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°