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

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1 Answer 1

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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.

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