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Lecture 14. Equilibrium Between Phases

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Lecture 14. Equilibrium Between Phases Lecture 14. Phases of Pure Substances (Ch.5) P Up to now we have dealt almost exclusively with systems consisting of a single phase. In this lecture, we will learn how more complicated, multi- phase systems can be treated by the methods of thermodynamics. The guiding principle will be the minimization of the Gibbs free energy in equilibrium for all systems, including the multi- phase ones. T The generic phase diagram of a substance in the P-T coordinates is shown above. Every point of this diagram is an equilibrium state; different states of the system in equilibrium are called phases. The lines dividing different phases are called the coexistence curves. Along these curves, the phases coexist in equilibrium, and the system is macroscopically inhomogeneous. All three coexistence curves can meet at the triple point – in this case, all three phases coexist at (Ttr , Ptr). The Coexistence Curves Along the coexistence curves, two different phases 1 and 2 coexist in equilibrium (e.g., ice and water coexist at T = 00C and P = 1bar). The system undergoes phase separation each time we cross the equilibrium curve (the system is spatially inhomogeneous along the equilibrium curves). Any system in contact with the thermal bath is governed by the minimum free energy principle. The shape of coexistence curves on the P-T phase diagram is determined by the condition: GPTGPT,,= and, since 1()2 ( ) GN= μ PTPTμ1( ,,) = μ2 ( ) - otherwise, the system would be able to decrease its Gibbs free energy by transforming the phase with a higher μ into the phase with lower μ. Two phases are in a state of diffusive equilibrium: there are as many molecules migrating from 1 to 2 as the molecules migrating from 2 to 1. - as for any two sub-systems in equilibrium Also for equilibrium TT1= 2 between the phases: - the phase boundary does not move PP1= 2 Though G is continuous across the transition, H demonstrates a step-like behavior: G N= Uμ = PV + −= TS− HHTSΔ TS= Δ (different phases have different values of the entropy) Example: the Gas-Liquid Transformation ⎛ ∂G ⎞ ⎛ ∂F ⎞ ⎛ ∂U ⎞ ⎛ ∂S ⎞ μ = ⎜ ⎟ = ⎜ ⎟=F[] ≡ U − TS⎜ = ⎟ −T⎜ ⎟ ⎝ ∂N ⎠TP, ⎝ ∂N ⎠TV, ⎝ ∂N ⎠TV, ⎝ ∂N ⎠TV, Gas: ∂U/∂N term is sall and positive (kinm), e. energy of a single moleculT(∂S/∂N) term is large and positive ⇒ μ is negative, and rapidly decreases with increasing T. Liquid: ∂U/∂N term is negative (ttraction abetween molecules), T(∂S/∂N) term is smaller than that in gas and positive ⇒ μ is also negative, and slowly increases with decreasing T . μ Table on page 404 (a very useful source of inforation) provides the values of mH and G for 0 liquid different phases of many substances. The data are provided per mole, at T=298 K and P=1 bar. For example, let’s check that at the boiling point, gas the values of G for liquid water and water vapor are equal to each other: phase T ⎛ ∂G ⎞ transforationm = −S GGTTS≈ − − ⎜ ⎟ ( PNT) ,, ( PNT) ,, 0 ( 0 ) ⎝ ∂T ⎠PN, 237G≈ − 10J × / mol3 75 − K 70 × J /( K ⋅ mol ) ≈3 J mo 242 ×l 10 / S(water) = 70 J/K ( liqTK) =373 S(vapor) = 189228 J/K .G≈ 6 − 10J × / mol3 75 − K × 189 J K /( ⋅ molJ ≈ )3 mol × 242 10 / ( vapTK) =373 Phases of Carbon The phase equilibrium on the P,T-plane is determined by GPTGPT1(),,= 2 ()or PTPTμ1( ,,) = μ2 ( ) At normal conditions, graphite is more stable than diamond: G(graphite) = 0, G(diamond) = 2.9 kJ (diamonds are not forever...). What happens at higher pressures? ⎛ ∂G ⎞ = V GGPPV≈ + − ⎜ ⎟ ( TN) , ( TNP) ,, 0 ( 0 ) ⎝ ∂P TN⎠ , - since the molar volume of graphite is greater than the molar volume of diamond, G(graphite) will grow with G pressure faster than G(diamond) [we neglected V = V(P)] ond diam te hi D. becomes more stable than G. only at P > 1.5 MPa ap 2.9 kJ gr T = 300K With increasing T, the pressure range of graphite 12 stability becomes wider: P (MPa) G diamo ⎛ ∂G ⎞ nd ⎜ ⎟ = −S GGTTS() ≈ ( ) − ( − ) 2.9 kJ PNT,, PNT,, 0 0 g ⎝ ∂T ⎠ rap PN, hite P = 1 bar S(graphite) = 5.74 J/K, S(diamond) = 2.38 J/K, 300 800 1300 T (K) The First-Order Transitions Because molecules aggregate differently in different phases, we have to provide (or remove) energy when crossing the coexistence curves. The energy difference is called the latent heat ; crossing the coexistence curve, the system releases (absorbs) a latent heat L. The entropy of the system changes abruptly: T1 The transitions which displays a δ Q L jump in entropy and a latent heat ΔS = = associated with this jump are T T S gas T=T1 = const called the first-order phase transitions. gas+liquid critical liquid S point