Trying to understand the derivation of the Gibbs adsorption isotherm

Question

I have looked at several derivations online, and the clearest I have found is that on the Wiki page.

The problem being considered is that of two phases, with an interface of a given area between them. The Gibbs adsorption isotherm relates the change in the surface tension to the change in chemical potential in equilibrium, via the excess surface concentrations.

The result for a two species system,

$-d\gamma = \Gamma_1 \mu_1 + \Gamma_2 \mu_2$,

where $\mu_i$ are the chemical potentials of the species at the surface, and thus also in each bulk phase in equilbirum. $\Gamma_i$ is an 'appropriately defined surface excess', the significance of which I don't quite get, but I see that it measures a number density giving the per unit area of the species.

A more familiar form might be

$\Gamma_i = -\frac{1}{RT}(\frac{\partial \gamma}{\partial \ln(C_i)})_{T,p}$.

The part of the derivation which I don't understand is right at the start, where the Gibbs free energy for a given phase is stated to be

$G=U-TS+PV+\Sigma_i \mu _i n_i.$

I thought that $U=TS-PV+\Sigma_i \mu_i n_i +\gamma A$, so with the above definition of $G$ we have an extra $\Sigma_i \mu _i n_i$ here! However the derivation does not seem to work without it.

Another similar problem is that the $\gamma A$ term seems to be given additionally to the Gibbs free energy, aside from that implicitly contained in $U$. Again, I am not quite sure why we are putting in these extra pieces.

Answer 1

I have trouble with this Wikipedia derivation, for the reasons you give, and also because it starts by assuming that it is possible to define the properties of the surface phase. As Rowlinson and Widom state in Molecular theory of capillarity, this can't be done unambiguously, unless one follows Gibbs' approach: use the difference between those of the whole system and the idealized two-phase system, taking the properties of the bulk phases to be unchanged up to a dividing surface. They also mention the issue of whether or not to include the $\gamma A$ term in the definition of $G$.

Their derivation uses Gibbs' approach, and is consistent with the Helmholtz free energies $$ F=-pV+\gamma A+\sum_i \mu_i n_i $$ for the whole system and $$ F^\alpha=-pV^\alpha +\sum_i \mu_i n_i^\alpha $$ for phase $\alpha$, and similarly for phase $\beta$. Surface excess properties, denoted $n_i^s$ etc, are then defined by equations such as $$ n_i^s+n_i^\alpha+n_i^\beta=n_i . $$ It then follows that $$ F^s=\gamma A + \sum_i \mu_i n_i^s . $$ The key equation, though, comes from the fundamental equation of thermodynamics $$ dU = TdS - pdV +\gamma dA + \sum_i \mu_i dn_i $$ and is $$ dF =-SdT - pdV +\gamma dA + \sum_i \mu_i dn_i $$ and similar equations, without the surface term, for the two bulk phases, which are subtracted from this one to give $$ dF^s = -S^s dT +\gamma dA + \sum_i \mu_i dn_i^s . $$ This is then compared with the total derivative of the equation for $F^s$ above to give a Gibbs-Duhem equation $$ S^s dT + A d\gamma + \sum_i n_i^s d\mu_i =0 . $$ Setting $dT=0$ and dividing by area gives the Gibbs adsorption isotherm $$ d\gamma + \sum_i \Gamma_i d\mu_i =0 $$ where the surface adsorption is define by $\Gamma_i=n_i^s/A$. This is your starting equation, but you missed some d's.

I've reproduced this in case you can't get hold of that book, but I do recommend it, if you are interested in the thermodynamics of inhomogeneous systems.