Surface tension, surface Helmholtz and Gibbs energies, and entropies

The surface tension is related to the derivative of the Helmholtz energy by the equations

$${\gamma} = ({\partial}A^{\sigma}/{\partial}A_s)_{T,n^{\sigma}_i} =({\partial}A^s/{\partial}A_s)_{T,V^s,n^s_i}$$

$$= ({\partial}A/{\partial}A_s)_{T,V^{\alpha},V^{\beta},n^{\alpha}_i,n^{\beta}_i,n^{\sigma}_i} =({\partial}A/{\partial}A_s)_{T,V,n_i},$$

where in the last equality $A$ is the Helmholtz energy of the whole system, and $V^{\alpha}$ and $V^{\beta}$ refer to the volumes of the bulk phases relative to the Gibbs surface; and to the derivative of the Gibbs energy by the equations

$${\gamma} = ({\partial}G^{\sigma}/{\partial}A_s)_{T,{\gamma},n^{\sigma}_i} =({\partial}G^s/{\partial}A_s)_{T,p^s,n^s_i}$$

$$= ({\partial}G/{\partial}A_s)_{T,p^s,n^{\alpha}_i,n^{\beta}_i,n^{\sigma}_i}.$$

Expressed in terms of integral quantities

$${\gamma}A_s=A^{\sigma}-{\sum}_in_i^{\sigma}{\mu}^{\sigma}_i=A^s-{\sum}_in^s_i{\mu}^s_i,$$

or

$${\gamma}=a^{\sigma}-{\sum}_i{\Gamma}_i^{\sigma}{\mu}^{\sigma}_i=a^s-{\sum}_i{\Gamma}^s_i{\mu}^s_i,$$

Under equilibrium conditions the superscripts $s$ and ${\sigma }$ attached to the ${\mu}_i$ terms may be omitted.

Note: Only when ${\sum}_i{\Gamma}_i{\mu}_i = 0$ is ${\gamma }$ equal to the surface (excess) Helmholtz energy per unit area. In general, for a multicomponent system it is not possible to define either an interfacial layer, or a Gibbs surface, for which this condition is satisfied. However, it is satisfied automatically when the system exhibits an adsorption azeotrope at which all the ${\Gamma}_i$ are zero.

For a one-component system, treated in terms of a Gibbs surface it is always possible to choose this surface so that ${\Gamma}_1 = 0$, so that the surface tension is equal to the value of $a^{\sigma}$ relative to this surface; on the other hand ${\Gamma}^s_i$ must always be positive for an interfacial layer so that $a^s$ and ${\gamma }$ can never be equated.

The surface excess entropy is given by

$$-\left(\frac{{\partial}A^{\sigma}}{{\partial}T}\right)_{A_s,n^{\sigma}_i}=S^{\sigma}.$$