MATERIALS SCIENCE & ENGINEERING:AN INTRO
MATERIALS SCIENCE & ENGINEERING:AN INTRO
10th Edition
ISBN: 9798203933584
Author: Callister
Publisher: ZYBOOKS
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Chapter 4, Problem 1QAP
To determine

The number of vacancies per cubic meter in Gold.

Expert Solution & Answer
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Answer to Problem 1QAP

The number of vacancies: NV=14.1×1023vacancies/m3.

Explanation of Solution

Given Information:

The equilibrium fraction of lattice sites at 800C

  =2.5×105 .

The density of gold: ρFe=18.45g/cm3 .

The temperature: T=800°C .

From the table, find the atomic weight of Gold i.e. AAu=196.97g/mol .

First, calculate the total number of lattice sites in gold using the relation:

  NAu=NAρAuAAuWhere,NAAvogadro'snumber = 6.022×1023atoms/mol

Calculation:

  NAu=NAρAuAAuWhere,NAAvogadro'snumber = 6.022×1023atoms/molSubstitutingvalues;NAu=6.022×1023atoms/mol18.45g/cm3196.97g/molNAu=5.64×1022atoms/cm3or,NAu=5.64×1028atoms/m3

Therefore, the number of vacancies in gold will be determined as:

  NV=2.5×105NAuor,NV=2.5×1055.64×1028atoms/m3NV=14.1×1023vacancies/m3

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