Chapter III: Statistical mechanics

Transcription

1 Chater III: Statistical mechanics Section 3: Temerature and heat caacity... Schottky model... Relation between heat and entroy...5 From fundamentals functions to exerimental measurable...6 What drive heat exchanges...9 Energy fluctuation and C V...0 Section 4: Chemical Equilibria...5 Conditions of equilibria...5 Partition functions for chemical reactions...6 More comlex equilibria...7 Finding ground state energies...8 Pressure-ased equilibrium constant... Standard state otential... Le Chatelier s Princile...4 van t Hoff equation...5 Gibbs-Helmholtz equation...7 Pressure deendence of equilibrium constant...8 Section 5: Equilibria between Liquids, solids and gases...3 What forces drive the change of hysical hase...3 Lattice model for liquids and solids...3 Cavities in liquids and solids...35 The Claeyron equation...36 Refrigerators and heat ums...4 Surface tension...43

2 Section 3: Temerature and heat caacity Temerature and heat caacity = easy to measure but not easy to concetualize Two simle models: ideal gas and Schottky two states The two-state model can have negative temeratures (lower than 0K ), ideal cannot Negative temeratures hel illuminating the meaning of temerature and heat caacity The temerature: T is a roerty of single object S Thermodynamic definition: = T U Schottky model Two energy levels: ground state with ε = 0 and excited state where ε = ε0 > 0 V, N For N articles, n are in excited state and N ( ( )) The entroy SU ( ) = S W n ( U) nin the ground state U (3.3.) U = nε 0 n = ε0 The multilicity: W N! = n!( N n )! The entroy (using Sterling aroximation): S n N n (3.3.) = lnw = nln ( N n) ln k N N Substituting (3.3.3) U n = a simle way to get T is to use the definition: ε 0 lnw lnw dn = k k T U = n du VN,

3 dn Since = du ε 0 k n N n k nn k U Nε 0 (3.3.4) = ln ln ln ln T ε + + = = 0 N N ε0 nn ε0 U Nε0 ut since (3.3.5) n f excited N = and ground N n = f we also have k f ground = ln T ε0 fexcited So the temerature of a two state system deends on the sacing ε0between the levels which is a roerty that distinguish one tye of material from another The temerature also deends on N and U through n The figure shows the entroy variation with energy S( U ) for a two-state system with N = 3 - the sloe of the curve is S U = T From left to right one can see that the sloe change sign the temerature becomes negative 3

4 Positive temerature: If an external source of energy (bath) is available the system will tend to absorb energy exciting more and more articles nd law: the system tend towards maximum entroy increase the multilicity The system absorb energy to increase the entroy Infinite temerature: ecause the entroy cannot grow, the system will not tend to absorb more energy from the bath This is the oint of the curve where S( U) is maximum Negative temerature: ecause shifting the few articles left in ground state to the excited state will not increase the multilicity the system will tend to loose energy to increase its entroy A system with T < 0 is hotter than a system with T > 0 U is low most articles are in the ground state fground fground > ln > 0 f f excites excites Multile states system, the oulations of the levels follow the oltzmann law lower energy levels are more oulated than the higher energy levels U is intermediate energy, like for a coin fli exeriment, W is maximum of the articles are in two states fground fground = ln = 0 T = f f excites excites Multile states system equal oulation on all state (similar to outcomes of unbiased dice) For a system with high energy, most articles are in excited states fground fground < ln < 0 T < 0 f f excites excites Multiles state oulations increase exonentially with level high energy levels more oulated than low energy ones 4

5 The two states model illustrate that reresents inclination of system to absorb energy T following the second law of thermodynamics: 0 T > Absorb energy 0 T = No trend 0 T < Give away energy Why T > 0 is more common thant < 0? Negative temerature occur only is saturated system finite number of states Ordinary materials infinite ladder of energy levels (translation + rotation + vibration + electronic) cannot saturate o In these cases, nd law of thermodynamics favor energy absortion System with T < 0 inversion of oulations (non thermal henomenon) o This cannot be achieved by equilibration with thermal bath the most one can achieve this way is T = o Inversion of oulation only ossible if articles excited by electromagnetic radiation (non equilibrium state) o Phenomena related to LASER or MASER in order to increase entroy the system will emit radiation Relation between heat and entroy q What is the microscoic justification behind the relationds =? T Consider a system with degrees of freedom ( U, V, N ) The nd law of thermodynamics S( U, V, N) is maximum at equilibrium If V and N constants, no work is done on the system the only way to increase U is through intake of heat du = q For a two-state system with T > 0 the heat intake fexcited q > 0 du > 0 increases f ground Since du = q = ds the multilicity increase with the intake of energy T T 5

