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I'm trying to derive a relationship between the mass-specific Helmholtz free energy,the pressure, the temperature, and the mass density. It is given by

$$p=\rho ^2\left ( \frac{\partial F_s}{\partial \rho } \right )_T$$

where $F_s$ is the mass-specific Helmholtz free energy. The steps that I carried out were:

$$dF(T,V(\rho ,T))=-SdT-pdV(\rho ,T)$$ $$=\left ( \frac{\partial F}{\partial T} \right )_VdT+\left ( \frac{\partial F}{\partial V} \right )_TdV(\rho ,T)$$ $$=\left ( \frac{\partial F}{\partial T} \right )_VdT+\left ( \frac{\partial F}{\partial V} \right )_T\left [ \left ( \frac{\partial V}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial V}{\partial T} \right )_\rho dT \right ]$$

$$\Rightarrow -pdV(\rho ,T)=\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial F}{\partial V} \right )_T\left ( \frac{\partial V}{\partial T} \right )_\rho dT$$

$$\left ( \frac{\partial V}{\partial T} \right )_\rho =0$$

$$\Rightarrow -pdV(\rho ,T)=\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho $$

$$\Rightarrow -p=\frac{\left ( \frac{\partial F}{\partial \rho } \right )_Td\rho }{\left [ \left ( \frac{\partial V}{\partial \rho } \right )_Td\rho +\left ( \frac{\partial V}{\partial T} \right )_\rho dT \right ] }$$

I don't see where to go from here, because carrying out the division would just yield that

$$-p=\left ( \frac{\partial F}{\partial V} \right )_T$$

which is where I started. The motivation for this is simply the fact that an expression for a mass-specific Helmholtz function given in the following pictures has been used to determine the properties of water. The first relation in part (a) in the second picture is what I'm trying to verify.

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    $\begingroup$ Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists. $\endgroup$
    – Poutnik
    Feb 12, 2022 at 22:37
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    $\begingroup$ Yup, plain text titles are always better. If you must use MathJax, please do it in inline style: $...$ rather than $$...$$ and also don't write fractions as \frac{a}{b}, just use a/b where possible. $\endgroup$
    – orthocresol
    Feb 12, 2022 at 23:07
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    $\begingroup$ There seems to be a dimensional inconsistency in the equation. The LHS (p) has units of E/V. If rho is, as you state, mass density, then the RHS has units of E*rho = E*(m/V). $\endgroup$
    – theorist
    Feb 13, 2022 at 1:17
  • $\begingroup$ @theorist the Helmholtz free energy considered is the mass-specific Helmholtz free energy, which has units of E/m. $\endgroup$
    – CalebWilliamsUIC
    Feb 13, 2022 at 15:17

2 Answers 2

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Starting with dF=-SdT-PdV, you have: $$\left(\frac{\partial F}{\partial V}\right)_T=-P$$where V is the molar volume. The molar volume is related to the molar density $\rho$ by $$V=\frac{1}{\rho}$$Therefore, $$\left(\frac{\partial F}{\partial V}\right)_T=-\rho^2\left(\frac{\partial F}{\partial \rho}\right)_T$$

and hence $$P=\rho^2\left(\frac{\partial F}{\partial \rho}\right)_T$$

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  • $\begingroup$ Thanks a lot, this makes perfect sense. $\endgroup$
    – CalebWilliamsUIC
    Feb 13, 2022 at 15:19
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$$V=\frac{m}{\rho }\Rightarrow \frac{\partial V}{\partial \rho }=-\frac{m}{\rho ^2}\Rightarrow -p\left ( \frac{\partial V}{\partial \rho } \right )_T=\left ( \frac{\partial F}{\partial \rho } \right )_T$$

$$-p\left (-\frac{m}{\rho ^2} \right )=p\frac{m}{\rho ^2}\Rightarrow p=\frac{\rho ^2}{m}\left ( \frac{\partial F}{\partial \rho } \right )_T=\rho^2\left ( \frac{\partial F_s}{\partial \rho } \right )_T$$

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