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The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

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 isHere $S$ is entropy, $V$ is volume, $T$ 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.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. 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 $x_i=n_i/n$: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 $0<x<1$,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 ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

An intensive quantity is unchanged under replication at the same thermodynamic equilibrium. Temperature, pressure, chemical potential and the mass density of a uniform substance are examples. Independence from sample size does not mean independence from state: a material's mass density can change with temperature or pressure.

The relation to extensive quantities is particularly elegant for a differentiable bulk fundamental relation. Put $\mathbf{x}=(S,V,n)$ and writeTaking a partial derivative with respect to $x_i$ and cancelling $\lambda$ givesThe equilibrium differential $dU=TdS-pdV+\mu dn$ identifies these derivatives with temperature, minus pressure and molar chemical potential. They are degree-zero homogeneous functions. The same degree-one assumption yields the Euler relation in thermodynamics, but a surface energy contribution requires extra care.

An intensive quantity generally has no meaningful sum across samples: two equal temperatures do not add to a doubled temperature. Neither is the final temperature after thermal contact generally an unweighted average; heat capacities and the first law of thermodynamics determine the balance. Ratios of additive inventories lead to the useful weighted rules in normalizing extensive quantities.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.