Eight Lectures on Theoretical Physics/II

In the lecture of yesterday I sought to make clear the fact that the essential, and therefore the final division of all processes occurring in nature, is into reversible and irreversible processes, and the characteristic difference between these two kinds of processes, as I have further separated them, is that in irreversible processes the entropy increases, while in all reversible processes it remains constant. Today I am constrained to speak of some of the consequences of this law which will illustrate its rich fruitfulness. They have to do with the question of the laws of thermodynamic equilibrium. Since in nature the entropy can only increase, it follows that the state of a physical configuration which is completely isolated, and in which the entropy of the system possesses an absolute maximum, is necessarily a state of stable equilibrium, since for it no further change is possible. How deeply this law underlies all physical and chemical relations has been shown by no one better and more completely than by John Willard Gibbs, whose name, not only in America, but in the whole world will be counted among those of the most famous theoretical physicists of all times; to whom, to my sorrow, it is no longer possible for me to tender personally my respects. It would be gratuitous for me, here in the land of his activity, to expatiate fully on the progress of his ideas, but you will perhaps permit me to speak in the lecture of today of some of the important applications in which thermodynamic research, based on Gibbs works, can be advanced beyond his results.

These applications refer to the theory of dilute solutions, and we shall occupy ourselves today with these, while I show you by a definite example what fruitfulness is inherent in thermodynamic theory. I shall first characterize the problem quite generally. It has to do with the state of equilibrium of a material system of any number of arbitrary constituents in an arbitrary number of phases, at a given temperature $T$ and given pressure $p$. If the system is completely isolated, and therefore guarded against all external thermal and mechanical actions, then in any ensuing change the entropy of the system will increase:

$\begin{array}{cccc}& .(00)& & dS>0.\end{array}$

But if, as we assume, the system stands in such relation to its surroundings that in any change which the system undergoes the temperature $T$ and the pressure $p$ are maintained constant, as, for instance, through its introduction into a calorimeter of great heat capacity and through loading with a piston of fixed weight, the inequality would suffer a change thereby. We must then take account of the fact that the surrounding bodies also, e. g., the calorimetric liquid, will be involved in the change. If we denote the entropy of the surrounding bodies by $S_0$, then the following more general equation holds:

$\begin{array}{cccc}& .(00)& & dS+d{S}_0>0.\end{array}$

In this equation

$\begin{array}{cccc}& .(00)& & d{S}_0=-\frac{Q}{T},\end{array}$

if $Q$ denote the heat which is given up in the change by the surroundings to the system. On the other hand, if $U$ denote the energy, $V$ the volume of the system, then, in accordance with the first law of thermodynamics,

$\begin{array}{cccc}& .(00)& & Q=dU+pdV.\end{array}$

Consequently, through substitution:

$\begin{array}{cccc}& .(00)& & dS-\frac{dU+pdV}{T}>0\end{array}$

or, since $p$ and $T$ are constant:

$\begin{array}{cccc}& .(00)& & d(S-\frac{U+pV}{T})>0.\end{array}$

If, therefore, we put:

$\begin{array}{cccc}& (1).0& & S-\frac{U+pV}{T}=\mathrm{\Phi },\end{array}$

then

$\begin{array}{cccc}& .(00)& & d\mathrm{\Phi }>0,\end{array}$

and we have the general law, that in every isothermal-isobaric ($T=\text{const.}$, $p=\text{const.}$) change of state of a physical system the quantity $\mathrm{\Phi }$ increases. The absolutely stable state of equilibrium of the system is therefore characterized through the maximum of $\mathrm{\Phi }$:

$\begin{array}{cccc}& (2).0& & \delta \mathrm{\Phi }=0.\end{array}$

If the system consist of numerous phases, then, because $\mathrm{\Phi }$, in accordance with $(1)$, is linear and homogeneous in $S$, $U$ and $V$, the quantity $\mathrm{\Phi }$ referring to the whole system is the sum of the quantities $\mathrm{\Phi }$ referring to the individual phases. If the expression for $\mathrm{\Phi }$ is known as a function of the independent variables for each phase of the system, then, from equation $(2)$, all questions concerning the conditions of stable equilibrium may be answered. Now, within limits, this is the case for dilute solutions. By “solution” in thermodynamics is meant each homogeneous phase, in whatever state of aggregation, which is composed of a series of different molecular complexes, each of which is represented by a definite molecular number. If the molecular number of a given complex is great with reference to all the remaining complexes, then the solution is called dilute, and the molecular complex in question is called the solvent; the remaining complexes are called the dissolved substances.

Let us now consider a dilute solution whose state is determined by the pressure $p$, the temperature $T$, and the molecular numbers $n_0$, $n_1$, $n_2$, $n_3$, $\cdots$, wherein the subscript zero refers to the solvent. Then the numbers $n_1$, $n_2$, $n_3$, $\cdots$ are all small with respect to $n_0$, and on this account the volume $V$ and the energy $U$ are linear functions of the molecular numbers:

$\begin{array}{cccc}& .(00)& V& =n_0{v}_0+n_1{v}_1+n_2{v}_2+\cdots ,\\ & & U& =n_0{u}_0+n_1{u}_1+n_2{u}_2+\cdots ,\end{array}$

wherein the $v$'s and $u$'s depend upon $p$ and $T$ only.

