Vapor-Liquid Equilibrium Basics

Concept

Vapor-Liquid Equilibrium (VLE) dictates how a chemical mixture distributes itself between a liquid phase and a vapor phase at a specific temperature and pressure. Understanding VLE is fundamental for designing separation processes like distillation. When a liquid mixture is in thermodynamic equilibrium with its vapor, the rate of evaporation equals the rate of condensation for every component. For ideal mixtures containing chemically similar compounds (e.g., benzene and toluene), the VLE can be accurately described by Raoult's Law.

Formula & Method

Raoult's Law states that the partial pressure ( $p_i$ ) of a component in the vapor phase is proportional to its mole fraction in the liquid phase ( $x_i$ ) and its pure component vapor pressure ( $P_i^*$ ) at the system temperature:

$p_i = x_i P_i^*$

By Dalton's Law, the total pressure ( $P$ ) of the vapor mixture is the sum of the partial pressures:

$P = \sum p_i = \sum (x_i P_i^*)$

The mole fraction of the component in the vapor phase ( $y_i$ ) is the ratio of its partial pressure to the total pressure:

$y_i = \frac{p_i}{P} = \frac{x_i P_i^*}{P}$

Variables & Units

$x_i$ = Mole fraction of component $i$ in the liquid phase (dimensionless).
$y_i$ = Mole fraction of component $i$ in the vapor phase (dimensionless).
$P_i^*$ = Vapor pressure of pure component $i$ at the given temperature (e.g., kPa, mmHg, atm).
$p_i$ = Partial pressure of component $i$ in the vapor mixture (same units as $P$ ).
$P$ = Total system pressure.

Worked Example

Problem: An ideal liquid mixture of component A and component B contains 40 mol% A ( $x_A = 0.40$ ). At the system temperature of $80^\circ\text{C}$ , the pure vapor pressure of A is $120 \text{ kPa}$ and B is $60 \text{ kPa}$ . Determine the total equilibrium pressure ( $P$ ) and the composition of the vapor phase ( $y_A, y_B$ ).

Calculation:

Identify variables: $x_A = 0.40$ , $x_B = 1 - 0.40 = 0.60$ , $P_A^* = 120 \text{ kPa}$ , $P_B^* = 60 \text{ kPa}$ .
Calculate partial pressures using Raoult's Law: $p_A = x_A P_A^* = 0.40 \times 120 = 48 \text{ kPa}$ $p_B = x_B P_B^* = 0.60 \times 60 = 36 \text{ kPa}$
Calculate total pressure $P$ : $P = p_A + p_B = 48 + 36 = 84 \text{ kPa}$
Calculate vapor mole fractions: $y_A = \frac{p_A}{P} = \frac{48}{84} \approx 0.571 \text{ (or 57.1 mol%)}$ $y_B = \frac{p_B}{P} = \frac{36}{84} \approx 0.429 \text{ (or 42.9 mol%)}$ (Note: $y_A + y_B = 1.0$ )

Engineering Meaning

The example demonstrates the core principle of distillation: the vapor phase is richer in the more volatile component ( $A$ , because $P_A^* > P_B^*$ ). While the liquid is 40% A, the vapor boils off at 57.1% A. By continuously boiling and condensing this mixture across multiple stages, engineers can achieve high-purity separation of chemicals.

Engineering Check

Raoult's Law is strictly valid only for ideal solutions. Always verify if the mixture consists of chemically similar, non-polar molecules of similar size. If the mixture is highly non-ideal (e.g., ethanol and water), Raoult's Law will yield significant errors, and an activity coefficient must be introduced. Ensure that $\sum x_i = 1$ and $\sum y_i = 1$ .

Explicit Exclusions

This foundational article excludes non-ideal VLE requiring activity coefficients (Modified Raoult's Law), fugacity coefficients, azeotrope formation, and equation-of-state methods.