# How to see the equivalence of two definitions of fluid isothermal compressibility?

Question: How can one show the equivalence of two definitions of fluid isothermal compressibility?

I see the isothermal compressibility, $\beta_T$, commonly defined as: $$\tag{1} \beta_T=-\frac{1}{V_o}\left(\frac{\partial V}{\partial p}\right)_T$$

where $V_o$ is the initial or starting volume, $\partial V$ is the change in volume ($=V_o-V_{new}$), and $\partial p$ is the change in pressure ($=p_o-p_{new}$).

I have also seen the isothermal compressibility defined as: $$\tag{2} \beta_T=\frac{1}{\rho_o}\left(\frac{\partial \rho}{\partial p}\right)_T$$ where $\rho_o$ is the initial or starting density, $\partial \rho$ is the change in density($=\rho_o-\rho_{new}$), and $\partial p$ is the change in pressure ($=p_o-p_{new}$).

If I use the relationship between volume and density: $V=m/\rho$, where $m$ is the mass of the fluid, and I assume it to be constant, I suppose I can re-write Eqn (1) as:

$$\tag{3} \beta_T=-\rho_o\left(\frac{\partial \rho^{-1}}{\partial p}\right)_T$$

If I can do this, I am having trouble seeing how I can make my next step to arrive at Eqn (2).

• The correct definitions should really be without the subscript 0's. See what you get then. Dec 21, 2017 at 16:35
• @ChesterMiller so to begin, I should say $\beta_T=-\frac{1}{V(p)}\left(\frac{\partial V(p)}{\partial p}\right)_T$, where $V(p)$ is some volume at pressure $p$, $\partial V(p)$ is the change in volume ($=V(p)-V(p+\partial p)$), and $\partial p$ is the change in pressure ($=p-\partial p$). Then using the reciprocal rule as Endulum has suggested, I would then have $$\beta_T=\frac{\rho(p)}{\rho(p)^2}\left(\frac{\partial \rho(p)}{\partial p}\right)_T=\frac{1}{\rho(p)}\left(\frac{\partial \rho(p)}{\partial p}\right)_T$$ which would be the correct definition for Eqn (2)? Dec 21, 2017 at 16:45
• Yes. That is correct. Dec 21, 2017 at 16:54

You are almost there. You need only compute the derivative in your Eq. (3). Use the fact that $\frac{\partial}{\partial x}\frac{1}{f(x)} = -\frac{1}{f^2}\frac{\partial f}{\partial x}$ and you will immediately get Eq. (2).
• is this the chain rule of differentiation? So $\frac{\partial}{\partial p} (\rho(p))^{-1} = - \frac{1}{\rho^2} \frac{\partial p}{\partial p}$? Dec 21, 2017 at 16:14
• OK, so you are saying I can use the [reciprocal rule][1]. If I use this, I would get:$$\beta_T=\frac{\rho_o}{\rho(p)^2}\left(\frac{\partial \rho(p)}{\partial p}\right)_T$$ What statement(s) can be made to be able to make this equation Eqn (2)? [1]: en.wikipedia.org/wiki/Reciprocal_rule Dec 21, 2017 at 16:32