## Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications (2013)

### 10. Describing the Behavior of Liquids: Polar Liquids

In the previous chapters we have discussed various models to describe liquids, but we have limited ourselves so far largely to nonpolar liquids. In this chapter we extend our discussion to polar liquids. First, we discuss macroscopic (linear) dielectric behavior. Second, the dielectric behavior of polar fluids is discussed in microscopic terms. Because H_{2}O is the most important example of a polar liquid and behaves rather anomalously, we deal thereafter with the structure of liquid water and the relation to its properties.

### 10.1. Basic Aspects

Let us first briefly describe macroscopic, (linear) dielectric behavior (see also __Appendix D__). In macroscopic terms, if we place in vacuum on a parallel plate capacitor__ ^{1)}__ with electrode area

*A*and electrode distance

*d*an electrical voltage difference

*ϕ*on the electrodes, a charge

(10.1)

will develop, where *I*(*t*) is the current during the period of charging. The capacitance

(10.2)

is the proportionality constant between the voltage applied and the charge developed, that is, *q*_{0} = *C*_{0}*ϕ*. The surface charge *σ*_{0} is defined by

(10.3)

where *E* = *ϕ*/*d* is the electric field strength and *ε*_{0} is the permittivity of the vacuum, which is a universal constant. The response to the applied electric field *E*, the surface charge *σ*_{0}, can for isotropic media be interpreted as the dipole moment per unit volume since its dimension *C* m^{−2} can be written as (*C* m)m^{−3}. In that case the response is denoted as the (di)electric displacement *D*, and so in vacuum we have *D* = *ε*_{0}*E*.

If we put a dielectric material in the capacitor the charge changes to *q* = *ε*_{r}*q*_{0}, with *ε*_{r} the relative permittivity and thus the capacitance becomes *C* = *ε*_{r}*ε*_{0}*A*/*d* = *εA*/*d* with *ε* ≡ *ε*_{r}*ε*_{0} the permittivity. The relative permittivity *ε*_{r} is a material-specific function, dependent on temperature and pressure, and possibly also on the electric field. We can also say that the surface charge changes from *σ*_{0} = *q*_{0}/*A* to

(10.4)

where *σ*_{pol} is the extra surface charge at the electrodes due to the polarization of the material put in the capacitor. In terms of the field *E*, the displacement *D* becomes

(10.5)

where *P _{E}* is the

*polarization*.

__Solving for__

^{2)}*P*we obtain

_{E}(10.6)

with *χ* the (electric) *susceptibility*, for which it always holds that *χ* ≥ 0.

The variables *E* and *D* are conjugated variables. To show this we calculate the work δ*W* required to transfer an element of charge d*q* from the negative to the positive electrode. Because *E* = *ϕ*/*d* and *q* = *AεE*, we obtain *ϕ* = *Ed* and d*q* = *A*d(*εE*). Therefore, the work δ*W* = *ϕ* d*q* = *V*_{cap}*E*d*D* with displacement *D* = *εE* and volume of the capacitor *V*_{cap} = *Ad*. The first law (for a one-component system at temperature *T* and pressure *P*) thus becomes

(10.7)

Applying a Legendre transform using *G* = *U* − *TS* + *PV* − *V*_{cap}*ED*, we obtain

__(10.8)__

For normal dielectrics at low field, the permittivity is not a function of the field *E*, and the integration of __Eq. (10.8)__ results in *G* = *G*^{(0)} − ½*εE*^{2}*V*_{cap}, where *G*^{(0)} represents the value of *G* at zero field at given temperature and pressure. From this expression, results for the entropy *S*, the volume *V*, and the energy *U* follow in the usual manner.

Problem 10.1

Show that the entropy *S*, volume *V*, and chemical potential energy *μ* for a normal dielectric, using temperature *T*, pressure *P* and number of moles *n*, are given by, respectively,

Problem 10.2

Show that the energy *U* for a normal dielectric, using temperature *T* and pressure *P*, is given by