From the general equation of entropy:

$\begin{array}{cccc}& .(00)& & dS=\frac{dU+pdV}{T},\end{array}$

in which the differentials depend only upon changes in $p$ and $T$, and not in the molecular numbers, there results therefore:

$\begin{array}{cccc}& .(00)& & dS=n_0\frac{d{u}_0+pd{v}_0}{T}+n_1\frac{d{u}_1+pd{v}_1}{T}+\cdots ,\end{array}$

and from this it follows that the expressions multiplied by $n_0$, $n_1$ $\cdots$, dependent upon $p$ and $T$ only, are complete differentials. We may therefore write:

$\begin{array}{cccc}& (3).0& & \frac{d{u}_0+pd{v}_0}{T}=d{s}_0,\frac{d{u}_1+pd{v}_1}{T}=d{s}_1,\cdots \end{array}$

and by integration obtain:

$\begin{array}{cccc}& .(00)& & S=n_0{s}_0+n_1{s}_1+n_2{s}_2+\cdots +C.\end{array}$

The constant $C$ of integration does not depend upon $p$ and $T$, but may depend upon the molecular numbers $n_0$, $n_1$, $n_2$, $\cdots$. In order to express this dependence generally, it suffices to know it for a special case, for fixed values of $p$ and $T$. Now every solution passes, through appropriate increase of temperature and decrease of pressure, into the state of a mixture of ideal gases, and for this case the entropy is fully known, the integration constant being, in accordance with Gibbs:

$\begin{array}{cccc}& .(00)& & C=-R(n_0\log c_0+n_1\log c_1+\cdots ),\end{array}$

wherein $R$ denotes the absolute gas constant and $c_0$, $c_1$, $c_2$, $\cdots$ denote the “molecular concentrations”:

$\begin{array}{cccc}& .(00)& & c_0=\frac{n_0}{n_0+n_1+n_2+\cdots },c_1=\frac{n_1}{n_0+n_1+n_2+\cdots },\cdots .\end{array}$

Consequently, quite in general, the entropy of a dilute solution is:

$\begin{array}{cccc}& .(00)& & S=n_0(s_0-R\log c_0)+n_1(s_1-R\log c_1)+\cdots ,\end{array}$

and, finally, from this it follows by substitution in equation $(1)$ that:

$\begin{array}{cccc}& (4).0& & \mathrm{\Phi }=n_0({\phi }_0-R\log c_0)+n_1({\phi }_1-R\log c_1)+\cdots ,\end{array}$

if we put for brevity:

$\begin{array}{cccc}& (5).0& & {\phi }_0=s_0-\frac{u_0+p{v}_0}{T},{\phi }_1=s_1-\frac{u_1+p{v}_1}{T},\cdots \end{array}$

all of which quantities depend only upon $p$ and $T$.

With the aid of the expression obtained for $\mathrm{\Phi }$ we are enabled through equation $(2)$ to answer the question with regard to thermodynamic equilibrium. We shall first find the general law of equilibrium and then apply it to a series of particularly interesting special cases.

Every material system consisting of an arbitrary number of homogeneous phases may be represented symbolically in the following way:

$\begin{array}{cccc}& .(00)& & n_0{m}_0,n_1{m}_1,\cdots \mid n_{0^{\prime }}{m}_{0^{\prime }},n_{1^{\prime }}{m}_{1^{\prime }},\cdots \mid n_{0^{″}}{m}_{0^{″}},n_{1^{″}}{m}_{1^{″}},\cdots \mid \cdots .\end{array}$

Here the molecular numbers are denoted by $n$, the molecular weights by $m$, and the individual phases are separated from one another by vertical lines. We shall now suppose that each phase represents a dilute solution. This will be the case when each phase contains only a single molecular complex and therefore represents an absolutely pure substance; for then the concentrations of all the dissolved substances will be zero.

If now an isobaric-isothermal change in the system of such kind is possible that the molecular numbers

$\begin{array}{cccc}& .(00)& & n_0,n_1,n_2,\cdots ,n_{0^{\prime }},n_{1^{\prime }},n_{2^{\prime }},\cdots ,n_{0^{″}},n_{1^{″}},n_{2^{″}},\cdots \end{array}$

change simultaneously by the amounts

$\begin{array}{cccc}& .(00)& & \delta n_0,\delta n_1,\delta n_2,\cdots ,\delta n_{0^{\prime }},\delta n_{1^{\prime }},\delta n_{2^{\prime }},\cdots ,\delta n_{0^{″}},\delta n_{1^{″}},\delta n_{2^{″}},\cdots \end{array}$

then, in accordance with equation $(2)$, equilibrium obtains with respect to the occurrence of this change if, when $T$ and $p$ are held constant, the function

$\begin{array}{cccc}& .(00)& & \mathrm{\Phi }+{\mathrm{\Phi }}^{\prime }+{\mathrm{\Phi }}^{″}+\cdots \end{array}$

is a maximum, or, in accordance with equation $(4)$:

