工程材料科学与设计原书后习题答案4-8章

1.FIND:Whatmaterialhasapropertythatishugelyaffectedbyasmallimpuritylevel?SOLUTION:Electricalconductivityspansawiderange.Incorporationofafewpartspermillionimpuritiescanchangeelectricalconductivityordersofmagnitude.Smallcracksinbrittlematerialsdecreasetheirtensilestrengthbyordersofmagnitude.Smalladditionsofimpuritycanchangethecolorofgems. COMMENTS:Thesearebutafewexamples.2.COMPUTE:Thetemperatureatwhichthevacancyconcentrationisonehalfthatof25oC.GIVEN:EQUATION:where Cv=vacancyconcentrationQfv=activationenergyforvacancyinformationR=gasconstant8.314J/mole-KT=absolutetemperatureInthepresentproblemand T1=35+273=308KT2=25+273=298Kalso Cv(35oC)=2Cv(25oC)Thus,SolvingforQfvwegetQfv=52893.5J/mole.UsingthisvalueofQfv,theCv(25oC)canbecalculatedTheproblemrequiresustocalculatethetemperatureatwhichthevacancyconcentrationis½Cv(25oC). ½Cv(25oC)=2.675x10-10ThusforsolvingT,weget:oC.3.COMPUTE:GIVEN:EQUATION:Dividing(1)by(2)weget:SolvingforQ,weget: Q=22033.56J/mole=exp(-7.511)=5.46x10-4TheproblemrequirescomputingatemperatureatwhichCv=3Cv(80oC). 3Cv(80oC)=3x5.46x10-4=1.63x10-3solvingforT,weget:oC4.5.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)AlinNiSize:r(Ni)=0.125nm;r(Al)=0.143nmdifference=14.4%Electronegativity:MostCommonValence:Al3+;Ni2+CrystalStructure:Al:FCC;Ni:FCCThecrystalstructureofAlandNiarethesameandthemostcommonvalenciesarealsocomparable.However,thesizedifferenceiscloseto15%andthedifferenceiselectronegativitiesisrathersignificant.Basedonthis,itappearsthatNiandAlwouldnotformasolidsolutionovertheentirecompositionalrange.(b)TiinNiSize:r(Ti)=0.147nm,r(Ni)=0.125nmdifference=17.6%Electronegativity:Valence: Ti4+;Ni2+CrystalStructure: Ti:HCP;NiFCCTiinNiwouldnotexhibitextensivesolidsolubility(c) ZninFeSize r(Zn)=0.133nm;r(Fe)-0.124nmdifference=7.25%Electronegativity:MostCommonValence:Zn2+;Fe2+CrystalStructure: An:HCP;Fe:BCCSinceelectronegativitiesandcrystalstructuresareverydifferent,Zn-Fewillnotexhibitextensivesolidsolubility.(d) SiinAlSizer(Si)=0.117nm;r(Al)=0.143nm;difference=22.2%Electronegativity:Valence:Si4+;A;3+CrystalStructure:Si:DiamondCubic;Al:FCCSincethesizedifferenceisgreaterthan15%,andthecrystalstructuresaredifferent,Si-Alwouldnotexhibitextensivesolidsolubility.(e) LiinAlSizer(li):0.152,r(Al):0.143;difference-6.29%Electronegativity:MostCommonValence:Li1+;Al3+CrystalStructure:Li:BCC;Al:FCCSinceelectronegativityandcrystalstructuresareverydifferent,Li-Alwillnotexhibitextensivesolidsolubility.(f) CuinAuSizer(Cu)=0.125nm;r(au)=0.144nm;difference=12.5%Electronegativity:MostCommonValence:Cu+;Au+CrystalStructure:Cu:FCC;Au:FCCCu-Auwillexhibitextensivesolidsolubility.(g) MninFeSizer(Mn)=0.112,r(Fe)=0.124difference=10.71%Electronegativity:MostCommonValence:Mn2+;Fe2+CrystalStructure:Mn:BCC;FeBCCThedifferenceinelectronegativityishighbutMn-Fedoesobeytheother3Hume-Rotheryrules.Therefore,itwillformsolidsolutionsbutnotovertheentirecompositionalrange.