Chapter 12, Problem 12.13QP

Interpretation Introduction

Interpretation:

The molar heat of vaporization of the liquid has to be calculated when its vapor pressure gets doubled during temperature rise from ${\text{75}}^{\text{°}}\text{C}$ to ${\text{100}}^{\text{°}}\text{C}$.

Concept Introduction:

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 as follows,

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