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°