Chapter 12.2, Problem 12.2.4SR

Interpretation Introduction

Interpretation:

From the given graph the molar heat of vaporization has to be calculated.

Concept Introduction:

Molar heat of vaporization:

Molar heat of vaporization of the liquid is the amount of heat required 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,

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