Chapter 12.2, Problem 1PPB

Interpretation Introduction

Interpretation:

• Molar heat of vaporization of the liquid has to be determined.
• The temperature at which the vapor pressure of the liquid will be five times the vapor pressure at $\text{85°C}$ has to be determined.

Concept Introduction:

Molar heat of vaporization of the liquid id the amount of heat require to vaporize one mole of the liquid at its boiling point.

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid by Clausius – Clapeyron equation as follows –

$\begin{array}{l}\text{ln P = }\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{\text{T}}\right]\text{+ C}\\ \text{Where P = vapor pressure ; R = Universal Gas constant}\\ {\text{ T = Temperature ; ΔH}}_{\text{vap}}\text{= molar heat of vaporization}\\ \text{ C = constant}\end{array}$

At two different temperatures, a substance has two different vapor pressure.  In this case the Clausius – Clapeyron equation becomes,

$\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{Where}{\text{P}}_{\text{1}}\text{=}\text{vapor}\text{pressure}\text{of}\text{the}\text{liquid}\text{}\text{at}\text{temperature}{\text{T}}_{\text{1}}\text{.}\\ {\text{P}}_{\text{2}}\text{=}\text{vapor}\text{pressure}\text{of}\text{the}\text{liquid}\text{}\text{at}\text{temperature}{\text{T}}_{\text{2}}\text{.}\\ {\text{ΔH}}_{\text{vap}}\text{=}\text{Molar}\text{heat}\text{of}\text{vaporization}\\ \text{R}\text{=}\text{Universal}\text{Gas constant}\text{.}\end{array}$