Chapter 5, Problem 5C.9E

Interpretation Introduction

Interpretation:

The vapor pressure of each component and total vapor pressure of the mixture at ${\text{25}}^{\text{o}}\text{C}$ has to be calculated.

Concept Introduction:

The equilibrium between a liquid and its vapor produces a characteristic vapor pressure for each substance that depends on the temperature.  The lowering of the vapor pressure is caused by a lesser ability of the solvent to evaporate, so equilibrium is reached with a smaller concentration of the solvent in the gas phase.  The vapor pressure of a solution is expressed using Raoult’s law:

  ${\text{P}}_{\text{solv}}{\text{=χ}}_{\text{solv}}{\text{P}}^{\text{o}}{}_{\text{solv}}$

The vapor pressure of the solvent $\left({\text{P}}_{\text{solv}}\right)$ above a dilute solution is equal to the mole fraction of the solvent ${\text{(χ}}_{\text{solv}}\text{)}$ times the vapor pressure of the pure solvent $\left({\text{P}}^{\text{o}}{}_{\text{solv}}\right).$

Mole fraction:  Mole fraction of a substance in a solution is the number of moles of that substance divided by the total number of moles of all substances present.  The formula is,

  $\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Moles}\text{of}\text{A (in mol) +Moles}\text{of}\text{B (in mol) +Moles}\text{of}\text{C (in mol) +}\mathrm{...}}\\ \\ {\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\end{array}$

Dalton’s Law:

The total pressure of a gas mixture is the sum of the partial pressures of its component gases.

  ${\text{P}}_{\text{total}}{\text{=P}}_{\text{a}}{\text{+P}}_{\text{b}}\text{+}\mathrm{...}$

Where ${\text{P}}_{\text{a,}}{\text{P}}_{\text{b}}$ is the partial pressure of the gases in the mixture.

The partial pressure of the gas can be obtained by multiplying the total pressure of the mixture with the percent of the gases present in the mixture.