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11Chapter2StructureandpropertyofmaterialsChapter4

ChemicalThermodynamicsofMaterials材料化學(xué)熱力學(xué)24.2

EllinghamDiagramsanditsapplication(埃靈罕姆圖的應(yīng)用)4.2.1EllinghamDiagramsΔG0-TlineLinearrelationshipB=-ΔS03stableunstable4Slope(斜率)ofG0-Tline:※Ifthenumberofgasdecreaseduringtheoxidationprocess,S0<0,(-S0)>0,slopispositive.(金屬氧化物形成過程中,反應(yīng)中有氣體,而生成物為固體,反應(yīng)結(jié)果是熵減少,斜率為正。即在高溫下)Metal(s)+O2(g)-----Metaloxide(s)※Ifthenumberofgasincreaseduringtheoxidationprocess,S0>0,(-S0)<0,slopisnegative※Ifthenumberofgasunchangeduringtheoxidationprocess,S0=0,(-S0)=0,slopiszero,thatistosay,

G0isunrelatedtotemperature.5

wecanstudythethermodynamicsandoxidativepropertiesofvariousmaterialsinalargetemperaturerangeaccordingtoEllinghamdiagrams,andthenprovideprovidenceanddataforthedevelopmentofnewmaterials.4.2ApplicationofEllinghamDiagrams(埃靈罕姆圖的應(yīng)用)61)EquilibriumandControlofoxideformation

(氧化物生成平衡及控制)

Underacertaintemperature,wecancontrolthedirectionoftheinteraction(相互作用)bymodifytheO2pressure.

72)Comparisonofstabilityofoxide

(氧化物穩(wěn)定性比較)MetaloxidewithG0-Tlineinthebeloworthe-G0hasbigger

negativevalue,theoxideismorestable.曲線越在下方,G0

負(fù)值越大,越穩(wěn)定Atagiventemperature,correspondingelementinthebelowlinemaymaketheelementreductionoftheaboveline.

在給定的溫度,在相應(yīng)元素線下面的可使上面的元素還原。MetaloxidelieabovetheH2OgeneratinglinewillbereducedbyH.

金屬氧化物在水生成線上的可被H還原ThedistancebetweentwolinesofoxidativeandreductivereactionsstandfortheG0。8i.e.ComparisonofTiO2andMnOTiO2generatinglineliesbelowtheMnOgeneratingline,sotheTiO2ismorestable.

1000℃,thedistancebetweentwolines:

G0<0,MnOwillbereducedbyTiatthestandardcondition.93)Reverseofreducingcapacity

還原能力的相互反轉(zhuǎn)Whenthetwogeneratinglinesconvergeatcertaintemperature(當(dāng)兩根氧化物生成線在特定溫度相交時(shí)),therelativereducingcapacityoftwoelementsmayreverse.SlopofCOgeneratinglineisnegative,soCOismorestablewiththeincreaseoftemperature.

Alltheoxidecouldbereducedaslongasthetemperatureishighenough.Bywhat?104.3PhasebalanceandphasediagramKeyterms:

Phasebalance,phasediagram,component,unitarysystem,binarysystem,ternarysystem,isomorphous,leverrule,eutecticphasediagram,peritecticphasediagram,monotectic11

f——isthenumberofdegreesoffreedom,whichmeansthenumberofpropertiessuchastemperatureorpressure,whichareindependentofothervariables.(自由度數(shù))

c

——

isthenumberofcomponent(獨(dú)立組元數(shù))

p——

isthenumberofphasesinthermodynamicequilibriumwitheachother(平衡相的數(shù)目)

2

——

temperatureandpressure;ifitisasolidsystem,itshouldbe1.GibbsPhaseRule

wasproposedbyJosiahWillardGibbsinthe1870sastheequality

f=c-p+2

Binaryphasediagrams

二元相圖c=2Condensedstatussystem:f=c-p+1=3-pMaximumnumberofdegreesoffreedom:f=3-1=2Componentandtemperature2Dplane

?132curves(liquidlineandsolidusline)2single-phaseregions1two-phaseregion1)Binaryisomorphousdiagramandleverrule

二元勻晶相圖與杠桿規(guī)則P103ABTATBLαL+αwB/%※Twocomponentshaveanalogouschemicalpropertiesandsamecrystalstructure.※Theydissolveeachotheratbothliquidandsolidusstate.※Theyforminfinite(successive)solidsolution.14LeverruleTACuNiC0CLCαTBT1LαL+αacb0100.wNi/%15Deduceofleverrule16PhaseanalysisBinaryisomorphousdiagram液相線固相線液相區(qū)L固相區(qū)α兩相共存區(qū)ABTATBL+αwB,%2curves(liquidlineandsolidusline)c=2,p=2,f=12single-phaseregionsc=2,p=1,f=21two-phaseregionc=2,p=2,f=1f=c-p+117m1818Atextremumpoint,itisnotaccordwithphaserule.Cshouldbeconsideredasaspecificcomponent.thephasediagramshouldbeconsideredasacombinationoftwobinaryisomorphousdiagram(ACandCB).Binaryisomorphousdiagramwithextremum極值MaximumpointMinimumpoint192)Eutecticphasediagram二元共晶相P105

