Result

Entropy of mixing

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Statement

For ideal gases or ideal solutions mixed at constant temperature and pressure, \(\Delta S_{\text{mix}}=-nR\sum_i x_i\ln x_i\), where \(x_i\) are mole fractions; since each \(x_i\) satisfies \(0\lt x_i\lt1\), \(\ln x_i\) is always negative, making \(\Delta S_{\text{mix}}\) always strictly positive for any genuine mixing. Because ideal mixing involves no new intermolecular interactions (\(\Delta H_{\text{mix}}=0\) exactly), \(\Delta G_{\text{mix}}=-T\Delta S_{\text{mix}}\lt0\) always — mixing of ideal substances is spontaneous purely from entropy increase, with no enthalpic driving force at all.

Why it matters

This result explains, from the same thermodynamic machinery already established in this unit, one of the most universally observed everyday phenomena: gases and miscible liquids spontaneously mix and never spontaneously separate on their own. It is a direct, elegant application of gibbs-equilibrium-constant's chemical potential formula, extended from chemical reactions and phase equilibrium to the mixing process itself, closing out this unit's progressive build from the first law through to a fully quantitative account of mixing.

Hypotheses

Mixing is ideal: no new intermolecular interactions form between the different components upon mixing.This is a genuine idealisation, directly analogous to the ideal-gas assumptions already flagged in van-der-waals-equation. Real mixtures generally do have some nonzero \(\Delta H_{\text{mix}}\) — positive (unfavourable) for components that interact less favourably with each other than with themselves, negative (favourable) for components with strong new interactions (e.g. hydrogen bonding) upon mixing — and this real-world departure from ideality is exactly what explains limited or zero miscibility for some substance pairs (Fails without). Mixing occurs at constant temperature and pressure.Consistent with the constant-\(T,P\) condition used throughout this unit for Gibbs free energy (gibbs-free-energy) and the equilibrium relations built on it.

Proof

\[ \mu_i(\text{mixture}) = \mu_i^\circ + RT\ln\frac{P}{P^\circ} + RT\ln x_i \]

Using gibbs-equilibrium-constant's chemical potential formula with the partial pressure \(P_i=x_iP\) (Dalton's law, for an ideal gas mixture at total pressure \(P\)) splits the logarithm into a term shared by every component (independent of \(x_i\)) and a term depending explicitly on mole fraction. A

\[ \Delta G_{\text{mix}} = G_{\text{mixture}} - \sum_i G_{i,\text{pure}} = RT\sum_i n_i\ln x_i \]

Comparing the total Gibbs free energy after mixing (summing Step 1 over all components, weighted by moles \(n_i\)) against the sum of each component's Gibbs free energy as a separate pure substance at the same total pressure \(P\): the shared, \(x_i\)-independent term is identical before and after mixing and cancels exactly, leaving only the mole-fraction-dependent contribution. A

\[ \Delta H_{\text{mix}}=0 \ (\text{ideal, Hypotheses}) \ \Longrightarrow\ \Delta G_{\text{mix}} = -T\Delta S_{\text{mix}} \]

Since ideal mixing has no enthalpy change, \(\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}\) (gibbs-free-energy) reduces to purely the entropy term. A

\[ \Delta S_{\text{mix}} = -\frac{\Delta G_{\text{mix}}}{T} = -R\sum_i n_i\ln x_i = -nR\sum_i x_i\ln x_i \]

Solving Step 3 for \(\Delta S_{\text{mix}}\) using Step 2's result, and writing \(n_i=x_in\) (total moles \(n\)) gives the standard entropy-of-mixing formula. Since each \(x_i\in(0,1)\), \(\ln x_i\lt0\), so each term \(-x_i\ln x_i>0\) — the sum, and hence \(\Delta S_{\text{mix}}\), is always strictly positive whenever more than one component is genuinely present. A

Result

\[ \Delta S_{\text{mix}} = -nR\sum_i x_i\ln x_i \ > 0\ \text{always}, \qquad \Delta G_{\text{mix}} = -T\Delta S_{\text{mix}} < 0\ \text{always (ideal mixing)} \]

Reading. Mixing distinguishable ideal components always increases entropy and always decreases Gibbs free energy, regardless of which specific substances are involved or in what proportion — spontaneous mixing is a purely statistical, entropy-driven consequence, not something requiring any favourable enthalpic interaction between the components.

Scope. Requires ideal mixing (Hypotheses); real, non-ideal mixtures can have \(\Delta H_{\text{mix}}\ne0\), which can, in extreme cases, overcome the entropy term and prevent spontaneous mixing altogether (Fails without).

