A solution containing two volatile liquids A and B obeys Raoult's law. Derive the expression for its total vapour pressure.

📘 Derivation of Total Vapour Pressure

🔵 Step 1: Consider a Binary Solution

Consider a solution containing two volatile liquids A and B.

Let:

• x_A = Mole fraction of A in the liquid solution
• x_B = Mole fraction of B in the liquid solution
• P_A° = Vapour pressure of pure A
• P_B° = Vapour pressure of pure B
• P_A = Partial vapour pressure of A
• P_B = Partial vapour pressure of B

Since it is a binary solution:

x_A + x_B = 1

🟢 Step 2: Apply Raoult's Law to A

According to Raoult's law, the partial vapour pressure of a volatile component is equal to the product of its mole fraction and vapour pressure in the pure state.

P_A = x_AP_A°

🟠 Step 3: Apply Raoult's Law to B

P_B = x_BP_B°

🟣 Step 4: Calculate Total Vapour Pressure

The total vapour pressure of the solution is equal to the sum of the partial vapour pressures of A and B.

P_total = P_A + P_B

Substituting the Raoult's law expressions:

P_total = x_AP_A° + x_BP_B°

🎬 3D-Style Molecular Animation

Both volatile liquids contribute to the vapour pressure.

A B A B A B

A contributes → P_A = x_AP_A°

B contributes → P_B = x_BP_B°

Total → P_total = P_A + P_B

🔴 Step 5: Expression in Terms of Number of Moles

If n_A and n_B are the number of moles of A and B, then:

x_A = n_A / (n_A + n_B)

x_B = n_B / (n_A + n_B)

Therefore:

P_total = [n_AP_A° + n_BP_B°] / [n_A + n_B]

✅ Final Derived Expression

P_total = x_AP_A° + x_BP_B°

Thus, the total vapour pressure of a binary solution of two volatile liquids is the sum of their partial vapour pressures.

🎯 Exam Formula

P_A = x_AP_A°
P_B = x_BP_B°

P_total = x_AP_A° + x_BP_B°

Explain the difference between ideal and non-ideal solutions with suitable examples.

📘 Ideal vs Non-Ideal Solutions

🔵 Step 1: Ideal Solution

An ideal solution is a solution that obeys Raoult's law over the entire range of composition and temperature.

P_i = x_iP_i°

In an ideal solution, the intermolecular attractions between A–A, B–B and A–B molecules are nearly equal.

ΔH_mix = 0

ΔV_mix = 0

Examples: Benzene + Toluene, n-Hexane + n-Heptane.

🟠 Step 2: Non-Ideal Solution

A non-ideal solution does not obey Raoult's law over the entire range of composition.

This occurs when the A–B intermolecular attractions are significantly different from A–A and B–B attractions.

Examples: Ethanol + Acetone, Chloroform + Acetone.

🟣 Step 3: Positive Deviation

When A–B attractions are weaker than A–A and B–B attractions, molecules escape more easily into the vapour phase.

Vapour pressure observed > Vapour pressure predicted by Raoult's law

This is called positive deviation.

Example: Ethanol + Acetone.

🔴 Step 4: Negative Deviation

When A–B attractions are stronger than A–A and B–B attractions, molecules escape less easily into the vapour phase.

Vapour pressure observed < Vapour pressure predicted by Raoult's law

This is called negative deviation.

Example: Chloroform + Acetone.

🎬 3D-Style Molecular Animation

Ideal solution: A–B attraction is approximately equal to A–A and B–B.

A B A B A B

📊 Step 5: Main Differences

Property	Ideal Solution	Non-Ideal Solution
Raoult's Law	Obeys completely	Shows deviation
A–B Interaction	Approximately equal to A–A and B–B	Different from A–A and B–B
ΔHmix	Zero	Non-zero
ΔVmix	Zero	Usually non-zero
Examples	Benzene + Toluene	Ethanol + Acetone

✅ Exam Summary

Ideal Solution → Raoult's Law obeyed

Non-Ideal Solution → Positive or Negative deviation

Positive deviation: A–B attraction weaker → higher vapour pressure.

Negative deviation: A–B attraction stronger → lower vapour pressure.

Derive Raoult's law for a solution containing volatile components.

📘 Derivation of Raoult's Law for a Solution of Volatile Components

🔵 Step 1: Consider a Binary Solution

Consider a binary solution containing two volatile components, A and B.

x_A + x_B = 1

Let:

• P_A° = Vapour pressure of pure A
• P_B° = Vapour pressure of pure B
• x_A = Mole fraction of A in liquid phase
• x_B = Mole fraction of B in liquid phase
• P_A = Partial vapour pressure of A
• P_B = Partial vapour pressure of B

🟢 Step 2: Vapour Pressure of Component A

According to Raoult's law, the partial vapour pressure of a volatile component is directly proportional to its mole fraction in the solution.

P_A ∝ x_A

Introducing the proportionality constant, which is the vapour pressure of pure A:

P_A = x_AP_A°

🟠 Step 3: Vapour Pressure of Component B

Similarly, for component B:

P_B = x_BP_B°

🟣 Step 4: Total Vapour Pressure

The total vapour pressure of the solution is the sum of the partial vapour pressures of A and B.

P = P_A + P_B

Substituting the expressions:

P = x_AP_A° + x_BP_B°

⭐ Raoult's Law for a Binary Solution

P = x_AP_A° + x_BP_B°

🔴 Step 5: Special Case

If component B is absent, then x_A = 1. Therefore:

P = P_A°

Thus, the vapour pressure of pure A is obtained when the mole fraction of A is unity.

🎬 3D-Style Raoult's Law Animation

Volatile molecules A and B escape from the liquid surface into the vapour phase.

A B A B A B

A → P_A = x_AP_A°

B → P_B = x_BP_B°

Total → P = P_A + P_B

📌 General Form

For a solution containing several volatile components:

P_total = Σ x_iP_i°

where x_i is the mole fraction and P_i° is the vapour pressure of the pure component.

✅ Final Result

P_A = x_AP_A°

P_B = x_BP_B°

P_total = x_AP_A° + x_BP_B°

🎯 Exam Point

Raoult's law: P_i = x_iP_i°

For a binary volatile solution: P_total = x_AP_A° + x_BP_B°