(h) CrinFeSizer(Cr)=0.125nm,Fe=0.144nmdifference=12.5%Electronegativity:MostCommonValence:Cr3+;Fe2+CrystalStructure:Cr:BCC;Fe:BCCCrinFewillexhibitextensivesolidsolubilitybutnotovertheentirecompositionalrangesinceitobeysonly3of4Hume-Rotheryrules.(i) NiinFeSizer(Ni)=0.125nm,r(Fe)=0.124nmdifference=0.8%Electronegativity:MostCommonValence:Ni3+;Fe3+CrystalStructure:Ni:FCC;Fe:BCCNiandFeobeys3ofthe4Hume-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) electronegativity valence 1+,2+ 3+ crystalstructure FCC FCCSOLUTION: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:Thevacancyconcentrationat727SYMBOL176\f"Symbol"C=1000Kis0.00022.SOLUTION:Weuseequation4.2-2tosolvethisproblem:Cv=exp(-Qfv/RT)SolvingforQfv:Qfv=-RTlnCv=-(8.31J/mole-K)(1000K)ln0.00022=7.0x104J/mole11.SHOW:ASchottkyandFrenkeldefectinMgF2structuresA2-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:NeglectinterstialvacanciesWehave2molesofNiOand98molesofSiO2.SinceNiOisa1:1compoundthereare2molesofNi2+ionsand2molesofO2-ionspresent.SiO2ontheotherhandisa1:2compound;therefore,thereare98molesofSi4+and196molesofO2-.ThetotalnumberofeachtypeofionisNNi=2molesNSi=98molesNO2=196molesThetotalnumberofmolesofionsinthesystemisNT=NNi+NSi+NO=2+98+196=196molesEachsubstitutionofanNi2+forSi4+resultsinalossof2positivecharges.Ifnointerstitialsarecreated,thislossofpositivechargeisbalancedbythecreationofanionvacancies.ChargeneutralityrequiresoneoxygenvacancycreatedforeveryNi2+ion.Therefore,thenumberofoxygenvacanciesisNOv=NNi=2molesThereare2molesofoxygenionvacanciescreatedwiththeadditionof2molesofNiOto98molesofSiO2.14. Calculatethenumberofdefectscreatedwhen1moleofMgOisaddedto99molesofAl2O3.MgOisa1:1compound,thereforethereis1moleofMg2+ionsand1moleofO2-ionsinthesystem.FromAl2O3,thereare198molesofAl3+ionsand297molesofO2-ionsinthesystem.EachsubstitutionofanMg2+ionforAl3+ionresultsinalossofonepositivecharge.Thislossofpositivechargeisbalancedbyoxygenvacancy.ChargeneutralityrequiresoneoxygenvacancytobecreatedforeverytwoMg2+ion3.Thereforethenumberofoxygenionvacanciescreatedis0.5molesofoxygenionvacanciesarecreatedbytheadditionof1moleofMgOto99molesofAl2O3.15.COMPUTE:Relativeconcentrationofcationvacancies,anionvacanciesandcationinterstitials.GIVEN: QCv=20kJ/moleQAv=40kJ/moleQCI=30kJ/moleASSUMPTION:assumeroomtemperature T=298KConcentrationofcationvacancies,CCvisgivenbySimilarlyforanionvacanciesandforcationinterstitials16. (a)DescribeaSchottkydefectinU2(b)WouldyouexpecttofindmorecationoranionFrenkeldefectsinthiscompound?Why?UO2hasafluoritestructurewithU4+ionsoccupyingFCClatticesitesandO2-occupyingtetrahedralinterstitialsites.(a)ASchottkydefectinUO2willinvolveoneU4+cationvacancyand2O2-anionvacancies.