Chapter 10.5, Problem 10.9P

Interpretation Introduction

Interpretation:

The boiling point of Benzene in $\text{°C}$ has to be calculated.

Concept introduction:

Equilibrium vapor pressure:

The pressure throw by vapor of the substance is equilibrium liquid or solid state of same substance is called the equilibrium vapor pressure, thus,

$\begin{array}{l}\text{ln}\left(\frac{{\text{P}}_{\text{vap}\text{1}}}{{\text{P}}_{\text{vap}\text{2}}}\right)\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{\text{R}}\left({\frac{\text{1}}{\text{T}}}_{\text{2}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\mathrm{......}(1)\\ {\text{P}}_{\text{vap}\text{1}}\text{and}{\text{P}}_{\text{vap}\text{2}}\text{are}\text{vapor pressures}\\ {\text{T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}\text{are}\text{temperatures}\\ {\text{ΔΗ}}_{\text{vap}}\text{is}\text{enthalpy of vaporization}\\ \text{R is }\text{constant}\end{array}$

The liquid is boiled when the vapor pressure of liquid is equal to atmospheric pressure.