Thermodynamics, Part 2

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Charles Burns
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1 hermodynamics, art November 1, hase equilibria Consider the three phase diagrams in Fig.(6.1) (a) CO, (b) water; and (c) helium. CO is a normal material with a typical and simple phase diagram. At 1 bar pressure and at 00K, CO (s) CO (g), carbon dioxide sublimes and in order to have liquid-gas coexistence, higher pressures are needed. Water s phase diagram is characterized by the multiplicity of solid phases and that the vs. line has a negative slope. he huge number of solid phases arises because of the tetrahedral, low pressure solid is susceptible to elevated pressures that induce structures with more than four nearest neighbors. Helium has but two electrons, is only weakly polarizable and becomes liquid at very low temperatures, near 4 K. In these circumstances of its low mass and the low temperatures needed to liquify, quantum wavelengths are large and hence quantum interference becomes important. He, I is a normal liquid, whereas He, II is a superfluid with a vanishing shear viscosity. he two conditions for two phase equilibrium are (ρ 1, ) = (ρ, ) µ(ρ 1, ) = µ(ρ, ) (1) viz., the chemical potential in phase 1 characterized by density ρ 1 must equal the chemical potential for that same species in phase at density ρ. Likewise a similar situation prevails for the pressure. he µ-condition implies that it must take equal work to insert a particle in phase 1 as it does to insert it into phase. he condition implies that the forces along the 1- interface must be equal otherwise the phase boundary line is on the move. So how could the pressure of an ideal gas ever equal that of a liquid? he attractive and repulsive forces operative at close intermolecular separations in the liquid phase must nearly cancel so that the pressure of the liquid is that of a nearly ideal gas. hese two equations, for µ and, when solved simultaneously give the densities of the coexisting phases. 1.1 hase boundary lines in, diagrams In an equilibrium between the liquid and solid phases, dµ L = S L d + V L d = dµ S = S S d + V S d () 1

2 where the over-bar denotes per mole. Rearrange to form the slope of vs. along the coexistence line, ( ) = S S S L V coex S V = S(L S) L V (L S) = H(L S) V (L S) For the freezing transition, V is small and negative, and since the process is exothermic, both V, H < 0. hus the vs. slope is positive and steep. For the L, S G transition, V is comparatively large wherein the slope, although still positive, is small. If the final state is a gas, (3) V (L, S G) V G R (4) and which integrates to ( ) = H(L, S G) (5) R coex d = d H(L, S G) ln( R / 1 ) = H(L, ( S G) 1 1 ) R 1 (6) his equation determines the shape of the vs. curve and determines the vapor pressure at one temperature, given a value at another. 1. and dependence of the chemical potential For a system with one component, dµ = Sd + V d (7) Hence the slope of µ vs. is the negative of the entropy and the slope of µ vs. is the molar volume. Shown in Fig.(6.a) is the variation of µ vs. for a single component substance with G, L and S phases. Suppose we apply an external pressure to the system, then because of dµ = V d, we shift the chemical potential phase boundaries as shown in Fig.(6.b). In Fig.(6.b) we implement V L > V S, the solid is more compact than the liquid, and hence the solid melts at a higher temperature (as determined by the intersection point of the chemical potential lines). For water, the molar volume of the liquid is smaller than the solid, and therefore by applying pressure to water, one shifts the melting point to lower temperatures, i.e., apply pressure to ice and it melts. 1.3 Ehrenfest and phase transitions In the 1960 s and 70 s there was a revival of interest in the properties of liquids and phase transitions. he molecular structure and dynamics of liquids is a classic many body problem

