Two-Component Liquid–Vapour Diagrams

Imagine boiling a flask of vodka. The vapour that escapes is not 40% ethanol like the liquid. It is far richer in alcohol. This gap between liquid and vapour composition is the entire reason distillation works. Two-component liquid–vapour diagrams give us the map for this behaviour. They plot how pressure or temperature changes with composition, and they show two distinct curves: one for the liquid, one for the vapour in equilibrium with it. Reading these diagrams correctly is the foundation for everything that follows in binary phase equilibria. This article builds the ideal p–x diagram from Raoult's law, then converts it into the more practically useful T–x diagram. You will learn why the two curves separate. You will learn how to read tie lines. You will see two fully worked numerical examples. The lever rule and fractional distillation, covered next, both depend on the diagrams built here.

Key takeaways

For an ideal binary mixture, Raoult's law gives a straight-line total vapour pressure versus liquid composition.
The vapour composition differs from the liquid composition whenever the two components have different vapour pressures.
Plotting vapour pressure against vapour composition instead of liquid composition curves the line into the "vapour curve."
A T–x diagram flips the picture: temperature rises as volatile component depletes, producing a lens-shaped two-phase region.
Any point inside the two-phase lens splits into a liquid point and a vapour point connected by a horizontal tie line.
Key idea: because liquid and vapour in equilibrium never share the same composition, plotting both curves on one diagram reveals a separation gap that distillation exploits.

The core insight

In an ideal binary mixture, Raoult's law fixes the total vapour pressure as a linear function of the liquid mole fraction, but Dalton's law converts that pressure into a vapour composition that is enriched in the more volatile component; plotting both the liquid line and the resulting vapour curve on the same axes is what makes a p–x or T–x phase diagram.

The physical picture

Consider a binary liquid mixture of components A and B that forms an ideal solution — meaning A–B interactions resemble A–A and B–B interactions closely enough that Raoult's law holds across the whole composition range (see Raoult's Law and Ideal Solutions). Each component contributes a partial vapour pressure proportional to its mole fraction in the liquid, and the total pressure is the sum. Because the two pure vapour pressures $p_A^*$ and $p_B^*$ are generally different, the vapour phase does not have the same composition as the liquid it came from. The vapour is always richer in whichever component has the higher vapour pressure — the more volatile one. This is the single fact from which the entire shape of the phase diagram follows. At fixed temperature, plotting total pressure against liquid mole fraction gives a straight line — the liquidus (in this context often just called the liquid line). Plotting total pressure against the vapour mole fraction, computed from Dalton's law, gives a curve that bows away from the liquid line, since $y_A \neq x_A$ except at the pure-component endpoints. Together the two lines bound a lens-shaped two-phase region: above it only liquid exists (high pressure, both phases squeezed to liquid), below it only vapour exists, and inside it liquid and vapour coexist. When we instead fix the total pressure (typically 1 atm, as in a distillation column open to the atmosphere) and plot the boiling temperature against composition, the diagram inverts: temperature is now the vertical axis, the liquidus curves downward toward the more volatile component, and the vapour curve lies below it. The two-phase lens still appears, but now bounded above by the vapour curve and below by the liquid curve, because higher temperature corresponds to more vapour. The Gibbs phase rule confirms why a two-dimensional diagram is exactly what we need here. For $C = 2$ components and $P$ phases, $F = C - P + 2$ . In the single-phase liquid or vapour regions, $P=1$ so $F=3$ ; but one variable (pressure or temperature) is already fixed as the diagram's other axis, leaving two degrees of freedom — composition and the remaining variable — matching the two-dimensional plot. In the two-phase region, $P=2$ , so $F=2$ ; at fixed pressure (T–x diagram) or fixed temperature (p–x diagram), only one degree of freedom remains, which is why the tie line at a given temperature (or pressure) connects only one specific liquid composition to one specific vapour composition — not a range.

Figure

p–x diagram for an ideal binary mixture at fixed temperature

Figure

T–x diagram at fixed pressure showing the two-phase lens

The core relations

p = x_A p_A^* + (1-x_A) p_B^* = p_B^* + (p_A^* - p_B^*)x_A

where: $p$ is total vapour pressure, $x_A$ liquid mole fraction of A, $p_A^*, p_B^*$ pure-component vapour pressures.

y_A = \frac{x_A p_A^*}{x_A p_A^* + (1-x_A) p_B^*}

where: $y_A$ is the mole fraction of A in the vapour, obtained from Dalton's law $y_A = p_A/p$ .

p = \frac{p_A^* p_B^*}{p_A^* + (p_B^* - p_A^*)y_A}

where: this rewrites $p$ as a function of vapour composition $y_A$ , tracing the vapour curve on the p–x diagram.

F = C - P + 2

where: Gibbs phase rule; $F$ degrees of freedom, $C$ components, $P$ phases (see The Gibbs Phase Rule).

Derivation AFrom Raoult's law to the vapour curve

Goal. Starting only from Raoult's law for each component, derive the vapour-composition expression $y_A(x_A)$ and the pressure-vs-vapour-composition curve $p(y_A)$ , showing why liquid and vapour compositions differ.

Step 1. Raoult's law states each component's partial vapour pressure is proportional to its liquid mole fraction:

p_A = x_A p_A^*, \qquad p_B = x_B p_B^* = (1-x_A)p_B^*

This holds across the whole range $0 \le x_A \le 1$ only for an ideal solution.

Step 2. Total pressure is the sum of partial pressures (Dalton's law applied to the vapour in equilibrium with the liquid):

p = x_A p_A^* + (1-x_A)p_B^*

This is linear in $x_A$ , so on a p–x plot the liquid line is straight, running from $p_B^*$ at $x_A=0$ to $p_A^*$ at $x_A=1$ .

Step 3. The mole fraction of A in the vapour is, by Dalton's law, the ratio of A's partial pressure to the total:

y_A = \frac{p_A}{p} = \frac{x_A p_A^*}{x_A p_A^* + (1-x_A)p_B^*}

Notice immediately that if $p_A^* > p_B^*$ (A more volatile), then $y_A > x_A$ for every intermediate composition — the vapour is enriched in A. This is the origin of the whole separation effect.

Step 4. To plot pressure against $y_A$ rather than $x_A$ , invert Step 3 for $x_A$ in terms of $y_A$ :

x_A = \frac{y_A p_B^*}{p_A^* + (p_B^* - p_A^*)y_A}

and substitute into the Step 2 expression for $p$ . After simplifying the algebra,

p = \frac{p_A^* p_B^*}{p_A^* + (p_B^* - p_A^*)y_A}

Step 5. This is a hyperbola, not a straight line, in $y_A$ . Because $p_A^* \ne p_B^*$ in general, the curve bows below the straight liquid line (on a p–x diagram, where higher $p$ is "more liquid-like"). The two curves meet only at the pure-component edges $x_A=y_A=0$ and $x_A=y_A=1$ , where there is nothing left to differ about.

\boxed{y_A = \dfrac{x_A p_A^*}{x_A p_A^* + (1-x_A)p_B^*}, \qquad p(y_A) = \dfrac{p_A^* p_B^*}{p_A^* + (p_B^*-p_A^*)y_A}}

Worked examples

Example 1 — Building a p–x point by point (benzene/toluene at 25 °C). Take a classic ideal pair: benzene (A) and toluene (B) at 25 °C, with $p_A^* = 12.8\ \text{kPa}$ and $p_B^* = 3.79\ \text{kPa}$ (benzene is more volatile, consistent with its lower boiling point of 80 °C versus toluene's 111 °C). Suppose the liquid mole fraction of benzene is $x_A = 0.40$ . Step 1: total pressure from Raoult's law.

p = (0.40)(12.8\ \text{kPa}) + (0.60)(3.79\ \text{kPa}) = 5.12 + 2.27 = 7.39\ \text{kPa}

