# Scientific Papers of Josiah Willard Gibbs, Volume 1/Chapter VIII

VIII.
SEMI-PERMEABLE FILMS AND OSMOTIC PRESSURE.
[Nature, vol. lv, pp. 461, 462, Mar. 18, 1897.]
Lord Kelvin's very interesting problem concerning molecules which differ only in their power of passing a diaphragm (see Nature for January 21, p. 272), seems only to require for its solution the relation between density and pressure for the fluid at the temperature of the experiment, when this relation for small densities becomes that of an ideal gas; in other cases, a single numerical constant in addition to the relation between density and pressure is sufficient.
This will, perhaps, appear most readily if we imagine each of the vessels ${\text{A}}$ and ${\text{B}}$ connected with a vertical column of the fluid which it contains, these columns extending upwards until the state of an ideal gas is reached. The equilibrium which we suppose to subsist will not be disturbed by communications between the columns at as many levels as we choose, if these communications are always made through the same kind of semi-permeable diaphragm as that which separates the vessels ${\text{A}}$ and ${\text{B}}$. It will be observed that the difference of level at which any same pressure is found in the two columns is a constant quantity, easily determined in the upper parts (where the fluids are in the ideal gaseous state) as a function of the composition of the fluid in the A-column, and giving at once the height above the vessel ${\text{A}}$, where in the A-column we find a pressure equal to that in the vessel ${\text{B}}$.
In fact, we have in either column
* $dp=-g\gamma dz,$
* ${\frac {1}{\gamma }}={\text{F}}'(p)$
* ${\text{F}}'(p)dp=-gdz.$
* ${\text{F}}(p_{\text{A}})=-g(z-{\text{C}}_{\text{A}}),$
* ${\text{F}}(p_{\text{B}})=-g(z-{\text{C}}_{\text{B}}).$
* ${\text{F}}(p_{\text{A}})-{\text{F}}(p_{\text{B}})=g({\text{C}}_{\text{A}}-{\text{C}}_{\text{B}}).$
​In the upper regions,
* ${\text{F}}'(p)={\frac {1}{\gamma }}={\frac {at}{p}};$ $\therefore {\text{F}}'(p)=at\log p,$
$t$
$a$
* $at\log p_{\text{A}}-at\log p_{\text{B}}=g({\text{C}}_{\text{A}}-{\text{C}}_{\text{B}}).$
$1\,:\,n$
* $\gamma _{\text{A}}=(1+n)\gamma _{\text{B}},$$p_{\text{A}}=(1+n)p_{\text{B}},$ $g({\text{C}}_{\text{A}}-{\text{C}}_{\text{B}})=at\log(1+n).$
* ${\text{F}}(p_{\text{A}})-{\text{F}}(p_{\text{B}})=at\log(1+n).$
${\text{A}}$
${\text{B}}$
${\text{A}}$
$n$
* ${\text{F}}'(p_{\text{A}})(p_{\text{A}}-p_{\text{B}})=atn$
* $p_{\text{A}}-p_{\text{B}}=atn\gamma _{\text{A}}.$
${\text{A}}$
To show that the case is substantially the same, at least for any one temperature, when the fluid is not volatile, we may suppose that we have many kinds of molecules, ${\text{A, B, C}}$, etc., which are identical in all properties except in regard to passing diaphragms. Let us imagine a row of vertical cylinders or tubes closed at both ends. Let the first contain A-molecules sufficient to give the pressure $p'$ at a certain level. Then let it be connected with the second cylinder through a diaphragm impermeable to B-molecules, freely permeable to all others. Let the second cylinder contain such quantities of A- and B-molecules as to be in equilibrium with the first cylinder, and to have a certain pressure $p''$ at the level of $p'$ in the first cylinder. At a higher level this second cylinder will have the pressure which we have called $p'$. There let it be connected with the third cylinder through a diaphragm impermeable to C-molecules, and to them alone. Let this third cylinder contain such quantities of A-, B-, and C-molecules as to be in equilibrium with the second cylinder, and have the pressure $p''$ at the diaphragm; and so on, the connections being so made, and the ​quantities of the several kinds of molecules so regulated, that the pressures at all the diaphragms shall have the same two values.