6 From fundamentals functions to exerimental measurable To untangle relationshi between SUTand,, C V follow grahical transformations egin with S( U ), the closest to rincile of equilibrium, and move toward ( T) the most directly measurable quantity CV We comare the ideal gas model with the two-state model for V and N constants S increases monotically without bounds (infinite ladder of energy) S reaches a maximum and then decreases 3 S = NklnU Taking the sloe of S( U) yields S = U T the driving force to absorb heat T always ositive: 3 Nk = T U The sloe change sign at high energies Higher for low energy (T small) and saturating towards high energy (T ) 6

7 The transformation y ' y = yields T ( U ) The temerature is roortional to the energy U T = 3 Nk Temerature increase more raidly when aroaching saturation f T ln excited f ground In general the temerature measure the relative oulation on different energy levels The transformations y = x and x y = yield U( T ) The energy deends on T : 3 U = NkT Legitimate the exression utting energy into a system : the higher the temerature of the bath and the higher the internal energy of the equiartitioned system absortion rocess is linear As long as T > 0, the energy limited to N ε 0, higher energies only when T < 0 utting thermal energy into a system : absortion is a saturated rocess at higher temerature the system will not absorb more energy from the bath 7

8 To obtain C takes the derivative of U( T) CV V U = T For ideal gas, C V is constant For two-states C V eaks near T = 0 No thermal energy is absorbed either for T > 0 or T < 0 when T is high What these series of grahs show is that CV is much more different than T is driving force to absorb energy T C is telling how much energy is soaked u as you change the driving force V 8

9 What drive heat exchanges Consider two -state systems: A: UA, ε Aand N A : U, εand N At low ositive temerature: In equilibrium: = T T (3.3.6) A U Nε and k 0 ln N ε = T ε U 0 N ε N ε ln = ln ε U ε U A A A A 0 If the two systems are made of the same material εa = εand contain the same amount of articles NA = Nthen the trend toward maximum entroy is a tendency to equalize the energy ut if NA N the exchange of energy will deend on the size of the systems 9

10 Energy fluctuation and C V Hold a system at a constant temerature utting it in contact with heat bath Although T = constant, U will fluctuate C V measures the magnitude of these fluctuations To see this we must consider the density of states: 3 Particle in cubic box V = a h (3.3.7) εtrans = εn,, ( ) x ny n = n z x + ny + nz 8ma For nx, ny, n z =,,3, counts the number of states that have energy between ε and ε + ε R ( n n n x y z ) = + + different values of n yields similar result ( hr) ε trans = 8ma R is the radius of sace integer Volume of sace integer: π R where 8 have n,, x ny nzall ositive (3.3.8) The number of ositive integer oints: To get W ( ε) π 8mε M π R V = = 83 6 h ε the number of ositive integer oints in shell of energy between dm ε and ε + ε we take the derivative W( ε) ε = M( ε + ε) M( ε) ε dε (3.3.9) W( ) 3 π 8m ε ε = Vε ε 4 h Where W ( ε ) = the number of states of single article in box of volume V energy between ε and ε + ε - Generally an extremely large number 3 = a that have 0

11 Examle 3.3.: Atom of Argon g m = 40 = mol At If W T = 300K, 6 a = cm and ε = 0.0ε ( ε) kg atom J K 300K kt J ε = = 3 6 kg π ε = ( ) ( ) 4 4 ( J s) 3-3 atom 0 m 6. 0 J states The above develoment focus on robabilities ( nx, ny, nz) articular microstate, say (,, x y z) n n n for the 3D article in a box βε = e that a article is in a q Focusing on ( ε) ( ε) = q W e βε that a system is in a articular energy level The artition function is the same: (3.3.0) ( ) states 3 βε βε π 8m βε energylevels 4 h 0 q = e = W ε e = Vε e dε The form of the integral is βx x e dx which has solution Γ ( ) 0 3 β 3 π where Γ ( 3) =, therefore the solution of the integral is ( kt ) 3 8m mkt π π π q = V kt = V 4 h h (3.3.) ( ) π This is the same result as before when focusing on states instead of energy

