When a pure liquid is kept in a closed vessel (in which no air is present) the liquid evaporates to give the vapours.
The molecules of the liquid which possess high kinetic energy have a tendency to change to vapour state.
After some time a dynamic equilibrium is established between liquid and vapours.
At Equilibrium
rate of evaporation = rate of condensation.
H_{2}O (ℓ) ⇌ H_{2}O (g)
The pressure that the vapours exert at equilibrium on the walls of the container or on the surface of the liquid is called the vapour pressure of the liquid at that temperature, Figure (a).
Case 1: What would happen to the vapour pressure of the liquid if we take a bigger container, Figure (b) or a wider container, Figure (c).
Key concept: When the liquid is in equilibrium with vapours,
Liquid ⇌ Vapours,
the K_{p} of the system is K_{p} = P_{v},
where P_{v} is the vapour pressure.
We know that K_{p} for a given chemical equilibrium is a constant and only depends on temperature. Therefore the vapour pressure of the liquid is a constant and does not depend on the nature of the vessel used and that it only depends on temperature.
Case 2: Take a vessel with the same liquid as in case 1 and cover it with a glass plate (of negligible weight) having a hole in it. Would the vapour pressure be the same now?
Key concept: Well, the liquid system is the same and the temperature is also same. Therefore, the vapour pressure should be same. The vapour pressure of a liquid is independent of the surface area exposed on top of the liquid surface provided the surface area exposed should not be zero.
The vapour pressure of liquid depends upon
(i) Nature of the Liquid: When the intermolecular forces of attractions are stronger then the vapour pressure will be low because less number of molecules can leave the liquid.
Out of C_{2}H_{5}OH, CH_{3}OCH_{3}, CH_{3}CHO the one with highest vapour pressure is CH_{3}OCH_{3} and the one with lowest vapour pressure is C_{2}H_{5}OH because in C_{2}H_{5}OH there is hydrogen bonding leading to strongest intermolecular forces where as in CH_{3}OCH_{3} there is Vander Waals forces leading to weakest intermolecular forces.
Decreasing order of vapour pressure of few liquids is
Fig: HBonding in Methanol
(ii) Temperature: Higher the temperature, greater would be the vapour pressure.
This is because when temperature is raised, kinetic energy of the molecules increase and therefore the number of molecules leaving surface of liquid, is large.
The Clausius Clapeyron equation tells very clearly that as temperature increases the right hand side remains highly and thereby vapour pressure increases.
This equation is,
which tells as to how vapour pressure changes with nature of liquid and temperature when vapour pressure is low then the temperature at which the liquid would boil will be high. In the above equation P_{1} and P_{2} are vapour pressure of liquid at T_{1} and T_{2} respectively, ΔHv, heat of vaporization and R is gas constant.
Fig: 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.
Some characteristics of the vapour pressure of liquid solutions are as follows:
Fig: Vapour pressure equilibrium in different solution
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 nonvolatile.
Vapour pressure can be evaluated in two cases which are:
Vapour Pressure of LiquidLiquid Solutions
To find out vapour pressure of the liquidliquid 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.
Applying Dalton’s law of partial pressure the total vapour pressure of the solution will be
P_{solution} = P_{A} + P_{B}
Where P_{A} and P_{B} are the partial vapour pressures of A and B.
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.
Raoult’s law states that:
View AnswerA solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
Mathematically, Raoult’s law equation is written as:
P_{solution} = Χ_{solvent }P^{0}_{solvent}
Where,
P_{solution} = vapour pressure of the solution
Χ_{solvent} = mole fraction of the solvent
P^{0}_{solvent} = vapour pressure of the pure solvent
Let's understand this by looking at the example below:
Let two volatile liquids A and B dissolve each other to form an ideal solution. Then the vapours above the solution will contain the vapours of A and B. Applying Dalton’s law of partial pressure the total vapour pressure of the solution will be
P_{solution} = P_{A} + P_{B}
Where,
P_{A} and P_{B} are the partial vapour pressures of A and B.
Solution being dilute (Condition for ideality) Raoult’s law can be applied.
Raoult’s law for binary solutions state that
The vapour pressure of any volatile constituent of a binary solution at any given temperature is equal to the product of the vapour pressure of pure constituent and its mole fraction
P_{A} = P^{º}_{A} X_{A}
P_{B} = P^{º}_{B} X_{B}
Where:
Hence, 𝑃 (Solution) 𝑜𝑟 P_{s} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}
Also, P_{s} = P^{0}_{A}(1X_{B}) + P^{0}_{B}X_{B}
⇒ P_{s} = P^{0}_{A}+(P^{0}_{B}  P^{0}_{A})X_{B}
This equation is of the form y = mx + c. The value of m (slope) may be (+ve) or (–ve) depending upon whether P^{0}_{B} > P^{0}_{A} (m = + ve) or P^{0}_{B} < P^{0}_{A} (m = –ve).
Hence a plot of a graph of P_{s} versus X_{B} will be a straight line with an intercept on the yaxis equal to p_{A} and slope equal to (P^{0}_{B}  P^{0}_{A})
Total Pressure of a Solution will be:
where y_{A} and y_{B} are mole fraction of ‘A’ and ‘B’ in the vapour phase and y_{A} + y_{B} = 1
SPECIAL NOTE:
x_{A} and x_{B} are mole fraction of ‘A’ and ‘B’ in the solution phase and x_{A} + x_{B} =1y_{A} and y_{B} are mole fraction of ‘A’ and ‘B’ in the vapour phase and y_{A} + y_{B} =1
There are a few limitations to Raoult’s law:
There is either a negative or a positive deviation. The negative deviation occurs when the vapour pressure is lower than expected from Raoult’s law. An example of negative deviation is a mixture of chloroform and acetone or a solution of water and hydrochloric acid.