$\begin{array}{cccc}& .(00)& & \sum ({\phi }_0-R\log c_0)\delta n_0+({\phi }_1-R\log c_1)\delta n_1+\cdots =0\end{array}$

(the summation $\sum$ being extended over all phases of the system). Since we are only concerned in this equation with the ratios of the $\delta n$'s, we put

$\begin{array}{cccc}& .(00)& & \delta n_0:\delta n_1:\cdots :\delta n_{0^{\prime }}:\delta n_{1^{\prime }}:\cdots :\delta n_{0^{″}}:\delta n_{1^{″}}:\cdots \\ & & & ={\nu }_0:{\nu }_1:\cdots :{\nu }_{0^{\prime }}:{\nu }_{1^{\prime }}:\cdots :{\nu }_{0^{″}}:{\nu }_{1^{″}}:\cdots ,\end{array}$

wherein we are to understand by the simultaneously changing $\nu$'s, in the variation considered, simple integer positive or negative numbers, according as the molecular complex under consideration is formed or disappears in the change. Then the condition for equilibrium is: Template:Anchor

$\begin{array}{cccc}& (6).0& & \sum {\nu }_0\log c_0+{\nu }_1\log c_1+\cdots =\frac{1}{R}\sum {\nu }_0{\phi }_0+{\nu }_1{\phi }_1+\cdots =\log K.\end{array}$

$K$ and the quantities ${\phi }_0$, ${\phi }_1$, ${\phi }_2$, $\cdots$ depend only upon $p$ and $T$, and this dependence is to be found from the equations:

$\begin{array}{cccc}& .(00)& \frac{\partial \log K}{\partial p}& =\frac{1}{R}\sum {\nu }_0\frac{\partial {\phi }_0}{\partial p}+{\nu }_1\frac{\partial {\phi }_1}{\partial p}+\cdots ,\\ & & \frac{\partial \log K}{\partial T}& =\frac{1}{R}\sum {\nu }_0\frac{\partial {\phi }_0}{\partial T}+{\nu }_1\frac{\partial {\phi }_1}{\partial T}+\cdots .\end{array}$

Now, in accordance with $(5)$, for any infinitely small change of $p$ and $T$:

$\begin{array}{cccc}& .(00)& & d{\phi }_0=d{s}_0-\frac{d{u}_0+pd{v}_0+v_{0}dp}{T}+\frac{u_0+p{v}_0}{T^2}\cdot dT,\end{array}$

and consequently, from $(3)$:

$\begin{array}{cccc}& .(00)& & d{\phi }_0=\frac{u_0+p{v}_0}{T^2}dT-\frac{v_{0}dp}{T},\end{array}$

and hence:

$\begin{array}{cccc}& .(00)& & \frac{\partial {\phi }_0}{\partial p}=-\frac{v_0}{T},\frac{\partial {\phi }_0}{\partial T}=\frac{u_0+p{v}_0}{T^2}.\end{array}$

Similar equations hold for the other $\phi$'s, and therefore we get:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log K}{\partial p}=-\frac{1}{RT}\sum {\nu }_0{v}_0+{\nu }_1{v}_1+\cdots ,\\ & & & \frac{\partial \log K}{\partial T}=-\frac{1}{R{T}^2}\sum {\nu }_0{u}_0+{\nu }_2{u}_2+\cdots +p({\nu }_0{v}_0+{\nu }_1{v}_1+\cdots )\end{array}$

or, more briefly:

$\begin{array}{cccc}& (7).0& & \frac{\partial \log K}{\partial p}=-\frac{1}{RT}\cdot \mathrm{\Delta }V,\frac{\partial \log K}{\partial T}=\frac{\mathrm{\Delta }Q}{R{T}^2},\end{array}$

if $\mathrm{\Delta }V$ denote the change in the total volume of the system and $\mathrm{\Delta }Q$ the heat which is communicated to it from outside, during the isobaric isothermal change considered. We shall now investigate the import of these relations in a series of important applications.

Template:C

The system consists of a single phase:

$\begin{array}{cccc}& .(00)& & n_0{H}_{2}O,n_1\stackrel{+}{H},n_2\stackrel{-}{HO}.\end{array}$

The transformation under consideration

$\begin{array}{cccc}& .(00)& & {\nu }_0:{\nu }_1:{\nu }_2=\delta n_0:\delta n_1:\delta n_2\end{array}$

consists in the dissociation of a molecule $H_{2}O$ into a molecule $\stackrel{+}{H}$ and a molecule $\stackrel{-}{HO}$, therefore:

$\begin{array}{cccc}& .(00)& & {\nu }_0=-1,{\nu }_1=1,{\nu }_2=1.\end{array}$

Hence, in accordance with $(6)$, for equilibrium:

$\begin{array}{cccc}& .(00)& & -\log c_0+\log c_1+\log c_2=\log K,\end{array}$

or, since $c_1=c_2$ and $c_0=1$, approximately:

$\begin{array}{cccc}& .(00)& & 2\log c_1=\log K.\end{array}$

The dependence of the concentration $c_1$ upon the temperature now follows from $(7)$:

