Derivation of the Equation $p=\rho^2 (\partial F_s/\partial \rho)_T$

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.


[![enter image description here](https://i.stack.imgur.com/hsHMo.jpg)](https://i.stack.imgur.com/hsHMo.jpg)


[![enter image description here](https://i.stack.imgur.com/lX8oj.jpg)](https://i.stack.imgur.com/lX8oj.jpg)




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


 4

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


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


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