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Friday, 23 May 2025

Vapour Pressure of Liquid Solutions

 

🌑️ Vapour Pressure of Liquid Solutions

Introduction:
Liquid solutions are formed when the solvent is a liquid. The solute in such solutions can be a gas, a liquid, or a solid. In this section, we will discuss binary solutions, i.e., solutions containing two components. These include:

  • (i) Solutions of liquids in liquids

  • (ii) Solutions of solids in liquids

Such solutions may contain one or more volatile components, with the solvent generally being volatile.


πŸ”Ή Vapour Pressure of Liquid–Liquid Solutions

Let us consider a binary solution made of two volatile liquids, denoted by components 1 and 2.

When this mixture is placed in a closed container, both liquids evaporate. Eventually, an equilibrium is established between the vapour phase and the liquid phase.

Let:

  • p1p_1 = Partial vapour pressure of component 1

  • p2p_2 = Partial vapour pressure of component 2

  • ptotalp_{\text{total}} = Total vapour pressure of the solution

  • x1x_1 = Mole fraction of component 1 in the liquid phase

  • x2x_2 = Mole fraction of component 2 in the liquid phase


πŸ“˜ Raoult’s Law:

Given by: FranΓ§ois-Marie Raoult (1886)

Statement: For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.

So,

p1x1p1=p10x1(1)p_1 \propto x_1 \quad \Rightarrow \quad p_1 = p_1^0 \cdot x_1 \tag{1.12}
p2x2p2=p20x2(2)p_2 \propto x_2 \quad \Rightarrow \quad p_2 = p_2^0 \cdot x_2 \tag{1.13}

Where:

  • p10p_1^0 and p20p_2^0 = Vapour pressures of pure components 1 and 2 respectively.


πŸ§ͺ Total Vapour Pressure:

According to Dalton’s Law of Partial Pressures, the total pressure is the sum of partial vapour pressures:

ptotal=p1+p2(3)p_{\text{total}} = p_1 + p_2 \tag{1.14}

Substituting from equations (1) and (2):

ptotal=x1p10+x2p20(4)p_{\text{total}} = x_1 p_1^0 + x_2 p_2^0 \tag{1.15}

Since x1+x2=1x_1 + x_2 = 1, we can write:

ptotal=p10+(p20p10)x2(5)p_{\text{total}} = p_1^0 + (p_2^0 - p_1^0)x_2 \tag{1.16}


πŸ“Š Interpretation of Equation :

  • (i) Total vapour pressure depends on mole fraction of any one component.

  • (ii) The relation is linear with respect to mole fraction x2.

  • (iii) If p10<p20p_1^0 < p_2^0, then total vapour pressure increases with increasing x2x_2.


🧩 Graphical Representation:

A plot of:

  • p1p_1 vs. x1x_1

  • p2p_2 vs. x2x_2

  • ptotalp_{\text{total}} vs. x2x_2

gives straight lines, intersecting at the points where mole fractions = 1 (i.e., pure components).

The line for ptotalp_{\text{total}} lies between those of p1p_1 and p2p_2.


🌫️ Composition of Vapour Phase at Equilibrium:

Let:

  • y1y_1 = Mole fraction of component 1 in vapour phase

  • y2y_2 = Mole fraction of component 2 in vapour phase

Using Dalton’s law:

p1=y1ptotal(6)p_1 = y_1 \cdot p_{\text{total}} \tag{1.17}
p2=y2ptotal(7)p_2 = y_2 \cdot p_{\text{total}} \tag{1.18}

In general, for any component ii:

pi=yiptotalp_i = y_i \cdot p_{\text{total}}

This shows that mole fraction in vapour phase ( yiy_i ) is not necessarily equal to mole fraction in liquid phase ( xix_i ).


πŸ” Important Points to Remember:

  • Raoult’s law applies to ideal solutions.

  • For non-ideal solutions, deviations occur due to intermolecular interactions.

  • Mole fractions in liquid and vapour phases differ unless the solution is ideal and forms an azeotrope.

  • Vapour pressure is a colligative property, depending on the number of particles, not their nature.


πŸ“˜ Application in Daily Life:

  • Perfumes and essential oils use volatile liquid mixtures.

  • Distillation techniques (like in petrochemicals) rely on vapour pressure principles.

  • Meteorology and humidity studies use vapour pressure data of water.


✍️ Practice Questions:

  1. State Raoult’s law and write its mathematical form.

  2. Derive the expression for total vapour pressure in a binary liquid solution.

  3. Explain why total vapour pressure changes with mole fraction.

  4. A solution has x1=0.6x_1 = 0.6, p10=80kPap_1^0 = 80 \, \text{kPa}, p20=60kPap_2^0 = 60 \, \text{kPa}. Calculate ptotalp_{\text{total}}.

  5. Define ideal and non-ideal solutions with examples.

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