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Vapour Pressure

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Vapour pressure is the force exerted by a liquid’s (or solid’s) vapour above the surface of the liquid. At a particular temperature and thermodynamic equilibrium, this pressure is formed in a closed container. The rate of liquid evaporation is controlled by the equilibrium vapour pressure. The vapour pressure increases with increasing temperature. When atmospheric pressure and vapour pressure are equal, a liquid is said to have reached its boiling point.

Characteristics of Vapour Pressure

Vapour pressure is defined as the pressure exerted by the liquid in the thermodynamic equilibrium state. Any substance that has a high vapour pressure at normal temperature is termed a volatile substance. The vapour pressure at the equilibrium state indicates the rate of evaporation in the liquid.

In the stage of equilibrium, the pressure exerted by the atmosphere is equal to the vapour of the liquid.

Factors Affecting Vapour Pressure

Vapour Pressure of the liquid depends on the various factor discussed below:

  • Nature of Liquid: Intermolecular forces between the atoms of the liquid explain the nature of the liquid. Vapour Pressure of the liquid changes according to different types of liquids.
  • Effect of Temperature: With the increase in temperature of the liquid, the kinetic energy of the atoms of the liquid also increases. Now, the liquid molecules with high kinetic energy are less likely to escape thus, the vapour pressure of the liquid increases. Hence we can say that vapour pressure is directly proportional to temperature.
  • Concentration of Solute: The existence of a solute in the liquid will significantly reduce the vapour pressure. And this fall in vapour pressure also differs with respect to the concentration of solute.

What is Raoult’s Law?

According to Raoult’s law, the vapour pressure of a pure component (liquid or solid) multiplied by its mole fraction in the mixture results in the partial pressure of that component in a perfect mixture of liquids. As a result, the mole fraction of solute in the solution equals the relative decrease in vapour pressure of a diluted solution of a non-volatile solute.
A solution that complies with Raoult’s law is referred to as the “perfect solution”. While some liquid combinations adhere to this law over a wide range of concentrations, this law only holds for diluted solutions. Perfect solutions are produced because the intermolecular forces between molecules of pure substances are the same as the forces between molecules of one material and molecules of another.

Raoult’s Law or Vapour Pressure Formula

When a solid is dissolved into a liquid, a solution is made. When the solute is added, the vapour pressure that results from this solution is reduced. Here is the formula for vapour pressure using Raoult’s law, which explains how the vapour pressure of a liquid changes when a solute is added.

Psolution = (Xsolvent)×(P° solvent)

where,

Psolution is the vapour pressure of a solution 
Xsolvent is the mole fraction of solvent in a solution
solvent is the vapour pressure of a solvent

Importance of Raoult’s law

Let’s say a volatile liquid A is placed inside a closed container. After some time, vapour particles will start to develop as a result of evaporation. The liquid particles on the surface will eventually be in dynamic equilibrium with the vapour particles of A as time goes on. As a result, the pressure that A’s vapour particles are exerting at any given temperature is known as A’s vapour pressure at that temperature. All solids and liquids display vapour pressure, which is solely dependent on the kind of liquid and temperature.

Vapour in liquid solution

 

The A particles will now fill the spaces between the B particles on the surface of the solution if liquid B is introduced to this container.
A portion of the molecules on the surface of any given liquid have enough energy to convert to the vapour phase.
Since there are currently fewer A particles on the surface, there will be fewer A vapour particles in the vapour phase. This will cause A’s vapour pressure to decrease.
If B is also volatile, there will be fewer B particles in the vapour phase than there would be if B were a pure liquid.

Liquid in Liquid solution

 

This new pressure, which is determined by Raoult’s equation and depends on the component concentration of each liquid phase, is the partial pressure of each A and B

PA ∝ XA = XAA  

PB ∝ XB  = XBB

P° represents the component of the mole fraction.

Vapour Pressure of Liquid-Liquid Solutions

Let’s use two volatile liquid solutions labelled 1 and 2 to calculate the vapour pressure of a liquid-liquid solution. An equilibrium between the binary solution’s liquid phase and vapour phase is created when the liquid components are added to a closed vessel.
Assume that PTotal is the total vapour pressure of the binary solution at equilibrium and P1 and P2 are the respective partial vapour pressures of components 1 and 2. Let the mole fractions of components 1 and 2 be X1 and X2, as the partial pressures are related to the mole fractions.

