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Vapour Pressure & Raoult's Law

Student can calculate total vapour pressure using Raoult's Law and predict deviations in non-ideal solutions.

8 lessons

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Theory

Vapour Pressure of Volatile Liquids

Introduction to Raoult's Law for binary solutions.

When two volatile liquids (Component 1 and Component 2) are mixed in a closed container, both evaporate. Eventually, a dynamic equilibrium is reached between the liquid and vapour phases.

Raoult's Law defines the relationship between the liquid composition and the vapour pressure. It states that at a given temperature, the partial vapour pressure of each component in a solution is directly proportional to its mole fraction in the liquid phase.

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Diagram

Raoult's Law Plot

Graph showing linear relationship of vapour pressure and mole fraction.

$A clean scientific coordinate graph showing Raoult's law for an ideal binary solution. The horizontal x-axis represents the mole fraction, starting from x1=1, x2=0 on the left and ending at x1=0, x2=1 on the right. The vertical y-axis represents vapour pressure. Two dashed diagonal lines represent the partial pressures p1 and p2, showing them increasing linearly with their respective mole fractions. A solid straight line connects the pure vapour pressure point p1^0 on the left y-axis to the higher pure vapour pressure point p2^0 on the right y-axis, representing the total vapour pressure.$

Click to zoom

Key Components

Individual Partial Pressures (Dashed Lines): These follow Raoult's Law, which states the partial pressure ( $p_i$ ) is proportional to the mole fraction ( $x_i$ ): $p_1 = x_1 p_1^0$ $p_2 = x_2 p_2^0$
Total Vapor Pressure (Solid Blue Line): This follows Dalton’s Law, representing the sum of the partial pressures: $P_{\text{total}} = p_1 + p_2 = x_1 p_1^0 + x_2 p_2^0$
Pure Components: The plot correctly terminates at $p_1^0$ (when $x_1 = 1$ ) and $p_2^0$ (when $x_2 = 1$ ).

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Practice

Example 1.5: Total Vapour Pressure

Calculates vapour pressure and vapour phase composition.

Problem

Vapour pressure of chloroform ( $CHCl_3$ ) and dichloromethane ( $CH_2Cl_2$ ) at 298 K are 200 mm Hg and 415 mm Hg respectively.

(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of $CHCl_3$ and 40 g of $CH_2Cl_2$ at 298 K. (ii) Calculate the mole fractions of each component in the vapour phase.

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Worksheet

Finding Composition

Finds liquid and vapour composition given total pressure.

0 of 2 blanks filled

We are given the pure vapour pressures $p_A^0 = 450$ mm Hg, $p_B^0 = 700$ mm Hg, and $p_{total} = 600$ mm Hg.

Using Raoult's law, we can write the equation for total pressure as $p_{total} = p_A^0 + (p_B^0 - p_A^0) x_B$ . Substituting the given values yields $600 = 450 + (250) x_B$ , which gives the liquid mole fraction $x_B = ␟blank1␟$ .

The mole fraction of component A in the liquid mixture is therefore $x_A = 1 - x_B = 0.40$ . To find the vapour phase composition, we first calculate the partial pressures using $p_A = x_A p_A^0$ , which equals $0.40 \times 450 = 180$ mm Hg.

The mole fraction of A in the vapour phase is then calculated as $y_A = p_A / p_{total} = 180 / 600 = 0.30$ .

Following the same logic, the partial pressure $p_B$ is $420$ mm Hg, making the vapour mole fraction $y_B = ␟blank2␟$ .

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Diagram

Deviation Curves

Graphs of positive and negative deviations from Raoult's Law.

$A side-by-side comparison of coordinate graphs showing deviations from Raoult's law. The left graph illustrates positive deviation, showing solid vapour pressure curves bowed upward, bulging above the straight dashed lines that represent ideal behavior. The right graph illustrates negative deviation, showing the solid vapour pressure curves bowed downward, sagging below the straight dashed lines of ideal behavior. Both graphs plot vapour pressure against mole fraction.$

Click to zoom

The graphs show vapour pressure versus mole fraction for a binary solution.

