   Chapter 17, Problem 84AE

Chapter
Section
Textbook Problem

# Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data.   Boilling point (1atm) ∆Hvap Methane 112K 8.20 KJ/mol Hexane 342K 28.9 KJ/mol Compare the molar volume of gaseous methane at 112 K with that of gaseous hexane at 342 K. How do the differences in molar volume affect the values of ∆Svap for these liquids?

Interpretation Introduction

Interpretation: The boiling point and ΔHvap is given for methane and hexane. The entropy change for the vaporization of liquid methane and liquid hexane is to be calculated from these data. The change in ΔSvap of these liquids due to the differences in their molar volume is to be stated.

Concept introduction: The change in entropy for the vaporization of liquid is expressed by the formula,

ΔSvap=ΔHvapTB.P

Molar volume of a gaseous substance is directly proportional to its boiling point.

Explanation

Explanation

Given

Boiling point of methane is 112K .

Boiling point of hexane is 342K .

The value of ΔHvap of methane is 8.20kJ/mol .

The value of ΔHvap of hexane is 28.9kJ/mol .

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 8.20kJ into joule is,

8.20kJ=(8.20×103)J=8200J

The conversion of 28.9kJ into joule is,

28.9kJ=(28.9×103)J=28900J

The change in entropy for the vaporization of liquid is expressed by the formula,

ΔSvap=ΔHvapTB.P (1)

Where,

• ΔSvap is the entropy change of vaporization.
• ΔHvap is the heat change of vaporization.
• TB.P is the boiling point.

Substitute the values of TB.P and ΔHvap for methane in equation (1).

ΔSvap=ΔHvapTB

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