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 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
will develop, where I(t) is the current during the period of charging. The capacitance
is the proportionality constant between the voltage applied and the charge developed, that is, q0 = C0ϕ. The surface charge σ0 is defined by
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 = εrq0, with εr the relative permittivity and thus the capacitance becomes C = εrε0A/d = εA/d with ε ≡ εrε0 the permittivity. The relative permittivity εr 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
where σpol 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
where PE is the polarization.2) Solving for PE we obtain
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 = VcapEdD with displacement D = εE and volume of the capacitor Vcap = Ad. The first law (for a one-component system at temperature T and pressure P) thus becomes
Applying a Legendre transform using G = U − TS + PV − VcapED, we obtain
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) − ½εE2Vcap, 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.
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,
Show that the energy U for a normal dielectric, using temperature T and pressure P, is given by