Question

Consider a binary mixture of two ideal volatile liquids A and B.
Which of the following represents the correct relation?

(Here total pressure of the solution is $\left(P_T\right)$, mole fraction of component A in vapour phase is $y_A$, vapour pressure of component A and B in pure states is $p_A^{\circ }and{p}_B^{\circ }$ respectively.)

• A. $P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$
• B. $P_T=\frac{p_A^{\circ }+p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$
• C. $P_T=\frac{p_A^{\circ }-p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }+p_A^{\circ })\times y_A}$
• D. $P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }+p_A^{\circ })\times y_A}$

Answer 1

The correct option is A $P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$

From raoult's Law we have:
$p_A=p_A^{\circ }{\chi }_A\Rightarrow {\chi }_A=\frac{p_A}{p_A^{\circ }}....eqn\left(i\right)p_B=p_B^{\circ }{\chi }_B\Rightarrow {\chi }_B=\frac{p_B}{p_B^{\circ }}....eqn\left(ii\right)$
where $p_A,p_B$ are the partial pressures of A and B repsectively
Also by dalton's law , we have :
$p_A=y_A\times P_T{p}_B=y_B\times P_T$

Put the values of $p_A,p_B$ in eqn (i) and eqn (ii)

${\chi }_A=\frac{y_A\times P_T}{p_A^{\circ }}$

${\chi }_B=\frac{y_B\times P_T}{p_B^{\circ }}$

Also ${\chi }_A+{\chi }_B=1$

So, $\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{y_B}{p_B^{\circ }}\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{1-y_A}{p_B^{\circ }}$
Solving,

$P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$

(c) is the correct answer