Step 2: vapour composition from Dalton's law.

y_A = \frac{(0.40)(12.8)}{7.39} = \frac{5.12}{7.39} = 0.693

Check: $y_A = 0.693 \gg x_A = 0.40$ , confirming strong enrichment in the volatile component, exactly as Step 3 of the derivation predicts since $p_A^*/p_B^* = 3.38$ is large. Step 3: locate this point on both curves. On the liquid line, the point $(x_A, p) = (0.40, 7.39\ \text{kPa})$ lies on the straight segment between $(0, 3.79)$ and $(1, 12.8)$ — check: $3.79 + 0.40(12.8-3.79) = 3.79+3.60=7.39$ kPa ✓. On the vapour curve, the same total pressure corresponds to $(y_A, p) = (0.693, 7.39\ \text{kPa})$ . These two points, at the same pressure, are joined by a horizontal tie line — the liquid and vapour compositions that coexist at $p = 7.39$ kPa. Sanity check: at $x_A = 0$ , $p = p_B^* = 3.79$ kPa and $y_A = 0$ ; at $x_A=1$ , $p=p_A^*=12.8$ kPa and $y_A=1$ . Both endpoints match the pure-component vapour pressures, as required.

Example 2 — Reading a T–x tie line and estimating enrichment per distillation stage. A binary mixture of heptane (A, more volatile, $T_A^* = 98.4\ ^\circ\text{C}$ ) and octane (B, $T_B^* = 125.6\ ^\circ\text{C}$ ) is charged into a still at 1 atm with overall liquid composition $x_A = 0.30$ . Treat the mixture as ideal with vapour pressures (in kPa, from the Antoine equation or tabulated data) at three trial temperatures:

$T$ (°C)	$p_A^*$ (kPa)	$p_B^*$ (kPa)
105	136.2	61.3
110	151.0	68.9
115	167.0	77.1

We need the boiling temperature of the $x_A = 0.30$ liquid at total pressure $p = 101.3$ kPa (1 atm), then the vapour composition in equilibrium with it. Step 1: at each trial $T$ , compute $p = x_A p_A^* + (1-x_A)p_B^*$ and compare to 101.3 kPa. At 105 °C: $p = 0.30(136.2) + 0.70(61.3) = 40.86 + 42.91 = 83.8$ kPa — too low (liquid hasn't reached its boiling point at 1 atm yet). At 110 °C: $p = 0.30(151.0) + 0.70(68.9) = 45.30 + 48.23 = 93.5$ kPa — still low. At 115 °C: $p = 0.30(167.0) + 0.70(77.1) = 50.10 + 53.97 = 104.1$ kPa — now slightly above 101.3 kPa. Interpolating linearly between 110 °C (93.5 kPa) and 115 °C (104.1 kPa) for target 101.3 kPa:

T_b \approx 110 + 5 \times \frac{101.3 - 93.5}{104.1-93.5} = 110 + 5\times\frac{7.8}{10.6} = 110 + 3.68 = 113.7\ ^\circ\text{C}

Step 2: at $T_b \approx 113.7\ ^\circ\text{C}$ , interpolate the pure vapour pressures similarly: $p_A^* \approx 151.0 + \tfrac{3.68}{5}(167.0-151.0) = 151.0+11.8=162.8$ kPa; $p_B^* \approx 68.9+\tfrac{3.68}{5}(77.1-68.9)=68.9+6.0=74.9$ kPa. Step 3: compute the vapour composition at the boiling point.

y_A = \frac{x_A p_A^*}{p} = \frac{(0.30)(162.8)}{101.3} = \frac{48.84}{101.3} = 0.482

So at 1 atm and 113.7 °C, a liquid with $x_A = 0.30$ boils to give a vapour with $y_A \approx 0.48$ — the tie line on the T–x diagram runs horizontally from $(0.30,\ 113.7\,^\circ\text{C})$ on the liquidus to $(0.48,\ 113.7\,^\circ\text{C})$ on the dew curve. Sanity check: $y_A = 0.48 > x_A = 0.30$ , correctly enriched in the more volatile heptane; and $T_b = 113.7\,^\circ\text{C}$ lies between the two pure boiling points (98.4 °C and 125.6 °C), as any mixture's boiling point must. This single-stage enrichment from 30% to 48% is exactly the mechanism that fractional distillation repeats stage after stage.

