Vapour Pressure of Liquid Solutions: A solution is formed when a solute is dissolved in a solvent. The solvent is generally a liquid. The solute can be a solid, liquid or gas. Additionally, a solution may contain one or more volatile components. Generally, the solvent which is in a liquid state is volatile. The solute dissolved in the solvent may or may not be volatile. A binary solution is formed if the solution consists of two components, namely- (i) liquids in liquids and (ii) solids in liquids.

The addition of a solute to a solvent affects the vapour pressure of the resultant solution. Let’s understand what vapour pressure is and how it is affected by adding a solute to a solvent.

What is Vapour Pressure?

When a liquid evaporates in a closed vessel, a part of the liquid changes into a gaseous state and fills the available space with vapours. As evaporation proceeds, the number of liquid molecules that get converted to the vapour phase increases. These gaseous molecules are in random motion with each other; some of these strike the surface of the liquid. As they strike the surface of the liquid, the vapours get condensed, and the process of condensation occurs in a direction opposite to the process of evaporation. Thus, evaporation and condensation processes go on simultaneously, and a stage is reached when the rate of evaporation becomes equal to the rate of condensation. An equilibrium gets established between the liquid and vapour phases.

At equilibrium, the pressure exerted by the vapours of the liquid on the liquid surface at a given temperature is called the vapour pressure.
The vapour pressure of a pure liquid is more compared to the vapour pressure of the same liquid to which a solute has been added. For example, water will have more vapour pressure compared to lemonade.

Learn About Lowering Of Vapour Pressure Here

Factors on which vapour pressure depends:

(1) Nature of the Liquid
The Vapour pressure of a liquid varies inversely to the forces of attraction that exist between molecules of a liquid. This means the liquids with weaker intermolecular forces tend to escape readily into the vapour phase and have a greater vapour pressure.

(2) Temperature
The vapour pressure of a liquid is directly proportional to its temperature. This means with an increase in temperature, and more molecules will have larger kinetic energies. An increased number of molecules will escape from the surface of the liquid to the vapour phase, resulting in higher vapour pressure.

Vapour pressure can be evaluated in two cases which are:
1. Vapour pressure of a binary solution in which the solute and solvent both are in the liquid phase.i.e liquid-liquid solution.
2. Vapour pressure of a binary solution in which solute is in the solid-state and the solvent is in the liquid phase, i.e., solid-liquid solutions.

Vapour Pressure of Liquid-Liquid Solutions

To determine the vapour pressure of a liquid-liquid solution, let us consider two volatile liquid solutions denoted as \(1\) and \(2.\) When the liquid components are taken in a closed vessel, an equilibrium is established between the liquid phase and vapour phase of the binary solution.
Suppose\({{\text{P}}_{{\text{Total}}}}\) is the overall vapour pressure of the binary solution at equilibrium and \({{\text{P}}_1}\) and \({{\text{P}}_2}\) be the partial vapour pressures of components \(1\) and \(2,\) respectively. The partial pressures are related to the mole fractions, so let the mole fractions of components \(1\) and \(2\) be \({{\text{X}}_1}\) and \({{\text{X}}_2}.\)
The relationship between partial pressures and mole fractions was given by the French chemist François-Marie Raoult in \(1886.\) This relationship is popularly known as Raoult’s Law.

Raoult’s Law

According to Raoult’s law, the partial vapour pressure of each component of a binary solution consisting of two volatile liquids is directly proportional to its mole fraction present in the solution.
Thus, for component \(1,\) we have-
\({{\text{P}}_1}\alpha {{\text{X}}_1}\)
\({{\text{P}}_{\text{1}}}{\text{ = P}}_1^0 \cdot {{\text{X}}_{\text{1}}}\)
Where \({\text{P}}_1^0\) represents the vapour pressure of component \(1\) in its pure form.
Similarly, for component \({\text{B}},\) we have-
\({{\text{P}}_2}{\text{ = P}}_2^0 \cdot {{\text{X}}_2}\)
Where \({\text{P}}_2^0\) is the vapour pressure of component \(2\) in its pure form.