Corollaries & converses

For a two-component mixture, \(\Delta S_{\text{mix}}\) is maximised at an equimolar composition (\(x_1=x_2=0.5\)), by the symmetry of \(-x\ln x-(1-x)\ln(1-x)\) about \(x=0.5\); highly unequal mixtures (e.g. \(x_1=0.9\), \(x_2=0.1\)) give a noticeably smaller entropy of mixing (Worked examples, Problems).
Mixing more distinct components generally increases the total entropy of mixing further, since additional terms (each individually positive) are added to the sum — a three-component equal mixture has a larger \(\Delta S_{\text{mix}}\) per mole than a two-component equal mixture.
Converse: since \(\Delta S_{\text{mix}}\) depends only on the number of distinguishable components and their mole fractions, not on which specific substances are involved, it says nothing at all about \(\Delta H_{\text{mix}}\) or real-world miscibility — a real mixture's actual spontaneity still requires \(\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}\lt0\) overall, not \(\Delta S_{\text{mix}}>0\) alone.

Fails without

Assume all substances mix spontaneously in all proportions, since \(\Delta S_{\text{mix}}\) is always positive (drop Hypotheses): oil and water are the classic real-world counterexample — their mutual intermolecular interactions are sufficiently unfavourable (a substantial positive \(\Delta H_{\text{mix}}\), driven by water's strong hydrogen-bonding network being disrupted) that \(\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}\) remains positive despite the always-favourable entropy term, and the two liquids remain largely immiscible at ordinary temperature.
Apply the formula naively to mixing two samples of the identical gas (indistinguishable particles): the formula as derived assumes distinguishable components; mixing identical particles involves no actual physical rearrangement that could be detected or reversed, and careful treatment (accounting for particle indistinguishability, a statistical-mechanics refinement beyond this introductory scope) shows the true entropy change in that specific case is exactly zero, not the positive value a naive application of the formula to "\(x_1=x_2=0.5\) of the same substance" would suggest — the well-known historical "Gibbs paradox" (Discussion).

Common errors

Dropping or mishandling the leading negative sign in the formula, since \(\ln x_i\) is always negative for \(0\lt x_i\lt1\); the negative sign is essential to obtain the correct, always-positive \(\Delta S_{\text{mix}}\).
Assuming a positive \(\Delta S_{\text{mix}}\) alone guarantees spontaneous mixing for any real substance pair, without checking \(\Delta H_{\text{mix}}\) (Fails without, first bullet).
Applying the distinguishable-components formula to mixing identical particles (Fails without, second bullet, the Gibbs paradox).
Using mass fraction or volume fraction in place of mole fraction \(x_i\) in the formula, which requires mole fractions specifically.

Discussion

Entropy of mixing became clearly understood once Gibbs's chemical potential framework (gibbs-free-energy's Discussion) was applied specifically to solutions and gas mixtures in the late 19th century. The formula's derivation also surfaces a genuinely famous historical puzzle, the "Gibbs paradox": naively applying the mixing-entropy formula to combining two samples of the identical gas predicts a nonzero entropy increase, even though physically nothing has actually changed — swapping the positions of indistinguishable particles produces no detectable difference at all. The resolution, only fully understood with the later development of statistical mechanics and quantum indistinguishability, is that the classical derivation implicitly (and incorrectly, for identical particles) treats every particle as distinguishable; correctly accounting for particle indistinguishability removes the paradoxical prediction entirely for the identical-gas case, while leaving the ordinary formula (Result) fully valid whenever the mixed components are genuinely distinct substances.

The always-positive nature of \(\Delta S_{\text{mix}}\) for ideal mixing is a direct, quantitative instance of the broader statistical-mechanical principle (briefly touched on in entropy-second-law's Hypotheses) that entropy measures the number of accessible microscopic arrangements: a mixed state has vastly more ways to arrange the same total number of particles among the same total volume than a fully separated state does, and this combinatorial abundance is precisely what the logarithmic mole-fraction terms in the Result quantify.

Common misconception: that entropy of mixing being always positive means all substances are always miscible. As the oil-and-water example demonstrates directly (Fails without), real-world miscibility is governed by the full \(\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}\), not by the entropy term in isolation; \(\Delta S_{\text{mix}}>0\) is a universal, always-favourable contribution toward mixing, not a guarantee of it.