12 Energy fluctuation The robability for a article to be in a articular energy E is: W E e Q (3.3.) ( E) = ( ) The function dominated by two terms: W( E) which grows extremely raidly, and e βε which shrinks extremely raidly function is highly eaked βε If E is any value of energy at one articular instant and U is the value at equilibrium then E is aroximately Gaussian with a width roortional to C near the eak ( ) V To see this, aroximate the function as a Taylor series: (3.3.3) ln ( ) ln ( ) ( ) ln E ln E E = U + ( E U) + ( E U) + E E E U ( ) = E= U The equilibrium = eak and fluctuation = deviation from it at eak E= U and S E S U T E = T of the heat bath ( ) = ( ) and ( ) 0 To evaluate ln ( E ) relace it by ( ) ln W E β E Q Where Q is not a function of E and does not contribute to derivative

13 Since S( E) klnw( E) (3.3.4) At the eak, = and S = E T E ( ) ( ) ( ) ( ) ln E lnw E Sk = β = β = E E E kt E kt ( ) ln E E ( ) T0 = 0 T E = Deriving a second time second derivative term of series: (3.3.5) ( ) ln E T = = E kt E kt C E= U 0 ( ) 0 V Substituting in the Taylor series and exonentiating (3.3.6) ( ) ( ) ( ) ( ) ( ) E U kt0cv U TSU 0 E U kt0 CV E = U e = e e Comaring with Gaussian exression: ( x) With variance: = e σ π (3.3.7) ( ) ( ) x σ E U E U kt0 CV σ = = = We see that CV determines the width of the energy fluctuation 3

14 Examle 3.3. very narrow width in general Useful dimensionless measure of width: U σ Since 3 U = NkT and ( kt CV ) CV 3 = Nk 3 kt N σ 3 = = N U 3 3 = NkT NkT For σ N ~0 0 U extremely shar eak for ( E ) 3 Fluctuations extremely small Excetions = hase transition and critical oints where CV and fluctuations are large CONSEQUENCES: If ( E) sharly eaked and U is of interest, then fixing T is tantamount to fixing U within narrow range Very few errors in using Microcanonical ensemble ( U, V, N) assuming reonderance of accessible states have same energy If fluctuation are of interest use Canonical ensemble where U fluctuate even if T is constant no fluctuations in U in Microcanonical ensemble 4

15 Section 4: Chemical Equilibria Major goal of chemistry = redict equilibria of chemical reactions + relative amounts of reactant and roducts from atomic masses + bond lengths + moment of inertia + other standard roerties Usually gas hases easier than liquids no interaction between molecules Conditions of equilibria Two state equilibrium: (3.4.) A K Ex. chemical isomerization, folding of bioolymers from oen to comact state, binding of ligand to surface or molecule, condensation of vaor to liquid or freezing of liquid to solid Equilibrium constant K : ratio of number or concentration of molecules in each state at equilibrium [ ] Arrow oints to the final state K = A Once direction determined signs of thermodynamic quantities is fixed [ ] The Chemical otentials redict the equilibrium At fixed T and, extremum function = Gibbs free energy dg = SdT + Vd+ µ dn + µ dn A A At equilibrium ( T and fixed): (3.4.) dg = µ dn + µ dn = 0 A A Since the total number of molecules is constant: (3.4.3) dn + dn = 0 dn = dn A A Substituting for the equilibrium condition: (3.4.4) ( µ µ ) dn = 0 A A 5

16 And since dn A 0 in general (3.4.5) µ A = µ What we need to do is to get µ Aand µ from microscoic model To do that relate µ to artition functions Partition functions for chemical reactions Consider the artition function: t j 0 ε kt (3.4.6) q = e = e + e + + e j= 0 ε kt ε kt εt kt The reduced artition function (factoring out the ground level) (3.4.7) ( ε ε ) kt ( ε ε ) q = e q = + e + + e ε 0 kt 0 t 0 kt Exressing the chemical otentials in terms of q q (3.4.8) µ ln A A = kt NA And q (3.4.9) µ ln = kt N At equilibrium: All the energy levels of both secies A and are accessible by any molecule Distributed according to oltzmann distribution with largest number of molecules in lowest energy states The number of molecules oulating a given state is determined by energy not whether molecule of secies A or q A q Substituting into equilibrium condition: µ A = µ kt ln = kt ln N N (3.4.0) N q q K = = = N q q A A A e ( ε ε ) 0 0 A kt A 6