3 where molecular order is neither chaotic as in the gas phase nor regular or crystalline or regular as in the solid phase. he emergence of short range order proved a fascinating topic, fueled in part by the importance of liquids as environments for nearly all chemical reactions of significance. An understanding of the phase transition from gas to liquid is key to the understanding the properties of liquids. In the study of phase transitions, particularly near the critical point, there is a degree of universality. If one could understand the transition from a paramagnetic solid to a ferromagnetic one, or the phase separation of a two component liquid into two separate immiscible phases, or the orientational and positional order in some liquid crystal phases, one could, at the same time, understand the liquid-gas critical point. Since the one size fits all approach is very appealing, we shall look at critical points. he language of phase transitions was influenced by Ehrenfest (one of Boltzmann s students) who introduced the criteria of first and second order transitions. Consider dµ = Sd + V d (8) A first order phase transition illustrates discontinuous changes in the first derivative of G or µ at a phase transition. For a melting transition or at the boiling point, S = H/ and the molar volume, V, change discontinuously. See Fig.(6.3a,b). For a second order transition, the second derivative of G diverges at the critical point. ( ) ( V ) µ = (9) ( ) S = C (10) In Fig.(6.4) we present plots of vs. an order parameter in the vicinity of the liquid-gas, binary liquid mixture and paramagnet-ferromagnet critical points. In the case of liquid-gas transitions, the order parameter is the density difference ρ L ρ G, and its phase diagram is a part of the phase diagram we have shown earlier. For the binary liquid mixture, the order parameter is the difference between the mole fraction of the mixture and that at the critical point, x x c. he phenol-water system is immiscible for < c and miscible for > c. In the magnetic case, the order parameter is the magnetization (M z ) (i.e.,the number of up electron spins minus down spins). At temperatures slightly below c, M z is non-zero and reflects the existence of a permanent magnetic moment. he curvature of the order parameter vs. temperature near the critical point is the same in all three cases (among others not mentioned as well) Critical exponents he critical exponents measure how various properties vanish or diverge as the critical point is reached. Each of the exponents is defined on a path to criticality, the α exponent, C p c α along ρ = ρ c α = 0.1 (11) 3

4 the β exponent ρ ρ c c β along the coexistence curve β = 0.35 (1) the δ exponent c ρ ρ c δ along = c δ = 4.5 (13) the γ exponent κ = c V ( ) V 1 c γ along ρ = ρ c, γ = (14) Sketched in Fig.(6.54) are representative behaviors of thermodynamic quantities as they approach the critical point. erfectly ordinary quantities, C p and κ, actually diverge at c, ρ c. Simple thermodynamic arguments capture some of the physics of the critical point, but they are not quantitative descriptors of the path to the critical point, as we shall see. 1.4 Landau theory predictions for critical exponents In the Landau approach, one assumes that a power series can describe the path to a critical point. Although this turns out to be wrong, it is so sensible and simple, that all transitions are usually analyzed by first quoting Landau theory (the so-called mean field theory) before using more sophisticated theories. hree exponents, β, δ and γ, are calculated using the equation of state. So, expand (, ρ) about the critical point, with ( ) ( = c ) V ( ) ( ( ρ) 3 ) + 6 ρ 3 ρ V,ρ ( ) + ρ + ( ) ( ρ + 1 ) ( ρ) (15) ρ ρ = c, ρ = ρ ρ c (16) At the critical point, the first two derivatives of with respect to ρ vanish, and then c = a 1 ( ρ) 3 + a + a 3 ρ (17) Along a critical isotherm, = 0 and = a 1 ( ρ) 3 or δ = 3. Along the ρ = ρ c path, and γ = 1. κ = 1 ρ ( ) ρ = 1 ( ) 1 ρ c 3a 1 ρ + a 3 1 (18) 4

5 he exponent associated with phase coexistence, β follows from the chemical potential. ndµ = V d, or ρdµ = d = ( ) dρ (19) ρ Since the chemical potentials of coexisting phases must be equal, then ρd( µ) = ( ρ ) dρ = ( 3a 1 ( ρ) + 3 ) ρ = 0 (0) Being that dρ 0, then the quantity in brackets must vanish and and β = 1. 3a 1 ( ρ) + a 3 = 0 ρ 1/ (1) hese predictions were made without the assumption of an equation of state, they are amazingly general and yet wrong. We need a new theory. he approach that works here is very different and does not yield a simply interpretable equation of state. Rather it focusses on the statistical fluctuations in the system that occur on all length scales. Emerging from this approach (called the renormalization group) is a scale invariance, and this scale invariance introduces an iterative transform that generates the power laws for the critical exponents. he issue of scale invariance arises at the critical point and outside the small critical domain, a Landau analysis of phase transitions is perfectly adequate. he exponents derived from the renormalization group explain all the fractional exponents derived from experiment. 1.5 Liquids in equilibrium with vapor Next, we turn our attention to liquids in equilibrium with their vapor. he chemical potentials of the liquid and the vapor must be equal, µ A,gas (, ) = µ 0 A( ) + R ln A = µ A,liq (, ) () where the standard state is, as before, a pure gas at one bar pressure. Whereas before we used Dalton s law A = y A (3) now we use Raoult s law A = x A 0 A (4) where x A is the mole fraction of A in the liquid and 0 A is the vapor pressure of pure liquid A. herefore µ A,liq (, ) = µ 0 A( ) + R ln 0 A + R ln x A = µ A,liq (, 0 A) + R ln x A (5) 5

