# Isothermal compressibility

How does one get from the thermodynamics definition :

$$\chi_T = -\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T$$

to the fluid dynamics definition :

$$\chi_T = \frac{1}{\rho} \left(\frac{\partial \rho}{\partial p}\right)_T$$

?

• Have you thought about what you could do to the equation to turn all your $V$'s into $\rho$'s? – tpg2114 Nov 14 '14 at 23:41
• I have, and I didn't find the answer to that (that's why I'm asking) – mwa1 Nov 15 '14 at 0:11
• Have you considered using an ideal gas and proving the relation is true? – Kyle Kanos Nov 15 '14 at 0:32
• Do you know the relationship between mass, density and volume? – tpg2114 Nov 15 '14 at 0:53
• Well I know that, dimensionally speaking, density is a mass by unit volume. But I get confused because density used to be defined as $\rho = \frac{m}{V}$ and now in Fluid Mechanics I often see $\rho = \frac{dm}{d^3r}$ so I'm not sure how I'm supposed to manipulate this. – mwa1 Nov 15 '14 at 1:08

## 1 Answer

If anyone is looking for the same thing, here is the solution :

\begin{align}m &= \rho V = \textrm{constant}\\ \Leftrightarrow~~~~~~~~~~~~~~~~ \rho~\mathrm dV + V~\mathrm d\rho &= 0\\\Leftrightarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\mathrm d\rho}{\rho} &= -~ \frac{\mathrm dV}{V}\\\Leftrightarrow~~~~ \chi_T = - ~\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T &= \frac{1}{\rho} \left(\frac{\partial \rho}{\partial p}\right)_T\end{align}

It's simple but not obvious if you don't know where to start...