Science, asked by devenkapoor199p1tl45, 1 year ago

why does vapour pressure is directly proportinal to mole fraction of that liquid?

Answers

Answered by akkipilania34
2
Raoult's law (/ˈrɑːuːlz/ law) is a law of thermodynamics established by French chemist François-Marie Raoult in 1887. [1] It states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. In other words it is stated as the relative lowering of vapour pressure of a dilute solution containing nonvolatile solute is equal to the mole fraction of solute in the solution.

Mathematically, Raoult's law for a single component in an ideal solution is stated as

{\displaystyle p_{i}=p_{i}^{\star }x_{i}},

where {\displaystyle p_{i}} is the partial vapor pressure of the component {\displaystyle i} in the gaseous mixture (above the solution), {\displaystyle p_{i}^{\star }} is the vapor pressure of the pure component {\displaystyle i}, and {\displaystyle x_{i}} is the mole fractionof the component {\displaystyle i} in the mixture (in the solution).[2]

Once the components in the solution have reached equilibrium, the total vapor pressure of the solution can be determined by combining Raoult's law with Dalton's law of partial pressures to give

{\displaystyle p=p_{\rm {A}}^{\star }x_{\rm {A}}+p_{\rm {B}}^{\star }x_{\rm {B}}+\cdots }.

If a non-volatile solute (zero vapor pressure, does not evaporate) is dissolved into a solvent to form an ideal solution, the vapor pressure of the final solution will be lower than that of the solvent. The decrease in vapor pressure is directly proportional to the mole fraction of solute in an ideal solution.

{\displaystyle p=p_{\rm {A}}^{\star }x_{\rm {A}}}{\displaystyle \Delta p=p_{\rm {A}}^{\star }-p=p_{\rm {A}}^{\star }(1-x_{\rm {A}})=p_{\rm {A}}^{\star }x_{\rm {B}}}.

PrincipleEdit



Vapor pressure of a binary solution that obeys Raoult's law. The black line shows the total vapor pressure as a function of the mole fraction of component B, and the two green lines are the partial pressures of the two components.

Raoult's law is a phenomenological law that assumes ideal behavior based on the simple microscopic assumption that intermolecular forces between unlike molecules are equal to those between similar molecules: the conditions of an ideal solution. This is analogous to the ideal gas law, which is a limiting law valid when the interactive forces between molecules approach zero, for example as the concentration approaches zero. Raoult's law is instead valid if the physical properties of the components are identical. The more similar the components are, the more their behavior approaches that described by Raoult's law. For example, if the two components differ only in isotopiccontent, then Raoult's law is essentially exact.

Comparing measured vapor pressures to predicted values from Raoult's law provides information about the true relative strength of intermolecular forces. If the vapor pressure is less than predicted (a negative deviation), fewer molecules of each component than expected have left the solution in the presence of the other component, indicating that the forces between unlike molecules are stronger. The converse is true for positive deviations.

For a solution of two liquids, A and B, Raoult's law predicts that if no other gases are present, then the total vapor pressure {\displaystyle p} above the solution is equal to the weighted sum of the "pure" vapor pressures of the two components, {\displaystyle p_{A}} and {\displaystyle p_{B}}. Thus the total pressure above the solution of A and B would be

{\displaystyle p=p_{\rm {A}}^{\star }x_{\rm {A}}+p_{\rm {B}}^{\star }x_{\rm {B}}.}

Since the sum of the mole fractions is equal to one,

{\displaystyle p=p_{\rm {A}}^{\star }(1-x_{\rm {B}})+p_{\rm {B}}^{\star }x_{\rm {B}}=p_{\rm {A}}^{\star }+(p_{\rm {B}}^{\star }-p_{\rm {A}}^{\star })x_{\rm {B}}}

Similar questions