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

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 H₂O 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 capacitor¹⁾ 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, q₀ = C₀ϕ. The surface charge σ₀ is defined by

(10.3)

where E = ϕ/d is the electric field strength and ε₀ is the permittivity of the vacuum, which is a universal constant. The response to the applied electric field E, the surface charge σ₀, 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 = ε₀E.

If we put a dielectric material in the capacitor the charge changes to q = ε_rq₀, with ε_r the relative permittivity and thus the capacitance becomes C = ε_rε₀A/d = εA/d with ε ≡ ε_rε₀ 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 σ₀ = q₀/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.²⁾ Solving for P_E we obtain

(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 δW required to transfer an element of charge dq from the negative to the positive electrode. Because E = ϕ/d and q = AεE, we obtain ϕ = Ed and dq = Ad(εE). Therefore, the work δW = ϕ dq = V_capEdD 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_capED, 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⁾ − ½εE²V_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