Worked examples

\[ 1.00\,\text{mol gas A} + 1.00\,\text{mol gas B (equimolar, } n=2.00\,\text{mol), at }298\,\text{K}: \quad \Delta S_{\text{mix}}=-(2.00)(8.314)\left[0.5\ln0.5+0.5\ln0.5\right]\approx11.53\,\text{J/K} \]

A direct application of the Result to the maximum-entropy (equimolar) two-component case; \(\Delta G_{\text{mix}}=-T\Delta S_{\text{mix}}=-(298)(11.53)\approx-3.43\,\text{kJ}\), confirming spontaneous mixing with zero enthalpic contribution, purely from the entropy term. A

\[ \text{Air-like mixture: } 3.00\,\text{mol N}_2 + 1.00\,\text{mol O}_2\ (x_{\text{N}_2}=0.75,\ x_{\text{O}_2}=0.25,\ n=4.00\,\text{mol}) \]

\(\Delta S_{\text{mix}}=-(4.00)(8.314)\left[0.75\ln0.75+0.25\ln0.25\right]\approx18.70\,\text{J/K}\) — larger in absolute terms than Worked example 1's equimolar case, since more total moles are involved, though this specific, unequal \(3{:}1\) ratio gives a smaller per-mole entropy of mixing than a perfectly equimolar mixture would (Corollaries, first bullet). A

\[ \text{Equimolar (1:1): } \Delta S_{\text{mix}}\approx11.53\,\text{J/K}; \qquad \text{Air-like (3:1), } n=4\,\text{mol: } \Delta S_{\text{mix}}\approx18.70\,\text{J/K} \]

Reading. The formula applies identically to any mole-fraction composition and any total amount of substance, always giving a positive entropy of mixing.

Scope. Both examples treat the gases as ideal, per the Result's scope restriction.

Further example. A sealed vessel contains \(5.00\,\text{mol}\) of helium gas and \(5.00\,\text{mol}\) of neon gas at \(298\,\text{K}\). Calculate the entropy of mixing \(\Delta S_{\text{mix}}\) and the Gibbs free energy of mixing \(\Delta G_{\text{mix}}\) for this ideal gas mixture.

\[ n_{\text{total}} = 5.00 + 5.00 = 10.00\,\text{mol}; \quad x_{\text{He}} = \frac{5.00}{10.00} = 0.500; \quad x_{\text{Ne}} = \frac{5.00}{10.00} = 0.500 \]

Determine the total moles and the mole fractions for each component, which are required inputs for the entropy of mixing formula. A

\[ \Delta S_{\text{mix}} = -n_{\text{total}} R \sum_i x_i \ln x_i = -(10.00)(8.314)\left[0.500\ln(0.500) + 0.500\ln(0.500)\right] \]

Substitute the values into the standard formula for ideal mixing. Since the mixture is equimolar, the sum simplifies to \(-\ln 2\). A

\[ \Delta S_{\text{mix}} = -(10.00)(8.314)\left[2 \times 0.500 \times (-0.6931)\right] = -(83.14)(-0.6931) \approx 57.62\,\text{J/K} \]

Perform the arithmetic. Note that this is exactly double the value in Worked example 1 because the total number of moles is doubled while the composition (equimolar) remains the same. A

\[ \Delta G_{\text{mix}} = -T\Delta S_{\text{mix}} = -(298\,\text{K})(57.62\,\text{J/K}) \approx -17171\,\text{J} = -17.2\,\text{kJ} \]

Calculate the Gibbs free energy change using the relation for ideal mixing where \(\Delta H_{\text{mix}}=0\). The negative sign confirms spontaneity. A

\[ \Delta S_{\text{mix}} \approx 57.6\,\text{J/K}; \qquad \Delta G_{\text{mix}} \approx -17.2\,\text{kJ} \]

Reading. The entropy of mixing scales linearly with the total amount of substance for a fixed composition. The large negative \(\Delta G\) indicates a strong thermodynamic driving force for mixing, entirely derived from the increase in configurational entropy.

Scope. Valid for ideal gases or ideal solutions where intermolecular interactions are identical before and after mixing (\(\Delta H_{\text{mix}}=0\)).

