= Gibbs-Duhem equation
{c}
{title2=$S\,dT-V\,dp+\sum_i n_i\,d\mu_i=0$}
{wiki=Gibbs–Duhem_equation}
The <Euler theorem for homogeneous functions> applied to an <extensive quantity>, together with $dE=T\,dS-p\,dV+\mu\,dN$ give
$$
E=TS-pV+\mu N,
\qquad
S\,dT-V\,dp+N\,d\mu=0.
$$
The <Gibbs-Duhem equation> constrains changes of the <intensive quantities> describing one uniform <thermodynamic phase>. For a <mixture> with $k$ independently specified component amounts, its molar form is
$$
S\,dT-V\,dp+\sum_{i=1}^k n_i\,d\mu_i=0.
$$
Here $S$ is <entropy>, $V$ is <volume (thermodynamics)>, $T$ is <temperature>, $p$ is <pressure>, $n_i$ is the <amount of substance> of component $i$, and $\mu_i$ is its molar <chemical potential>. This assumes a <simple compressible system> whose bulk <internal energy> is an <extensive quantity>; additional <work> coordinates require additional terms.<Thermodynamics and Chemistry by Howard DeVoe>[Howard DeVoe, Thermodynamics and Chemistry, second edition, version 10, §9.2.8, printed p. 240, equations 9.2.42–9.2.44]{ref}
The proof exposes the source of the constraint. For <internal energy> $E(S,V,n_1,\ldots,n_k)$, replication of the whole <thermodynamic phase> by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\ldots,\lambda n_k)=\lambda E(S,V,n_1,\ldots,n_k)$. The <Euler theorem for homogeneous functions> and the fundamental <differential> of <internal energy> therefore give
$$
E=TS-pV+\sum_i\mu_i n_i,
\qquad
dE=T\,dS-p\,dV+\sum_i\mu_i\,dn_i.
$$
Taking the <differential> of the first equation produces all the terms in the second, plus $S\,dT-V\,dp+\sum_i n_i\,d\mu_i$. Subtraction proves the <Gibbs-Duhem equation>. Thus the constraint comes from two descriptions of the same <state function>, rather than an assumed microscopic interaction model.
At fixed <temperature> and <pressure>, divide by $n=\sum_i n_i>0$ and introduce the <mole fractions> $x_i=n_i/n$:
$$
\sum_i x_i\,d\mu_i=0.
$$
The <chemical potentials> cannot be varied independently. In a two-component <mixture>, put $x_A=x$ and $x_B=1-x$. Within a smooth single-<thermodynamic phase> region with $0<x<1$,
$$
x\left(\frac{\partial\mu_A}{\partial x}\right)_{T,p}
+(1-x)\left(\frac{\partial\mu_B}{\partial x}\right)_{T,p}=0,
\qquad
d\mu_B=-\frac{x}{1-x}\,d\mu_A.
$$
One <chemical potential> fixes the other's change, but determining its absolute value still requires a reference value. This does not say $\mu_A=\mu_B$: equality at <phase coexistence> concerns the same <chemical species> in different <thermodynamic phases>. Nor does the constraint say $dG=0$ whenever <temperature> and <pressure> are fixed; changing composition still contributes $\sum_i\mu_i\,dn_i$ to the <Gibbs free energy>.
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