S liquid+solid gas beyond critical solid mixed point, gas is phase P indistinguishable from liquid The “evaporation” L is generally greater than the “melting” L (the Sliquid disorder introduced by evaporation is greater than that TTC temperature introduced by melting). Q: Can the critical point exist along the melting coexistence curve? The First-Order Transitions (cont.) Note that in the first-order transitions, the G(T) curves G have a real meaning even beyond the intersection point, this results in metastability and hysteresis. There is usually an energy T barrier that prevents a S G P,N = const transition occurring from the higher μ to the lower μ phase (e.g., gas, being cooled below ΔS = L/T solid Ttr does not immediately liquid condense, since surface energy makes the formation of T gas very small droplets ⎛ ∂S ⎞ C = T energetically unfavorable). P ⎜ ⎟ T ⎝ ∂T ⎠P,N (Pr. 5.9). L. water can exist at T far On the graph G(T) at P,N = const, lower than the freezing CP the slope dG/dT is always temperature: water in organic negative: cells can avoid freezing down ⎛ ∂G ⎞ 0 S = −⎜ ⎟ to –20 C in insects and down 0 ⎝ ∂T ⎠P,N to –47 C in plants. T Problem The entropy of water at atmospheric pressure and 1000C is 1.3 J/g·K, and the entropy of water vapor at the same T and P is 7.4 J/g K. (a) What is the heat of vaporization at this temperature? (b) The enthalpy of vapor under these conditions is 2680 J/g. Calculate the enthalpy of water under these conditions. (c) Compute the Gibbs free energies of water and steam under these conditions. (a) The heat of vaporization: L = TΔS = 373K×6.1 J/g·K=2275 J/g (b) The differential of enthalpy dH = TdS+VdP. Hence, Hwater = Hvapor –TdS=Hvapor –L= (2680-2275)J/g = 405 J/g (c) Since G = H-TS, Gwater = Hwater –TSwater = 405J/g - 373K × 1.3J/g·K = -80J/g Gvapor = Hvapor –TSvapor = 2680J/g - 373K × 7.4J/g·K = -80J/g The Second Order Transitions Second-order transition The vaporization coexistence curve ends at a point called the G critical point (TC, PC). As one moves along the coexistence curve toward the critical point, the distinction between the liquid phase on one side and the gas phase on the other gradually decreases and finally disappears at (TC, PC). The T-driven T phase transition that occurs exactly at the critical point is S called a second-order phase transition. Unlike the 1st- order transitions, the 2nd-order transition does not require ΔS=0 any latent heat (L=0). In the higher-order transitions (order-disorder transitions or critical phenomena) the entropy is continuous across the transition. The specific heat T C =T(δS/δT) diverges at the transition (a cusp-like “λ” ⎛ ∂S ⎞ P P = TC ∞→ singularity). P ⎜ ⎟ ⎝ ∂T ⎠ ,NP Whereas in the 1st-order transitions the G(T) curves have a real meaning even beyond the intersection point, nothing of CP the sort can occur for a 2nd-order transition – the Gibbs free energy is a continuous function around the critical temperature. T The Clausius-Clapeyron Relation GPTGPTP 1( ,,) = 2 ( ) Along the phase boundary: phase ΔP boundary GPTGPT1(),,= 2 () PTPTμ1( ,,) = μ2 ( ) Consider two distinct displacements along the coexistence curve, one immediately above the curve (in phase 1), the other immediately below the curve, in phase 2. Because the ΔT T chemical potentials remain equal along the curve, dμ1 = dμ2 For the slope of the boundary we have: S dT− V1 + dP1 = − S2 + dT2 V dP - the slope is determined by the entropies and volumes of the SPTSPTdP ()(),,− = 1 2 two phases. The larger the difference in entropy between the VPTVPTdT 1()(),,− 2 phases – the steeper the coexistence curve, the larger the difference in molar volumes – the shallower the curve. (compare the slopes of melting and vaporization curves) dP LT( ) Since S1 - S2 = L/T (L is the latent heat), we = (applies to all arrive at the Clausius-Clapeyron Relation : dTTVTΔ () coexistence curves) Example: -since Vgas > Vliq , and L > 0 for the “liquid→gas” transformation, the boiling temperature increases with pressure. The “freezing” temperature with increasing TVTVTdT [ gas ()− liq ()] = pressure either increases or decreases, depending on the dP LT() 3 sign Vliq – Vsolid (exception – He). Problem 1 kg of water at 200C is converted into ice at -100C (all this happens at P = 1 bar). The latent heat of ice melting Ltmel = 334 kJ/kg, the heat capacity of water at constant pressure is 4.2 kJ/(kg·K) and that of ice 2.1 kJ/(kg·K). (a) What is the total change in entropy of the water-ice system? (b) If the density of water at 00C is taken as 10% greater than that of ice, what is the slope of the melting curve of ice at this temperature? Give both sign and size.
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