$\begin{array}{cccc}& .(00)& & 2\frac{\partial \log c_1}{\partial T}=\frac{\mathrm{\Delta }Q}{R{T}^2}.\end{array}$

$\mathrm{\Delta }Q$, the quantity of heat which it is necessary to supply for the dissociation of a molecule of $H_{2}O$ into the ions $\stackrel{+}{H}$ and $\stackrel{-}{HO}$, is, in accordance with Arrhenius, equal to the heat of ionization in the neutralization of a strong univalent base and acid in a dilute aqueous solution, and, therefore, in accordance with the recent measurements of Wörmann,

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }Q=27,857-48.5T\text{gr}.\text{cal}.\end{array}$

Using the number $1.985$ for the ratio of the absolute gas constant $R$ to the mechanical equivalent of heat, it follows that:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log c_1}{\partial T}=\frac{1}{2\cdot 1.985}(\frac{27,857}{T^2}-\frac{48.5}{T}),\end{array}$

and by integration:

$\begin{array}{cccc}& .(00)& & \stackrel{10}{\log }{c}_1=-\frac{3047.3}{T}-12.125\stackrel{10}{\log }T+\text{const.}\end{array}$

This dependence of the degree of dissociation upon the temperature agrees very well with the measurements of the electric conductivity of water at different temperatures by Kohlrausch and Heydweiller, Noyes, and Lundén.

Template:C

Let the system consists of an aqueous solution of acetic acid:

$\begin{array}{cccc}& .(00)& & n_0{H}_{2}O,n_1{H}_4{C}_2{O}_2,n_2\stackrel{+}{H},n_3\stackrel{-}{H_3{C}_2{O}_2}.\end{array}$

The change under consideration consists in the dissociation of a molecule $H_4{C}_2{O}_2$ into its two ions, therefore

$\begin{array}{cccc}& .(00)& & {\nu }_0=0,{\nu }_1=-1,{\nu }_2=1,{\nu }_3=1.\end{array}$

Hence, for the state of equilibrium, in accordance with $(6)$:

$\begin{array}{cccc}& .(00)& & -\log c_1+\log c_2+\log c_3=\log K,\end{array}$

or, since $c_2=c_3$:

$\begin{array}{cccc}& .(00)& & \frac{{c_2}^2}{c_1}=K.\end{array}$

Now the sum $c_1+c_2=c$ is to be regarded as known, since the total number of the undissociated and dissociated acid molecules is independent of the degree of dissociation. Therefore $c_1$ and $c_2$ may be calculated from $K$ and $c$. An experimental test of the equation of equilibrium is possible on account of the connection between the degree of dissociation and electrical conductivity of the solution. In accordance with the electrolytic dissociation theory of Arrhenius, the ratio of the molecular conductivity $\lambda$ of the solution in any dilution to the molecular conductivity ${\lambda }_{\mathrm{\infty }}$ of the solution in infinite dilution is:

$\begin{array}{cccc}& .(00)& & \frac{\lambda }{{\lambda }_{\mathrm{\infty }}}=\frac{c_2}{c_1+c_2}=\frac{c_2}{c},\end{array}$

since electric conduction is accounted for by the dissociated molecules only. It follows then, with the aid of the last equation, that:

$\begin{array}{cccc}& .(00)& & \frac{{\lambda }^{2}c}{{\lambda }_{\mathrm{\infty }}-\lambda }=K\cdot {\lambda }_{\mathrm{\infty }}=\text{const.}\end{array}$

With unlimited decreasing $c$, $\lambda$ increases to ${\lambda }_{\mathrm{\infty }}$. This “law of dilution” for binary electrolytes, first enunciated by Ostwald, has been confirmed in numerous cases by experiment, as in the case of acetic acid.

Also, the dependence of the degree of dissociation upon the temperature is indicated here in quite an analogous manner to that in the example considered above, of the dissociation of water.

Template:C

In equilibrium the system consists of two phases, one liquid, and one gaseous or solid:

$\begin{array}{cccc}& .(00)& & n_0{m}_0\mid n_{0^{\prime }}{m}_{0^{\prime }}.\end{array}$

Each phase contains only a single molecular complex (the solvent), but the molecules in both phases do not need to be the same. Now, if a liquid molecule evaporates or solidifies, then in our notation

$\begin{array}{cccc}& .(00)& & {\nu }_0=-1,{\nu }_{0^{\prime }}=\frac{m_0}{m_{0^{\prime }}},c_0=1,c_{0^{\prime }}=1,\end{array}$

and consequently the condition for equilibrium, in accordance with $(6)$, is:

$\begin{array}{cccc}& (8).0& & 0=\log K.\end{array}$

Since $K$ depends only upon $p$ and $T$, this equation therefore expresses a definite relation between $p$ and $T$: the law of dependence of the pressure of vaporization (or melting pressure) upon the temperature, or vice versa. The import of this law is obtained through the consideration of the dependence of the quantity $K$ upon $p$ and $T$. If we form the complete differential of the last equation, there results:

$\begin{array}{cccc}& .(00)& & 0=\frac{\partial \log K}{\partial p}dp+\frac{\partial \log K}{\partial T}dT,\end{array}$

or, in accordance with $(7)$:

$\begin{array}{cccc}& .(00)& & 0=-\frac{\mathrm{\Delta }V}{T}dp+\frac{\mathrm{\Delta }Q}{T^2}dT.\end{array}$

If $v_0$ and $v_0\text{'}$ denote the molecular volumes of the two phases, then:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }V=\frac{m_0{v}_{0^{\prime }}}{m_{0^{\prime }}}-v_0,\end{array}$

consequently:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }Q=T(\frac{m_0{v}_{0^{\prime }}}{m_{0^{\prime }}}-v_0)\frac{dp}{dT},\end{array}$

or, referred to unit mass:

$\begin{array}{cccc}& .(00)& & \frac{\mathrm{\Delta }Q}{m_0}=T(\frac{v_{0^{\prime }}}{m_{0^{\prime }}}-\frac{v_0}{m_0})\cdot \frac{dp}{dT},\end{array}$

the well-known formula of Carnot and Clapeyron.

Template:C

Most aqueous salt solutions afford examples. The symbol of the system in this case is, since the second phase (gaseous or solid) contains only a single molecular complex:

$\begin{array}{cccc}& .(00)& & n_0{m}_0,n_1{m}_1,n_2{m}_2,\cdots \mid n_{0^{\prime }}{m}_{0^{\prime }}.\end{array}$

The change is represented by:

$\begin{array}{cccc}& .(00)& & {\nu }_0=-1,{\nu }_1=0,{\nu }_2=0,\cdots {\nu }_{0^{\prime }}=\frac{m_0}{m_{0^{\prime }}},\end{array}$

and hence the condition of equilibrium, in accordance with $(6)$, is:

$\begin{array}{cccc}& .(00)& & -\log c_0=\log K,\end{array}$

or, since to small quantities of higher order:

$\begin{array}{cccc}& .(00)& c_0=\frac{n_0}{n_0+n_1+n_2+\cdots }& =1-\frac{n_1+n_2+\cdots }{n_0},\\ & (9)& \frac{n_1+n_2+\cdots }{n_0}& =\log K.\end{array}$

A comparison with formula $(8)$, found in example III, shows that through the solution of a foreign substance there is involved in the total concentration a small proportionate departure from the law of vaporization or solidification which holds for the pure solvent. One can express this, either by saying: at a fixed pressure $p$, the boiling point or the freezing point $T$ of the solution is different than that ($T_0$) for the pure solvent, or: Template:Anchorat a fixed temperature $T$ the vapor pressure or solidification pressure $p$ of the solution is different from that ($p_0$) of the pure solvent. Let us calculate the departure in both cases.

1. If $T_0$ be the boiling (or freezing temperature) of the pure solvent at the pressure $p$, then, in accordance with $(8)$:

$\begin{array}{cccc}& .(00)& & (\log K{)}_{T=T_0}=0,\end{array}$

and by subtraction of $(9)$ there results:

$\begin{array}{cccc}& .(00)& & \log K-(\log K{)}_{T=T_0}=\frac{n_1+n_2+\cdots }{n_0}.\end{array}$

Now, since $T$ is little different from $T_0$, we may write in place of this equation, with the aid of $(7)$:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log K}{\partial T}(T-T_0)=\frac{\mathrm{\Delta }Q}{R{T}_0^2}(T-T_0)=\frac{n_1+n_2+\cdots }{n_0},\end{array}$

and from this it follows that:

$\begin{array}{cccc}& (10).& & T-T_0=\frac{n_1+n_2+\cdots }{n_0}\cdot \frac{R{T}_0^2}{\mathrm{\Delta }Q}.\end{array}$

This is the law for the raising of the boiling point or for the lowering of the freezing point, first derived by van't Hoff: in the case of freezing $\mathrm{\Delta }Q$ (the heat taken from the surroundings during the freezing of a liquid molecule) is negative. Since $n_0$ and $\mathrm{\Delta }Q$ occur only as a product, it is not possible to infer anything from this formula with regard to the molecular number of the liquid solvent.

2. If $p_0$ be the vapor pressure of the pure solvent at the temperature $T$, then, in accordance with $(8)$:

$\begin{array}{cccc}& .(00)& & (\log K{)}_{p=p_0}=0,\end{array}$

and by subtraction of $(9)$ there results:

$\begin{array}{cccc}& .(00)& & \log K-(\log K{)}_{p=p_0}=\frac{n_1+n_2+\cdots }{n_0}.\end{array}$

Now, since $p$ and $p_0$ are nearly equal, with the aid of $(7)$ we may write:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log K}{\partial p}(p-p_0)=-\frac{\mathrm{\Delta }V}{RT}(p-p_0)=\frac{n_1+n_2+\cdots }{n_0},\end{array}$

and from this it follows, if $\mathrm{\Delta }V$ be placed equal to the volume of the gaseous molecule produced in the vaporization of a liquid molecule:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }V=\frac{m_0}{m_{0^{\prime }}}\frac{RT}{p},\\ & & & \frac{p_0-p}{p}=\frac{m_{0^{\prime }}}{m_0}\cdot \frac{n_1+n_2+\cdots }{n_0}.\end{array}$

This is the law of relative depression of the vapor pressure, first derived by van't Hoff. Since $n_0$ and $m_0$ occur only as a product, it is not possible to infer from this formula anything with regard to the molecular weight of the liquid solvent. Frequently the factor $m_0\text{'}/m_0$ is left out in this formula; but this is not allowable when $m_0$ and $m_0\text{'}$ are unequal (as, e. g., in the case of water).

