[PDF] Lecture 9: Phase Transitions Here are some example phase





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The triple point occurs at 175 5 K Sketch (not to scale) the single-component phase diagram for methanol Strategy Consider Figure 17 4 which shows a typical phase diagram for a one-component system Remember that the normal melting and boiling points correspond to a pressure of 1 atm = 1 013 bar Solution

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MatthewSchwartz

StatisticalMechanics,Spring2019

Lecture9:PhaseTransitions

1Introduction

Figure1.Twoofthephasesofsolidcarbon

•A⎷haseisauniformstateofmatter.

wantittobe.Wewant"gas"tobethephase.

Amoretechnicallyprecisedefinitionis

ofP;V;Tetc). importantareaofphysics. 1 ofthephasetransformationarerelated.

2Solids,liquidsandgases

T=¡1

V@V @P T (1) gaslawPV=NkBT,weseethatforagasT=∞

P=/0.Compressibilityisanexampleofanorder

=G

Aiscalled

thesurfacetension:2Section2

Atohavinga

surface,with thesurfacetension. (e.g. water=73mN mcomparedtosay,

CO2=17mN

m),butnotthelargest(mercuryhas

Hg=486mN

m). ofmetal. temperatureandpressure. k BT3 fora

2mkBp;foramoregeneralideal

gas=kBTlnP k BT equilibriumthatwearediscussinghere). energyG.Recallthat dG=VdP¡SdT+1dN1+2dN2(2) gas,ittakestheform=kBTlnN V

Melting:transitionfromsolidtoliquid.

Condensation:transitionfromgastoliquid.

Sublimation:transitionfromsolidtogas.

Nfixed.Then,=G

Nandso@

@T P =1 N@G @T P =-S N(4) infinitesimallybelowit∂G ∂T P ∂T P =-Sgas.Inapure solid=liquid=gas.

3Phaseboundaries

dominantes,andNisfixed(dN=0).SinceG

N=,thendG=NdandfromEq.(2),dG=

VdP-SdT,wefind

d=V NdP-S

NdT(5)

whichgives V 1

N1dP-S1

N1dT=V2

N2dP-S2

N2dT(6)

Thatis

dP dT=S N V

N(7)Phaseboundaries5

3.1Latentheat

N.RecallthatG=H-TSso

μ=G

N=H N-TS

N=·H

Natthephaseboundary.Sowecan

canalsowritetheClapeyronequationas dP dT=∞ TL

·¡1

n(8) wheren=N

Visthenumberdensityand

L=·H

N (9) iscalledthelatentheat.

T.Since·S=H

Twe putinpermoleculetochangethephase. -286kJ molandforwatervaporis·fH=-242kJ ofvaporizationofwaterat1atm:Lvap=44kJ

Notethat44kJ

molK, molofenergy.That L fuse=6.0kJ thatLfuse>0sinceittakesheattomeltice. L densethantheirliquidforms.So·¡1 n&0.ThereforebyEq.(8),dP dTisgenerallyverylargeand 2

Figure5.Phasediagramforwater.

dT<0.SinceL>0andT>0this mustmean·-1 n n=∞ n gas¡∞ n liquid∞ n gas=Vgas N gas=RT P(10) dP dT=PL

RT2(11)

PdP=L

RT2dT(12)

wecanintegratebothsidestogive

P=Cexp

¡L RT (13) wethenhave

P=Pexp

¡L

R∞

T¡∞

T (14)Phaseboundaries7 molandR=8.3J mol, weget T=1

T¡R

LlnP

P¡1

=370.1K(15) whichisthreedegreeslower.

3.2Vaporpressure

T =373K,P=1atmandL=42kJ w(P0;T0)=gas(P0;T0)(16) wmix(P;T)=w(P;T)¡kBTNs

Nw(17)

P

0bywritingP=P0+Pweget

wmixed(P;T0)=w(P0+P;T0)¡kBT0Ns

Nw=w(P0;T0)+P@w

@P T

¡kBT0Ns

Nw(18)8Section3

Similarly,

gas(P;T0)=gas(P0;T0)+P@gas @P T (19) Now,@ @P T =V PVw

Nw¡Vgas

Ngas =kBT0Ns

Nw(20)

Vismuchlarger),

sowecandropVw N wcomparedtoVgas N gas.Usingtheidealgaslaw,Vgas N gas=kBT0 P

0wethenhave

(P¡P0)

¡kBT0

P0 =kBT0Ns

Nw(21)

orP=¡Ns

NwP0(22)

atalowervaporpressure.

ClapeyronequationdP

dT=PL

RT2,Eq.(11).RecallthatdP

dTistheslopeofthephaseboundary.For smallPandTwecanusedP dT=P pressure,so

T=¡PRT2

P 0L=Ns

NwRT02

L(23) vaporizationofwaterL=42kJ mol,about1/7thofthelatentheatof vaporization.Then

T=¡8.7

558.3J

molK(273K)2 6.0kJ mol=¡16K(24)

3.3Chemicalpotentialphasediagrams

whichissometimesuseful. athighT.Wealsoknowthat@ @T P =¡S

N<0(25)

solidlines. bothexpansions.Tryityourself!10Section3 then@ @P T =V N=1 n>0(26) inqualitativeagreementwithEq.(14).

Here"sanotherexample

Figure8.Phasediagramforsubatomicmatter

4Generalphasetransitions

∂T P changesdiscontinuouslyatthe

Firstorderphasetransition:∂G

∂T P changesdiscontinuouslyatthephaseboundary ∂T P tobecontinuous, nthorderphasetransition:∂nG ∂Tn P intoone,andthelatentheatvanishes. theentropyS=¡@G @T P Vis equivalenttousingV=@F @P T

4.1Paramagnetism

anexampleofaphasetransition. phase. =2N F magentic=E¡TS=¡N"(27) andthefreeenergyofthedisorderedstateis F stateoccursat k BTc" ln2(29) T cisknownastheCurietemperature. transitionissecondorder. magneticfieldisavectorM,andwecandefineM=

M.AboveTc,M=0exactly.AsTis

symmetryisspontaneouslybroken.

4.2Criticalphenomena

properties.Generalphasetransitions13

N.Thegreenregionhas

liquidandgas. ∂v=0.Afterthephasetransitionis coexistenceregion. volume,v=1 ∂v? T =0.Atthecriticalpoint,thelengthofthe ∂v2? T =0aswell:thecritical14Section4 thederivativesofPvanish,∂nP ∂vn T

It"snotjustthederivatives∂nP

∂vn T diagramintheT-vplane:

Figure11.TVdiagramforwater

∂vn P =0,soT(v)isanon-analytic function. ∂vn T =∂nT ∂vn P =0,allthedimensionfulphysical andeverything surfacetension thecriticalvalues: T ^=T T c;P^=P P c;v^=v v c;n^=1 v^=n n c= c(30) n^?(T^)=1+3

4(1-T^)+7

4(1-T^)1/3;n^g(T^)=1+3

4(1-T^)-7

4(1-T^)1/3(32)

pointn^=1,T^(k)(1)=0.16Section4quotesdbs_dbs14.pdfusesText_20
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