17 The last equation does not take into account the interactions between molecules alies only to isolated articles such as in a gas hase More comlex equilibria (3.4.) K aa+ b cc Where abc,, are the stoichiometries of secies Aand, C For T and constants, the condition of equilibria (3.4.) dg = µ dn + µ dn + µ dn = 0 A A C C With two stoichiometric constraints N (3.4.3) A N + C = constant and N + N C = constant a c b c The logic behind these constraints: NA is the number of article in secies A a is the number of A tye article required for each stoichiometric conversion to C N A a C is the number of stoichiometric conversions consuming A N articles N A N + is the total number of ossible stoichiometric conversion of all A tye a c articles that were originally ut into reaction into the vessel To find the conditions for equilibria take the differential of elements in constraints a b (3.4.4) dn A = dnc and dn = dnc c c Substituting in free energy equation: (3.4.5) µ a b C a b dnc 0 c µ c µ = 7

18 At equilibrium (3.4.6) a b µ C = µ a + µ b c c Substituting the definition of chemical otential c a b q C q A q q C q A q (3.4.7) c kt ln = a ktln + b kt ln = = NC NA N NC NA N Rearranging (3.4.8) c ( qc ) b ( q ) ( q ) N q K = = = e N N c c C C a b a a b A qq A A ( ε0c ε0a ε0 ) c a b kt To determine K from exeriments we need to Detect the number of articles of tyes ACat,, equilibrium Know the stoichiometric coefficients abc,, Finding ground state energies We also need ε0 = cε aε bε the difference in ground state energies 0C 0A 0 For A both states must contain same atoms, so the fully dissociated state of A must be identical to the fully dissociated state of common zero energy = fully dissociated state 8

19 To define ground state energy need to resolve vibrational ground state Vibrational energy: ε v = v+ hv where ε v = 0 at the bottom of the energy well ut since there is no energy, the system cannot access this state cannot be measured exerimentally Latest energy level accessible ε 0 = hν this is the zero oint level The vibrational artition function: (3.4.9) q e = e hν kt vib hν kt Switch to zero oint instead of well bottom Dissociation energy: D= ε0 hν Can be determined from sectroscoic or calorimetric exeriments on the dissociation of molecules extraolating to T = 0K Combining vibrational artition function with D hν kt DkT DkT hν kt e D+ kt ε 0 (3.4.0) qvibe = e = e = q h kt h kt vze ν ν e e Where qvz is the vibrational artition function defined with zero of energy at zero oint (3.4.) q vz = h kt e ν To use the exression for K (eq ) relace roducts of vibrational artition 0 kt function that aear in q, and the oltzman factor qvibe ε with roducts qvze each secies in the chemical equilibrium DkT for 9

20 Examle 3.4.: K in charge symmetry reaction Gas hase exchange reaction: (3.4.) H + D K HD The equilibrium constant: (3.4.3) q K = q q HD H D e DRT Where D is the difference between molar dissociation energies of all roducts and reactants We use RT instead of kt to get oltzmann factors in dimensionless exonents Dissociation energies: H, kj 43.8, D, mol kj 439. mol and HD, kj 435. mol (3.4.4) DRT So e 0.79 at T = 300K D = D D D = kj 0.6 HD H D mol Equilibrium constant = roduct of comonents: translation t, rotation r, vibration v K = KKKe t r v DRT For translation: (3.4.5) K For rotation: (3.4.6) K r t 3 ( π mhdkth ) ( πmh kth ) ( πmdkth ) 3 m ( 3 HD ) 3 3 mh m 4 D = = =.9 8π I kt HD σ ( 6.3 HDh σ H σ ) D I HD I HD IH I H kt π I D kt σ D H h σ D h = = = π 8 σ Where σ = σ = and σ = H D HD 0

21 ecause vibrational energies are large for all secies at low temerature hν kt then we have q v = (3.4.7) Combining everything: DRT K = KKKe = t r v As T, K 4 K v hν HD kt e = = hνh kt hν D kt e e The change in rotational symmetry is the main contributor gain in entroy due to rotational asymmetry of roduct Pressure-ased equilibrium constant ecause ressure is easier to measure than article number for gases more convenient K V Use ideal gas equation: N = kt V C c c N C kt qc (3.4.8) K = = = a b N N V qq A V kt kt Multily both terms by V kt a b a b A A a+ b c C (3.4.9) = = ( ) Where q q π mkt = = V h 0 c c ( q0 ) ( ) ( ) K kt e e DkT c c a b C DkT a b a b Ab q0a q0 3 qq is the artition function with volume factored out r v

