _{2}\left({\frac {d\mu _{2}}{dx}}+c_{2}{\frac {dV}{dx}}\right)=At{\frac {k_{2}}{c_{2}M_{2}}}{\frac {d\gamma _{2}}{dx}}+k_{2}{\frac {dV}{dx}}\gamma _{2},$

(4)

where
$k_{1},k_{2}$ are constants ('migration velocities') depending on the solvent, the temperature, and the ion.
^{[18]} Now whatever the number of ions the flux of electricity is given by the equation
$\phi =\textstyle \sum \displaystyle \pm c_{1}\phi _{1},$^{[19]} 

where the upper sign is for cations and the lower for anions, and the summation for all ions. This gives
$\phi =At\textstyle \sum \displaystyle \mp {\frac {k_{1}}{M_{1}}}{\frac {d\gamma _{1}}{dx}}{\frac {dV}{dx}}\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}.$


That is,
$\phi {\frac {dx}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}=At{\frac {\textstyle \sum \displaystyle \mp {\frac {k_{1}}{M_{1}}}d\gamma _{1}}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}dV.$


The form of this equation shows that since
$\phi$ is the current,
${\frac {dx}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}$ is the "resistance" of an elementary slice of the cell, and the next term the (internal) electromotive force of that slice.
Integrating from one point to another in the electrolyte,
$\phi \int {\frac {dx}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}=At\int {\frac {\textstyle \sum \displaystyle \mp {\frac {k_{1}}{M_{1}}}d\gamma _{1}}{\textstyle \sum \displaystyle c_{1}k_{1}\gamma _{1}}}+V'V''.$


The evaluation of these integrals which denote the resistance and electromotive force for a finite part of the electrolyte depends on the distribution of the ions in the cell. For one salt with varying concentration,
$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {{\frac {k_{1}}{M_{1}}}d\gamma _{1}+{\frac {k_{2}}{M_{2}}}d\gamma _{2}}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}dV,$


or, since
$c_{1}\gamma _{1}=c_{2}\gamma _{2}$ and
$c_{1}d\gamma _{1}=c_{2}d\gamma _{2}$,
$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {{\frac {k_{1}}{c_{1}M_{1}}}+{\frac {k_{2}}{c_{2}M_{2}}}}{k_{1}+k_{2}}}{\frac {d\gamma _{1}}{\gamma _{1}}}dV,$


$\phi {\frac {dx}{c_{1}k_{1}\gamma _{1}+c_{2}k_{2}\gamma _{2}}}=At{\frac {{\frac {k_{1}}{c_{1}M_{1}}}+{\frac {k_{2}}{c_{2}M_{2}}}}{k_{1}+k_{2}}}\log {\frac {\gamma _{1}''}{\gamma _{1}'}}+V'V''.$


The resistance depends on the concentration throughout the part of the cell considered, but the electromotive force depends only on the concentration at the terminal points (
$'$ and
$''$).
For $c_{1}M_{1}$ and $c_{2}M_{2}$ we may write ${\frac {v_{1}}{a_{\text{H}}}}$ and ${\frac {v_{2}}{a_{\text{H}}}}$, where $v_{1}$ and $v_{2}$ are the "valencies" of the molecules. This gives
$V''V'=a_{\text{H}}At{\frac {{\frac {k_{1}}{v_{1}}}{\frac {k_{1}}{v_{1}}}}{k_{1}+k_{2}}}\log {\frac {\gamma _{1}'}{\gamma _{1}''}},$for$\phi =0$(circuit open).


I think this is identical with your equation (
$V$) when your ions have the same valency.
Planck's problem is less simple.^{[20]} We may regard it as relating to a tube connecting the two great reservoirs filled with different electrolytes of same concentration, i.e., $\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}'=\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}''$. I use (_{0}) for an y ion, (_{1}) for any cation, (_{2}) for any anion. [The accents ($'$) and ($''$) refer to the two reservoirs.]
The tube is supposed to have reached a stationary state and dissociation is complete. The number of ions is immaterial, but they all must have the same valency $v$.
Now by equations (3) and (4), since $c_{0}M_{0}={\frac {v}{a_{\text{H}}}}$,
$\phi _{0}={\frac {aAt}{v}}k_{0}{\frac {d\gamma _{0}}{dx}}\mp k_{0}{\frac {dV}{dx}}\gamma _{0},$


or, writing
$T$ for the constant
${\frac {aAt}{v}},$
${\begin{aligned}\phi _{0}&=Tk_{0}{\frac {d\gamma _{0}}{dx}}\mp k_{0}\gamma _{0}{\frac {dV}{dx}},\\{\frac {c_{0}\phi _{0}}{k_{0}}}&=Tc_{0}{\frac {d\gamma _{0}}{dx}}\mp c_{0}\gamma _{0}{\frac {dV}{dx}},\\\textstyle \sum _{0}\displaystyle {\frac {c_{0}\phi _{0}}{k_{0}}}&=T\textstyle \sum _{0}\displaystyle {\frac {c_{0}d\gamma _{0}}{dx}}\end{aligned}}$


[The terms
$\mp {\frac {dV}{dx}}k_{0}\gamma _{0}$ disappear in the algebraic sum since
$\textstyle \sum \displaystyle c_{1}\gamma _{1}=\textstyle \sum \displaystyle c_{2}\gamma _{2}$. For a similar reason]
$\textstyle \sum _{0}\displaystyle \mp {\frac {c_{0}\phi _{0}}{k_{0}}}={\frac {dV}{dx}}\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}.$


The first equation makes
${\frac {\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}}{dx}}$ constant throughout the tube, and since
$\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}''=\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}',$ $\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}$ must be constant throughout the tube. The second equation then makes
${\frac {dV}{dx}}$ constant throughout the tube. Let
$X={\frac {dV}{dx}}\cdot$
Our original equation is
$\phi _{0}=Tk_{0}{\frac {d\gamma _{0}}{dx}}\pm Xk_{0}\gamma _{0}.$


Now with $X$ constant this is easily integrated.
${\begin{aligned}{\frac {\phi _{0}}{Xk}}&={\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}\pm \gamma _{0},\\\pm {\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}&=\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}},\\{\frac {d\gamma _{0}}{\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}}}&=\pm {\frac {X}{T}}dx,\\\log \left(\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}\right)&=\pm {\frac {X}{T}}x+\log H_{0},\\\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}&=H_{0}e^{\pm {\frac {X}{T}}x}.\end{aligned}}$


To determine
$H_{0}$ we have
$\gamma _{0}''\gamma _{0}'=H_{0}\left(e^{\mp {\frac {X}{T}}x''}e^{\pm {\frac {X}{T}}x'}\right).$


If we put the origin of coordinates in the middle of the tube we have
$x'=x''.$


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


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


The condition of no electrical current gives
$\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x}}{PP^{1}}}\cdot$


Apply to both ends and add,
${\begin{aligned}\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}''+\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}'&=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {e^{\pm {\frac {X}{T}}x''}+e^{\pm {\frac {X}{T}}x'}}{PP^{1}}},\\&=\textstyle \sum _{0}\displaystyle \pm c_{0}k_{0}\Delta _{0}{\frac {P+P^{1}}{PP^{1}}}\cdot \end{aligned}}$


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


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.
I regret that I have been obliged to delay my writing so long. I presume that you would have preferred to have me reply more promptly and more briefly. But the matter did not seem to be capable of being dispatched in few words.
One might easily economize in letters in the formulae by referring densities ($\gamma$) and potentials ($\mu$) to equivalent or molecular weights, as you have done, but I thought I was more sure to be understood with the notations which I have used. Moreover, since the molecular weight is often the doubtful point in the whole problem, there is a certain advantage in bringing it in explicitly rather than implicitly, so that we can see at a glance how a change in our assumptions in regard to the molecules will affect the measurable quantities.