Problems

Compute \(\Delta S_{\text{mix}}\) for an equimolar three-component mixture (\(1.00\,\text{mol}\) each of three different ideal gases, \(x_i=1/3\) each, \(n=3.00\,\text{mol}\) total) at \(298\,\text{K}\), and compare with the two-component equimolar result from Worked example 1.
Solution
\(\Delta S_{\text{mix}}=-(3.00)(8.314)\left[3\times\left(\tfrac13\ln\tfrac13\right)\right]\approx27.40\,\text{J/K}\), substantially larger than the two-component equimolar result (\(11.53\,\text{J/K}\)), consistent with the Corollaries' claim that mixing more distinct components increases total entropy of mixing further.
Compute \(\Delta S_{\text{mix}}\) for an unequal, \(0.900{:}0.100\) mole-fraction two-component mixture (\(n=2.00\,\text{mol}\) total), and compare with Worked example 1's equimolar (\(0.500{:}0.500\)) result to confirm the equimolar case gives the maximum entropy of mixing.
Solution
\(\Delta S_{\text{mix}}=-(2.00)(8.314)\left[0.900\ln0.900+0.100\ln0.100\right]\approx5.41\,\text{J/K}\), noticeably smaller than the equimolar result (\(11.53\,\text{J/K}\)), confirming numerically that a highly unequal composition gives a smaller entropy of mixing than the equimolar (\(x=0.5\)) case, consistent with the Corollaries' maximum-at-equimolar claim.
Explain why oil and water do not spontaneously mix into a single homogeneous phase, despite \(\Delta S_{\text{mix}}\) being positive for any composition, using the Result and Fails without.
Solution
\(\Delta S_{\text{mix}}>0\) is a genuine, universal contribution favouring mixing for any substances, but real spontaneity is governed by the full \(\Delta G_{\text{mix}}=\Delta H_{\text{mix}}-T\Delta S_{\text{mix}}\), not the entropy term alone. Oil and water have a substantially positive \(\Delta H_{\text{mix}}\), since mixing them requires disrupting water's extensive hydrogen-bonding network without forming comparably favourable new interactions with the largely nonpolar oil molecules; this unfavourable enthalpy term is large enough to make \(\Delta G_{\text{mix}}>0\) overall despite the favourable entropy contribution, so the two liquids remain largely immiscible — a direct, real-world instance of Fails without's first bullet.
Explain, using the Discussion's account of the Gibbs paradox, why mixing two samples of the exact same pure gas does not actually increase entropy, even though naively substituting \(x_1=x_2=0.5\) into the Result's formula (as if the two samples were distinguishable) would suggest a positive \(\Delta S_{\text{mix}}\).
Solution
The Result's derivation (Step 2 onward) implicitly assumes the mixed components are distinguishable, different substances; the entropy increase it predicts comes from the vastly larger number of distinguishable arrangements available once different types of particles can occupy the combined volume. When the two "components" being mixed are actually identical particles, there is no genuine new distinguishable arrangement created by combining them — swapping identical particles' positions produces no physically detectable difference at all — so the true entropy change is exactly zero, not the positive value the naive formula would give if (incorrectly) treated as if the two samples were distinguishable. This resolution, requiring careful accounting of particle indistinguishability, is precisely the historical Gibbs paradox.
Calculate the entropy of mixing \(\Delta S_{\text{mix}}\) for a mixture containing \(2.00\,\text{mol}\) of argon and \(1.00\,\text{mol}\) of krypton at \(298\,\text{K}\). Assume ideal behavior.
Solution
First, determine the total moles and mole fractions: \[ n_{\text{total}} = 2.00 + 1.00 = 3.00\,\text{mol} \] \[ x_{\text{Ar}} = \frac{2.00}{3.00} = \frac{2}{3} \approx 0.6667 \] \[ x_{\text{Kr}} = \frac{1.00}{3.00} = \frac{1}{3} \approx 0.3333 \] Next, apply the entropy of mixing formula: \[ \Delta S_{\text{mix}} = -n_{\text{total}} R (x_{\text{Ar}} \ln x_{\text{Ar}} + x_{\text{Kr}} \ln x_{\text{Kr}}) \] \[ \Delta S_{\text{mix}} = -(3.00)(8.314) \left[ \left(\frac{2}{3}\right)\ln\left(\frac{2}{3}\right) + \left(\frac{1}{3}\right)\ln\left(\frac{1}{3}\right) \right] \] Calculate the term in the brackets: \[ \ln(2/3) \approx -0.4055 \] \[ \ln(1/3) \approx -1.0986 \] \[ \text{Sum} = \left(\frac{2}{3}\right)(-0.4055) + \left(\frac{1}{3}\right)(-1.0986) \] \[ \text{Sum} = -0.2703 - 0.3662 = -0.6365 \] Finally, compute \(\Delta S_{\text{mix}}\): \[ \Delta S_{\text{mix}} = -(24.942)(-0.6365) \approx 15.88\,\text{J/K} \] Rounding to 3 significant figures: \[ \Delta S_{\text{mix}} \approx 15.9\,\text{J/K} \]