Consider a pure liquid in a beaker is covered with a jar. Some molecules on the liquid surface evaporate and fill the vacant space available to them. The molecules in the vapour phase move randomly in the vacant space, during this motion, they strike the surface of the liquid and condensed. This process of evaporation and condensation go on and an equilibrium is established between evaporation and condensation. The pressure exerted by vapours over the liquid surface at equilibrium is called vapour pressure of the liquid.

If solute is non volatile solid or liquid the vapour pressure of solution is equal to partial vapour pressure of solvent in the solution and if the solute is volatile solid or liquid, then vapour pressure will be equal to the sum of partial vapour pressure of solute and that of solvent.

__Raoult’s Law :__

This law is applied for a solution of liquid in liquids and can be stated as follows.

” The partial vapour pressure of any component in the solution is directly proportional to its mole fraction ” .

P_{A} ∝ X_{A} Where, P_{A} = Partial vapour pressure of A

P_{A} = KX_{A} Where X_{A} = Mole fraction of A in solution

For pure liquids X_{A} = 1

Then K = P°_{A} Where P°_{A} is the vapour pressure of component A in pure state

Hence, P_{A} = X_{A}P°_{A}

Similarly for component B , P_{B} = X_{B}P°_{B}

__Raoult’s Law in Combination with Dalton’s Law of Partial Pressure:__

Assuming that vapours of a liquid is behaving like an ideal gas, then according to Dalton’s law of partial pressure the total pressure PT is given by

P_{T} = P_{A} + P_{B}

or P_{T} = X_{A} P°_{A} + X_{B} P°_{B} = P°_{B} + (P°_{A} − P°_{B}) X_{A}

**Illustration :** The vapour pressures of pure CCl_{4} (M_{m} = 154 g mol^{–1}) and SnCl_{4 }(M_{m} = 170 g mol^{–1}) at 25°C are 114.9 and 238.9 torr, respectively. Assuming ideal behaviour , calculate the total vapour pressure of a solution containing 10 g of CCl_{4} and 15 g of SnCl_{4}.

**Solution: **Let x_{A} be the mole fraction of CCl_{4} and X_{B} that of SnCl_{4+}

$\large X_A = \frac{10 g/(154 g mol^{-1})}{10 g/(154 g mol^{-1}) + 15 g/(170 g mol^{-1})} $

= 0.424

x_{B} = 1 – x_{A} = 0.576

According to Raoult’s law, the partial pressures are:

P_{A} = x_{A}P°_{A} = 0.424 × 114.9 torr = 48.72 torr

P_{B} = x_{B}P°_{B} = 0.576 × 238.3 torr = 137.26 torr

Total vapour pressure = 48.72 + 137.26 = 185.98 torr

__Composition of the Vapour:__

The composition of the liquid and vapour that are in mutual equilibrium are not necessarily the same, the common sense suggest that the vapour pressure should be richer in the more volatile component. This expectation can be confirmed as follows:

Let the mole fractions of A and B in vapour phase be YA and YB then from Dalton’s law,

$\large Y_A = \frac{P_A}{P_T}$ …(i)

and , $\large Y_B = \frac{P_B}{P_T}$ …..(ii)

Provided the mixture of vapours behaves as an ideal gas

Rewriting equation (i)

$\large Y_A = \frac{X_A P_A^o}{P_B^o + (P_A^o – P_B^o)X_A}$

and , Y_{B} = 1 − Y_{A}

**Illustration :** The vapour pressures of pure benzene and toluene at 40°C are 184.0 torr and 59.0 torr , respectively.

Calculate the partial pressures of benzene and toluene, the total vapour pressure of the solution and the mole fraction of benzene in the vapour above the solution that has 0.40 mole fraction of benzene. Assume that the solution is ideal.

**Solution: **Let us designate benzene by subscript A and toluene by subscript B.

In solution x_{A} = 0.40 and hence X_{B} = 0.6

P_{A} = x_{A}P°_{A} = (0.40) (184.0 torr) = 73.6 torr

P_{B} = x_{B}P°_{B} = (0.60) (59.0 torr) = 35.4 torr

P = P_{A} + P_{B} = 73.6 torr + 35.4 torr = 109.0 torr

X_{A , vap} = P_{A}/P = P_{A}/(P_{A} + P_{B}) = 73.6 torr/109.0 torr = 0.675

Condensation of vapours of solution: When the vapours of solution (containing liquids A and B) is condensed, the composition of liquids A and B in the condensate remains same.

The vapours over condensate can again recondensed and the composition of A and B in condenstate (2) remains same as it was in vapour phase over condensate (2).