(b)IngeneralcationFrenkeldefectsaremorecommonthananionFrenkeldefectsbecausecationsareusuallysmaller.Inthiscase,theradiiofU4+is0.106nmandthatofO2-is0.132nm.TheU4+cationissmallerthantheO2-anion.However,thesizedifferenceisnotveryhigh.Still,cationFrenkeldefectsareexpectedtobemore.17. IoniccompoundLi2O(a)DescribeaSchottkydefect(b)DescribeaFrenkeldefectLi2Ohasanantifluoritestructure.O2-ionsoccupyFCClatticesitesandLi+occupiestetrahedralinterstitialsites.(a)ASchottkydefectinLi2Oinvolves2Li2+cationvacanciesandoneO2-anionvacancy(b)TheionicradiiofLi+andO2-are0.078nmand0.132nmrespectively.ThismaterialismostlikelytoexhibitcationFrenkeldefectsincethesizeofthecationismuchsmallerthantheanion.18.DETERMINE:(a)InterstitialNa+ions(b)InterstitialO2-ions(c)VacantNa+sites(d)VacantO2-sitesinNa2OGIVEN:r(Na+r(O2-Na2Ostructureissimilartoantifluoritestructure.Na+ionsoccupytetrahedralinterstitialsitesandO2-ionsoccupyFCClatticesites.SincetheratioofNa:0is2:1forthismaterials,aSchottkydefectresultsin2cationvacanciesforeveryoneanionvacancy.no.ofvacantNa+sites=2xno.ofvacantO2-sitesAcationFrenkeldefectismorelikelytooccurinthismaterial(a)InterstitialNa+ions=1(b)InterstitialO2-ions=0(c)VacantNa+sites=2(d)VacantO2-sites=119.SOLVENT:AuSOLUTE:N,AgorCsDETERMINE:(a)whichelementismostlikelytoformaninterstitialsolidsolution.(b)whichelementismostlikelytoformasubstitutionalsolidsolution.(a)BasedonatomicradiiNismostlikelytoformareinterstitialsolidsolutionwithAuassolvent.(b)AgismostlikelytoformasubstitutionalsolidsolutionbecausethesizedifferencebetweenAu&NandAu&Csismorethan15%.Inaddition,AuandAghavesimilarvalence,andcrystalstructure.Theelectronegativitiesarenotquitesimilar,butsinceAg-Ausystemobeys3outof4oftheHume-Rotheryrules,AgisthemostlikelyelementwithwhichAuformsasubstitutionalsolidsolution.Section4.4Diffusion20. UnderwhatconditioncanFick’sfirstlawbeusedtosolvediffusionproblems.TheFick’sfirstlawcanbeusedtosolvediffusionproblemsprovidedtheconcentrationgradientdoesnotchangewithtime.21.GIVEN:1wt%BisaddedtoFe.FIND:(a)ifBwouldbepresentasaninterstitialimpurityorsubstitutionalimpurity,(b)fractionofsitesoccupiedbyBatoms,(c)ifFecontainingBweretobegascarburized,wouldtheprocessbefasterorslowerthanforFewhichhasnoB?Explain.(a)BasedontheatomicradiiBwouldbepresentasaninterstitialimpurity(b)amountofBpresent=1wt%Asabasisofcalculationassume100gmsofmaterial.Determinetheno.ofmolesofFeandBpresent.Totalno.ofmolesofFeandB=1.773+0.092=1.865moles.FractionofsitesoccupiedbyBatoms=Thus,Broughlyoccupies5%ofthesites.(c)IfFecontainingBweretobegascarburizedtheprocesswouldbeslowerthanforFewhichhasnoBsimplybecausethepresenceofBatomsalreadyininterstitialsitesleavefewersitesforinterstitialCtodiffusethrough.22. Determinewhichtypeofdiffusionwouldbeeasier(a)CinHCPTi(b)NinBCCTi(c)TiinBCCTir(C)<<r(Ti)sowecanpredictthatdiffusionoccursviaaninterstitialmechanismr(N)<<r(Ti).Inthiscasethediffusionalsooccursviainterstitialmechanism.TiinBCCTiisacaseofself-