22 We can also use the artial ressure to exress the chemical otential: using the definition kt ln q NkT µ =, factoring out the volume q = qv 0 and using V = N Standard state otential qkt 0 (3.4.30) µ = kt ln = kt ln = µ + kt ln Where int = qkt 0 and µ = kt ln qkt 0 The last equation is useful because divides chemical otential into a art that deends on ressure and a art that is indeendent of ressure Examle 3.4.: Dissociation reaction for Iodine int I K At I T = 000K, with m I kg 0.7 mol kg = 3 molecules mol The ground state electronic degeneracies: q ei, = 4 and q ei =, The rotational and vibrational temeratures: θ rot = Kand θ vib = 308K The difference in dissociation energies: D = 35.6kcal, remembering that cal= 4.86J (3.4.3) ( q ) 0I DkT K = kt e q0i Where kt RT = = N 3 molecules AV mol matm K mol 000K matm 3 And ( ) DkT e = e Factoring the artition function (3.4.3) ( ) 0I ti ei q q q = q q q q q 0I ti ri vzi ei

23 For rotation: σ = and q ri T 000 = = 930 θ rot For vibration: = = 3.77 e e q vzi θvib T For the electronic term: q q ei ei = 6 For the translation: with m = m I 3 h qti πmkt I 3 q ti π m h I kt 3 π mkt = = = h I 5 3 (.09 0 kg)(.38 0 J K )( 000K) 34 ( J s) I π = m Combining all the terms: K = matm m 3. 0 atm Comarison with exeriments very good K highly deendent on T due to exonential term Increasing the temerature dissociate Iinto iodine 3

24 Le Chatelier s Princile Any erturbation away from stable equilibrium state increase the free energy of the system The system resonds by moving towards the equilibrium Consider the two-state equilibrium: K A Suose the fluctuation change the number of molecules by (3.4.33) dg = ( µ µ ) dn A dn Defines reaction coordinate: ξ = system roceeded to N A N + N describes the fractional degree to which the Since the total number of articles is fixed NA + N = N N = ξ N and dn = Ndξ Substituting in free energy equation: dg = µ µ Ndξ (3.4.34) ( ) To move back towards equilibrium 0 different signs A dg < therefore ( µ µ ) and dξ must have If fluctuation haens when µ > µ Athen dξ < 0and N will decrease A This is why µ escaing trend - the higher the value of to escae from to A µ and the greater the tendency Le Chatelier s rincile: term that refers to the tendency of a system to return to equilibrium by moving in a direction oosite to that caused by external fluctuations 4

25 van t Hoff equation Describes the deendence of equilibrium on temerature If you measure K( T) you can learn something about the enthaly and entroy Useful to test microscoic models Consider the two-state equilibrium: A K At constant T and the condition of equilibrium At equilibrium: µ + ktln = µ + kt ln (3.4.35) ln K A A Where µ only deends on T not on µ = µ K = A ( A) µ µ µ = ln = = kt kt A A In terms of enthaly and entroy: µ = h T s The temerature deendence: ln K (3.4.36) µ h T s = = T T kt T kt If h = h h and A s = s s A are indeendent of T ln K (3.4.37) h = T kt This is van t Hoff relation - rovides useful way to lot data to obtain h Since d = T T dt (3.4.38) ln K ( T) h = k 5

26 In the grah of ln K vs. h the sloe is T k For dissociation h > 0characteristic of bond breaking rocess dissociation driven by entroy Common, but not universal, feature of van t Hoff lot h indeendent of T Integrating the relation: (3.4.39) ( ) ( ) K T h ln = K T k T T This relation can be used to find K from h or vice versa 6

27 Examle 3.4.3: getting Takes two oints T T h,ln K T from the lot: 4 = 500K K and ( ) T 4 = 57K K and ( ) T lnk T = 3.47 lnk T = 6.4 ( ) ( ) ( ) lnk T ln K T kj h = R = 8.34J K mol ( ) K mol T T At T = 500K µ = RTlnK = 8.34J K mol 500K ( 3.447) 64 kj mol And ( 49 64) kj mol h µ s = = 56.7kJ K mol T 500K The two equations and are general; aly to any two states in equilibrium Gibbs-Helmholtz equation When h deends on temerature Generalizing for any deendence on free energy: G = H TS (3.4.40) H = G + TS Substitute (3.4.4) G S = T G H = G T T u vu uv Using the differential rule: d =, with v= T and u= G v v ( GT) G G G (3.4.4) = G T T T T = T T T 7

