# Thermodynamic relations from Gibbs-Duhem

Given the Gibbs-Duhem relation $V dp = N d \mu + S dT$, I am having trouble deriving the following identity:

$\ \left(\frac{\partial N}{\partial \mu}\right)_{V,T} = N \left(\frac{\partial \rho}{\partial p}\right)_T$

The problem is that the variables $N$ and $\rho$ don't appear as infinitisemals in the equation. How can I proceed?

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Start by noting that the density is inversely proportional to volume: ρ = m / V, and also that the mass m equals the substance's molar mass M multiplied by the number of moles N: m = M N. Together we have, ρ = M (N / V) (M will be carried through the rest of the derivation as an arbitrary constant with dimensions kg / mol.) – David H Nov 27 '12 at 21:16

## 1 Answer

$\left( \frac {\partial N} {\partial \mu} \right)_{V,T} = \left( \frac {\partial (\rho V) } {\partial \mu} \right)_{V,T}$ since $V$ constant, this is the same as: $\left(V \frac {\partial \rho} {\partial \mu} \right)_{V,T}$.

From there you can put: $\left(V \frac {\partial \rho} {\partial p} \frac {\partial p} {\partial \mu} \right)_{V,T}$ where $\frac {\partial p} {\partial \mu} = \frac {dp} {d\mu}= \frac {N} {V}$ under $T$ constant, as can be deduced from the Gibbs-Duhem relation. So you get:
$\left( \frac {\partial N} {\partial \mu} \right)_{V,T} = \left(V \frac {\partial \rho} {\partial p} \frac {\partial p} {\partial \mu} \right)_{V,T} = \left(V \frac {\partial \rho} {\partial p} \frac {N} {V} \right)_{V,T} = N \left(\frac {\partial \rho} {\partial p} \right)_{T}$

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