What are three different cases for raoult’s law
Answers
Answer:
Raoult’s law states that a 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;
Psolution = ΧsolventP0solvent
Where,
Psolution = vapour pressure of the solution
Χsolvent = mole fraction of the solvent
P0solvent = vapour pressure of the pure solvent
We will further understand the principle behind the law by looking at the example below.
Consider a solution of volatile liquids A and B in a container. Because A and B are both volatile, there would be both particles of A and B in the vapour phase.
Hence, the vapour particles of both A and B exert partial pressure which contributes to the total pressure above the solution.
Solution of volatile liquids A and B
Raoult’s Law further states that at equilibrium,
{{P}_{A}}=P_{A}^{{}^\circ }{{x}_{A}},{{P}_{B}}=P_{B}^{{}^\circ }{{x}_{B}}P
A
=P
A
∘
x
A
,P
B
=P
B
∘
x
B
Where PA is the partial pressure of A.
P_{A}^{{}^\circ }P
A
∘
is vapour pressure of pure A at that temperature.
{{x}_{A}}x
A
is mole fraction of A in the liquid phase.
Similarly {{P}_{B}},P_{B}^{{}^\circ }{{x}_{B}}P
B
,P
B
∘
x
B
Hence {{P}_{T}}={{P}_{A}}+{{P}_{B}}\left( Dalton’s\,Law \right)P
T
=P
A
+P
B
(Dalton’sLaw) =P_{A}^{{}^\circ }{{x}_{A}}+P_{B}^{{}^\circ }{{x}_{B}}=P
A
∘
x
A
+P
B
∘
x
B
=P_{A}^{{}^\circ }+{{x}_{B}}\left( P_{B}^{{}^\circ }-P_{A}^{{}^\circ } \right)=P
A
∘
+x
B
(P
B
∘
−P
A
∘
)
Explanation:
Answer:
law of thermodynamics.
We will further take an in-depth look at Raoult’s law and understand the principle behind the law as well as its application and limitations in this lesson.
Table of Content
What is Raoult’s Law?
Importance of Raoult’s law
Raoult’s Law and its Relationship with Other Laws
Limitations of Raoult’s Law
FAQs
What is Raoult’s Law?
Raoult’s law states that a 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;
Psolution = ΧsolventP0solvent
Where,
Psolution = vapour pressure of the solution
Χsolvent = mole fraction of the solvent
P0solvent = vapour pressure of the pure solvent
We will further understand the principle behind the law by looking at the example below.
Consider a solution of volatile liquids A and B in a container. Because A and B are both volatile, there would be both particles of A and B in the vapour phase.
Hence, the vapour particles of both A and B exert partial pressure which contributes to the total pressure above the solution.
What is the Importance of Raoult’s law?
Assume that we have a closed container filled with a volatile liquid A. After some time, due to evaporation, vapour particles of A will start to form. Then as time passes, the vapour particles of A will be in dynamic equilibrium with the liquid particles (on the surface). The pressure exerted by the vapour particles of A at any particular temperature is called the
vapour pressure of A at that temperature.
Vapour pressure is exhibited by all solids and liquids and depends only on the type of liquid and temperature.
Now imagine we are adding another liquid B (solute) to this container. This will result in B particles occupying the space between A particles on the surface of the solution.
For any given liquid there are a fraction of molecules on the surface which will have sufficient energy to escape to the vapour phase.
Since now we have a lesser number of A particles on the surface, the number of vapour particles of A in the vapour phase will be lesser. This will result in lower vapour pressure of A.
Now if we assume that B is volatile as well, we will have lesser number of B particles in the vapour phase as compared to pure liquid B.