工程材料科学与设计原书第2版课后习题答案48章复习过程Word下载.docx
《工程材料科学与设计原书第2版课后习题答案48章复习过程Word下载.docx》由会员分享，可在线阅读，更多相关《工程材料科学与设计原书第2版课后习题答案48章复习过程Word下载.docx（125页珍藏版）》请在冰点文库上搜索。
Thus,
SolvingforQfvwegetQfv=52893.5J/mole.
UsingthisvalueofQfv,theCv（25oC）canbecalculated
Theproblemrequiresustocalculatethetemperatureatwhichthevacancyconcentrationis½
Cv（25oC）.
½
Cv（25oC）=2.675x10-10
Thus
forsolvingT,weget:
T=288.63Kor15.63oC.
3.COMPUTE:
Dividing
（1）by
（2）weget:
SolvingforQ,weget:
Q=22033.56J/mole
=exp（-7.511）
=5.46x10-4
TheproblemrequirescomputingatemperatureatwhichCv=3Cv（80oC）.
3Cv（80oC）=3x5.46x10-4
=1.63x10-3
solvingforT,weget:
T=413.05Kor140.05oC
4.
5.FIND:
AreAlandZncompletelysolubleinsolidsolution?
IfAl-ZnsystemobeysalltheHume-Rotheryrules.Thenitisexpectedtoshowcompletesolubility.
（i）TheatomicradiiofAlandZnare0.143nmand0.133nmrespectively.Thedifferenceintheirradiiis7.5%whichislessthan15%.
（ii）TheelectronegativitiesofAlandZnare1.61an1.65respectivelywhicharealsoverysimilar.
（iii）ThemostcommonvalenceofAlis+3and+2forZn.
（iv）AlhasanFCCstructurewhereZnhasaHCPstructure.
ItappearsthatAl-Znsystemobeys3outof4Hume-Rotheryrules.Inthiscasetheyarenotexpectedtobecompletelysoluble.
6.SHOW:
TheextentofsolidsolutionformationinthefollowingsystemsusingHume-RotheryRules.
（a）AlinNi
Size:
r（Ni）=0.125nm;
r（Al）=0.143nmdifference=14.4%
Electronegativity:
Al=1.61;
Ni=1.91
MostCommonValence:
Al3+;
Ni2+
CrystalStructure:
Al:
FCC;
Ni:
FCC
ThecrystalstructureofAlandNiarethesameandthemostcommonvalenciesarealsocomparable.However,thesizedifferenceiscloseto15%andthedifferenceiselectronegativitiesisrathersignificant.
Basedonthis,itappearsthatNiandAlwouldnotformasolidsolutionovertheentirecompositionalrange.
（b）TiinNi
r（Ti）=0.147nm,r（Ni）=0.125nmdifference=17.6%
Ti:
1.54;
1.91
Valence:
Ti4+;
HCP;
NiFCC
TiinNiwouldnotexhibitextensivesolidsolubility
（c）ZninFe
Sizer（Zn）=0.133nm;
r（Fe）-0.124nmdifference=7.25%
Zn=1.65;
Fe=1.83
Zn2+;
Fe2+
An:
HCP;
Fe:
BCC
Sinceelectronegativitiesandcrystalstructuresareverydifferent,Zn-Fewillnotexhibitextensivesolidsolubility.
（d）SiinAl
Sizer（Si）=0.117nm;
r（Al）=0.143nm;
difference=22.2%
（Si）=1.90;
Al=1.61
Si4+;
A;
3+
Si:
DiamondCubic;
FCC
Sincethesizedifferenceisgreaterthan15%,andthecrystalstructuresaredifferent,Si-Alwouldnotexhibitextensivesolidsolubility.
（e）LiinAl
Sizer（li）:
0.152,r（Al）:
0.143;
difference-6.29%
Li:
0.98;
1.61
Li1+;
Al3+
BCC;
Sinceelectronegativityandcrystalstructuresareverydifferent,Li-Alwillnotexhibitextensivesolidsolubility.
（f）CuinAu
Sizer（Cu）=0.125nm;
r（au）=0.144nm;
difference=12.5%
Cu=1.90;
Au=1.93
Cu+;
Au+
Cu:
FCC;
Au:
Cu-Auwillexhibitextensivesolidsolubility.
（g）MninFe
Sizer（Mn）=0.112,r（Fe）=0.124difference=10.71%
Mn1.55;
Fe1.83
Mn2+;
Mn:
FeBCC
ThedifferenceinelectronegativityishighbutMn-Fedoesobeytheother3Hume-Rotheryrules.Therefore,itwillformsolidsolutionsbutnotovertheentirecompositionalrange.
（h）CrinFe
Sizer（Cr）=0.125nm,Fe=0.144nmdifference=12.5%
Cr=1.66;
Cr3+;
Cr:
BCC
CrinFewillexhibitextensivesolidsolubilitybutnotovertheentirecompositionalrangesinceitobeysonly3of4Hume-Rotheryrules.
（i）NiinFe
Sizer（Ni）=0.125nm,r（Fe）=0.124nmdifference=0.8%
1.91;
Ni3+;
Fe3+
NiandFeobeys3ofthe4Hume-Rotheryrulestherefore,extensivesolidsolutionwillbeexhibitedbutnotovertheentirecompositionalrange.
7.（a）WhenoneattemptstoaddasmallamountofNitoCu,NiisthesoluteandCuisthesolvent.
