   Chapter 18, Problem 12PS

Chapter
Section
Textbook Problem

Calculate the entropy change that occurs when 1.00 mol of steam is converted to liquid water at 100 °C in a reversible process. (qvap = 40. 7 kJ/mol)

Interpretation Introduction

Interpretation:

The entropy change that occurs when 1.00 mol of steam is converted to liquid water at 100 °C in a reversible process should be calculated.

Concept introduction:

The second law of thermodynamics relate the entropy of the system to the heat absorbed by the system. The change in entropy is related to the heat absorbed by the formula,

ΔS=qrevT

Here, ΔS is the change in entropy, qrev is the heat absorbed and T is the temperature. The formula is applicable if the work done is reversible.

Explanation

Given:

1molof steamT =100oCqvap= 40.7 kJ/mol

The given reaction is depicted as follows since steam gets converted into water at given temperature.

H2O(g)H2O(l)

The entropy change that occurs when 1.00 mol of steam is converted to liquid water at 100 °C in a reversible process is calculated as follows,

The molar mass is 18.015 g/mol. The given vaporization value is positive which then the conversion of steam into liquid should have negative of given vaporization value.

Thus,

qrev(- 40

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