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Vapour Pressure of Liquid Solutions
Vapour Pressure of Liquid in Liquid Solution
Raoult’s Law
Conclusions on Vapour Pressure relation in Liquid-Liquid Solution
Vapour Pressure of Solid in Liquid Solution
Image 1: Example of liquid in liquid solution
Liquid Solution is formed when we dissolve a solid, liquid or gas in a particular liquid solvent. Vapour Pressure of liquid solutions is defined as the pressure exerted by the vapours on the liquid solvent when kept in equilibrium and a certain temperature. It varies with the nature of liquid and temperature of the surroundings. Some characteristics of the vapour pressure of liquid solutions are as follows:
The pure liquid has more vapour pressure as compared to liquid’s solution. For Example, take two beaker one filled with water and other with lemon juice and water, you’ll find that the beaker filled with water has more vapour pressure
Vapour pressure is inversely proportional to forces of attraction between molecules of a liquid
Vapour pressure increases with increase in temperature, as molecules gain kinetic energy and vapourise briskly
Image 2: Vapour pressure equilibrium in different solutions
Liquid solution can contain volatile solute and solvent as well. In most cases, the solvent is volatile in nature and the solute left out to be non-volatile. Vapour pressure can be evaluated in two cases which are:
Vapour pressure of liquid-liquid solutionsthat is, solute and solvent both are in liquid phase
Vapour pressure of solid- liquid solutions that is, solute is in solid state and solvent is in liquid phase
To find out vapour pressure of the liquid-liquid solution, we take two volatile liquid solutions and name their liquid component as A and B. When we place the volatile liquid with their components in a closed vessel, we find that equilibrium is established between the liquid phase and vapour phase.
Suppose P_{total} is the overall vapour pressure at an equilibrium state and let P_{A} and P_{B} be the partial vapour pressures of components A and B respectively. Adding further, the mole fraction of respective components is x_{A} and x_{B} respectively.
To evaluate vapour pressure of volatile liquids, we use Raoult’s Law.
In 1986, a French Chemist named Francois Marte Raoult proposed a quantitative relation between partial pressure and mole fraction of volatile liquids. The law states that mole fraction of the solute component is directly proportional to its partial pressure.
So, according to Raoult’s Law, the partial pressure of A will be
P_{A} ∝ x_{A}
P_{A} = P_{A}^{0} x_{A}
where P_{A}^{0} is the vapour pressure of pure liquid component A. Similarly partial pressure of B will be
P_{B} ∝ x_{B}
P_{B} = P_{B}^{0 }x_{B}
where P_{B}^{0} is the vapour pressure of pure liquid component B.
Now, by Dalton’s law of partial pressures, this states that total pressure ( P_{total}) the solution placed in a container is the sum of partial pressures of its respective components. That is
P_{total} = P_{A} + P_{B}
P_{total} = P_{A}^{0} x_{A} + P_{B}^{0 }x_{B}
Also since, x_{A} + x_{B} = 1 the relation can also be re-written as
P_{total} = P_{A}^{0} + (P_{B}^{0} - P_{A}^{0}) x_{B}
Image 3: Graph between mole fraction and vapour pressure in liquid-liquid solution
Following conclusions are observed from the following relation.
Total vapour pressure relies on mole fraction of any one of the component that is, either A or B. For Example, the equation can also be written as
P_{total} = P_{B}^{0} + (P_{A}^{0} – P_{B}^{0}) X_{A}
We can draw a linear graph between total pressure and mole fraction of any of the following component
Total vapour pressure increases or decreases with respect to mole fraction of component A, depending on pure vapour pressures of components A and B
If the components are in vapour form, the total vapour pressure can be calculated as
P_{A} = y_{A} P_{total}
P_{B} = y_{B} P_{total}
where y_{A} and y_{B} are mole fraction of components A and B in the vapour phase.
Example: The vapour pressures of pure chloroform CHCl_{3} and dichloromethane CH_{2}Cl_{2} are 200 mm Hg and 415 mm Hg. If we add 30 g of CHCl_{3} and 42 g of CH_{2}Cl_{2} at some normal temperature 298 K, then what will be the total vapour pressure of solution? Also, find out mole fraction of both the components in the vapour phase.
So, in this problem, we have two liquid solutions chloroform ( CHCl_{3}) and dichloromethane CH_{2}Cl_{2}. To find out total vapour pressure, we need mole fraction of both the components.
We know that mole fraction of component A = moles of A / Total moles of the solution
And moles of component is mass of component/ molar mass of component
Moles of chloroform CHCl_{3} = 30/ 119.5 = 0.25
Moles of dichloromethane CH_{2}Cl_{2} = 42/ 85 = 0.49
Let A component be CHCl_{3} and B component be CH_{2}Cl_{2} , then
x_{A }= 0.25 / ( 0.25 + 0.49) = 0.25 / 0.74 = 0.33
x_{B} = 0.49 / ( 0.25 + 0.49) = 0.49 / 0.74 = 0.66
Also, we have P_{A}^{0} as 200 mm Hg and P_{B}^{0} as 415 mm Hg
Then total vapour pressure P equals
P_{total} = 200 × 0.33 + 415 × 0.66
P_{total} = 339.9 mm Hg
Let the mole fraction of CHCl_{3} chloroform y_{A} and CH_{2}Cl_{2} be y_{B} in vapour phase, then we know that
P_{A}^{0} x_{A }= y_{A} P_{total}
200 × 0.33 = y_{A} P_{total}
66 / 339.9 = y_{A}
0.19 = y_{A}
Also, y_{A} + y_{B} = 1
Then y_{B} = 1 – y_{A} = 1 – 0.19 = 0.81
So, the total vapour pressure of the solution is 339.3 mm Hg and mole fraction of CHCl_{3} chloroform is 0.19 and that of CH_{2}Cl_{2} dichloromethane is 0.81 respectively.
Another type of solutions is solids in liquid solution, in which we take solid as the solute and liquid as the solvent. For example, when we dissolve glucose, sugar or salt in water, we get solids in liquids type of solution. Generally, the solute is non-volatile in nature and the vapour pressure is less than the pure vapour pressure of the solution.
Why Vapour Pressure decreases when we add a non-volatile solute in the solvent?
The decrease in vapour pressure is due to:
As evaporation is a surface phenomenon, the more the surface, the greater the evaporation and hence more the vapour pressure. In a pure liquid, there is more surface area available for the molecules to vapourise, thereby have more vapour pressure. On the other hand, when we add a non-volatile solute, the solvent molecules get less surface to escape and hence experience low vapour pressure
The number of molecules evaporating or leaving the surface is much greater in pure liquid solutions to that of non-volatile solute in solvent
Image 4: Pure solvent has more number of molecules on the surface as compared to non-volatile solute- solvent solution
To find out overall vapour pressure of the solid-liquid solution, we consider a solution in which lets say A is solvent and B is solute. According to Raoult’s law, we know that partial vapour pressure of individual component (solute/solvent) is directly proportional to its mole fraction.
Now, when we add a non-volatile solute, it is obvious that vapour pressure will only come from the solvent part, as they are an only available component in the vapour phase. Hence, if P_{A} is the vapour pressure of the solvent, x_{A} is its mole-fraction and P_{A}^{0} is the vapour pressure of the pure solvent, then by Raoult’s law, the relation will come out to be:
Image 5: Graph between Vapour Pressure and mole fraction in case of solid in liquid solution.
When we plot a graph between mole fraction of solvent and vapour pressure, we find its nature to be linear.
Watch this Video for more reference
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Vapour Pressure of Liquid Solution
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