Dashed lines show ideal Raoult’s law behaviour. Solid curves show actual vapour pressure.

(a) Positive deviation: vapour pressure is higher than ideal because unlike molecules attract weakly, so evaporation is easier.

(b) Negative deviation: vapour pressure is lower than ideal because unlike molecules attract strongly, so evaporation is harder.

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Theory

Ideal and Non-Ideal Solutions

Defines ideal solutions and positive/negative deviations from Raoult's law.

Not all mixtures play by the rules. While Raoult's law provides a great mathematical baseline, real-world molecular interactions often cause deviations.

An Ideal Solution obeys Raoult's law over the entire concentration range. This happens when the intermolecular attractive forces between the mixed components (A-B) are nearly identical to the pure components (A-A and B-B).

For ideal solutions:

$\Delta H_{mix} = 0$ (No heat is absorbed or released)
$\Delta V_{mix} = 0$ (Total volume perfectly matches the sum of individual volumes)
Examples: n-hexane & n-heptane, benzene & toluene.

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Comparing Deviations

Summary matrix of Ideal, Positive, and Negative deviations.

Ideal Solution

• Obeys Raoult's Law
• $Δ H_{mi x} = 0$
• A-B forces = pure forces
• Never forms azeotropes

Positive Deviation

• Vapour pressure > Ideal
• $Δ H_{mi x} > 0$ Endothermic
• A-B forces are weaker
• Minimum boiling azeotrope

Negative Deviation

• Vapour pressure < Ideal
• $Δ H_{mi x} < 0$ Exothermic
• A-B forces are stronger
• Maximum boiling azeotrope

Diagram

Raoult's Law Plot

Graph showing linear relationship of vapour pressure and mole fraction.

Click to zoom

Key Components

Individual Partial Pressures (Dashed Lines): These follow Raoult's Law, which states the partial pressure ( $p_i$ ) is proportional to the mole fraction ( $x_i$ ): $p_1 = x_1 p_1^0$ $p_2 = x_2 p_2^0$
Total Vapor Pressure (Solid Blue Line): This follows Dalton’s Law, representing the sum of the partial pressures: $P_{\text{total}} = p_1 + p_2 = x_1 p_1^0 + x_2 p_2^0$
Pure Components: The plot correctly terminates at $p_1^0$ (when $x_1 = 1$ ) and $p_2^0$ (when $x_2 = 1$ ).

Practice

Example 1.5: Total Vapour Pressure

Calculates vapour pressure and vapour phase composition.

Problem

Vapour pressure of chloroform ( $CHCl_3$ ) and dichloromethane ( $CH_2Cl_2$ ) at 298 K are 200 mm Hg and 415 mm Hg respectively.

(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of $CHCl_3$ and 40 g of $CH_2Cl_2$ at 298 K. (ii) Calculate the mole fractions of each component in the vapour phase.

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Worksheet

Finding Composition

Finds liquid and vapour composition given total pressure.

0 of 2 blanks filled

We are given the pure vapour pressures $p_A^0 = 450$ mm Hg, $p_B^0 = 700$ mm Hg, and $p_{total} = 600$ mm Hg.

The mole fraction of A in the vapour phase is then calculated as $y_A = p_A / p_{total} = 180 / 600 = 0.30$ .

Following the same logic, the partial pressure $p_B$ is $420$ mm Hg, making the vapour mole fraction $y_B = ␟blank2␟$ .

Theory

Ideal and Non-Ideal Solutions

Defines ideal solutions and positive/negative deviations from Raoult's law.

Not all mixtures play by the rules. While Raoult's law provides a great mathematical baseline, real-world molecular interactions often cause deviations.

For ideal solutions:

$\Delta H_{mix} = 0$ (No heat is absorbed or released)
$\Delta V_{mix} = 0$ (Total volume perfectly matches the sum of individual volumes)
Examples: n-hexane & n-heptane, benzene & toluene.

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