工程材料科学与设计原书第2版课后习题答案48章.docx
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工程材料科学与设计原书第2版课后习题答案48章.docx
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工程材料科学与设计原书第2版课后习题答案48章
SolutionstoChapter4
1.FIND:
Whatmaterialhasapropertythatishugelyaffectedbyasmallimpuritylevel?
SOLUTION:
Electricalconductivityspansawiderange.Incorporationofafewpartspermillionimpuritiescanchangeelectricalconductivityordersofmagnitude.Smallcracksinbrittlematerialsdecreasetheirtensilestrengthbyordersofmagnitude.Smalladditionsofimpuritycanchangethecolorofgems.
COMMENTS:
Thesearebutafewexamples.
2.COMPUTE:
Thetemperatureatwhichthevacancyconcentrationisonehalfthatof25oC.
GIVEN:
EQUATION:
whereCv=vacancyconcentration
Qfv=activationenergyforvacancyinformation
R=gasconstant8.314J/mole-K
T=absolutetemperature
Inthepresentproblem
andT1=35+273=308K
T2=25+273=298K
alsoCv(35oC)=2Cv(25oC)
Thus,
SolvingforQfvwegetQfv=52893.5J/mole.
UsingthisvalueofQfv,theCv(25oC)canbecalculated
Theproblemrequiresustocalculatethetemperatureatwhichthevacancyconcentrationis½Cv(25oC).
½Cv(25oC)=2.675x10-10
Thus
forsolvingT,weget:
T=288.63Kor15.63oC.
3.COMPUTE:
GIVEN:
EQUATION:
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
Size:
r(Ti)=0.147nm,r(Ni)=0.125nmdifference=17.6%
Electronegativity:
Ti:
1.54;Ni:
1.91
Valence:
Ti4+;Ni2+
CrystalStructure:
Ti:
HCP;NiFCC
TiinNiwouldnotexhibitextensivesolidsolubility
(c)ZninFe
Sizer(Zn)=0.133nm;r(Fe)-0.124nmdifference=7.25%
Electronegativity:
Zn=1.65;Fe=1.83
MostCommonValence:
Zn2+;Fe2+
CrystalStructure:
An:
HCP;Fe:
BCC
Sinceelectronegativitiesandcrystalstructuresareverydifferent,Zn-Fewillnotexhibitextensivesolidsolubility.
(d)SiinAl
Sizer(Si)=0.117nm;r(Al)=0.143nm;difference=22.2%
Electronegativity:
(Si)=1.90;Al=1.61
Valence:
Si4+;A;3+
CrystalStructure:
Si:
DiamondCubic;Al:
FCC
Sincethesizedifferenceisgreaterthan15%,andthecrystalstructuresaredifferent,Si-Alwouldnotexhibitextensivesolidsolubility.
(e)LiinAl
Sizer(li):
0.152,r(Al):
0.143;difference-6.29%
Electronegativity:
Li:
0.98;Al:
1.61
MostCommonValence:
Li1+;Al3+
CrystalStructure:
Li:
BCC;Al:
FCC
Sinceelectronegativityandcrystalstructuresareverydifferent,Li-Alwillnotexhibitextensivesolidsolubility.
(f)CuinAu
Sizer(Cu)=0.125nm;r(au)=0.144nm;difference=12.5%
Electronegativity:
Cu=1.90;Au=1.93
MostCommonValence:
Cu+;Au+
CrystalStructure:
Cu:
FCC;Au:
FCC
Cu-Auwillexhibitextensivesolidsolubility.
(g)MninFe
Sizer(Mn)=0.112,r(Fe)=0.124difference=10.71%
Electronegativity:
Mn1.55;Fe1.83
MostCommonValence:
Mn2+;Fe2+
CrystalStructure:
Mn:
BCC;FeBCC
ThedifferenceinelectronegativityishighbutMn-Fedoesobeytheother3Hume-Rotheryrules.Therefore,itwillformsolidsolutionsbutnotovertheentirecompositionalrange.
(h)CrinFe
Sizer(Cr)=0.125nm,Fe=0.144nmdifference=12.5%
Electronegativity:
Cr=1.66;Fe=1.83
MostCommonValence:
Cr3+;Fe2+
CrystalStructure:
Cr:
BCC;Fe:
BCC
CrinFewillexhibitextensivesolidsolubilitybutnotovertheentirecompositionalrangesinceitobeysonly3of4Hume-Rotheryrules.
(i)NiinFe
Sizer(Ni)=0.125nm,r(Fe)=0.124nmdifference=0.8%
Electronegativity:
Ni:
1.91;Fe1.83
MostCommonValence:
Ni3+;Fe3+
CrystalStructure:
Ni:
FCC;Fe:
BCC
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
SOLUTION:
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.
GIVEN:
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.
GIVEN:
QCv=20kJ/mole
QAv=40kJ/mole
QCI=30kJ/mole
ASSUMPTION:
assumeroomtemperature
T=298K
Concentrationofcationvacancies,CCvisgivenby
Similarlyforanionvacancies
andforcationinterstitials
16.(a)DescribeaSchottkydefectinU2
(b)WouldyouexpecttofindmorecationoranionFrenkeldefec
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