Energetics - Describing Liquids: Structure and Energetics - Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications (2013)

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

6. Describing Liquids: Structure and Energetics

6.5. Energetics

If we assume that the correlation function g(r) is known and the pairwise additivity (of spherical potentials) holds for the potential energy, that is, Φ(r) = ½Σijϕij(r), we can write down for the internal energy c6-math-5028 almost by inspection the energy equation

(6.32) c6-math-0032

where the first term is the kinetic contribution. More formally, the canonical average, as given by Eq. (6.9), results in

(6.33) c6-math-0033

and using Φ = ½Σijϕij leads to N(N − 1)/2 identical terms so that

(6.34) c6-math-0034

to which the kinetic term has to be added. In the thermodynamic limit (N → ∞, V → ∞ but N/Vρ), the factor (N − 1)/V becomes ρ, so that the result is Eq. (6.32).

Similarly, though in a somewhat more complex fashion, an expression for the pressure P can be obtained. We start with the previously derived expression (see Chapter 5)

(6.35) c6-math-0035

with the thermal wavelength Λ ≡ (h2/2πmkT)1/2. Before differentiating, we consider that the pressure is independent of the shape of the container, so that we can take a cube and employ the coordinate transformation ri = V1/3ri′. The expression for QN then becomes

(6.36) c6-math-0036

while for the intermolecular distance rij = [(xixj)2 + (yiyj)2 + (zizj)2]1/2 we obtain the expression c6-math-5048. Therefore ∂QN/∂V becomes

(6.37) c6-math-0037

(6.38) c6-math-0038

Transforming back, noting that we have N(N − 1)/2 identical terms and taking the thermodynamic limit, one obtains the pressure (or virial) equation (see Section 3.8)

(6.39) c6-math-0039

In Eqs (6.32) and (6.39), the first terms are due to the momenta (i.e., the kinetics), while the second term is due to the coordinates (i.e., the configuration). Thus, both equations in principle relate a thermodynamic quantity to the structure, as represented by g(r).

For future reference we note that the compressibility κT can be calculated using the grand canonical ensemble (see Justification 6.1). The final result

(6.40) c6-math-0040

is known as the compressibility equation, and this expression is derived without assuming any pairwise additivity for the potential energy of the system, and is thus generally valid.

Justification 6.1: The compressibility equation*

The derivation of the compressibility equation requires the grand canonical ensemble. We recall first the definition of the isothermal compressibility


From PV = kT lnΞ we have, using c6-math-5030,


We further need the absolute activity given by λ = exp(βμ), and note that ∂lnΞ/∂ρ = (∂lnΞ/∂λ)/(∂ρ/∂λ). For the numerator ∂lnΞ/∂λ we obtain, defining IN ≡ ∫ … ∫exp(−βΦ)dr1 … drN,


where the distribution as defined for the grand canonical ensemble


is used. The denominator ∂ρ/∂λ is given by


so that


where for the last step the definition g(r1,r2) = ρ(2)(r1,r2)/ρ(1)(r1)ρ(2)(r2) = ρ(2)(r1,r2)/ρ2 is used. Combining the expressions for ∂lnΞ/∂λ and ∂ρ/∂λ yields


The isothermal compressibility is then given by


Note that the structure factor S(s) for s = 0 reduces to


and therefore S(0) = ρkTκT, a fact that can be used for reconstructing g(r).

One might wonder whether the compressibility equation and the pressure equation yield the same result for the pressure. On integrating Eq. (6.40), one obtains

(6.41) c6-math-0041

to be compared with PP as obtained from Eq. (6.39). For the ideal gas g(r) = 1, leading for both expressions to the ideal gas EoS P = ρkT. This is no longer true for a nonideal fluid for which ϕ(r) ≠ 0, and thus g(r) is unknown. Using approximations for g(r) usually leads to different values for PP and Pκ, indicating a loss of thermodynamic consistency (see Section 6.3).

In order to obtain a complete description of the thermodynamic state, some extra information is required. This is probably most clear from the Gibbs equation

(6.42) c6-math-0042

which shows that information is required about the energy U, the pressure P, and the chemical potential μ to obtain a complete description of the thermodynamic state of the system.

If the correlation function g(r) were to be known over a wide temperature range, we could use the energy Eq. (6.32) and integrate the Gibbs–Helmholtz equation c6-math-5040 and so obtain the required information, since the chemical potential μ can be calculated from the Helmholtz energy F. Similarly, if g(r) were to be known over a wide density range, we could use the pressure Eq. (6.39) and integrate ∂(F/N)/∂(1/ρ) = ρ2P (as obtained from dF = −PdVSdT + μdN). Alas, neither the temperature nor density dependence of g(r) is usually available.

As an alternative we introduce the coupling parameter approach, also known as thermodynamic integration. A coupling parameter ξ is defined, varying from 0 to 1, which controls the interaction between the reference molecule 1 and another, say j, by replacing ϕ1j by ξϕ1j. The potential energy then becomes

(6.43) c6-math-0043

and a molecule is added to the system when ξ varies from 0 to 1. The partition function ZN(ρ,T) reads now ZN(ρ,T,ξ) while the correlation function g(r;ρ,T) becomes g(r;ρ,T,ξ). Because the number of particles N is very large, we can write

(6.44) c6-math-0044

From F = −kTlnZ, we obtain −βF = ln QN − ln N! − 3N ln Λ, and hence we get

(6.45) c6-math-0045

Further, we have QN(ξ = 1) = QN and QN(ξ = 0) = VQN−1. Therefore,

(6.46) c6-math-0046

(6.47) c6-math-0047

Calculating ∂QN(ξ)/∂ξ using Eq. (6.43) results in

(6.48) c6-math-0048

Further, employing ρ(N)(ξ) = N!n(N)(ξ) with n(N)(ξ) = QN(ξ)−1exp[−βΦ(ξ)], and recognizing that we have N − 1 identical integrals, we obtain

(6.49) c6-math-0049

where in the last step ρ(2)(ξ) = ρ2g(ξ) is employed. Substitution in Eq. (6.45) yields the required result

(6.50) c6-math-0050

From c6-math-5041, P and μ, we can calculate all of the thermodynamic properties if g(r;ξ) is known. The theoretical calculation of g(r;ξ) is the topic of Chapter 7.

Problem 6.7*

Check the steps necessary to obtain Eqs (6.39), (6.40), and (6.50).