Question

The mole fraction of component A in the vapour phase is ${\chi }_1$ and mole fraction of component A in liquid mixture is ${\chi }_2$ ($P_A^o$ = vapour pressure of pure A; $P_B^o$ = vapour pressure of pure B). Then the total vapour pressure of the liquid mixture is:

• A. $\frac{P_A^o{\chi }_2}{{\chi }_1}$
• B. $\frac{P_A^o{\chi }_1}{{\chi }_2}$
• C. $\frac{P_B^o{\chi }_1}{{\chi }_2}$
• D. $\frac{P_B^o{\chi }_2}{{\chi }_1}$

Answer 1

The correct option is A $\frac{P_A^o{\chi }_2}{{\chi }_1}$

ANS: - Mole fraction of component A in vapor phase - $x_1$

In Liquid Phase $x_2$

Vapor Pressure of pure $A=P_A^{\circ }$

Vaporer pressure of pure $B=P_B^{\circ }$

Mole fraction of component A in the vapor phase is given by

$x_1=\frac{P_A}{P_A+P_B}$

$P_A+P_B=\frac{P_A}{X_1}$

$P_B=\frac{P_A}{X_1}-P_A$(1)

According to Dalton's low

$P_T=P_A+PB$(2)

Where $P_A=P_A^{\circ }{X}_2$(3)

putting (1) and (3) in (2)

$P_T=P_A^0{X}_2+\frac{P_A}{x_1}-P_A$

$=P_A^0{X}_2+\frac{P_A^P{X}_2}{X_1}-P_A^{\sigma }{x}_2$

$P_T=\frac{P_A^0{X}_2}{x_1}$