28 Substituting 3.4.4, we obtain the Gibbs-Helmholtz equation: (3.4.43) ( GT) H ( T) = T T Similarly for the Helmholtz free energy: (3.4.44) ( FT) U( T) = T V T Examle 3.4.4: obtaining ST ( ) and H( T) from GT ( ) Suose GT ( ) 3 = at, where a = constant G T Then ST ( ) = = 3aT and H( T) = G T = at 3aT = at G T Consistent with ( ) H( T) d at dt = at = T This examle shows that the Gibbs-Helmholtz equation does not mean that the temerature deendence of the Gibbs free energy is solely due to enthaly Pressure deendence of equilibrium constant Deduction similar than for the temerature K reflects a volume change (3.4.45) ( ) ( A) ln K µ µ µ = = kt kt To eliminate the ressure deendence µ ( ) relations use Gibbs-Duhem equation or Maxwell s 8

29 Using the Maxwell relation: (3.4.46) µ V = = v N T, N T, Where v = V moleculeor er mole deending on units of µ Substituting (3.4.47) ( µ µ A) = v va = v T Relacing in the equation for the equilibrium constant: ln K( ) v (3.4.48) = kt If is the state with smaller volume, v va < 0, then increasing the ressure will shift the equilibrium from A to 9

30 Examle 3.4.5: ressure affects -state equilibrium Common anesthetic drug molecule = halothane (-homo--chloro-,,-trifluorethane) Mode of action = artitioning from water ( A) into liid bilayer membranes ( ) We choose two oints (,ln K ) in the grah above: ( 0,7.84) and ( 80,7.6 ) v = v v bilayer ln K water lnk ( )( ) ( ) v= RT = matom K mol 300K. 0 mmol Or N AV 80atm cm cm Å Å. 0 mmol m mol cm molecule Therefore the anesthetic occuies more volume in the liids bilayer than in the water a ressure increase force the anesthetic to go into water where the volume is smaller 30

31 Section 5: Equilibria between Liquids, solids and gases What forces drive the change of hysical hase Chemical otential also describes tendency to move articles from one hase to another Surface tension describe tendency for articles to move between the interior (bulk) and surface of material Why do liquids boil? At normal temerature and ressure water is liquid + few articles in vaor hase Above boiling temerature water vaorizes o Vaor material is liquid at normal temerature and ressure o Gas material is gaseous at normal temerature and ressure What is the effect of T and on vaorization rocess? alance of two forces F = U T S o Attraction between molecules in liquid hase o Gain in translational entroy in vaor hase Increasing T entroy at high temerature greater than loss of bond energies (second terms dominates) vaor hase is favored At lower temerature first term dominates liquid hase favored Consider a beaker of liquid (c hase) in equilibrium with vaor (v hase) hold T and constants Nc is number of articles in condensed hase, hase Free energy change must be exressed in terms of chemical otentials (3.5.) Since dt = d = 0 Since dg = SdT + Vd+ µ dn = µ dn + µ dn N + N = N = constant v c (3.5.) dnc = dnv Substituting, the condition of equilibrium yields dg = µ µ dn = (3.5.3) ( ) 0 j= Nv is number of articles in vaor j j v v c c v c v Or equivalently (3.5.4) µ v = µ c µ c and µ v 3

32 To comute the vaorization equilibrium we need µ v and µ c Assuming the vaor hase is dilute enough ideal gas (3.5.5) µ v = kt ln kt Where int = describes the internal degrees of freedom in the molecule 3 Λ indeendent of T To determine µ c need a model of liquid and solid hase int Lattice model for liquids and solids Particles in solids and liquids significant interactions between articles ρg ρl While 000 we have 0.99 both can be treated by same model ρ ρ L Infinite lattice each site of lattice occuied by a article S Solid have long range order satial distribution of molecules is more regular than in liquids same amount of neighbor atoms or molecules More variations in liquids however, can be well defined on average More imortant insight of lattice model = most imortant energetic interactions for holding liquids and solids molecules together are short range interactions between near neighbors only weak interactions with more distant neighbors 3

