...Thermodynamics. If Clausius Clapeyron fails. l T (v 2 v 1 ) = 0/0 Second order phase transition ( S, v = 0)

Transcription

1 If Clausus Clapeyron fals ( ) dp dt pb =...Thermodynamcs l T (v 2 v 1 ) = 0/0 Second order phase transton ( S, v = 0) ( ) dp = c P,1 c P,2 dt Tv(β 1 β 2 ) Two phases ntermngled Ferromagnet (Excess spn-up atoms) pb Superflud Helum (Many partcles n quantum ground state) Superconductor (Many pared electrons n same quantum state) Lecture 17 November 16, / 21

2 Open systems How to deal wth partcle exchange? Extra partcles N N + dn Energy ganed by addng a partcle? du = TdS PdV + µdn wth µ = ( U ) N S,V Mathematcally µdn equvalent to ncludng an extra type of work. Lecture 17 November 16, / 21

3 Chemcal Potental µ and thermodynamc potentals Defntons unchanged F = U TS H = U + PV, G = H TS = F + PV Therefore dg = SdT + VdP + µdn etc. µ = Also, for many speces µ = ( U N ) ( U ) ( F ) ( G ) N = S,V N = T,V N T,P S,V,N j = ( F N ) µ s useful wth ANY boundary condton T,V,N j = ( G N )P,T,N j Lecture 17 November 16, / 21

4 Chemcal potental and Gbbs Free Energy Consder the two ways to wrte the free energy of αn partcles: αg(p, T, N) = G(P, T, αn) Take the dervatve of both sdes wth respect to α: ( ) G(P, T, αn) G = α P,T ( ) G(P, T, αn) = N (αn) = N ( ) G(P, T, αn) α N ( ) G(P, T, N) = N N P,T P,T P,T G = Nµ for a pure substance Lecture 17 November 16, / 21

5 State functons n terms of µ G = Nµ for a pure substance The chemcal potental for one speces s the specfc Gbbs free energy! µ = G(T, P, N)/N we fnd that: dµ = sdt + vdp µ can be wrtten as a functon of P and T only for a pure substance. s = ( µ/ T ) p,n v = ( µ/ p) T,N ; for a pure substance Lecture 17 November 16, / 21

6 Contrbuton to µ Be careful wth extensve ( N) and ntensve quattes. dg = N(vdP sdt ) + µdn d(g/n) = dg = dµ = (vdp sdt ) The fnal term vanshed for large N! µ s typcally ncreased by rasng the pressure, or lowerng the temperature. These are sngle-phase relatons: G s also reduced by spontaneous chemcal reactons. Lecture 17 November 16, / 21

7 A closed system equlbrates Closed system, two parts, A and B, ntally out of equlbrum. Va Ua Na Vb Ub Nb du A + du B = 0, dn A + dn B = 0, dv A + dv B = 0. Second law: ds A + ds B 0 ds(u, V, N) = du T + P T dv µ T dn Whch, when appled to the two-part system gves, ( 1 1 ) ( PA du A + P ) ( B µa dv A µ ) B dn A 0 T A T B T A T B T A T B Hold that thought... Lecture 17 November 16, / 21

8 Flow of heat and partcles From prevous slde... Va Ua Na Vb Ub Nb ( 1 1 ) ( PA du A + P ) ( B µa dv A µ ) B dn A 0 T A T B T A T B T A T B At equlbrum T A = T B, P A = P B and µ A = µ B. energy flows from hot to cold untl T A = T B. volume moves from hgh to low pressure (for equal T) partcles flow from hgh to low chemcal potental (for equal T). Partcles (mass) flow along gradents of the chemcal potental. Lecture 17 November 16, / 21

9 Phase separaton n Planets Chemcal potental ncludes gravty: µ = u Ts + Pv + mg r h. Heavy atoms fall to the bottom: can be drawn up f soluble (u) Lecture 17 November 16, / 21

10 Chemstry Do I really have to do all these? Lecture 17 November 16, / 21

11 Chemstry of an solated system Consder Isolated system : du = 0, dv = 0, S(U, V, N 1, N 2...) TdS = du + PdV µ 1 dn 1 µ 2 dn 2 µ 3 dn 3... Equlbrum (maxmum entropy): ds = 0 µ dn = 0 solated system at equlbrum Lecture 17 November 16, / 21