Dalton’s law of partial pressures states that the total pressure \(\left( {{{\text{P}}_{{\text{Total}}}}} \right)\) over the solution phase in the container will be the sum of the partial pressures of the solution’s individual components.
Hence,
\({{\text{P}}_{{\text{Total}}}} = {{\text{P}}_1} + {{\text{P}}_2}\)
\({{\text{P}}_{{\text{Total}}}} = {\text{P}}_1^0 \cdot {{\text{X}}_1} + {\text{P}}_2^0 \cdot {{\text{X}}_2}\)
The total mole fraction of a solution is always equal to \(1.\) So,
\({{\text{X}}_1} + {{\text{X}}_2} = 1\)
\({{\text{X}}_1} = 1 – {{\text{X}}_2}\)
Hence, total pressure \({\text{P}}\) is given by-
\({{\text{P}}_{{\text{Total}}}} = \left( {1 – {{\text{X}}_2}} \right){\text{P}}_1^0 + {\text{P}}_2^0 \cdot {{\text{X}}_2}\)
\({{\text{P}}_{{\text{Total}}}} = {\text{P}}_1^0 – {{\text{X}}_2}\left( {{\text{P}}_1^0 – {\text{P}}_2^0} \right)\)
From the above equation, we can conclude that-
(i) Total vapour pressure over the resulting solution can be related to the mole fraction of any one of the dissolved components, i.e. either \(1\) or \(2.\) For example, the above equation can also be written as-
\({{\text{P}}_{{\text{Total}}}} = {\text{P}}_2^0 + {{\text{X}}_1}\left( {{\text{P}}_1^0 – {\text{P}}_2^0} \right)\)
(ii) Total vapour pressure over the resulting solution varies linearly with \({{\text{X}}_2}\) i.e., the mole fraction of component \(2.\)
(iii) The total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component \(1\) or \(2.\)
On plotting, \({{\text{P}}_1}\) or \({{\text{P}}_2}\) versus the mole fractions \({{\text{X}}_1}\) and \({{\text{X}}_2}\) for a binary solution consisting of two volatile liquids, a linear graph is obtained. The dashed lines give the partial pressure of the two components \(1\) or \(2\) versus their composition \({{\text{X}}_1}\) or \({{\text{X}}_2}\) and the solid line gives the total vapour pressure of the binary solution.
When \({{\text{X}}_1} = 1,\) i.e. the liquid is pure \(1\)
\({{\text{P}}_1} = {\text{P}}_1^0 \times 1 = {\text{P}}_1^0\)
When \({{\text{X}}_1} = 0,\) i.e. the liquid is pure \(2\)
\({{\text{P}}_1} = {\text{P}}_1^0 \times 0 = 0\)
Thus, the plot of \({{\text{P}}_1}\) against \({{\text{X}}_1}\) should give a straight line passing through unity (when \({{\text{X}}_1} = 1\)) and \(0\) (when \({{\text{X}}_1} = 0\))
When component \(1\) is pure \(\left( {{{\text{X}}_1} = 1} \right),\) its vapour pressure is equal to \({\text{P}}_1^0.\) As component \(2\) is added to component \(1\) (\({{{\text{X}}_1}}\)decreases), the vapour pressure decreases till it becomes zero \(\left( {{{\text{X}}_1} = 0} \right).\) This is represented by line I in the above graph.
The variation of partial pressure of component \(2\left( {{{\text{P}}_2}} \right)\) with its mole fraction \(\left( {{{\text{X}}_2}} \right)\) is represented by the plot from \({{\text{X}}_2} = 0\left( {{{\text{P}}_2} = {\text{P}}_2^0 \times 0 = 0} \right)\) to \({{\text{X}}_2} = 1\left( {{{\text{P}}_2} = {\text{P}}_2^0 \times 1 = {\text{P}}_2^0} \right).\)
This is represented by line II in the above graph.
The total vapour pressure \({{\text{P}}_{{\text{Total}}}}\) exerted by the solution as a whole at any composition is given by the sum of partial vapour pressures of components \(1\) and \(2.\) This is represented by line III in the above graph.
The minimum value of \({{\text{P}}_{{\text{Total}}}}\) is \({\text{P}}_{\text{1}}^{\text{0}}\) and the maximum value is \({\text{P}}_2^{\text{0}},\) assuming that component \(1\) is less volatile than component \(2,\) i.e., \({\text{P}}_1^{\text{0}} < {\text{P}}_2^0.\)
If \({{\text{Y}}_1}\) and \({{\text{Y}}_2}\) are the mole fractions of the components \(1\) and \(2\) respectively in the vapour phase then, using Dalton’s law of partial pressures:
\({{\text{P}}_1} = {{\text{Y}}_{\text{1}}}{{\text{P}}_{{\text{Total}}}}\)
\({{\text{P}}_2} = {{\text{Y}}_2}{{\text{P}}_{{\text{Total}}}}\)
In general,
\({{\text{P}}_{\text{i}}} = {{\text{Y}}_{\text{i}}}{{\text{P}}_{{\text{Total}}}}\)

