Vapour Pressure

At a constant temperature, the pressure exerted by the vapours of a liquid on its surface when they (liquid and its vapours) are in equilibrium, is known as vapour pressure.

Factors affecting Vapour Pressure:

The process of evaporation depends on different factors:-

(a) Nature of the liquid

(b) Effect of temperature

1.0Vapour Pressure of Liquid Solution

Raoult's Law

According to this law, the partial pressure of any volatile constituents of a solution at a constant temperature is directly proportional to the mole fraction of that constituent in the solution.

Vapour Pressure of Liquid – Liquid Solution 

Let P_A and P_B be the partial vapour pressures of two constituents A and B in solution and P_A⁰ and P_B⁰ the vapour pressures in pure state respectivity.

At constant temperature partial vapour pressure of the component is directly proportional to the mole fraction of the component in solution.

Let P_A and P_B be the partial pressures of two constituents A and B in solution and P_A⁰ and P_B⁰  the vapour pressures in pure state respectively.

Thus, according to Raoult's law [for volatile liquids] 

$P_A=\frac{n_A}{n_A+n_B}{P}_A^0$...(1)                    

Partial pressure of A = mole fraction of

$A\times P_A^0=X_A{P}_A^0$

and

$P_B=\frac{n_B}{n_A+n_B}{P}_B^0$...(2)

Partial pressure of B = mole fraction of

$B\times P_B^0=X_B{P}_B^0$

If total pressure is P_S, then

According to Dalton's law of partial pressure given below: (If A is more volatile than B (P_A⁰<P_B⁰ )

If total pressure be P_S, then                          

$P_s=P_A+P_B=X_A{P}_A^0+X_B{P}_B^0$

$P_S=X_{A_{}}{P}_A^0+(1–X_A)P_B^0$

$[X_A+X_B=1]$

$P_S=X_A{P}_A^0–X_A{P}_A^0+P_B^0$

$P_S=X_A[P_A^0–P_B^0]+P_B^0$

$X_B=4/5$

Following conclusions can be drawn from equation 

(i) Total vapour pressure over the solution can be related to the mole fraction of any one component.

(ii) Total vapour pressure over the solution varies linearly with the mole fraction of component A.

2.0Related Questionnaire 

1 mole heptane (V.P. = 92 mm of Hg) is mixed with 4 mole Octane (V.P. = 31 mm of Hg), forming an ideal solution. Find out the vapour pressure of the solution.

Sol. Total mole = 1 + 4 = 5

Mole fraction of heptane =

$X_A=1/5$

Mole fraction of octane =

$P_S=X_A{P}_A^0+X_B{P}_B^0=\frac{1}{5}\times 92+\frac{4}{5}\times 31$ =43.2 mm of Hg. 

3.0Also Read