12 Thermodynamcs n Chemstry Molecules react to form other molecules. A chemcal potental can P be defned for each. du = TdS PdV + P µ dn dg = SdT + VdP + µ dn Total Gbbs, G s also the sum of the chemcal potentals X X G= µ N = dg = N dµ + µ dn Equatng these expressons for dg yelds the Gbbs-Duhem relaton: X N dµ = S dt + V dp Ths gves balance of concentraton of components Lecture 17 November 16, / 21

13 Chemcal equlbrum of a reactng system Closed system n equlbrum wth a T, P reservor. Amounts of each component N are all nternal degrees of freedom. Mnmse G (set dg = 0) dg = µ dn = 0 closed system at equlbrum, fxed T The dn are constraned by the reacton equaton b dn = 0 e.g. H O 2 H 2 O leads to dn H dn O 2 dn H2 O = 0 Combnng mnmsaton of G and constrant: chemcal process equlbrum: b µ = 0 e.g. µ H µ O 2 = µ H2 O Reacton contnues untl chemcal potentals are equal. Lecture 17 November 16, / 21

14 Chemcal Potental for sngle component deal gas dg = VdP SdT + µdn For constant N, ntegratng g from T=0, P=P 0, we get dµ = dg = ( ) d(g/n) dt P,N ( ) d(g/n) dt + dp T,N dp = (S/N)dT + (V /N)dP = c P dt + RT P dp µ = c P T + RT ln(p/p 0 ) Where we used the Thrd Law for S(T=0)=0. Lecture 17 November 16, / 21

15 Statng the obvous Non-nteractng objects n the same system can be treated as ndependent deal gases, thanks to... Dalton s Law Total pressure s the sum of partal pressures P = p Raoult s Law Partal pressure s proportonal to concentraton p V = N RT e.g. dlute chemcals n soluton, photons n a cavty etc. Lecture 17 November 16, / 21

16 Example: Solublty Surroundngs, ncl. external chemcal potental T 0, P 0, µ (0), Epecfc enthalpy of soluton δh (bndng to solvent). Fnd concentraton, c, of n the system at equlbrum? Equlbrum system T = T 0, P 0 = p N/N, µ = µ (0) µ (0) Usng Raoult s Law (p = P 0 N /N = c P 0 ) = µ = δh + RT ln p /P 0 Rearrangng: c = exp(µ (0) δh)/rt If nsoluble: δh µ (0) there s stll some n soluton, If soluble, δh < µ (0) ( s soluble ) the concentraton s larger than 1, (not an deal gas) Lecture 17 November 16, / 21

17 Reacton between deal gases at fxed T, N Dlute soluton Ideal Gas = Ideal Soluton. Ideal gas mxture: P = p Raoult s Law: p V = N RT For sothermal reacton dg = V dp SdT + µ dn dg dµ = RT dp p Get change n µ wth pressure by ntegratng from reference state µ 0, µ = µ 0 + RT ln[p /p 0 ] Substtutng nto equlbrum condton b µ = 0 Lecture 17 November 16, / 21

18 Equlbrum constant...from prevous b (µ 0 + RT ln[p /p 0 ]) = 0 Each reacton has an Equlbrum constant (whch depends on T). [ ] ln(k) ln (p /p 0 ) b = b µ 0 RT = K(T ) (p /p 0 ) b = exp[ b µ 0 ] reference state) RT If we consder T=0, Gbbs/µ s the same as Enthalpy At T=0, b µ s just the enthalpy of reacton. K(T) looks lke the Boltzmann factor exp E/kT. Lecture 17 November 16, / 21

19 Measurng K(T) wthout a chemcalpotentalometer Dfferentatng wrt T and usng s = µ/ T P ( (ln K) ) T p = 1 RT 2 b (µ 0 + Ts 0 ) = b H 0 RT 2 b H 0 s just the molar enthalpy of reacton, whch s readly measurable. Smlar to the Gbbs-Helmholtz Equaton (handn queston). Lecture 17 November 16, / 21

20 Chemstry made easy Need only measure Chemcal Potental µ for each component (N measurements), not K for every possble reacton (N! measurements) Lecture 17 November 16, / 21

21 The Chemcal Potental s... The extra energy from addng a partcle du = TdS PdV + µdn The specfc Gbbs µ = g for a pure substance. The quantty whch drves partcle flow. The quantty whch defnes chemcal equlbrum All of the above! Lecture 17 November 16, / 21