Raoult’s Law as a Special Case of Henry’s Law

In a binary liquid-liquid solution, one of the components is so volatile that it exists as a gas whose solubility is governed by Henry’s Law. According to Henry’s Law, the partial pressure of the gas in the vapour phase is proportional to the mole fraction of the gas in the solution” and is expressed as: \({\text{P}} = {{\text{K}}_{\text{H}}} \times {\text{X}}\)
Where \({\text{P}}\) represents the partial pressure of the solution and \({\text{X}}\) is the mole fraction of the solvent. \({{\text{K}}_{\text{H}}}\) is the proportionality constant (Henry’s constant)
The Raoult’s Law is given by:
\({\text{P}} = {{\text{P}}^0} \cdot {\text{X}}\)
Where \({\text{P}}\) represents the partial pressure of the solution and \({\text{X}}\) is the mole fraction of the solvent. \({{\text{P}}^0}\) represents the vapour pressure of the pure solvent.
Comparing the equations of Henry’s law and Raoult’s law, we get-
\({{\text{K}}_{\text{H}}} = {{\text{P}}^0}\)
Hence, Raoult’s Law is a special case of Henry’s Law.

Vapour Pressure of Solids in Liquids solutions

While making a sugar solution, we add sugar to water. The sugar in this solution acts as the non-volatile solute, and water acts as the solvent. On evaporating this sugar solution, the vapour phase consists of vapours of the solvent (i.e., of water) only because the solute is non-volatile. However, the vapour pressure of this solution is found to be less than that of the pure solvent (water).

The vapour pressure of a solution consisting of a non-volatile solute and a solvent depends on the rate at which the solvent molecules escape from the liquid’s surface.
Hence, the vapour pressure of the solution (sugar \(+\) water) will be less than that of the pure solvent (water), or there will be a lowering in vapour pressure of the solution due to the addition of a non-volatile solute. As the concentration of sugar increases in the solution, the vapour pressure of the solution will be further lowered.
The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution and not on its nature.

Why Vapour Pressure Decreases when a Non-Volatile Solute is Added to the Solvent?

The decrease in vapour pressure is due to:
1. Evaporation is a surface phenomenon; hence, the more the surface, the higher the rate of evaporation and the more is the vapour pressure. The non-volatile solute occupies a certain surface area of the entire solution. As a result lesser number of solvent molecules are available on the surface that will escape as vapours.
2. The number of molecules evaporating or leaving the surface is much greater in pure liquid solutions than that of a non-volatile solute solution.

Pure solvent has more number of molecules on the surface (a) as compared to (b) that has a non-volatile solute added to the solution

Raoult’s Law of Solutions of Solids in Liquids

According to Raoult’s law, the partial vapour pressure of a volatile component in the solution is directly proportional to its mole fraction.

When the solute is non-volatile, only the solvent molecules will be present in the vapour phase. Therefore, the vapour pressure of the entire solution will be the vapour pressure due to solvent only.

