Chapter 10.5, Problem 6PE

Interpretation Introduction

Interpretation: The vapour pressures of pentane at two different temperatures are given. The heat of vaporisation of pentane is to be calculated.

Concept introduction: The heat of vaporisation is calculated by the formula,

$\ln \left(\frac{\text{Vapour}\text{pressure}\text{at}{\text{T}}_{\text{1}}}{\text{Vapour}\text{pressure}\text{at}{\text{T}}_2}\right)=\frac{\text{Heat}\text{of}\text{vaporisation}}{\text{Gas}\text{constant}}\left(\frac{1}{{\text{T}}_{\text{1}}}-\frac{1}{{\text{T}}_2}\right)$

Where, ${\text{T}}_{\text{1}}$ and ${\text{T}}_2$ are the two absolute temperatures.

To determine: The heat of vaporisation of pentane.