20SoliduslinewB21Twosolidusphasewereprecipitatedsimultaneouslyfromaliquidphase.Accordingtophaserule,whenthreephaseinaequilibriumstate,f=c-p+1=2-3+1=0,sointhiscase,thecomponentandtemperaturearecertain,itappearaplateau(平臺).CEDthree-phaselineCharacteristicsofeutecticreaction(共晶反應(yīng))22eutecticpoint共晶點(diǎn),thelocationisdefinedbyeutecticcomponentandtemperature(E).eutectictemperature共晶溫度eutecticcomposition共晶成分Eutecticreactionline2324Pb-SnAlloy1ABEFP113/8.吉布斯相律通常為f=c-p+2,為什么在固體材料的研究中,相律一般可表達(dá)成f=c-p+1?Gibbs

phase

rule

is

usually

expressed

as

f

=

c-p+2,

why

in

the

study

of

solid

materials,

the

phase

rule

be

expressed

as

f

=

c-p+1?

答:在固體材料的研究中,壓力對固相反應(yīng)的影響很小,通??梢院雎?,所以非成分的變量只有溫度這一項(xiàng),所以相律一般可表達(dá)成f=c-p+1。In

the

study

of

solid

materials,

pressure

has

little

effect

on

the

solid

phase

reaction,

usually

it

can

be

ignored,

so

temperature

is

the

only

non-composition

variable,

so

the

phase

rule

generally

be

expressed

as

f

=

c-p+1.重點(diǎn)題目P113/9.一合金之成分為90Pb-10Sn。(a)請問此合金在100℃、200℃、300℃時(shí)含有哪幾種相?(b)在哪個(gè)溫度范圍內(nèi)將只有一相存在?Compositionofanalloyis90Pb-10Sn.(a)Whichphasesiscontainedat100℃,200℃,300℃?(b)Inwhichtemperaturerangewillbeonlyonephaseexists?(a)Compositionofalloyis90Pb-10Snalloywhichcorrespondingtotheredverticalline,itintersectwiththetemperaturelevellineata,bandcpoints.So,therearetwophasesat100℃,αandβ;thereisonlyonephaseat200℃,α;pointcliesontheliquidusline,sotherearetwophasesat300℃,Landα.(b)Drawaperpendicularlineacrosstheingredientlineof90Pb-10Snalloy,wecangetthreepoints,d,eandf,thendrawhorizontallinestogetthecorrespondingtemperature.Sowhenthetemperatureishigherthan300℃,liquidphaseLexists.Whenthetemperatureisbetween148~268℃,thereisonlyαphase.答:(a)成分為90Pb-10Sn的合金對應(yīng)的為紅色的垂直線,分別在100℃、200℃、300℃作水平線得到交點(diǎn)a,b,c,所以

100℃時(shí),交點(diǎn)a位于α+β相區(qū),所以有α和β兩相。

200℃時(shí),交點(diǎn)b位于α單相區(qū),所有只有α一相。

300℃時(shí),交點(diǎn)c位于液相線上,此為兩相共存線,所以有液相L和α兩相。(b)過成分為90Pb-10Sn的線作垂直線,可得到交點(diǎn)d和e,然后作水平線得到相應(yīng)的溫度。所以在溫度大于300℃時(shí),只有液相L存在。在溫度為148~268℃時(shí),只有α相存在。

P113/10.固溶體合金的相圖如下圖所示,試根據(jù)相圖確定:Phasediagramofsolidsolutionalloyisshowninthefigure,pleasedeterminethefollowingquestionsaccordingtoit.成分為40%B的合金首先凝固出來的固體成分是什么?若首先凝固出來的固體成分含60%B,合金的成分是什么?成分為70%B的合金最后凝固的液體成分是什么?合金成為為50%B,凝固到某溫度時(shí)液相含有40%B,固體含有80%B,此時(shí)液體和固體各占多少分?jǐn)?shù)?a)Whatisthecompositionofthesolidfirstfrozenfromalloycontain40%B?b)Ifthesolidfirstfrozencontaining60%B,pleasedeterminecompositionofthealloy.c)Whatisthecompositionoftheliquidfinalsolidifiedfromalloycontaining70%B?d)Analloycontaining50%B,whenitarriveatsometemperature,theliquidphasecontaining40%B,andthesolidphasecontaining80%B,pleasedeterminethefractionoftheliquidandsolid.Thecompositionofthesolidfirstfrozenfromalloycontain40%Bcorrespondtothepointa’,whichcontaining80%Band20%A.b)Ifthesolidfirstfrozencontaining60%B,wecangetpoin

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