Question

raoult law derive it's mathematical expression​

Answer 1

We have to derive the mathematical expression of Raoult's law.

Rauolt's law :

• This law is applicable for two or more than two volatile liquids are available in a solution.
• It is just a special case of Henry's law.
• At a given temperature, the partial pressure of a volatile liquid in a solution is the product of mole fraction of that liquid and vapor pressure of that liquid in pure state.

We know, a simple solution has two things ; solute and solvent.

Let solute is A and solvent is B.

$\text{partial pressure of solute, A}=P_A\\\\\text{mole fraction of solute, A}=x_A\\\\\text{vapor pressure of solute, A in pure state} =P^{\circ}_A\\\\\text{partial pressure of solvent, B}=P_B\\\\\text{mole fraction of solvent, B}=x_B\\\\\text{vapor pressure of solvent, B in pure state} =P^{\circ}_B$

From Rauolt's law,

$\implies P_A=P^{\circ}_Ax_A\\\\\implies P_B=P^{\circ}_Bx_B$

Therefore total pressure of solution,

$P_T=P_A+P_B\\\\=P^{\circ}_Ax_A+P^{\circ}_Bx_B\\\\=P^{\circ}_Ax_A+P^{\circ}_B(1-x_A)\\\\=P^{\circ}_B+(P^{\circ}_A-P^{\circ}_B)x_B$

Therefore the mathematical expression of Rauolt's law for pressure of solution is given by,

$\bold{P_T=P^{\circ}_B+(P^{\circ}_A-P^{\circ}_B)x_A}$

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