Limiting cases & common traps

Situation	What happens	Why
$p_A^* = p_B^*$	Liquid and vapour curves coincide; $y_A = x_A$ everywhere	No volatility difference means no separation possible; azeotrope-like degeneracy
$x_A \to 0$ or $1$	Liquid and vapour curves meet	Only one component present, so compositions must agree trivially
Reading a p–x diagram at fixed $T$	Higher pressure = more liquid	Compressing the vapour condenses it; opposite sense to T–x diagrams
Reading a T–x diagram at fixed $p$	Higher temperature = more vapour	Heating boils the liquid; do not transfer p–x intuition directly
Non-ideal mixtures	Curves can show a maximum or minimum (azeotrope)	Raoult's law fails when A–B interactions differ strongly from A–A, B–B

The chemist behind it

The p–x and T–x diagram framework rests directly on François-Marie Raoult's empirical law from the 1880s, but the graphical machinery for reading multi-phase equilibria — tie lines, the lever rule, and the phase rule that guarantees these diagrams have exactly the right number of independent variables — comes from Josiah Willard Gibbs. Gibbs published his phase rule in the 1870s, decades before chemists such as Ernest Sorel began applying it explicitly to distillation column design in the 1890s. It was really the synthesis of Raoult's linear pressure law with Gibbs's rule for counting degrees of freedom that turned distillation from a craft into a predictive science, letting engineers calculate how many theoretical plates a column needs before ever building one.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist whose phase rule underlies every liquid–vapour diagram

Ask the tutor: How do I use the lever rule on one of these tie lines to find the relative moles of liquid and vapour present?

Why does the vapour curve bow away from the straight liquid line on a p–x diagram?

The liquid line is straight because Raoult's law makes total pressure a linear function of liquid mole fraction $x_A$ . But the vapour curve plots the same pressure against the vapour mole fraction $y_A$ , which is a different, nonlinear function of $x_A$ (derived from Dalton's law). Substituting the inverse relation turns the straight line into a hyperbola in $y_A$ , always bowing toward lower pressure except at the pure endpoints where $x_A = y_A$ .

Why do T–x and p–x diagrams look "upside down" relative to each other?

On a p–x diagram at fixed temperature, high pressure favors liquid, so the liquid region sits above the two-phase lens. On a T–x diagram at fixed pressure, high temperature favors vapour, so the vapour region sits above instead. The physical variable driving each axis (compression versus heating) has an opposite effect on which phase is favored, flipping the diagram's vertical sense.

What happens to these diagrams if the mixture is not ideal?

Raoult's law no longer holds exactly, so the liquid line is no longer straight, and the deviations can be strong enough that the liquidus and vapour curve touch at an interior composition rather than only at $x_A = 0,1$ . That touching point is an azeotrope, where liquid and vapour share the same composition and simple distillation cannot separate the components further.

How does this diagram connect to the lever rule and fractional distillation?

Any point inside the two-phase lens represents an overall mixture that splits into a specific liquid point and vapour point joined by a horizontal tie line. The lever rule (covered next) tells you the relative amounts of liquid and vapour from the lengths of the two tie-line segments. Fractional distillation then repeats the boil-condense cycle — following successive tie lines up the T–x diagram — to push the vapour composition ever closer to the pure volatile component.

Why is the T–x diagram usually more useful in the laboratory than the p–x diagram?

Most distillations run at a fixed, known pressure — typically atmospheric — while temperature is what actually changes as the process proceeds. The T–x diagram directly answers the practical question, "at what temperature will this composition boil, and what vapour will it produce," which is exactly the operating condition of a real column or still.