Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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Chapter 4, Problem 4.1P
Interpretation Introduction

Interpretation:

The fraction of atomic sites which are vacant in gold at given conditions is to be determined.

Concept introduction:

The number of vacancies per cm3 is calculated by the formula;

  nv=n×exp(QvRT) ...... (1)

Here, n is the number of atoms per cm3, Qv is the energy required to deliver one mole of vacancy in cal/mol, T is the temperature in Kelvin, and R is the universalgas constant.

Expert Solution & Answer
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Answer to Problem 4.1P

The fraction of atomic sites which are vacant in gold at given conditions is 0.00091.

Explanation of Solution

Given information:

At 300 K, the number of vacancies in gold at equilibrium is 5.82×108 vacancies/cm3. Fraction of vacant atomic sites are to be calculated at 600 K.

The lattice parameter for gold with Face-centered cubic crystal structure is 4.065 A˙.

For 1 cell, the number of gold atoms per cm3 is calculated as shown below if there are 4 gold atoms per cell is:

  n=4 atoms/cell ( 4.065× 10 8  cm )3=5.95×1022 atoms/cm3

The temperature is given as 300 K .Universal gas constant to be used is 1.987 calmolK.

Using equation (1) to calculate the activation energy for gold as:

  nv=n×exp( Q v RT)5.82×108 vacancies cm3=(5.95× 10 22  atoms cm 3 )exp( Q v ( 1.987  cal molK )( 300 K ))0.978×1014=exp( Q v ( 1.987 )( 300 ))Qv=8351.159 cal/mol

Using this calculated value of activation energy, calculate the fraction of vacant atomic sites (nv/n) in gold at 600 K as:

  nvv=exp( Q v RT)=exp( Q v RT)=exp( 8351.159 cal/mol ( 1.987  cal molK )( 600 K ))=0.00091

Conclusion

The fraction of atomic sites which are vacant in gold at 600 K is calculated to be 0.00091.

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Essentials of Materials Science and Engineering, SI Edition

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