33 Since solids and liquids mostly incomressible we can use either ( T,, N ) or ( TVN,, ) If is constant so is V For free energy we can use: F = U TS The translational energy of a lattice must be zero because when airs of articles trade laces we can t dis tinguish new arrangement from original one W = and S = klnw = 0 At equilibrium, exchange of articles between L and G hase involves no changes in internal quantum state: rotation + vibration + electronic Energies relevant for vaorization are those between airs of articles in lattice Interactions are attractive bond energy between two A tye articles w AA < 0 Aly to any airs in neighborhood of a article Energies of uncharged atoms and molecules in solids and liquids very weak comared to covalent or ionic bonds that hold atoms in molecules A few kcal/mol comared to a few 0-00 kcal/mol First aroximation is that bond energies are indeendent of T To comute the interaction energy counts the number of bonds among the N articles and multilies by w AA Each oints have z neighbors (coordination number of the lattice) However, we cannot use U = Nzw AA because we count twice z the bonds use Number of bonds (edge contribute z negligibly) m = N (3.5.6) U = mw = AA Nzw AA And the free energy: (3.5.7) Nzw F = U TS = AA 33

34 The chemical otential for the condensed hase is therefore F zwaa (3.5.8) µ c = = N TV, For the vaor ressure, since at equilibrium we have µ c = µ v zw kt ln = AA (3.5.9) int zwaa kt = inte This is the vaor ressure of the molecules that escae the liquid N Also the density of molecules in vaor hase since = kt V When the bonds are strong ( w AA more negative) the vaor ressure decreases because the molecules refer to stay in liquid form If waais fixed, increasing T increase the ressure heating of lake + air above it increase vaorization when temerature decreases condensation occurs fog or rain 34

35 Cavities in liquids and solids Molecular rocesses involve creating or filling cavities The energy cost when a molecule leave the medium is: U = zw (3.5.0) removed AA This is assuming the cavity does not disturb or adjust after art removal Energy goes u (less negative) because it involves bonds breaking Of the z broken, z associated to molecule that left and z to the molecules that line the cavity Subsequent closure of the cavity energy cost, U removed+ closure, is equal to the difference between the energy of a system withn article and the original system with N articles: zwaa (3.5.) Uremoved+ closure = U( N ) U( N) = Therefore, the energy costs of oening and closing a cavity are: zwaa zwaa (3.5.) Uoen = and Uclose = These equations will be useful for solvation and artitioning rocesses Note that since µ is the same at the surface than in the bulk there is no secial treatment for the molecules at the surface 35

36 The Claeyron equation Describes T ( ) at hase equilibrium Consider the following hase diagram: the lines T, at which hases are stable are set of oints ( ) Solid-gas (sublimation) Solid-liquid (melting-freezing) Liquid-gas (boiling or condensation) The second grah shows a hase diagram for water Water boils at T = 00 C also boil at T = 5 Cand and = atm = 0.03atm In general increasing the ressure imlies increasing the boiling temerature but can If = atm and T = 5 C, water stays in liquid form When the ressure = vaor ressure water change hase When ressure is greater than water ressure water is liquid + small amount of vaor 36

37 Positive sloe of T ( ) is a general feature of many tye of hase equilibrium Derivation of sloe: Choose two oints ( T, ) and (, ) T at equilibrium L G Since we have equilibrium µ L T, = µ G T, (3.5.3) µ T, = µ T, L ( ) ( ) ( ) ( ) G The chemical otential at (, ) at ( T, ) µ G( T, ) = µ G ( T, ) + dµ G( T, ) µ ( T, ) = µ ( T, ) + dµ ( T, ) (3.5.4) T is the result of a small erturbation of the value L L L Considering the conditions at equilibrium this simlify to dµ T, = dµ T, (3.5.5) ( ) ( ) Since µ is function of ( T, ) we get for the differential: (3.5.6) µ (, ) G µ µ d T = dt + d T L N, T, N If we use free energy/mol instead of /molecules then the differential of the chemical otential is equal to the artial molar entroy and volume: µ S = = s T N, N T, (3.5.7) µ V = = v N T, T, N 37