diffusionandself-diffusionoccursviaavacancymechanism.Ingeneraltheactivationenergyforselfdiffusionishigherthaninterstitialmechanismbecausevacancymechanisminvolvestwosteps.Oneistocreateavacancyandsecondistopromoteavacancy/atomexchange.ThusTiinBCCTiwillbetheslowest.Theactivationenergyfordiffusionviainterstitialmechanismisjusttheenergynecessarytomoveanatomintoaneighboringinterstitialsite.Anopencrystalstructure,asopposedtoadensestructure,shouldhavealoweractivationenergy.BetweenBCCTiandHCPTi,BCCTihasamoreopenstructure(lowerAPF)thanHCPTi.Thus,NinBCCTidiffusionwouldbetheeasiestbyvirtueofitslowestactivationenergy.23.GIVEN:C1=0.19at%atsurfaceC2=0.18at%atbelowthesurfaceD=4x10-14mao=4.049AoCOMPUTE:Fluxofcopperatomsfromsurfacetointerior.Wemustfirstcalculatetheconcentrationgradientintermsof[copperatoms/cm3/cm].Itcanbecalculatedasfollows:Theconcentrationgradientisthen24.FIND:Predictwhetherdiffusionisfasterinvitreousorcrystallinesilica.GIVEN:Diffusionisthemovementofatomsthroughthematerialonestepatatime.Theeaseofmovementisinpartdeterminedbytheamountofspacethatsurroundseachatom.Inmoreopenorlessdensestructures,atomshaveanincreasedchanceofbeingabletosqueezepastaneighborintoanewposition.SOLUTION:DiffusioncanbethoughtofasanArrheniusprocess.Theactivationenergyisthatrequiredtomoveanatomfromonepositiontoanother,asshowninFig.2.3-2.Inacrystaltheactivationenergywillbegreaterthaninaglass,sincethedensityishigherandthereislessfree,orunoccupied,volume.Thus,weexpectdiffusiontobeslowerincrystalthaninglassesatthesametemperature.COMMENTS:Whenanoncrystallinematerialisraisedtoatemperatureabovetheglasstransitiontemperature,diffusionincreasesenormously.Inmetalsthisbringsaboutrapidcrystallization.Insomeceramicandpolymersystems,crystallizationmaybesloworabsent.25.FIND:Dotextiledyesmorereadilypenetratecrystallineornoncrystallineregions?GIVEN:Mosttextilefibersaresemicrystalline,containingbothcrystallineandnoncrystallineregions.Thedensityofthenoncrystallineregionsislessthanthatofthecrystallineregions.Oftendyeingisconductedatatemperatureatwhichthenoncrystallineregionsareabovetheirglasstransitiontemperature.SOLUTION:Dyepenetrationthroughtheglasswillbegreaterthanthatthroughthecrystal;however,therateofdyeingisnotsufficientlyhightobecommerciallyfeasible.Thetemperaturemustberaisedsothatthenoncrystallinepolymerisintherubberstate.Diffusionbecomesrapid(radiallyinward)intothesmallfibers.COMMENTS:Oneofthekeylessonsthatdyehouseslearnisthatasufficientamountofnoncrystallinepoorlyorientedpolymermustbepresentinthefiber.Thetemperatureofthedyebathneedstobeabovetheglasstransitiontemperature.Sometimeswaterandcarriersareusedtoswellthenoncrystallineregionstogetyetagreaterdiffusionrate.Thedyesmayattachtothepolymerusingionicbondsorcovalentbonds.Unattacheddyemaywashoutlater.26.CALCULATE:ThefactorbywhichthediffusioncoefficientofAlinAl2O3changewhentemperatureisincreasedfrom1800oCto2000oCGIVEN: T1=1800oC=2073KT2=2000oC=2273KEQUATION:dividing(1)by(2),wegetfromtable4.4-1ofthetextQ=477kJ/moleandR=8.31J/mole-KThus,thediffusioncoefficientofAlinAl2O3changesbyafactorof11.43whenthetemperatureisincreasedfrom1800oCto2000oC.27.FIND:TemperatureatwhichaspecimenofFemustbecarburizedfortwohourstoachievethesamediffusionresultasat900oCfor15hrs.GIVEN: T1=900oC=1173K;Q=84000J/molet1=15hrs;Do=2.00x10-6mt2=2hrs;R=8.31J/mole-KThevalueoffluxJisinunitsofcm2persec.Fluxpercm2Jf=Jxtime(1)Weneedthesameresultin2hours.dividing(1)by(2).28.GIVEN: D=4x10-4m2/s@20C1=2.2x10-3kmol/m3wallthickness=3mm,diameter=50cmheight=10cmCOMPUTE:Initialrateofmasslossthroughcylinder.InitiallytheconcentrationofHeoutsidethecylinder,C2,iszero.First,weneedtoconverttheconcentrationofHefromkmol/m3into(atoms/cm3)/cm.C1=2.2x10-3kmol/m33=2.2x10-6mol/cm3=2.2x10-6mole/cm3Intermsof(atoms/cm3)/cmTheconcentrationgradientisThefluxofatomspersecondpercm2isobtainedbyusingFick’sfirstlawofdiffusionTherateofmasslossis1.766x1019atoms/cm2sec.Thetotalsurfaceareaofthecylinderis2r(r+h)wherer=radiusandh=height.Totalsurfacearea=2x25(25+10)=5497.79cm2TherateofmasslosspersecondNote:(i)Thesteadystatemasslossiscalculatedbecausetheinitialrateofmassloss(i.e.,rateofmasslossattimet=0)is0.(ii)ItisassumedthatthecurvatureofthecylinderislargeenoughtocalculateJusingtheexpressionforplategeometry.29. Diffusionacrossapolymermembranedependsnotonlyonsizeofthediffusingspeciesbutalsothepolarityofthediffusingspecies.Apolarmembranemaypassnonpolarspeciesbutserveasabarriertopolarspecies.Saranwrapcontainshighlypolaratomsmakingitapolarmembranewhichservesasabarriertowaterwhichisapolarcompound.thus,thereisnodiffusionofwaterthroughthepackageunlikepolyethylene,whichisanonpolarmembraneandallowsdiffusionofwatermoleculeswhichformice.30.COMPUTE:Temperaturerequiredtoyieldacarboncontentof0.5%atadepthof0.4mnbelowthesurfaceoftherodin48hours.GIVEN:Basematerial:HCPTiEQUATION: Inthisproblem c(x,t)=0.5wt%co=0.2wt%cs=1.0wt%Fromfigure4.4-11,whenFromMetalsHandbook,DeskEdition,Pg.28.66forCdiffusioninTi,Do=3.02x10-3cm231. Thediffusionprocessthroughvacancy-interchangemechanismdependsoncreationofvacanciesandvacancy/atominterchange.Atcomparablehomologoustemperatures,forGeandCuthediffusioncoefficientforthatmaterialwhichhasahighervacancyconcentrationwouldbehigher.Acovalentbondasopposedtoametallicbondisstrongeranddirectional.Itisalsodifficulttocreatevacanciesinacovalentlybondedmaterialduetoitsstrongbonding.Therefore,theactivationenergyforvacancycreationinacovalentlybondedmaterialsuchasGeislargerthanCuwhichhasaweakmetallicbond.Thedirectionalnatureofacovalentbondplacesgeometricalrestrictionsonthevacancyatominterchangewhichagainresultsinanincreaseintheactivationenergy.Therefore,atcomparabletemperaturesthediffusioncoefficientforGewillbelarger.32. FIND:-5a,b,andc.SOLUTION:Theorderinpartaishigh.Thematerialsisperfect.Thereisonlyonewaytoarrangetheatomsinsuchasystem.Theentropyislow.Inpartbthereislessorder,moredisorder,andtheentropyhasincreased.Partcisnearlyrandom.Ithasloworderandhighentropy.Energycontainsacontributionfromentropy:E=H-TS,whereEisenergy,Tisabsolutetemperature,andSisentropy.Assumingallothercontributionstoenergychangenegligibly(TandH),theenergyofpartcisthelow,partaishighandpartbisintermediate. COMMENTS:Whatisshowningoingfromatocistheentropyofmixing.33.GIVEN:After10hrsat550oCanoxidelayerofthickness8misformed.COMPUTE:Thicknessafter100hrs.Usingthedefinitionofeffectivepenetrationdistanceandequation4.4-11oftext,with=2wehave.Inthiscase34.GIVEN: Dw=1.0x10-12mDdc=1.0x10(dyecarrier)Dd=1.0x10-14mCOMPUTE:(a)Timesrequiredforthewater,dyeandcarriertopenetratetothecenterofthefiber.(b)Sameas(a)butfiberdiameterdoubles(c)IfthermaldiffusivityofPETishowlongwillittakefortheheattopenetratetothecenterofa50mdiameterfiber.(a)usingequation4.4-11oftextwith=2.forwater,fordyecarriersimilarlyfordye t=6.25secs.(b)Ifthediameterfiberisdoubledxeff=50x10-6forwater,similarlyfordyecarrier t=6250secsandfordye t=6.25x104secs(c)SubstitutingDwithDth,wecanusethesameequationtocalculatethetimerequiredforheattopenetratethecenteroffiberdiameter=50m.Note:Theunitsofthermaldiffusivityism2/secandnotasprintedintext35.FIND:Howlongwillittaketocasecarburizeasteelchaintoadepthof1/16inch?GIVEN:Itrequires4hourstocarburizeaplateofsimilarcompositiontoadepthof1/16inch.ASSUMPTIONS:Allcarburizationconditionsarethesameinbothtreatments.SOLUTION:Equation4.4-11isusedtosolvetheproblem:xeff=SYMBOL103\f"Symbol"(Dt)1/2.Inthissituationwesetuparatiofortheplate(1)andthechain(2):,whereSYMBOL103\f"Symbol"is1fortheplateand2forthechain.SubscriptswereomittedfortheD's,sincediffusionisthesameinthetwocases.Similarly,x1=x2=1/16inch.Reducingtheequationgives:36.GIVEN: Dm=1x10-11mdw=25mdc=2musingequation4.4-11oftextweneedtofindtrequiredtoreachequilibriumWoolCotton(b)foratightlypackedcubicbaleoffiberwithsidelengthof1meter.Acubehassixsidesanddiffusionisexpectedtooccurthroughallsixsides.ThetimerequiredtoreachequilibriumHerexeff=Note:Thissolutionoverestimatestime.Amoreprecisesolutioncanbeobtainedbysettinguptheproblemin3DandsolvingtheFick’ssecondlawforacubegeometry.37.EXPLAIN:Whyfizzreductionishigherinplasticbottlescomparedtoglassbottlesandmetalcans.Plasticbottlesaresoftandcanbesqueezed.Uponsqueezingtheairinsidethebottleisdrivenoutandtomaintaintheequilibriumpressureoverthesoda,CO2comesoutoftheliquid.Suchisnotthecaseforbottlesandmetalcans.Itisforthisreason,thefizzreductionishighinplasticbottles.Comparethediffusioncoefficientofmethaneinrubberat239KtothediffusionofCuinAgatthesametemperature.Thediffusioncoefficientofmethaneinrubberat293Kis1.515x10-6m2/secwhereasthatofCuinAgis1.20x10-4m39.FIND:MakeaschematicplotofDvs.Tinnaturalrubber.GIVEN:NaturalrubberhasaTgofabout0SYMBOL176\f"Symbol"C.SKETCH:SOLUTION:Thesketchaboveisdiagrammatic.WeknowthatbeforeandaftertheglasstransitiontemperaturethediffusioncoefficientbehavesasanArrheniusfunction:D=Doexp(-Q/RT),sothatlnDislinearlyproportionalto1/T.QislargebelowTg.40.EXPLAIN:WillPTFEworkasamembranetoseparatewatervaporfrombenzene?PTFEisTeflon.BasedonthestructureofPTFE(Teflon),itappearstobeanon-polarmembrane.Non-polarmembranesallowpassageofpolarmoleculessuchaswater.Furthermore,thesizedifferencebetweenwatervapormoleculeandbenzenealsoplaysanimportantrole.AwatervapormoleculeismuchsmallerthanabenzenemoleculeandcaneasilypassthroughaTeflonmembranewhereasbenzenebeingalargemoleculewillnotpass.Thus,PTFEwillworkasaneffectivemembranetoseparatewatervaporandbenzene.ThesolutiontoFick’ssecondlawforathickplateiswhereC(x,t)carbonconcentrationatadistancexbelowthesurfaceattimetCscarbonconcentrationatthesurfaceCoinitialbulkcarbonconcentrationinstellttimeDDiffusivity(a)ThediffusivityDofcarboninsteeldependsontemperature.Soastemperatureincreases,Dincreasesanddecreases.Assumingallothervariablesremainconstant,anincreaseinD,increasesthecarbonconcentration(C(x,t))belowthesurface.(b)Whentemperatureisconstant,anincreaseincarbonconcentrationC(x,t)belowthesurfacecanbeincreasedbyincreasingthecarboncontentinthefurnace(i.e.,byincreasingCs).(c)Timehasineffectonthecarbonconcentrationwhichissimilartothatoftemperature.Asthetimethesteelpartspendsinthefurnaceincreases,thecarbonconcentration,C(x,t)increases.FIND:(a)timerequiredforCuatomtodiffuseadistanceof1nmthroughacrystallattice.(b)distanceaCuatomcandiffusethroughaCucrystallatticein1hourat1000oC.(c)commentondiffusioninsolidsatroomtemperatureversushightemperature.EQUATION:substitutingforDinsolvingfort,yields t=1.769x1021secs(b)Since1nmistheapproximatesizeofasingleunitcell,theCuatomdiffusesthrough24260unitcellsinonehourat1000oC.(c)ComparingthediffusionratesofCuatroomtemperatureandthatathightemperaturesuggeststhatweneednotbeconcernedaboutdiffusioninsolidsatroomtemperaturebutdiffusioninsolidsathightemperatureisextremelyimportant.FIND:thetimerequiredtoobtainacarboncontentof0.832%atabelowthesurfaceifthefurnaceatmosphereisfreeofcarbonandmaintainedat1200oC.GIVEN: T=1200oC=1473KCs=0CoC(x,t)=0.832%x=t=?EQUATION:WefirstneedtocalculateD.Do=2.00x10-6mQ=84000J/moleT=1273KR=8.31J/mole-K=2.093x10-9mNow,fromgraphoferffunctionintext(fig4.4-11)44.FIND:(a)TimerequiredforanAlatomtodiffuseadistanceof9xthroughAl2O3crystal.(b)TemperatureatwhichthediffusionrateofAlthroughAl2O3willbe10timesthatat1000oC.GIVEN: t13hrsT1=1000oC=1273KDo=2.8x10-3mQ=477000J/moleR=8.31J/mole-KEQUATION:WeusethefirstequationtocalculateDat1000oCfort=3hrs=3x3600secs,theAlatomdiffusesadistanceDividing(1)by(2),ButD2=D(T)=10D(1000oC)=10DT1SubstitutingforQ,R,&T,weget45.FIND:ThetemperaturerequiredforaZnatomtodiffuse