（b）BasedontherelativesizesofNiandCu,radiusofNi=0.128nm,radiusofCu=0.125nm,thesetwoareexpectedtoformsubstitutionalsolidsolutions.
（c）NiandCuwillbecompletelysolubleineachotherbecausetheyobeyallfourHume-Rotheryrules.
8.FIND:
PredicthowCudissolvesinAl.
DATA:
CuAl
atomicradius（A）1.281.43
electronegativity1.901.61
valence1+,2+3+
crystalstructureFCCFCC
AllofHume-Rothery'
srulesmustbefollowedforasubstitutionalsolution.Inthiscase,thevalencesdonotmatch.CuwillnotgointosubstitutionalpositionsinAltoalargeextent.
COMMENTS:
ThisprincipleisoftenusedtoprecipitationhardenAlusingCu.
9.WhattypeofsolidsolutionisexpectedtoformwhenCisaddedtoFe?
Theradiusofcarbonatomis0.077nmandthatofanFeatomis0.124nm.Thesizedifferencebetweenthesetwois~61%whichismuchgraterthan~15%.Thus,thesetwoarenotexpectedtoformsubstitutionalsolidsolution.
IfwecomparethesizeratioofCtoFeatomswiththesizeoftetrahedralandoctahedralinterstitialsitesinBCCiron,wefindthatCdoesnoteasilyfitintoeithertypeofinterstitialposition.C,however,formsaninterstitialsolidsolutionwithFebutthesolubilityislimited.
10.FIND:
CalculatetheactivationforvacancyformationinFe.
GIVEN:
Thevacancyconcentrationat727C=1000Kis0.00022.
SOLUTION:
Weuseequation4.2-2tosolvethisproblem:
Cv=exp（-Qfv/RT）
SolvingforQfv:
Qfv=-RTlnCv=-（8.31J/mole-K）（1000K）ln0.00022=7.0x104J/mole
11.SHOW:
ASchottkyandFrenkeldefectinMgF2structures
A2-DrepresentationoftheMgF2structurecontainingaSchottkydefectandaFrenkeldefectisshownbelow.
12.Explainwhythefollowingstatementisincorrect:
Inionicsolidsthenumberofcationvacanciesisequaltothenumberofanionvacancies.
Inioniccrystals,eveninthepresenceofvacancies,thechargeneutralitymustbemaintained.Therefore,singlevacanciesdonotoccurinioniccrystalssinceremovalofasingleionwouldleadtochargeimbalance.Insteadthevacanciesoccurinamannersuchthattheanion:
cationvacancyratiorenderthesolidelectricallyneutral.This,however,doesnotmeanthattheanionvacanciesareequaltocationvacancies.Forexample,aSchottkydefectinMgCl2orMgF2involvestwoCl-orF-cationvacanciesforeveryMg2+anionvacancytomaintainelectricalneutrality.
ThenumberofcationvacanciesequalsthenumberofanionvacanciesonlyforthelimitingcasewherethechemicalformulaofthecompoundisMX.
13.Calculatethenumberofdefectscreatedwhen2molesofNiOareaddedto98molesofSiO2.Also,determinethetypeofdefectcreated.
Neglectinterstialvacancies
Wehave2molesofNiOand98molesofSiO2.SinceNiOisa1:
1compoundthereare2molesofNi2+ionsand2molesofO2-ionspresent.SiO2ontheotherhandisa1:
2compound;
therefore,thereare98molesofSi4+and196molesofO2-.Thetotalnumberofeachtypeofionis
NNi=2moles
NSi=98moles
NO2=196moles
Thetotalnumberofmolesofionsinthesystemis
NT=NNi+NSi+NO=2+98+196=196moles
EachsubstitutionofanNi2+forSi4+resultsinalossof2positivecharges.Ifnointerstitialsarecreated,thislossofpositivechargeisbalancedbythecreationofanionvacancies.ChargeneutralityrequiresoneoxygenvacancycreatedforeveryNi2+ion.Therefore,thenumberofoxygenvacanciesis
NOv=NNi=2moles
Thereare2molesofoxygenionvacanciescreatedwiththeadditionof2molesofNiOto98molesofSiO2.
14.Calculatethenumberofdefectscreatedwhen1moleofMgOisaddedto99molesofAl2O3.
MgOisa1:
1compound,thereforethereis1moleofMg2+ionsand1moleofO2-ionsinthesystem.
FromAl2O3,thereare198molesofAl3+ionsand297molesofO2-ionsinthesystem.
EachsubstitutionofanMg2+ionforAl3+ionresultsinalossofonepositivecharge.Thislossofpositivechargeisbalancedbyoxygenvacancy.ChargeneutralityrequiresoneoxygenvacancytobecreatedforeverytwoMg2+ion3.Thereforethenumberofoxygenionvacanciescreatedis
0.5molesofoxygenionvacanciesarecreatedbytheadditionof1moleofMgOto99molesofAl2O3.
15.COMPUTE:
Relativeconcentrationofcationvacancies,anionvacanciesandcationinterstitials.
QCv=20kJ/mole
QAv=40kJ/mole
QCI=30kJ/mole
ASSUMPTION:
assumeroomtemperature
T=298K
Concentrationofcationvacancies,CCvisgivenby
Similarlyforanionvacancies
andforcationinterstitials
16.（a）DescribeaSchottkydefectinU2
（b）WouldyouexpecttofindmorecationoranionFrenkeld
升级会员 