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In 2 lines this book says that the second derivative of the thermodynamic Helmholtz free energy density $a\left(\rho,T\right)$ with respect to density of a one-component fluid, $\rho$, when we approach the critical point of the gas-liquid transition along the critical isochore ($\rho=\rho_{c}$) from $T>T_{c}$ can be written as:

$$ x=\left(\frac{\partial^{2}a\left(\rho,T\right)}{\partial\rho^{2}}\right)_{T,\rho=\rho_{c}}=\left(\frac{\partial\mu}{\partial\rho}\right)_{T,\rho=\rho_{c}}=\frac{1}{\rho_{c}^{2}\chi_{T}} $$

How can this be shown? What is the meaning of $x$? Any hint would be helpful. Thanks.

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Since you have tagged this as "homework-and-exercises", and asked for "any hint", I hope that an answer consisting of hints will be acceptable! Physics StackExchange discourages people from giving complete answers to questions of this kind.

The expressions you quote are standard thermodynamic relations for a one-component system, valid everywhere in the phase diagram, not just near a critical point. The main issue is converting from formulae which may be more familiar when expressed in terms of extensive variables, to intensive variables.

The first expression is essentially the definition of the chemical potential $$ \mu = \left(\frac{\partial A}{\partial N}\right)_{VT} $$ Just consider a system at fixed $V$, density $\rho=N/V$, so the derivative can easily be converted into one with respect to $\rho$. [Incidentally, physicists will be more accustomed to using $F$ and $f$ for the Helmholtz free energy and free energy density, respectively; you and I are using the notation familiar to chemists.]

The second equality is more subtle, and involves the isothermal compressibility which is usually defined $$ \chi_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = \frac{1}{\rho}\left(\frac{\partial \rho}{\partial P}\right)_T $$ So, your problem boils down to showing that $$ \left(\frac{\partial \mu}{\partial \rho}\right)_T =\frac{1}{\rho}\left(\frac{\partial P}{\partial \rho}\right)_T $$ or the equivalent, if we define $v=V/N=1/\rho$, $$ \left(\frac{\partial \mu}{\partial v}\right)_T =v\left(\frac{\partial P}{\partial v}\right)_T $$ The hint for this is to start from the expression for the total differential of the Gibbs free energy $G$, for a one-component system, and write it as $d\mu=$ an expression involving $dT$, $dP$, the entropy per mole $s=S/N$, and $v=V/N$.

As for the "meaning" of $x$, second derivatives of free energies are often related to the equilibrium fluctuations of quantities. In this case, close to the critical point, the isothermal compressibility diverges $\chi_T\rightarrow\infty$, and this is associated with density fluctuations becoming macroscopically large. So, if you like the (very rough!) analogy of $a(\rho)$ as being a function of $\rho$ like a harmonic oscillator, with the system lying near the bottom of a potential well, which dictates the magnitude of natural fluctuations in $\rho$, this second derivative (essentially the spring constant of the oscillator) will tend to zero as $\chi_T$ becomes infinite, and the density fluctuations will diverge.

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  • $\begingroup$ I though $\chi_{T}$ was the isothermal susceptibility and $\kappa_{T}$ the isothermal compressibility. $\endgroup$ – Igaturtle Aug 30 '18 at 11:16
  • $\begingroup$ $\kappa_T$ is often used for the same quantity: I have seen both symbols in use. As for the terminology, I believe that "isothermal susceptibility" is much more rarely used; "isothermal compressibility" is much more common. $\endgroup$ – user197851 Aug 30 '18 at 11:38
  • $\begingroup$ One more thing: is it true that $a=A/V$ and $g=G/V$? $\endgroup$ – Igaturtle Aug 30 '18 at 12:01
  • $\begingroup$ You have to be a bit careful, because sometimes it is useful to divide by $N$ instead of $V$ in defining intensive quantities. In this case I believe that the Helmholtz free energy density is indeed $a=A/V$. And rather than make use of $g=G/V$, it may be more useful to use $\mu=G/N$ which is true for one component systems. (Sometimes people define $g=G/N$, hence the need to look closely!) $\endgroup$ – user197851 Aug 30 '18 at 12:06
  • $\begingroup$ You should get $d\mu=$ a sum of two differentials. One of them, $-s\,dT$ will disappear if $T$ is constant. The other, $+v\,dP$ gives you exactly what you want, if you are considering how both sides vary with a small change $dv$ at constant $T$. Are you used to manipulations of this kind in thermodynamics? $\endgroup$ – user197851 Aug 30 '18 at 21:56

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