# How does compressibility depend on pressure?

I want to calculate the density distribution inside a column of a solid material of which I know all properties (density, compressibility) at the top (room pressure). Using the definition of isothermal compressibility as: $$\beta_T = \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T$$ and the definition of density as $\rho = \frac{m}{V}$, I arrive at $$\left( \frac{\partial \rho}{\partial P} \right)_T = \rho \beta_T$$ Integrating this differential equation I get an exponential function for density $\rho = \rho_0 \exp(\beta P)$. However, doing so I assumed that $\beta_T$ does not depend on pressure. But if I write: $$\frac{\partial \beta_T}{\partial P} = \frac{\partial}{\partial P} \left( \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T \right)$$ it looks to me like you could find an expression for this involving only stuff we already know. It might be a bit nit-picky, since the pressure depence might be small, but if the pressure becomes large enough, it could make a difference?

• You ask about $\partial\beta/\partial P$ but your title suggests a derivative w.r.t. temperature. Commented Jan 11, 2016 at 17:44
• I changed the title to "... depend on pressure". Thanks for pointing this out! Commented Jan 12, 2016 at 8:29

If you want to consider what happens to the pressure-density relationship at higher pressures when a linear relationship no longer holds, then the first thing you need to do is to discard the assumption that the isothermal compressibility $\beta_T$ has no pressure dependence. Instead, you have to use something like the non-linear Birch-Murnaghan equation of state, which is used by scientists studying the equations of state of materials under extreme pressure conditions using either diamond anvil cells or shock waves.