Ideal Solutions and Raoult's Law

Introduction

Dissolving a second liquid or a nonvolatile solute generally changes the vapour pressure above a solvent. The simplest model is an ideal solution, in which each component's chemical potential contains a logarithm of its liquid mole fraction. Equating liquid and vapour chemical potentials then yields Raoult's law. The route explains why lowering solvent mole fraction lowers its equilibrium vapour pressure.

Core explanation

For component i in an ideal liquid solution at T and p, μ i(liquid) = μ i (liquid) + RT ln x i, where μ i is the pure-liquid chemical potential at the same T and p. Since x i ≤ 1, the mixing term is zero for the pure component and negative when diluted. This expression is a model assumption about mixture behaviour, not a universal law for all solutions. An ideal solution has zero enthalpy and volume changes of mixing in the idealised thermodynamic description at specified conditions, and its Gibbs-energy lowering comes from mixing entropy.

At liquid–vapour equilibrium, μ i(liquid) = μ i(vapour). Treat the vapour as ideal and compare with equilibrium over the pure liquid, whose vapour pressure is p i . For the solution, μ i(vapour) = μ i° + RT ln(p i/p°), while for the pure liquid at the same temperature μ i = μ i° + RT ln(p i /p°), neglecting or consistently accounting for pressure effects on the liquid. Subtracting gives RT ln x i = RT ln(p i/p i ), so p i = x i p i . This is Raoult's law for each component of a fully ideal mixture.

For a nonvolatile solute B, essentially only solvent A contributes to vapour pressure, giving p A = x A p A . The lowering is Δp = p A − p A = x B p A for a binary ideal solution because x A + x B = 1. At dilute x B, the relative lowering Δp/p A equals x B. This relationship depends on counting dissolved particles appropriately; if a solute dissociates or associates, the relevant particle mole fraction is altered.

Real solutions can deviate because unlike and like molecular interactions differ. A more general expression is μ i = μ i + RT ln a i, where activity a i replaces x i. For a Raoult-standard component, a i = γ i x i and γ i approaches one in the appropriate pure-component limit. The ideal law remains a useful baseline for interpreting deviations and colligative properties.

Step-by-step reasoning

Choose the solvent and calculate liquid mole fractions from amounts, not masses. Write its ideal liquid μ formula. Equate that μ to its vapour μ and compare with the pure-solvent equilibrium at the same T. The standard-state terms cancel, giving p i = x i p i . If a nonvolatile solute is present, set only its vapour contribution negligible; do not set its liquid mole fraction to zero.

Visual explanation

Plot solvent vapour pressure p A against x A. The ideal Raoult line runs straight from zero at x A = 0 to p A at x A = 1. Draw a point below the pure-solvent endpoint after solute is added and label the vertical drop x B p A . A second plot of μ A against ln x A shows the logarithmic liquid chemical-potential decrease.

Real-world analogy

If fewer solvent molecules occupy a fraction of the positions at a liquid surface, fewer can contribute to the equilibrium vapour. That picture helps remember the direction of vapour-pressure lowering. The actual thermodynamic derivation uses equality of chemical potentials; simple surface coverage alone is insufficient for many real solutions.

Real-world example

Adding a small amount of a nonvolatile molecular solute to water lowers water's vapour pressure at a given temperature. This lowering contributes to boiling-point elevation and freezing-point depression. The ideal-solution formula estimates the change when the solution is sufficiently close to ideal and the solute remains as counted particles.

Why?

Mixing lowers the solvent's liquid chemical potential by RT ln x A. At the original pure-solvent vapour pressure, the vapour chemical potential would now be too high for equilibrium with the diluted liquid. Vapour pressure must fall until its logarithmic μ decreases by the same amount. Equating the two logarithms produces the proportionality p A = x Ap A .

Common misconception

Raoult's law uses liquid mole fraction, not mass fraction or vapour mole fraction. It also does not state that every real solution obeys a straight pressure–composition line. A nonvolatile solute has negligible vapour pressure but still changes x A and hence the solvent's pressure. Finally, lowering p A does not mean the solvent has stopped evaporating; evaporation and condensation still occur at equilibrium.

Worked example

Mix 9.0 mol of solvent A with 1.0 mol of nonvolatile solute B. Then x A = 9.0/10.0 = 0.90. If pure A has vapour pressure p A = 40.0 kPa at the temperature, ideal Raoult behaviour predicts p A = 0.90 × 40.0 = 36.0 kPa. The lowering is 4.0 kPa, or 10% of the pure pressure, equal to x B = 0.10. This does not predict the solute's own vapour peak because it is assumed nonvolatile.

Quick check

1. If an ideal solvent's liquid mole fraction changes from 1.00 to 0.80 at fixed T, what happens to its vapour pressure? Answer: It becomes 0.80 of the pure-solvent vapour pressure: p A = 0.80p A . The relative lowering is 0.20 under the ideal, nonvolatile-solute model.

Exam focus

Use x i from liquid amounts and p i at the same temperature. Derive Raoult's law by equating chemical potentials if requested, rather than relying only on a surface story. For a binary nonvolatile solute, Δp/p A = x B. Mention activity and deviations when the mixture is real or concentrated.

Advanced insight

An ideal binary solution with two volatile components gives p total = x Ap A + x Bp B . Its vapour composition is y i = p i/p total and usually differs from liquid composition x i. This difference underlies distillation even when the liquid solution itself follows Raoult's law closely.

Summary

An ideal liquid component has μ i = μ i + RT ln x i. Equilibrium with an ideal vapour gives Raoult's law p i = x i p i . A nonvolatile solute lowers solvent pressure in proportion to the solvent mole-fraction decrease. Real solutions require activities and may deviate from the ideal straight-line relation.

Practice questions

1. A binary ideal liquid has x A = 0.30, p A = 20 kPa and p B = 50 kPa. Find both partial vapour pressures and total pressure. Answer: x B = 0.70, so p A = 0.30(20) = 6 kPa and p B = 0.70(50) = 35 kPa. Total pressure is 41 kPa. 2. Why is p A lower over an ideal solution than over pure A at the same T when x A < 1? Answer: Mixing lowers liquid μ A by RT ln x A. The equilibrium vapour must have lower μ A too, which requires lower p A through its logarithmic pressure dependence. 3. A student uses 10 g A and 10 g B to set x A = 0.5. What information is missing? Answer: Mole fractions require mole amounts, so the molar masses of A and B are needed. Equal masses do not generally mean equal numbers of molecules.