Vapour Pressure of Solutions of Solids in Liquids

We mix sugar and water to create a sugar solution. The non-volatile solute in this mixture is sugar, while the solvent is water. Due to the non-volatility of the solute, when this sugar solution evaporates, the vapour phase only consists of water vapours. However, it is found that this solution’s vapour pressure is lower than that of the pure solvent (water).

The rate at which the solvent molecules escape from the surface of the liquid determines the vapour pressure of a solution made up of a non-volatile solute and a solvent. As a result, the vapour pressure of the solution (sugar and water) will be lower than that of the pure solvent (water), or the addition of a non-volatile solute will cause the vapour pressure of the solution to decrease. The solution’s vapour pressure will decrease more when the concentration of sugar in the solution rises.
The amount of non-volatile solute present in the solution, not the type of non-volatile solute, determines how much of the solvent’s vapour pressure is reduced.

What is a boiling point?

Vapour pressure of any liquid increases with the increase in temperature as the temperature increases there comes a stage at which the vapour pressure of the liquid is equal to the atmospheric pressure. It reaches a stage where the vapour pressure of the liquid becomes equal to the atmospheric pressure. In this stage, the vapour of liquid escapes to the atmosphere and the temperature of the liquid at this stage is called the boiling point of the liquid.

For the standard value of the boiling point of liquid pressure is taken to be,

Pressure = 1 atm = 102325 Pa = 1 bar

What is the Heat of Vapourization?

The heat required by 1 mole of liquid to change its phase from liquid to vapour form is called the Heat of Vapourization. The heat of Vapourization only changes the phase of the liquid to gaseous form it does not increase the temperature of the liquid.

Raoult’s Law and its Relationship with Other Laws

Raoult’s Law is one of the basic laws and its relation with other laws can be stated as,

  • Ideal gas law and Raoult’s law resemble one other quite a bit. Raoult’s law does not apply to solutions, which is the single exception. If you’ve read about the ideal gas law, you know that it implies that gases would behave in an ideal manner, with zero or no intermolecular forces between molecules of different types. Meanwhile, Raoult’s law makes the hypothesis that the intermolecular forces between unlike and comparable molecules are equal.
  • Non-ideal solutions are similarly subject to Raoult’s law. The interactions between molecules of various substances must be taken into account when incorporating several components, though.
    If we use an ideal system made up of an ideal liquid and an ideal vapour, we may also deduce a very
  • By combining Raoult’s law and Dalton’s law, we may further obtain a highly helpful equation for an ideal system made up of an ideal liquid and an ideal vapour.
  • This equation tells us that each component of an ideal solution made up of pure substances will have a different vapour pressure. Additionally, the component will have a higher pure vapour pressure in the gas phase compared to the solution, which will have a lower pure vapour pressure.

Limitations of Raoult’s Law

Raoult’s Law has limitations which are discussed below,

  • Raoult’s law is especially relevant since it refers to ideal solutions, i.e., those in which the gas phase has thermodynamic properties similar to a combination of ideal gases. 
  • The only drawback is that they are hard to find and rare. Different chemical components must be chemically equivalent. 
  • Several solutions deviate from Raoult’s law because various attractive forces exist in many liquid mixtures. So, do not properly obey it.

Also, Check,

Solved Examples on Vapour Pressure Formula

Example 1: At 25 °C, an aqueous solution’s vapour pressure is measured to be 23.80 mmHg. What fraction of the solute by a mole in this solution? At 25 °C, water has a vapour pressure of 25.756 mm Hg.

Solution:

Vapour pressure of the solution, P Solution = 23.80

Vapour pressure of solvent, P° Solvent= 25.756

Using vapour pressure formula:

Psolution = (Xsolvent) (P°solvent)

23.80= (XSolvent) 25.756

XSolvent = 0.92405

Mole fraction of Solvent = 0.92405

Mole fraction of Solute = 1 – 0.92405

                                      = 0.07595

Example 2: At 25°C, an aqueous solution’s vapour pressure is measured at being 20mmHg. What is the mole fraction of the solvent in this solution? At 25 °C, water has a vapour pressure of 60 mm Hg.

Solution: 

Vapour pressure of the solution, P Solution = 20

Vapour pressure of solvent, P° Solvent= 60

Using vapour pressure formula:

Psolution = (Xsolvent) (Posolvent)

20= (XSolvent) 60

XSolvent = 0.333

Mole fraction of solvent = 0.333

Example 3: What is the vapour pressure of the solution if the mole fraction of the solute is 0.3? Water has a 16.358 mmHg vapour pressure at 23 °C.

