Vapor pressure of liquid solutions

 1. Vapour pressure of liquid-liquid solutions:

Raoult’s Law:- The partial pressure of each component in a binary volatile liquid-liquid solution at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state at same temperature.

P_A = [n_A/ (n_A + n_B)] x P°_A = X_A * P°_A

P_B = ([n_B/ (n_A + n_B)] x P°_B = X_B * P°_B

P_solution = P_A + P_B = X_A x P°_A + X_B x P°_B

= (1 - X_B) x P°_A + X_B x P°_B

= P°_A + (P°_B - P°_A) * X_B

i.e., the total pressure varies linearly with mole fractions of the components

⇒ If component A is less volatile than component B i.e., P°_A < P°_B the minimum of P_total is P°_A and maximum is P°_B

⇒ If y_A and y_B are mole fractions of component A and B in vapour phase then according to Dalton’s law of partial pressures

P_A = y_A x P_total

P_B = y_B x P_total

Raoult’s law as a special case of Henry’s law:

⇒ According to Raoult’s law

P_A = P°_A * X_A … (i) 

⇒ According to Henry’s law for a gas in liquid solution the partial pressure of the volatile component, i.e., gas, is directly proportional to its mole fraction p α K_H x … (ii)

⇒ In equation (i) and (ii), only the proportionality constant differs. Thus, Raoult’s law is a special case of Henry’s Law where K_H = P°_A

2. Vapour pressure of solid in liquid solution:

⇒ Non-volatile solute is added to a solvent then the vapour pressure is only due to the solvent. The vapour pressure of the solution decreases after adding solute.

⇒ The decrease in vapour pressure depends on the quantity of solute added (moles of solute).

According to Raoult’s law:

P_A = P°_A * X_A

Where, A is the solvent.

P°_A = vapour pressure of pure solvent.

The plot of P_A and X_A is a straight line.