Template:C

The system, consisting of two phases, is represented by the following symbol:

$\begin{array}{cccc}& .(00)& & n_0{m}_0,n_1{m}_1,n_2{m}_2,\cdots \mid n_{0^{\prime }}{m}_{0^{\prime }},n_{1^{\prime }}{m}_{1^{\prime }},n_{2^{\prime }}{m}_{2^{\prime }},\cdots ,\end{array}$

wherein, as above, the figure $0$ refers to the solvent and the figures $1$, $2$, $3$ $\cdots$ refer to the various molecular complexes of the dissolved substances. By the addition of primes in the case of the molecular weights ($m_0\text{'}$, $m_1\text{'}$, $m_2\text{'}$ $\cdots$) the possibility is left open that the various molecular complexes in the vapor may possess a different molecular weight than in the liquid.

Since the system here considered may experience various sorts of changes, there are also various conditions of equilibrium to fulfill, each of which relates to a definite sort of transformation. Let us consider first that change which consists in the vaporization of the solvent. In accordance with our scheme of notation, the following conditions hold:Template:Anchor

$\begin{array}{cccc}& .(00)& & {\nu }_0=-1,{\nu }_1=0,{\nu }_2=0,\cdots {\nu }_{0^{\prime }}=\frac{m_0}{m_{0^{\prime }}},{\nu }_{1^{\prime }}=0,{\nu }_{2^{\prime }}=0,\cdots ,\end{array}$

and, therefore, the condition of equilibrium $(6)$ becomes:

$\begin{array}{cccc}& .(00)& & -\log c_0+\frac{m_0}{m_{0^{\prime }}}\log c_{0^{\prime }}=\log K,\end{array}$

or, if one substitutes:

$\begin{array}{cccc}& .(00)& & c_0=1-\frac{n_1+n_2+\cdots }{n_0}\text{and}{c}_{0^{\prime }}=1-\frac{n_{1^{\prime }}+n_{2^{\prime }}+\cdots }{n_{0^{\prime }}},\\ & & & \frac{n_1+n_2+\cdots }{n_0}-\frac{m_0}{m_{0^{\prime }}}\cdot \frac{n_{1^{\prime }}+n_{2^{\prime }}+\cdots }{n_{0^{\prime }}}=\log K.\end{array}$

If we treat this equation upon equation $(9)$ as a model, there results an equation similar to $(10)$:

$\begin{array}{cccc}& .(00)& & T-T_0=(\frac{n_1+n_2+\cdots }{n_0{m}_0}-\frac{n_{1^{\prime }}+n_{2^{\prime }}+\cdots }{n_{0^{\prime }}{m}_{0^{\prime }}})\frac{R{T}_0^2{m}_0}{\mathrm{\Delta }Q}.\end{array}$

Here $\mathrm{\Delta }Q$ is the heat effect in the vaporization of one molecule of the solvent and, therefore, $\mathrm{\Delta }Q/m_0$ is the heat effect in the vaporization of a unit mass of the solvent.

We remark, once more, that the solvent always occurs in the formula through the mass only, and not through the molecular number or the molecular weight, while, on the other hand, in the case of the dissolved substances, the molecular state is characteristic on account of their influence upon vaporization. Finally, the formula contains a generalization of the law of van't Hoff, stated above, for the raising of the boiling point, in that here in place of the number of dissolved molecules in the liquid, the difference between the number of dissolved molecules in unit mass of the liquid and in unit mass of the vapor appears. According as the unit mass of liquid or the unit mass of vapor contains more dissolved molecules, there results for the solution a raising or lowering of the boiling point; in the limiting case, when both quantities are equal, and the mixture therefore boils without changing, the change in boiling point becomes equal to zero. Of course, there are corresponding laws holding for the change in the vapor pressure.

Let us consider now a change which consists in the vaporization of a dissolved molecule. For this case we have in our notation

$\begin{array}{cccc}& .(00)& & {\nu }_0=0,{\nu }_1=-1,{\nu }_2=0\cdots ,{\nu }_{0^{\prime }}=0,{\nu }_{1^{\prime }}=\frac{m_1}{m_{1^{\prime }}},{\nu }_{2^{\prime }}=0,\cdots \end{array}$

and, in accordance with $(6)$, for the condition of equilibrium:

$\begin{array}{cccc}& .(00)& & -\log c_1+\frac{m_1}{m_{1^{\prime }}}\log c_{1^{\prime }}=\log K\end{array}$

or:

$\begin{array}{cccc}& .(00)& & \frac{{c_{1^{\prime }}}^{\frac{m_1}{m_{1^{\prime }}}}}{c_1}=K.\end{array}$

This equation expresses the Nernst law of distribution. If the dissolved substance possesses in both phases the same molecular weight ($m_1=m_{1^{\prime }}$), then, in a state of equilibrium a fixed ratio of the concentrations $c_1$ and $c_1\text{'}$ in the liquid and in the vapor exists, which depends only upon the pressure and temperature. But, if the dissolved substance polymerises somewhat in the liquid, then the relation demanded in the last equation appears in place of the simple ratio.

