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Thermodynamics, Part 2

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Thermodynamics, Part 2 Thermodynamics, Part 2 November 12, 2014 1 Phase equilibria Consider the three phase diagrams in Fig.(6.1) (a) CO2, (b) water; and (c) helium. CO2 is a 'normal' material with a typical and simple phase diagram. At 1 bar pressure and at 200K, CO2(s) )* CO2(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 P vs: T line has a negative slope. The 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. The two conditions for two phase equilibrium are P (ρ1;T ) = P (ρ2;T ) µ(ρ1;T ) = µ(ρ2;T ) (1) viz., the chemical potential in phase 1 characterized by density ρ1 must equal the chemical potential for that same species in phase 2 at density ρ2. Likewise a similar situation prevails for the pressure. The µ-condition implies that it must take equal work to insert a particle in phase 1 as it does to insert it into phase 2. The P condition implies that the forces along the 1-2 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? The 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. These two equations, for µ and P , when solved simultaneously give the densities of the coexisting phases. 1.1 Phase boundary lines in P, T diagrams In an equilibrium between the liquid and solid phases, ¯ ¯ ¯ ¯ dµL = −SLdT + VLdP = dµS = −SSdT + VSdP (2) 1 where the over-bar denotes per mole. Rearrange to form the slope of P vs. T along the coexistence line, ! ¯ ¯ ¯ ¯ @P SS − SL ∆S(L ! S) ∆H(L ! S) = ¯ − ¯ = ¯ ! = ¯ ! (3) @T coex VS VL ∆V (L S) T ∆V (L S) For the freezing transition, ∆V is small and negative, and since the process is exothermic, both ∆V; ∆H < 0. Thus the P vs. T 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, RT ∆V¯ (L; S ! G) ' V¯ ' (4) G P and ! @P ∆H¯ (L; S ! G)P = 2 (5) @T coex RT which integrates to ¯ ! ¯ ! ( ) dP dT ∆H(L; S G) ∆H(L; S G) 1 − 1 = 2 ln(P2=P1) = (6) P RT R T1 T2 This equation determines the shape of the P vs: T curve and determines the vapor pressure at one temperature, given a value at another. 1.2 T and P dependence of the chemical potential For a system with one component, dµ = −SdT¯ + V¯ dP (7) Hence the slope of µ vs: T is the negative of the entropy and the slope of µ vs: P is the molar volume. Shown in Fig.(6.2a) is the variation of µ vs: T 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¯ dP , we shift the chemical potential phase boundaries as shown in Fig.(6.2b). In ¯ ¯ Fig.(6.2b) we implement VL > VS, 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. The molecular structure and dynamics of liquids is a classic many body problem 2 where molecular order is neither chaotic as in the gas phase nor regular or crystalline or regular as in the solid phase. The 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. The 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 = −SdT¯ + V¯ dP (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=T 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 @2µ = ! 1 (9) @P @P 2 !T T @S C = P ! 1 (10) @T P T In Fig.(6.4) we present plots of T 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 ρL − ρG, 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 − xc. The phenol-water system is immiscible for T < Tc and miscible for T > Tc. In the magnetic case, the order parameter is the magnetization (Mz) (i.e.,the number of up electron spins minus down spins). At temperatures slightly below Tc, Mz is non-zero and reflects the existence of a permanent magnetic moment. The curvature of the order parameter vs. temperature near the critical point is the same in all three cases (among others not mentioned as well). 1.3.1 Critical exponents The 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, −α Cp ' jT − Tcj along ρ = ρc α = 0:1 (11) 3 • the β exponent β jρ − ρcj ' jT − Tcj along the coexistence curve β = 0:35 (12) • the δ exponent δ P − Pc ' jρ − ρcj along T = Tc δ = 4:5 (13) • the γ exponent ! P @V 1 κ = − c ' along ρ = ρ ; γ = 1:1 ··· 1:4 (14) T j − jγ c V @P T T Tc Sketched in Fig.(6.54) are representative behaviors of thermodynamic quantities as they approach the critical point. Perfectly ordinary quantities, Cp and κT , actually diverge at Tc; ρc. 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. Three exponents, β; δ and γ, are calculated using the equation of state. So, expand P (T; ρ) about the critical point, ! ! ! ! @P @2P @P @2P P = P + ∆T + 1 (∆T )2 + ∆ρ + 1 (∆ρ)2 (15) c @T 2 @T 2 @ρ 2 @ρ2 ! V ! V T T @3P @2P 1 3 ··· + 6 3 (∆ρ) + ∆ρ∆T + @ρ T @T @ρ T,ρ with ∆T = T − Tc; ∆ρ = ρ − ρc (16) At the critical point, the first two derivatives of P with respect to ρ vanish, and then 3 P − Pc = a1(∆ρ) + a2∆T + a3∆ρ∆T (17) 3 • Along a critical isotherm, ∆T = 0 and ∆P = a1(∆ρ) or δ = 3. • Along the ρ = ρc path, ! ! 1 @ρ 1 1 ! 1 κT = = 2 (18) ρ @P T ρc 3a1∆ρ + a3∆T ∆T and γ = 1. 4 • The exponent associated with phase coexistence, β follows from the chemical potential. ! @P ndµ = V dP; or ρdµ = dP = dρ (19) @ρ T Since the chemical potentials of coexisting phases must be equal, then ! ( ) @∆P 2 ρd(∆µ) = dρ = 3a1(∆ρ) + 3∆T ∆ρ = 0 (20) @ρ T Being that dρ =6 0, then the quantity in brackets must vanish and 2 1=2 3a1(∆ρ) + a3∆T = 0 ) ∆ρ ' ∆T (21) 1 and β = 2 .
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