Solution:

Mole Fraction of solute = 0.3

Mole Fraction of solvent, XSolvent = 1 – 0.3

                                       = 0.7

Now,

Vapour pressure of solvent, P°Solvent = 16.358

Mole Fraction of solvent = 0.7

Using vapour pressure formula:

Psolution = (Xsolvent) (P°solvent)

            = (0.7)(16.358)

            = 11.4506

Vapour pressure of the solution is 11.4506

Example 4: 200 grams of sucrose dissolved in 300 grams of water at 30 degrees. How much vapour pressure does this solution have? At 30.0 °C, the vapour pressure of water is 41.62 mmHg.

Solution: 

First, we need to find the fraction of mole in a solvent.

Now, the amount of mole in solute

Moles of Sucrose = wt (gm)/ molecular mass              (molecular weight of sucrose = 342.2948)
                             = 200 / 342.2948g
                             = 0.5843 

Moles of Water = 300 / 18.015                                     (molecular weight of water = 18.015)
                          = 16.6528 

Total mole of solution = 0.5843+16.6528 = 17.2371

Mole fraction of solvent (water) = Mole of water / Mole of solution

Mole fraction of solvent, X Solvent = 16.6528/(16.6528 + 0.5843)

                                                     = 16.6528/17.2371

                                                     = 0.9661

vapour pressure of solvent, P°Solvent = 41.62

Using vapour pressure formula:

Psolution = (Xsolvent) (P°solvent)

           = (0.9661)(41.62)

           = 40.2090

Vapour pressure of the solution is 40.2090.

Example 5: What is the vapour pressure of 400 g of propanol (molecular weight = 60 g/mol) and 130 g of acetone (molecular weight = 58 g/mol)? Propanol and acetone have vapour pressures of 11 and 20 mmHg, respectively, at 25 °C.

Solution:

Calculating Molar fractions: 

Number of moles in acetone, nacetone = 130g / 58g/mol = 2.24 mol

Number of moles in propanol, npropanol = 400g / 60g/mol = 6.66 mol

ntotal = nacetone + npropanol = 2.24 + 6.66 = 8.9 mol

Molar fraction of acetone, xacetone= 2.24/8.9 = 0.2516

Molar fraction of propanol, xpropanol = 6.66/8.9 = 0.7483

Vapour pressure of acetone, Pacetone = 20

Vapour pressure of Propanol, P°Propanol= 11

Now, partial pressure can be calculated as:

Pacetone = (xacetone) ( acetone) = 0.2516 × 20 = 5.03 mmHg

PPropanol = (xPropanol) ( Propanol) = 0.7483 × 11 = 8.2313 mmHg

vapour Pressure of the Mixture

Pmixture = Pacetone + PPropanol 
            = 5.03 + 8.2313 
            = 13.2613 mmHg

FAQs on Vapour Pressure Formula

Question 1: What is Raoult’s law and its application?

Answer:

Raoult’s law states that the partial pressure of a pure component (liquid or solid) in a perfect mixture of liquids is equal to the vapour pressure of that component multiplied by the mole fraction in the mixture.

Raoult’s law is applicable:

  • To evaluate the reduction in non-volatile solute vapour pressure.
  • To evaluate the liquids’ ability to bind strongly.

Question 2: Write the Vapour Pressure formula.

Answer:

A solution is created when a solid is dissolved in a liquid. The resultant vapour pressure of this solution is decreased when the solute is introduced. Raoult’s law, which describes how the vapour pressure of a liquid changes when a solute is added, is used to calculate vapour pressure.

Psolution = (Xsolvent)×(P° solvent)

where,

Psolution is the vapour pressure of a solution 
Xsolvent is the mole fraction of solvent in a solution
solvent is the vapour pressure of a solvent

Question 3: For what type of solutions, Raoult’s law is valid?

Answer:

Raoult’s law is only valid for ideal solutions. The solvent-solute interaction is the same as a solvent-solvent or solute-solute interaction in an ideal solution. This suggests that both the solvent and the solute expend the same amount of energy to reach the vapour phase as they do in their pure states.

Question 4: Write the limitations of Raoult’s law.

Answer:

There are limitations of Raoult’s law that are mentioned below:

Since Raoult’s law applies to ideal solutions, i.e., those in which the gas phase exhibits thermodynamic features resembling a combination of ideal gases, it is particularly pertinent. The fact that they are uncommon and hard to find is the only negative. Because distinct attractive forces exist in many liquid mixes, certain solutions depart from Raoult’s equation.



Last Updated : 20 Dec, 2023
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