Alternatively, positive deviation takes place when the cohesion between similar molecules is greater or that it exceeds adhesion between unlike or dissimilar molecules. Both components of the mixture can easily escape from the solution. An example of positive deviation includes the mixtures of benzene and methanol or ethanol and chloroform.
Example 1. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate the total vapour pressure of the solution.
Solution.
Number of moles of ethanol = 60/40 = 1.5
Number of moles of methanol = 40/32 = 1.25
X_{A }= 1.25/1.25 + 1.3 = 0.4545 and X_{B} = 1 – 0.4545 = 0.545
Let A = CH_{3}OH, B = C_{2}H_{5}OH
Total pressure of the solution
P_{T} = X_{A}P^{0}_{A} + X_{B}P^{0}B
= 0.4545 × 88.7 + 0.545 × 44.5 = 40.31 + 24.27
= 64.58 mm Hg
Example 2. The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If X_{A} and Y_{A} are the mole fraction of A in the liquid and vapour, respectively find the value of X_{A} for which Y_{A} − X_{A }has a minimum. What is the value of the pressure at this composition?
Solution.
Subtracting x_{A} from both sides, we get
Now differentiating w.r.t. x_{A}, we get
The value of x_{A} at which y_{A} − x_{A} has a minimum value can be obtained by putting the above derivative equal to zero. Thus we have
Solving for x_{A}, we get
hence
Example 3. A very small amount of nonvolatile solute (that does not dissociate) is dissolved in 56.8 cm^{3} of benzene (density 0.889 g cm^{3}). At room temperature, the vapour pressure of this solution is 99.88 mm Hg while that of benzene is 100mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degrees lower than that of benzene. What is the value of the molal freezing point depression constant of benzene?
Solution.
Also,
ΔT_{f} = K_{f} × molality f
∴ 0.73 = K × 0.1452
f_{K} = 5.028 K. molality
Example 4. What is the composition of the vapour which is in equilibrium at 30°C with a benzenetoluene solution with a mole fraction of benzene of 0.400?
(P^{o}_{B} = 119 torr and P_{T}^{o} = 37.0 torr)
Solution.
Total pressure of the solution is given by
P_{T} = X_{B} P^{0}_{B} + X_{T}P^{0}_{T}
= 0.4 × 119 + 0.6 × 37
= 47.6 + 22.2
= 69.8 torr
Applying Dalton’s law for mole fraction in the vapour phase.
Y_{B} = P_{B}/P_{T} = 47.6/62.4
= 0.763
Y_{T} = 1– 0.763
= 0.237
Raoult’s Law states that the mole fraction of the solute component is directly proportional to its partial pressure.
On the basis of Raoult’s Law, liquidliquid solutions are classified into two types of solutions, they are:
• Ideal Solutions
• Nonideal Solutions
An ideal solution is a mixture in which the molecules of different species are distinguishable, however, unlike the ideal gas, the molecules in ideal solution exert forces on one another. When those forces are the same for all molecules independent of species then a solution is said to be ideal.
Note: Perfectly ideal solutions are rare in nature, only some solutions show some ideal behaviour.
The solutions which don’t obey Raoult’s law at every range of concentration and at all temperatures are called NonIdeal Solutions. Nonideal solutions deviate from ideal solutions and are also known as NonIdeal Solutions.
Types of NonIdeal Solutions
Example: consider two components A and B to form nonideal solutions.
Let the vapour pressure, pure vapour pressure and mole fraction of component A be P_{A} , P_{A}^{0} and x_{A} respectively and that of component B be P_{B} , P_{B}^{0} and x_{B} respectively.
These liquids will show positive deviation when Raoult’s Law when
Examples: Following are examples of solutions showing positive deviation from Raoult’s Law
Examples: Following are examples of solutions showing negative deviation from Raoult’s Law
Azeotropes are defined as a mixture of two liquids which has a constant composition in the liquid and vapour phase at all temperatures.
Each azeotrope has a characteristic boiling point.
The boiling point of an azeotrope is either less than the boiling point temperatures of any of its constituents (a positive azeotrope), or greater than the boiling point of any of its constituents (a negative azeotrope).
1. Maximum Boiling Azeotrope
Maximum Boiling Azeotrope is formed when we mix two nonideal solutions at some specific composition, showing a large negative deviation from Raoult’s law.
Examples:
2. Minimum Boiling Azeotrope
Minimum Boiling Azeotrope is formed when we mix two nonideal solutions at some specific composition, which shows a large positive deviation from Raoult’s Law.
Example: Ethanol ( 95.5%) and water (4.5%) form minimum boiling azeotrope at a boiling temperature of 351.5 K
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1. What is vapor pressure? 
2. How is vapor pressure affected by temperature? 
3. How does the presence of solutes affect the vapor pressure of a liquid solution? 
4. How does the concentration of solutes affect the vapor pressure of a liquid solution? 
5. How is the vapor pressure of a liquid solution related to boiling point elevation? 

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