38 Substituting in 3.5.5: (3.5.8) dµ G = sdt G + vd G = dµ L = sdt L + vd L Rearranging the terms d sg sl s (3.5.9) = = dt v v v Where s = sg sl is the artial molar change in entroy and v= vg vl is the artial molar change in volume At hase transition µ = h T s = 0 (3.5.0) h = T s Where h = hg hlis the enthaly difference between the two hases Substituting we get the Claeyron equation: d h (3.5.) = dt T v Since molar volume of gas hase much higher than for liquid or solid v= vg vl vgand using the ideal gas law RT volume = while mole G kt volume = molecules The Clausius-Claeyron equation: dln h (3.5.) = dt RT Integrating with h indeendent of and T : (3.5.3) ln h = R T T L RT v where T h d ln = dt R T T 38

39 39

40 Examle 3.5.: oiling water at high altitude At low altitude = atm, water boils at T = 373K If we go u to = atm at what temerature will water boil? Since kj h va = we find that mol T R 8.34JK mol T hva Jmol = ln = ln 354K~8C Other hase equilibria: can be comuted relacing enthaly of vaorization by adequate one for the rocess From the relation d s d = and since 0 dt v dt > we deduce that the hase with greater entroy as usually greater volume: s > 0 v > 0 if s > 0 v d One interesting and imortant excetion is water: between 4C and 0C 0 dt < and s< 0 v> 0 ; because liquid has greater entroy than Ice, the volume of Ice is bigger (lower density) than that of water and Ice float over water Comaring the vaorization enthaly with the lattice model zwaa (3.5.4) hva = The enthaly of vaorization is the result of removing one article from the liquid and closing the cavity 40

41 Refrigerators and heat ums Illustrates the rinciles of vaor-liquid equilibria Refrigerator (heat um) use external energy to absorb heat from low temerature environment en dum it in high temerature environment Working fluid umed around system of tubes and undergoing reeated thermodynamic cycle of condensation and vaorization Choice of fluid based on ability to boil at low temerature and aroriate ressure o Tyical fluid,,,-tetrafluorethane with T =.6 C boiling o Older fluid = FREON containing chlorine which reacts and deletes ozone layer of atmoshere Two rinciles: ) oiling store energy (break bonds) and condensation get it back ) Fluid can be boiled at low T and condensed at high T by controlling the ressure 4-ste rocess: the grah of enthaly vs. ressure shows when non covalent bonds of working fluid break (high enthaly) and reform (low enthaly) Ste : absorbing heat efore ste, the working fluid is in liquid state at low ressure, oised to boil The working fluid flows through the coil inside the box absorbing heat The heat utake boils the fluid breaking liquid-liquid bonds increasing the enthaly Ste : comression low ressure gas is comressed at high ressure oised to condense into liquid hase 4

42 Ste 3: duming heat Working gas flows to coils outside the box giving heat to surrounding Condensation of gas into liquid decrease the enthaly (reformed liquid-liquid bond) Ste 4: reducing the ressure Fluid flows through exansion valve where ressure is reduced without much enthaly change Low ressure liquid oised to vaorized 4

43 Surface tension Describes equilibria between molecules at surface and in the bulk Surface = boundary between condensed hase and gas or vaor hase In general, interface = boundary between any two media Surface tension = free energy cost of increasing surface area of the system Water drolet (any liquid) sherical because yields smallest ratio surface volume Deviation away from sherical shae is oosed by surface tension Model lattice condensed hase of N molecules If n is the number of molecules in surface then N nis the number of molecules in the bulk ulk articles have z nearest neighbors Surface articles have z nearest neighbors Total energy of surface + bulk zw ( z ) w AA AA waa U = N n + n = ( Nz n) (3.5.5) ( ) Surface tension is the derivative of the free energy with resect to total area A of F the surface: γ = A T, V, N Since the lattice has entroy S = 0 and F = U then F F dn U dn (3.5.6) γ = A = = T, V, N nda nda Where U w (3.5.7) = AA n 43

44 And because A (3.5.8) ecause w AA < 0 γ > 0 = na where a is the area er article w γ = AA a dn da = a The surface tension describes the free energy cost of transferring one article from the bulk to the surface Greater for materials with strongest inter molecular bonds Very strong in liquid metals ex. Silver, Iron, Mercury in which free flowing electrons bind atoms tightly together o Exlained why Mercury drolet are so round Water dyn γ = 7.8 (where cm Lower in n-hexane 5 dyn 0 N = ) dyn γ = 8.4 where binding due to disersion forces cm 44

45 Adding surfactants to water decreases surface tension increase surface area ex. foam in dishwacher and beer Different measures relate to each other: since hva and γ are both linear functions of w then there must be a relation between hva and γ AA From the lattice model we get h γ = va er unit of area z 45