   Chapter 13.5, Problem 1.4ACP

Chapter
Section
Textbook Problem

The vapor pressure of pure heptane is 361.5 mm Hg at 75.0 °C and its normal boiling point is 98.4 °C. Use the Clausius-Clapeyron equation (page 512) to determine the enthalpy of vaporization of heptane.

Interpretation Introduction

Interpretation: The enthalpy of vaporization of heptane has to be determined using Clausius-Clapeyron equation.

Concept introduction:

• Enthalpy of vaporization (ΔHvap ) is the amount of energy required to change one mole of a substance from the liquid phase to the gas phase at constant temperature and pressure.
• Clausius-Clapeyron equation is used to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

ln(P2P1)=ΔHvapR(1T2-1T1)

ΔHvap is the enthalpy of vaporization.

R is the gas constant (8.3145Jmol1K1 )

P1andP2 are the vapor pressures at two temperatures T1andT2.

Explanation

Given data:

Vaporpressureofheptane=361.5mmHgT1=98.4°C=371.55KT2=75°C=348.15K

Clausius-Clapeyron equation is used to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

ln(P2P1)=ΔHvapR(1T2-1T1)

Substituting the given values of pressure and temperatures in the above equation,

ln(361

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts 