Equation of State for a Liquid with Double Exponential Potential

Abstract

We propose a simple method for the approximation of the integrand in the integral representation of the free energy of a system of particles with two-particle interactions admitting a Fourier expansion, based on which the equation of state of a with a double exponential potential liquid is obtained without laborious calculations. The found temperature dependences of the equilibrium thermodynamic properties appropriately describe the corresponding experimental data. The effectiveness of the approach is demonstrated on the example of a model with double Yukawa potential.

APPENDIX

To prove the inequality

$$\frac{{\ln \left( {1 + xD} \right)}}{{\ln \left( {1 + xd} \right)}} > \frac{D}{d},$$

(22)

in which 0 < D < 1, 0 < d < 1, D < d, in region G₁, x > 0, let us consider the function

$$\varphi \left( \nu \right) = {{\left( {1 + p} \right)}^{\nu }} - \left( {1 + p\nu } \right),$$

(23)

where 0 ≤ ν ≤ 1, p > 0.

Differentiating function (23) twice with respect to ν, we get

$$\varphi {\kern 1pt} ''\left( \nu \right) = {{\left( {1 + p} \right)}^{\nu }}{{\left[ {\ln \left( {1 + p} \right)} \right]}^{2}} > 0.$$

It follows from this equation that the curve φ(ν) is concave, and, since φ(0) = φ(1) = 0, then φ(ν) < 0 at 0 < ν < 1, or, as follows from (23):

$${{\left( {1 + p} \right)}^{\nu }} < \left( {1 + p\nu } \right).$$

(24)

By assuming in (24) ν = D/d and p = xd, we arrive at the inequality

$${{\left( {1 + xd} \right)}^{{{D \mathord{\left/ {\vphantom {D d}} \right. \kern-0em} d}}}} < \left( {1 + xD} \right).$$

(25)

Taking the logarithm of (25), we obtain inequality (22), from which inequality (9) follows. In conclusion, it should be noted that the values used in the proof, ν = 0 (D = 0) and ν = 1 (d = 1), correspond to the upper and lower boundaries of the region G₁; however, the actual values of δ and ε are contained inside G₁, where D ∈ (0; 1) and d ∈ (0; 1).