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Given the Carnahan-Starling equation of state for a solution of hard-spheres, $$ Z = \frac{P}{\rho k_BT} = \frac{1 + \eta + \eta^2 - \eta^3}{(1-\eta)^3}$$ where $\rho = N/V$ is the number density and $\eta = \frac{\pi}{6}\sigma^3 \rho$ is the sphere packing fraction, one can use the relationship between the pressure and Helmholtz free energy $F$, $$ P = -\left ( \frac{\partial F}{\partial V} \right )_{T,V} = -\left ( \frac{\partial F}{\partial (N/\rho)} \right )_{T,V} = -\left ( \frac{\partial F}{\partial (N\rho/\rho^2 )} \right )_{T,V} = -\frac{\rho^2}{N} \left ( \frac{\partial F}{\partial V} \right )_{T,V} $$ and integrate this expression with respect to density to obtain the Helmholtz free energy, \begin{align*} F &= -\int_0^{\rho'} \frac{N}{\rho^2}P \ d\rho \\ &= -Nk_BT \int_0^{\rho'} \frac{\rho}{\rho^2}\frac{1 + \eta + \eta^2 - \eta^3}{(1-\eta)^3} \ d\rho \\ &= -Nk_BT \int_0^{\rho'} \frac{1}{\rho}\frac{1 + \eta + \eta^2 - \eta^3}{(1-\eta)^3} \ d\rho \\ &= -Nk_BT \int_0^{\eta'} \frac{1}{\eta}\frac{1 + \eta + \eta^2 - \eta^3}{(1-\eta)^3} \ d\eta \end{align*} where I have used the substitution $\rho = \frac{6}{\pi\sigma^3}\eta$. However in this last step, Attard ($\textit{Thermodynamics and Statistical Mechanics: Equilibrium by Entropy Maximisation}$, p. 202), arrives at

$$ Nk_BT \int_0^{\eta'} \frac{1}{\eta}\left (\frac{1 + \eta + \eta^2 - \eta^3}{(1-\eta)^3} - 1 \right ) \ d\eta $$

which appears critical to actually evaluating the integral to obtain the correct form of $F$. I cannot understand where the $-1$ term comes from or how Attard gets rid of the negative sign in front of the integral. I feel as if I am missing something obvious. Where have I gone wrong?

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  • $\begingroup$ Great question (and answer), I was just asking myself the same thing! +1 $\endgroup$ – Jxx Dec 6 '18 at 22:21
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That's two questions!

  1. What about the minus sign? That goes away because $\frac{\partial V}{\partial \rho} = -\frac{\rho^2}{N}$, which means that $P = \frac{\rho^2}{N}\left( \frac{\partial F}{\partial \rho}\right)_{T,V}$ (without the - sign).
  2. What about the $-1$? Well, Attard is interested in the excess free energy, that is, the total free energy minus the ideal gas contribution. However, the Carnahan-Starling expression for $Z$ includes the ideal gas part. By adding $-1$ to it, that contribution is taken away (since $Z = 1$ for the ideal gas), so that we are left with just the excess free energy.
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