Chapter 11, Problem 11.147QP

Interpretation Introduction

Interpretation:

Vapor pressure of methanol at ${\text{25}}^{°}\text{C}$ has to be calculated.

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 by “Clausius – Clapeyron” equation as follows –

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

At two different temperature and pressure, the equation is rewritten as –

$\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\text{+}\text{C}\\ \text{Where}{\text{T}}_1{\text{ and T}}_2\text{ are two different Temperature}\\ {\text{ P}}_1{\text{ and P}}_2\text{ are two different Pressure}\\ {\text{ΔH}}_{\text{vap}}\text{=}\text{Molar}\text{heat}\text{of}\text{vaporization}\\ \text{R}\text{=}\text{Universal}\text{Gas constant}\text{.}\end{array}$