Template:C

This case is in a certain sense a special case of the one preceding. To it belongs that of the solubility of a slightly soluble salt, first investigated by van't Hoff, e. g., succinic acid in water. The symbol of this system is:

$\begin{array}{cccc}& .(00)& & n_0{H}_{2}O,n_1{H}_6{C}_4{O}_4\mid n_{0^{\prime }}{H}_6{C}_4{O}_4,\end{array}$

in which we disregard the small dissociation of the acid solution. The concentrations of the individual molecular complexes are:

$\begin{array}{cccc}& .(00)& & c_0=\frac{n_0}{n_0+n_1},c_1=\frac{n_1}{n_0+n_1},c_{0^{\prime }}=\frac{n_{0^{\prime }}}{n_{0^{\prime }}}=1.\end{array}$

For the precipitation of solid succinic acid we have:

$\begin{array}{cccc}& .(00)& & {\nu }_0=0,{\nu }_1=-1,{\nu }_{0^{\prime }}=1,\end{array}$

and, therefore, from the condition of equilibrium $(6)$:

$\begin{array}{cccc}& .(00)& & -\log c_1=\log K,\end{array}$

hence, from $(7)$:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }Q=-R{T}^2\frac{\partial \log c_1}{\partial T}.\end{array}$

By means of this equation van't Hoff calculated the heat of solution $\mathrm{\Delta }Q$ from the solubility of succinic acid at $0^{\circ }$ and at $8.5^{\circ }$ C. The corresponding numbers were $2.88$ and $4.22$ in an arbitrary unit. Approximately, then:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log c_1}{\partial T}=\frac{\stackrel{e}{\log }4.22-\stackrel{e}{\log }2.88}{8.5}=0.04494,\end{array}$

from which for $T=273$:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }Q=-1.98\cdot 273^2\cdot 0.04494=-6,600\text{cal}.,\end{array}$

that is, in the precipitation of a molecule of succinic acid, $6,600\text{cal}.$ are given out to the surroundings. Berthelot found, however, through direct measurement, $6,700$ calories for the heat of solution.

The absorption of a gas also comes under this head, e. g. carbonic acid, in a liquid of relatively unnoticeable smaller vapor pressure, e. g., water at not too high a temperature. The symbol of the system is then

$\begin{array}{cccc}& .(00)& & n_0{H}_{2}O,n_{1}C{O}_2\mid n_{0^{\prime }}C{O}_2.\end{array}$

The vaporization of a molecule $C{O}_2$ corresponds to the values

$\begin{array}{cccc}& .(00)& & {\nu }_0=0,{\nu }_1=-1,{\nu }_{0^{\prime }}=1.\end{array}$

The condition of equilibrium is therefore again:

$\begin{array}{cccc}& .(00)& & -\log c_1=\log K,\end{array}$

i. e., at a fixed temperature and a fixed pressure the concentration $c_1$ of the gas in the solution is constant. The change of the concentration with $p$ and $T$ is obtained through substitution in equation $(7)$. It follows from this that:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log c_1}{\partial p}=\frac{\mathrm{\Delta }V}{RT},\frac{\partial \log c_1}{\partial T}=-\frac{\mathrm{\Delta }Q}{R{T}^2}.\end{array}$

$\mathrm{\Delta }V$ is the change in volume of the system which occurs in the isobaric-isothermal vaporization of a molecule of $C{O}_2$, $\mathrm{\Delta }Q$ the quantity of heat absorbed in the process from outside. Now, since $\mathrm{\Delta }V$ represents approximately the volume of a molecule of gaseous carbonic acid, we may put approximately:

$\begin{array}{cccc}& .(00)& & \mathrm{\Delta }V=\frac{RT}{p},\end{array}$

and the equation gives:

$\begin{array}{cccc}& .(00)& & \frac{\partial \log c_1}{\partial p}=\frac{1}{p},\end{array}$

which integrated, gives:

$\begin{array}{cccc}& .(00)& & \log c_1=\log p+\text{const.},c_1=C\cdot p,\end{array}$

i. e., the concentration of the dissolved gas is proportional to the pressure of the free gas above the solution (law of Henry and Bunsen). The factor of proportionality $C$, which furnishes a measure of the solubility of the gas, depends upon the heat effect in quite the same manner as in the example previously considered.

A number of no less important relations are easily derived as by-products of those found above, e. g., the Nernst laws concerning the influence of solubility, the Arrhenius theory of isohydric solutions, etc. All such may be obtained through the application of the general condition of equilibrium $(6)$. In conclusion, there is one other case that I desire to treat here. In the historical development of the theory this has played a particularly important rôle.