Hence,
\({{\text{P}}_1}\alpha {{\text{X}}_1}\)
\({{\text{P}}_1} = {\text{P}}_1^0 \cdot {{\text{X}}_1}\)
Where \({\text{P}}_1^0\) is the vapour pressure of the solvent in its pure form, \({{\text{P}}_1}\) is the vapour pressure of the solution consisting of a non-volatile solute and volatile solvent, \({{\text{X}}_1}\) is the mole fraction of the solvent.
Rearranging the above equation, we get-
\(\frac{{{{\text{P}}_{\text{1}}}}}{{{\text{P}}_{\text{1}}^{\text{0}}}} = {{\text{X}}_1}\)
Substracting each side of the equation from \(1,\) we get-
\(1 – \frac{{{{\text{P}}_{\text{1}}}}}{{{\text{P}}_{\text{1}}^{\text{0}}}} = 1 – {{\text{X}}_1}\)
\(\frac{{{\text{P}}_{\text{1}}^{\text{0}} – {{\text{P}}_{\text{1}}}}}{{{\text{P}}_{\text{1}}^{\text{0}}}} = 1 – {{\text{X}}_1}\)
\(\frac{{{\text{P}}_{\text{1}}^{\text{0}} – {{\text{P}}_{\text{1}}}}}{{{\text{P}}_{\text{1}}^{\text{0}}}} = {{\text{X}}_2}\)
\(\frac{{{\text{P}}_{{\text{Solvent}}}^{\text{0}} – {{\text{P}}_{{\text{Solution}}}}}}{{{\text{P}}_{{\text{Solvent}}}^{\text{0}}}} = {{\text{X}}_{{\text{Solute}}}}\)
\({{\text{P}}_{{\text{Solvent}}}^{\text{0}} – {{\text{P}}_{{\text{Solution}}}}}\) represents the lowering of vapour pressure and \(\frac{{{\text{P}}_{{\text{Solvent}}}^{\text{0}} – {{\text{P}}_{{\text{Solution}}}}}}{{{\text{P}}_{{\text{Solvent}}}^{\text{0}}}}\) represents the relative lowering of vapour pressure due to the addition of a non-volatile solute.

Summary

The presence of a non-volatile solute plays an important role in determining the vapour pressure of a solution. The law which governs the vapour pressure of a solution is known as Raoult’s law. This law gives the relationship between the partial pressure of a substance dissolved in a solvent to its mole fraction. In this article, we learned about vapour pressure and how it varies for liquid-liquid and solid-liquid solutions. We also learnt Raoult’s law, lowering and relative lowering of the vapour pressure of a solution.

FAQs on Vapour Pressure of Liquid Solutions

Q.1. What is vapour pressure?
Ans: At equilibrium, the pressure exerted by the vapours of the liquid on the liquid surface at a given temperature is called the vapour pressure.

Q.2. How do you find the vapour pressure of a liquid?
Ans: The vapour pressure of a liquid can be determined by using Raoult’s law. The expression given by Raoult’s law is as follows-
\({{\text{P}}_1} = {\text{P}}_1^0 \cdot {{\text{X}}_1}\)
Where \({\text{P}}_1^0\) is the vapour pressure of the solvent in its pure form, \({{\text{P}}_1}\) is the vapour pressure of the solution consisting of a non-volatile solute and volatile solvent, \({{\text{X}}_1}\) is the mole fraction of the solvent.

Q.3. Which solution should have the highest vapour pressure?
Ans: At room temperature, the substance that has the lowest boiling point will have the highest vapour pressure and the substance with the highest boiling point will have the lowest vapour pressure.

Q.4. How will temperature affect the vapour pressure of a solution?
Ans: The vapour pressure of a liquid is directly proportional to its temperature, i.e. it increases with an increase in temperature. This is because, with an increase in temperature, more molecules will possess larger kinetic energies. Therefore, a larger number of molecules will escape from the surface of the liquid to the vapour phase resulting in higher vapour pressure.

Q.5. Does the vapour pressure of a solution depend on the nature of the non-volatile solute?
Ans: The vapour pressure of a liquid does not depend on the nature of the non-volatile solute. However, it depends on the quantity of the non-volatile solute used. For example, the decrease in the vapour pressure of water by adding \(1.0\,{\text{mol}}\) of glucose to one kg of water is almost similar to that produced by adding \(1.0\,{\text{mol}}\) of urea to the same quantity of water at the same temperature.

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