It is evident that the vertical distance between successive connections must be everywhere the same, say $l$; also, that at all the diaphragms, on the side of the greater pressure, the proportion of molecules which can and which cannot pass the diaphragm must be the same. Let the ratio be $1\,:\,n$. If we write $\gamma _{\text{A}},\gamma _{\text{B}}$, etc., for the densities of the several kinds of molecules, and $\gamma$ for the total density, we have for the second cylinder
* ${\frac {\gamma _{\text{A}}+\gamma _{\text{B}}}{\gamma _{\text{A}}}}=1+n.$
* ${\frac {\gamma _{\text{A}}+\gamma _{\text{B}}+\gamma _{\text{C}}}{\gamma _{\text{A}}+\gamma _{\text{B}}}}=1+n,$
* ${\frac {\gamma _{\text{A}}+\gamma _{\text{B}}+\gamma _{\text{C}}}{\gamma _{\text{A}}}}=(1+n)^{2}.$
* ${\frac {\gamma }{\gamma _{\text{A}}}}=(1+n)^{r-1}.$
* $(r-1)l.$
$\gamma /\gamma _{\text{A}}$
$\log(\gamma /\gamma _{\text{A}})$
$at/g$
The following demonstration of van't Hoff's law, which is intended to apply to existing substances, requires only that the solutum, i.e., dissolved substance, should be capable of the ideal gaseous state, and that its molecules, as they occur in the gas, should not be broken up in the solution, nor united to one another in more complex molecules.
It will be convenient to use certain quantities which may be called the potentials of the solvent and of the solutum, the term being thus defined: In any sensibly homogeneous mass, the potential of any independently variable component substance is the differential coefficient of the thermodynamic energy of the mass taken with respect ​to that component, the entropy and volume of the mass and the quantities of its other components remaining constant. The advantage of using such potentials in the theory of semi-permeable diaphragms consists pirtly in the convenient form of the conditions of equilibrium, the potential for any substance to which a diaphragm is freely permeable having the same value on both sides of the diaphragm, and partly in our ability to express van't Hoff's law as a relation between the quantities characterising the state of the solution, without reference to any experimental arrangement (see Transactions of the Connecticut Academy, vol. iii, pp. 116, 138, 148, 194) [this vol., pp. 63, 83, 92, 135].
Let there be three reservoirs, ${\text{R}}',{\text{R}}'',{\text{R}}'''$, of which the first contains the solvent alone, maintained in a constant state of temperature and pressure, the second the solution, and the third the solutum alone. Let ${\text{R}}'$ and ${\text{R}}''$ be connected through a diaphragm freely permeable to the solvent, but impermeable to the solutum, and let ${\text{R}}''$ and ${\text{R}}'''$ be connected through a diaphragm impermeable to the solvent, but freely permeable to the solutum. We have then, if we write $\mu _{1}$ and $\mu _{2}$ for the potentials of the solvent and the solutum, and distinguish by accents quantities relating to the several reservoirs,
* $\mu _{1}''=\mu _{1}'={\text{const.,}}$$\mu _{2}''=\mu _{2}'''.$
${\text{R}}''$
${\text{R}}'''$
$\gamma _{2}'',\gamma _{2}'''$
${\text{R}}''$
${\text{R}}'''$
${\text{R}}''$
${\text{R}}'''$
$1:{\text{B}}$
${\text{R}}'$
${\text{R}}''$
$1:{\text{B}}$
${\text{R}}''$
${\text{R}}'''$
* $\gamma _{2}'''={\text{B}}\gamma _{2}''.$
${\text{R}}'''$
* $\mu _{2}'''=a_{2}t\log \gamma _{2}'''+{\text{C}},$
$a_{2}$
${\text{C}}$
* $\mu _{2}''=a_{2}t\log({\text{B}}\gamma _{2}'')+{\text{C}}.$
${\text{B}}$
${\text{C}}$
* $vdp=\eta dt+m_{1}d\mu _{1}+m_{2}d\mu _{2},$
$v$
$\eta$
$m_{1}$
$m_{2}$
$t$
$\mu _{1}$
* $dp''=\gamma _{2}''d\mu _{2}''.$
$d\mu _{2}''$
* $dp''=a_{2}td\gamma _{2}'',$
$\gamma _{2}''=0$
$p''=p'$
* $p''-p'=a_{2}t\gamma _{2}'',$
We may extend this proof to cases in which the solutum is not volatile by supposing that we give to its molecules mutually repulsive molecular forces, which, however, are entirely inoperative with respect to any other kind of molecules. In this way we may make the solutum capable of the ideal gaseous state. But the relations pertaining to the contents of ${\text{R}}''$ will not be affected by these new forces, since we suppose that only a negligible part of the molecules of the solutum are within the range of such forces. Therefore these relations cannot depend on the new forces, and must exist without them.
To give up the condition that the molecules of the solutum shall not be broken up in the solution, nor united to one another in more complex molecules, would involve the consideration of a good many cases, which it would be difficult to unite in a brief demonstration. The result, however, seems to be that the increase of pressure is to be estimated by Avogadro's law from the number of molecules in the solution which contain any part of the solutum, without reference to the quantity in each.
J. Willard Gibbs.
New Haven, Connecticut, February 18.