Class 12 Chemistry


Vapor Pressure of Liquid Solutions

Let us consider a binary solution of two volatile liquids with two components 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually and equilibrium would be reached between vapor phase and liquid phase. At this stage, let us assume the total vapor pressure is ptotal and p1 and p2 are partial vapor pressures of the two components 1 and 2. These partial pressures are related to mole fractions x1 and x2 of the two components in that order.

Roult Law: For a solution of volatile liquids, the partial vapor pressure of each component of the solution is directly proportional to its mole fraction present in the solution.

Thus, for component 1

`p_1 ∝ x_1`

And `p_1=p_1^0x_1` ……………(3)

Here, `p_1^0` is the vapor pressure of pure component 1 at the same temperature.

Similarly, for component 2

`p_2=p_2^0x_2` ………….(4)

According to Dalton’s Law of Partial Pressures, the total pressure (ptotal) over the solution phase in the container will be the sum of partial pressures of the components of the solution.

Hence, `p_text(total)=x_1p_1^0+x_2p_2^0` …………….(5)


`=p_1^0+(p_2^0-p_1^0)x_2` ………………(6)

Following conclusions can be drawn from equation (6)

graph showing vapor pressure and mole fraction

A graph of p1 or p2 versus mole fractions x1 and x2 gives a linear plot. The lines I and II pass through the points for which x1 and x2 are equal to unity. The minimum value of ptotal is p10 and the maximum value is p20.

The composition of vapor phase in equilibrium with the solution is determined by partial pressures of the components. If y1 and y2 are the mole fractions of components 1 and 2 in vapor phase then, as per Dalton’s law of partial pressures:



In general

`p_1=y_1p_text(total)` ………….(7)

Example: Vapor pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapor pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298 K and (ii) mole fractions of each component in vapor phase.

Answer: Molar mass of CH2Cl2 `12xx1+x_1xx2+35.5xx2=85` g mol-1

Molar mass of CHCl3 `=12xx1+1xx1+35.5xx3=11.5` g mol-1

Moles of CH2Cl2 `=(40g)/(85g\text(mol)^(-1))=0.47` mol

Moles of CHCl3 `=(25.5g)/(119.5g\text(mol)^(-1))=0.213` mol

Total number of moles `=0.47+0.213=0.683` mol



Vapor pressure of solution can be calculated as follows:



`=200+147.9=347.9` mm Hg

Now, mole fraction of components in vapor phase can be calculated as follows:

`p_(CH_2Cl_2)=0.688xx415 = 285.5` mm Hg

`p_(CHCl_3)=0.312xx200=62.4` mm Hg



Raoult’s Law as a special case of Henry’s Law

We know that as per Rault’s law, the vapor pressure of a volatile component in a given solution is given as follows:


In case of a solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Solubility of this is given by Henry’s law which is expressed as follows:

`p = K_H\X`

In both the equations, partial pressure of the volatile component is directly proportional to its mole fraction in solution. Thus, Rault’s law becomes a special case of Henry’s law in which KH becomes p10.