Chapter 9, Problem 13QAP

Interpretation Introduction

Interpretation:

The vapor pressure of octane is obtained by the following data-

Vp(mmHg)	10	40	100	400
t (°C)	19.2	45.1	65.7	104.0

The graph between ln(vp) and 1/T is to be plotted. The heat of vaporization of octane is to be estimated by this graph.

Concept introduction:

The relation between pressure, temperature and change in enthalpy of vaporization is as follows:

$\mathrm{lnP}=A-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T}\right)$

Here,

P = pressure

T = temperature

$\Delta H_{vap}$ = heat of vaporization = slope

A = y-intercept from the graph.