# Expressing isothermal compressibility in terms of particle density and chemical potential

As listed on Wikipedia, isothermal compressibility is usually expressed as,

$$\beta=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T,N},$$

where $$V$$ is volume and $$p$$ is pressure. However, I am looking for the expression in terms of particle density and chemical potential,

$$\beta=\frac{1}{n^2}\frac{\partial n}{\partial \mu},$$

where $$n$$ is particle density ($$n=N/V$$).

I tried to use the Gibbs-Duhem equation (we only have 1 type of particle):

\begin{align}N d\mu &= - S dT + V dp\\ N d\mu &= V dp \quad\quad \text{Because T is constant, i.e. isothermal case}\\ dp&=\frac{N}{V}d\mu\\ &=n\ d\mu \end{align} Besides that fact, I recognize we should probably use $$-\frac{\partial}{\partial x}\frac{1}{f(x)}=\frac{1}{f^2}\frac{\partial f}{\partial x}$$, hence we are left with showing that \begin{align} \frac{\partial V}{\partial p}=V\frac{\partial}{\partial \mu} \frac{1}{n}. \end{align} However, using my $$dp$$ from earlier, I am instead stuck at: \begin{align} \frac{\partial V}{\partial p}=\frac{V}{N}\frac{\partial V}{\partial \mu}. \end{align} What is $$dV$$ and how do I reach the final form of $$\beta=\frac{1}{n^2}\frac{\partial n}{\partial \mu}$$?

Continuing from where you started, we have $$\frac{\partial V}{\partial p}= \frac{V}{N} \frac{\partial V}{\partial \mu}$$. But, as you stated $$V= \frac{N}{n}$$. So, plugging that in the expression of $$\beta$$, we get $$\beta = - \frac{1}{N} \frac{\partial V}{\partial \mu} = -\frac{\partial \frac{1}{n}}{\partial \mu}$$ which gives the quoted result.