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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).

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    $\begingroup$ The correct definitions should really be without the subscript 0's. See what you get then. $\endgroup$ – Chet Miller Dec 21 '17 at 16:35
  • $\begingroup$ @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)? $\endgroup$ – Armadillo Dec 21 '17 at 16:45
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    $\begingroup$ Yes. That is correct. $\endgroup$ – Chet Miller Dec 21 '17 at 16:54
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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).

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  • $\begingroup$ is this the chain rule of differentiation? So $\frac{\partial}{\partial p} (\rho(p))^{-1} = - \frac{1}{\rho^2} \frac{\partial p}{\partial p}$? $\endgroup$ – Armadillo Dec 21 '17 at 16:14
  • $\begingroup$ 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 $\endgroup$ – Armadillo Dec 21 '17 at 16:32
  • $\begingroup$ It looks like you got things sorted out up in the main comments. Just to reiterate my own way: in these definitions, there are no "start" and "end" values for the pressure, density, etc. They are differential formulas, so they only refer to the "current" pressure and density. You might recall from a past physics course that the average velocity and the instantaneous velocity are different in general. It's the same principle here. $\endgroup$ – Endulum Dec 21 '17 at 17:55

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