Template:C

We consider now a dilute solution separated by a membrane (permeable with regard to the solvent but impermeable as regards the dissolved substance) from the pure solvent (in the same state of aggregation), and inquire as to the condition of equilibrium. The symbol of the system considered we may again take as

$\begin{array}{cccc}& .(00)& & n_0{m}_0,n_1{m}_1,n_2{m}_2,\cdots \mid n_{0^{\prime }}{m}_0.\end{array}$

The condition of equilibrium is also here again expressed by equation $(6)$, valid for a change of state in which the temperature and the pressure in each phase is maintained constant. The only difference with respect to the cases treated earlier is this, that here, in the presence of a separating membrane between two phases, the pressure $p$ in the first phase may be different from the pressure $p^{\prime }$ in the second phase, whereby by “pressure,” as always, is to be understood the ordinary hydrostatic or manometric pressure.

The proof of the applicability of equation $(6)$ is found in the same way as this equation was derived above, proceeding from the principle of increase of entropy. One has but to remember that, in the somewhat more general case here considered, the external work in a given change is represented by the sum $pdV+p^{\prime }d{V}^{\prime }$, where $V$ and $V^{\prime }$ denote the volumes of the two individual phases, while before $V$ denoted the total volume of all phases. Accordingly, we use, instead of $(7)$, to express the dependence of the constant $K$ in $(6)$ upon the pressure:

$\begin{array}{cccc}& (11).& & \frac{\partial \log K}{\partial p}=-\frac{\mathrm{\Delta }V}{RT},\frac{\partial \log K}{\partial p^{\prime }}=-\frac{\mathrm{\Delta }{V}^{\prime }}{RT}.\end{array}$

We have here to do with the following change:

$\begin{array}{cccc}& .(00)& & {\nu }_0=-1,{\nu }_1=0,{\nu }_2=0,\cdots ,{\nu }_{0^{\prime }}=1,\end{array}$

whereby is expressed, that a molecule of the solvent passes out of the solution through the membrane into the pure solvent. Hence, in accordance with $(6)$:

$\begin{array}{cccc}& .(00)& & -\log c_0=\log K,\end{array}$

or, since

$\begin{array}{cccc}& .(00)& & c_0=1-\frac{n_1+n_2+\cdots }{n_0},\frac{n_1+n_2+\cdots }{n_0}=\log K.\end{array}$

Here $K$ depends only upon $T$, $p$ and $p^{\prime }$. If a pure solvent were present upon both sides of the membrane, we should have $c_0=1$, and $p=p^{\prime }$; consequently:

$\begin{array}{cccc}& .(00)& & (\log K{)}_{p=p^{\prime }}=0,\end{array}$

and by subtraction of the last two equations:

$\begin{array}{cccc}& .(00)& & \frac{n_1+n_2+\cdots }{n_0}=\log K-(\log K{)}_{p=p^{\prime }}=\frac{\partial \log K}{\partial p}(p-p^{\prime })\end{array}$

and in accordance with $(11)$:

$\begin{array}{cccc}& .(00)& & \frac{n_1+n_2+\cdots }{n_0}=-(p-p^{\prime })\cdot \frac{\mathrm{\Delta }V}{RT}.\end{array}$

Here $\mathrm{\Delta }V$ denotes the change in volume of the solution due to the loss of a molecule of the solvent (${\nu }_0=-1$). Approximately then:

$\begin{array}{cccc}& .(00)& & -\mathrm{\Delta }V\cdot n_0=V,\end{array}$

the volume of the whole solution, and

$\begin{array}{cccc}& .(00)& & \frac{n_1+n_2+\cdots }{n_0}=(p-p^{\prime })\cdot \frac{V}{RT}.\end{array}$

If we call the difference $p-p^{\prime }$, the osmotic pressure of the solution, this equation contains the well known law of osmotic pressure, due to van't Hoff.

The equations here derived, which easily permit of multiplication and generalization, have, of course, for the most part not been derived in the ways described above, but have been derived, either directly from experiment, or theoretically from the consideration of special reversible isothermal cycles to which the thermodynamic law was applied, that in such a cyclic process not only the algebraic sum of the work produced and the heat produced, but that also each of these two quantities separately, is equal to zero (first lecture). The employment of a cyclic process has the advantage over the procedure here proposed, that in it the connection between the directly measurable quantities and the requirements of the laws of thermodynamics succinctly appears in each case; but for each individual case a satisfactory cyclic process must be imagined, and one has not always the certain assurance that the thermodynamic realization of the cyclic process also actually supplies all the conditions of equilibrium. Furthermore, in the process of calculation certain terms of considerable weight frequently appear as empty ballast, since they disappear at the end in the summation over the individual phases of the process.

On the other hand, the significance of the process here employed consists therein, that the necessary and sufficient conditions of equilibrium for each individually considered case appear collectively in the single equation $(6)$, and that they are derived collectively from it in a direct manner through an unambiguous procedure. The more complicated the systems considered are, the more apparent becomes the advantage of this method, and there is no doubt in my mind that in chemical circles it will be more and more employed, especially, since in general it is now the custom to deal directly with the energies, and not with cyclic processes, in the calculation of heat effects in chemical changes.