6 In the second equation given immediately above, µ A,liq (, 0 A) is the chemical potential of pure liquid A, or alternatively the chemical potential of the gas above pure liquid A. Since the total vapor pressure is the sum of the partial pressures from A and B, = x A 0 A + x B 0 B (6) his is a nice result, but rarely true, since in a liquid that is primarily A, it is not clear why the vapor pressure of B should arise entirely from the unlikely B B interactions, instead of the more common A B interactions. Instead we find that the dilute component satisfies Henry s law B = x B k H,B (7) and k H,B is the Henry s law constant. For solutions dilute in B, with = x A 0 A + k H,B x B (8) µ B,liq = µ 0 B( ) + R ln k H,B + R ln x B (9) Shown in Fig.(6.6a) is the variation of the vapor pressure of component A over the entire range of liquid mole fractions; in Fig.(6.6b) components A and B are shown along with the total pressure (the sum of the A and B parts). he dashed line is the purely linear Raoult s law of ideal solutions, and the combined curve shows a positive deviation from Raoult s law. 1.6 Colligative properties In principle the colligative properties (boiling point elevation, freezing point depression and osmotic pressure) arise from the number of solute particles in solution rather than their identity, and to the lowest level of approximation that is true. Of course, as we look more carefully non-ideality enters in just the same way as the second and third virial coefficients complicated the treatment gases originally considered to be ideal. Analysis of the coligative properties starts with the chemical potential, dµ = Sd + V d (30) he thermal effects follow from the S-term and the osmotic pressure from the V-term. We begin with the thermal effects. Consider the equations for phase equilibrium of the solvent, species A, in the presence of a small amount of solute, µ A,gas (, ) = µ A,liq (, 0 A) + R ln x A = µ A,solid (, ) (31) Boiling point elevation is concerned with gas liquid equilibrium, the left hand side of the above equation, and liquid solid with freezing point depression. he change in boiling and freezing points, owing to the presence of solute, is shown in Fig.(6.7) by the shift of intersections of the chemical potential lines for the liquid mixture with that of the pure A solid, and the pure A gas. 6

7 1.6.1 Boiling point elevation and freezing point depression Here we consider only boiling point elevation since freezing point depression follows by relabeling a single quantity. Write µ gas (, ) µ liquid (, ) = R ln x A (3) Expand the left hand side in a power series around the normal boiling point of the solvent, B µ gas (, ) µ liquid (, ) ( ) ( ) µg (, ) µ L (, ) (µg µ L )/ +( B ) B At the normal boiling point, the first term vanishes, and being that then ( ) (G/ ) B = R B ln x A H(L G) = R ln x A (33) = H/ (34) R B x B H(L G) k Bm B (35) We have approximated the log term for small x B (ln(x A = 1 x B ) x B ) and changed variable to molality (moles of solute per kilogram of solvent). When we do the same for freezing point depression, we find F = R F x B H(L S) = k fm B (36) he minus sign arises because the L,S transition is exothermic. For water, k f = 1.86K(kg/mol), k b = 0.51K(kg/mol) Osmotic pressure Osmotic pressure is the additional pressure which must be exerted on a solution to prevent the flow of pure solvent through a solvent permeable membrane to further dilute the solution. Here we shall calculate the osmotic pressure and suggest how it is used to determine a molecular weight. Consider the arrangement shown in Fig.(6.8). If the pressure exerted on the solution, equals the pressure exerted on the solvent, 0, then since µ(,, x 1 ) < µ(, ), solvent will flow in the direction of the smaller chemical potential, further diluting the solution. o stop the flow, we must increase the pressure on the solution to increase its chemical potential. 7

