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For a nonideal chemical solution, write its thermodynamic activities on a mole fraction basis as $a_A=x{\gamma }_A$ and $a_B=(1-x){\gamma }_B$. Use composition-independent pure-component reference chemical potentials at the same temperature and pressure:The activity coefficients describe departure from the ideal solution under that reference convention.^[ref]

At fixed temperature and pressure, insert these expressions into the Gibbs-Duhem equation. The ideal solution terms cancel because $xd\ln x+(1-x)d\ln (1-x)=dx-dx=0$. What remains isThus two independent fits to measured activity coefficients can each look smooth while failing to describe one possible Gibbs free energy. This is a necessary consistency condition; satisfying it alone does not establish experimental accuracy or thermodynamic stability.

For an original algebraic example, suppose at one fixed temperature and pressure that $\ln {\gamma }_A=\alpha (1-x{)}^2$, with dimensionless constant $\alpha$. The constraint requires $d\ln {\gamma }_B/dx=2\alpha x$. With the pure-B normalization ${\gamma }_B(0)=1$, integration givesFor $\alpha =1/2$ and $x=1/4$, the derivatives of the two logarithms are $-3/4$ and $1/4$; their weighted sum is $-3/16+3/16=0$. Choosing $\ln {\gamma }_B=\alpha x$ instead would give a weighted sum $3/16$ at that composition, contradicting the Gibbs-Duhem equation. These are invented model functions, with no claim to fit a particular mixture.

The fixed-temperature, fixed-pressure restriction matters. Along a path with changing temperature or pressure, the general equation contains $SdT-Vdp$, and the reference chemical potentials and $RT$ also change. Across phase coexistence, apply the relation separately within each smooth thermodynamic phase; do not differentiate a discontinuous switch between phases as if it were a single composition curve.

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.

The Gibbs free energy of mixing compares the final mixture's Gibbs free energy with that of its initially separate components at the same temperature and pressure. For an ideal solution, subtract the pure-component reference from $G={\sum }_i{n}_i{\mu }_i$:For one mole of the chosen A/B ideal solution at $300\mathrm{K}$ and $x_A=0.4$,The individual chemical potential changes are $RT\ln 0.4\approx -2.28554\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for A and $RT\ln 0.6\approx -1.27417\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for B. Weighting these by $0.4\mathrm{mol}$ and $0.6\mathrm{mol}$ reproduces the total Gibbs free energy of mixing.

If ideality holds in a neighborhood of the specified temperature and pressure, differentiation at fixed component amounts givesThus the ideal mixing Gibbs free energy decrease comes entirely from the entropy increase; the mixing enthalpy and volume changes vanish. Ideality at a single isolated state alone does not determine temperature or pressure derivatives.^[ref]

The same formula rules out spontaneous separation into two liquid thermodynamic phases within this model. Per mole, the mixing Gibbs free energy $g_{\mathrm{mix}}=RT[x\ln x+(1-x)\ln (1-x)]$ hasAdding the pure-component reference contributions changes only a term linear in $x$. The total molar Gibbs free energy is therefore a strictly convex function: a separated pair of different compositions has a larger composition-weighted Gibbs free energy than the uniform sample with the same inventory. Such phase separation requires physics beyond this ideal solution model.

An ideal solution has molar chemical potentialswhere $R$ is the molar gas constant and the stars denote pure-component reference values at the same temperature and pressure.^[ref]

Hold temperature and pressure fixed. The derivatives with respect to $x$ are $RT/x$ and $-RT/(1-x)$, soEach chemical potential changes, yet their composition-weighted slopes cancel exactly. At $x=1/4$, the two slopes are $4RT$ and $-4RT/3$; multiplying by $1/4$ and $3/4$ gives $RT$ and $-RT$. Near either pure-component endpoint the dilute component's slope diverges, but its weighted contribution remains finite. The endpoint chemical potential itself requires a limiting interpretation because its logarithm diverges.

The same constraint has a geometric explanation. Subtract the pure-component reference contribution from the molar Gibbs free energy and divide by $RT$ to obtainIts derivative is $f^{\prime }(x)=\ln [x/(1-x)]$. At fixed total amount of substance, exchanging a little B for A changes the Gibbs free energy with slope ${\mu }_A-{\mu }_B$. Consequently the tangent to $f$ at $x$ has interceptsat horizontal coordinates $0$ and $1$, respectively. The two chemical potentials come from the same curve: choosing two unrelated functions would generally destroy this tangent construction.

An ideal solution is a chemical solution whose component chemical potentials depend on composition throughHere $n_i$ is an amount of substance, $x_i$ its mole fraction, $R$ the molar gas constant, and ${\mu }_i^{\ast }$ the pure-component liquid reference at the same temperature $T$ and pressure $P$. This model applies to every component across the composition range; it is stronger than an approximation valid only near a pure solvent.^[ref]

With this pure-liquid standard state convention, the thermodynamic activity satisfies ${\mu }_i={\mu }_i^{\ast }+RT\ln a_i$, so an ideal solution has $a_i=x_i$, or activity coefficient ${\gamma }_i=a_i/x_i=1$.^[ref] This statement concerns the liquid. Assuming that a coexisting vapor is an ideal gas mixture is an additional approximation. Its mole fractions $y_i$ generally differ from the liquid mole fractions $x_i$.

For an original two-component example, stipulate $T=300\mathrm{K}$ and pure-component vapor pressures $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. A and B are invented distinguishable components; these numbers are chosen model inputs, not material data. Combine an ideal solution with an ideal gas vapor, neglect the pure-liquid reference's pressure correction, and exclude other gases. At liquid mole fractions $x_A=0.4$ and $x_B=0.6$, Raoult's law and addition of partial pressures giveThus the coexisting vapor contains about $72.73\%$ A by amount of substance, although the liquid contains only $40\%$. The total pressure is neither pure-component vapor pressure, and $x_A$ cannot be substituted for $y_A$. This enrichment is the basis of distillation.

The usefulness of the ideal solution is that one composition law connects chemical potentials, vapor-liquid equilibrium, entropy of mixing and Gibbs free energy of mixing. Its liquid components can interact strongly: the defining condition is their composition dependence, rather than the absence of all interactions as in an ideal gas model.

In the ideal solution and ideal gas vapor approximation, each component's partial pressure equals its liquid mole fraction times its pure-component vapor pressure at the same temperature:These pressure formulas neglect the change in the pure-liquid reference between its saturation pressure and $P$; appreciable real gas behavior also requires a more general treatment. The ideal solution definition in terms of chemical potentials does not require either approximation.^[ref]

A short derivation explains the reference choice. At vapor-liquid equilibrium, each component has the same chemical potential in the liquid and vapor. For an ideal gas mixture, chemical potential of an ideal gas giveswhere $p^{\circ }$ is a reference pressure. Approximate ${\mu }_i^{\ast }(T,P)$ by ${\mu }_i^{\ast }(T,p_i^{\ast })$ and subtract the pure-component equation from the mixture equation:Taking the exponential function gives Raoult's law. Summing partial pressures and normalizing gives $P$ and $y_i$; this last step uses the ideal gas vapor assumption separately from the liquid model.

The law is named after François-Marie Raoult, who measured vapor pressures and freezing behavior of chemical solutions.^[ref]