8 π = 0 is the extra pressure that equalizes the chemical potential in both compartments. We want to achieve the equilibrium, µ(,, x) = µ(, 0 + π, x 1 ) = µ(, 0 ) (37) but, ( ) µ µ(, 0 + π, x) = µ(, 0 ) + d + R ln x 1 = µ(, 0 ) (38) 0 where ( ) µ = V 1 is the partial molar volume of the pure solvent. After performing the integral, and under the assumption that the solution is dilute, π V 1 + R ln x 1 = 0 π R x V 1 R n n 1 n 1 V R c (39) where c is the molar concentration of solute, x n /n 1, V 1 V/n 1. his expression for the osmotic pressure is known as the van t Hoff law. Since the pressure is measurable and the concentration can be determined in moles/liter, given the mass of substance, the molecular weight follows. 1.7 Regular solutions Roughly ninety years ago, J. H. Hildebrand of U. C. Berkeley provided a simple modification of the Gibbs energy of mixing of liquid solutions that could account for phase separation and critical phenomena. His derivation had myriad weak points, but in the final analysis, given what we know now from Density Functional heory, Hildebrand s ideas have proven to be insightful and exceedingly useful. His theory was called the theory of regular solutions, obviously not a Madison avenue label. Without derivation we present the Gibbs energy of mixing as G = G mix (, ) G pure (, ) = n (wx 1 x + R (x 1 ln x 1 + x ln x )) (40) G pure = n 1 (µ 0 1( ) + R ln 0 1 ) + n ( ) (41) where µ 0 1( ) is the standard state for a gas at one bar pressure, 1 0 is the vapor pressure of pure liquid 1, x 1 is the mole fraction of component 1 in the liquid phase. Essentially, the G consists of the entropy of mixing term S = nr i x i ln x i (4) plus an enthalpy term, H = nx 1 x w (43) where w is an interaction term to be discussed later. When we derive the chemical potential of component 1 in the mixture, we find that ( ) G µ 1 (, ) = = µ 0 n 1( ) + R ln ( 1 0 x 1 exp(wx /R ) ) (44) 1,,n 8

9 so that we can associate the pressure of component 1 in the mixture to be 1 = x exp(x w/r ) (45) Note: as x 0, 1 x 1 1 0, Raoult s law emerges, whereas as x 1, 1 x exp(w/r ) = k H,1 x 1 Henry s law is reclaimed. Consider Fig.(6.9) where we sketch G vs. liquid phase mole fraction x nr 1. For W < 1, R there is a minimum at x 1 = 1 corresponding to complete mixing. his location maximizes w the entropy of mixing. However, for = 1, the curve has a flat spot at x = 1. For w > 1, R R two minima develop. hese two minima correspond to a phase separation with two solutions having mole fractions greater than and less than one-half. Not surprisingly, the temperature at which w = 1 or = w is the critical temperature, for temperatures below R R c = w R give rise to phase separation. By adding a simple enthalpy correction to the free energy of mixing, we have opened the avenue to critical phenomena and liquid immiscibility Analysis of the critical domain For simplicity, select 0 1 = 0 = 0 1 wherein the total vapor pressure of the solution becomes ( x1 exp(wx /R ) + x exp(wx 1/R ) ) (46) he critical point is defined by two criteria, ( ) x 1 n, = ( ) x 1 n, = 0 (47) which yields two solutions, x = x c = 1 c = w R which we anticipated from Fig.(6.8). Next, expand g = G nr in a power series about x 1 = x c, ( g(x, ) = g(x c, c ) + 1 c (48) ) (x x c ) (x x c) 4 + (49) When we minimize g with respect to x, the so-called order parameter, we find ( ) g x n, ( = 4 1 c ) (x x c ) (x x c) 3 = 0 (50) When > c the solution to this equation is x = x c = 1, entropy of mixing dominates. he lowest free energy occurs for a 50:50 mixture. When < c, i.e., below the critical temperature, two new solutions are possible from ( 1 ) ( ) c (x x c) c = 0 or x x c = ± 4 (51) 9

10 and these solutions for x correspond to the phase separation in immiscible phases with x x c. Furthermore, when we compare the above equation with that for gas-liquid phase coexistence in a Landau theory, with our new result ρ β β = 1 (5) x β (53) we find the same β, and this is the universality we suggested previously. So what is w? w follows from statistical mechanics and there we find we can associate it with an energy difference, w = E 1 E 11 E (54) If the interaction between molecules 1 and is sufficiently weak, and weaker than the sum of the 1, 1 and, interactions, phase separation can occur to keep 1 s with their partners and s with their partners. Entropy always favors complete mixing and miscibility, but if solvation effects are unfavorable, phase separation occurs. hase separation in regular solutions is accomplished through one parameter, w, and the simplicity of this treatment is amazing. 10

5.4 Liquid Mixtures. G i. + n B. = n A. )+ n B. + RT ln x A. + RT ln x B. G = nrt ( x A. ln x A. Δ mix. + x B S = nr( x A

5.4 Liquid Mixtures. G i. + n B. = n A. )+ n B. + RT ln x A. + RT ln x B. G = nrt ( x A. ln x A. Δ mix. + x B S = nr( x A 5.4 Liquid Mixtures Key points 1. The Gibbs energy of mixing of two liquids to form an ideal solution is calculated in the same way as for two perfect gases 2. A regular solution is one in which the entropy