Phase transition. Asaf Pe er Background
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1 Phase transition Asaf Pe er 1 November 18, Background A hase is a region of sace, throughout which all hysical roerties (density, magnetization, etc.) of a material (or thermodynamic system) are essentially uniform. Well known examles are gaseous hase, liquid hase and solid hase. During a hase transition of a given medium certain roerties of the medium change, often discontinuously, as a result of some external condition, such as temerature, ressure, and others. For examle, a liquid may become gas uon heating to the boiling oint. We are interested in understanding the conditions under which a material undergoes a hase transition. In other words, we are interested in finding the equilibrium conditions between two hases of a medium. 2. Equilibrium conditions Let us look at an isolated system, which contains a medium which is in an equilibrium state between two hases. Being isolated, the energy, volume and number of articles are conserved, E 1 +E 2 = E V 1 +V 2 = V (1) N 1 +N 2 = N where the subscrits 1 and 2 reresent the two hases (see Figure 1). Sincethesystemisisolated,theequilibriumisachievedwhentheentroyS = S 1 (E 1,V 1,N 1 )+ S 2 (E 2,V 2,N 2 ) is maximized. ( S1 ds = ds 1 +ds 2 = de 1 + S ) ( 2 S1 de 2 + dv 1 + S ) ( 2 S1 dv 2 + dn 1 + S ) 2 dn 2 = 0 E 1 E 2 V 1 V 2 N 1 N 2 (2) 1 Physics De., University College Cork
2 2 Fig. 1. Phase transition. andsincee,v andn areconstants,de 1 = de 2,etc.,andthustheconditionforequilibrium is ( S1 ds = 0 = S ) ( 2 S1 de 1 + S ) ( 2 S1 dv 1 + S ) 2 dn 1 (3) E 1 E 2 V 1 V 2 N 1 N 2 Since E 1, V 1 and N 1 are indeendent variables, each of the terms in arenthesis must vanish for the system to be in equilibrium. he first two are familiar: they follow immediately from equilibrating the temerature and ressure using the entroy-based definitions: ( ) 1 S, (4) E hus the requirement 1 = 2 is written as V,N 1 = 2 S 1 E 1 = S 2 E 2 (5) Similarly, equilibrating the ressure using the entroy-based definition of the ressure, ( ) S, (6) V leads to hus, we are left with the 3 rd condition, E,N 1 = 2 S 1 V 1 = S 2 V 2 (7) S 1 N 1 = S 2 N 2 (8)
3 3 Its hysical meaning is that there is no assing of molecules between one hase and the other. We can use this to define the chemical otential µ i of each hase by ( ) Si µ i i (9) N i E i,v i hus, the condition in Equation 8 is simly µ 1 = µ 2 (10) In equilibrium, the chemical otential of the two hases must be the same. Note that the chemical otential is really a measure of the energy that is associated with the change in number of molecules in the system. Such a change can result from chemical reactions, hence the name Extension of the thermodynamic relations Let us leave for the moment the two-hase system, and look at a single hase only. We can write S = S(E,V,N). hus, ( ) ( ) ( ) S S S ds = de + dv + dn (11) E V,N V E,N N E,V or his can be re-arranged to write ds = 1 de + 1 dv 1 µdn (12) de = ds dv +µdn (13) Equation 13 is a generalization of the fundamental thermodynamic relation for a system in which the article number can change. Using this result in the definitions of Helmholtz and Gibbs free energies, We can write F = E S G = E +V S df = de ds Sd dg = de +dv +V ds Sd df = Sd dv +µdn dg = Sd +V+µdN. (14) (15)
4 4 We can thus exress the chemical otential in various ways deending on which of the variables (, or V) are held constant: ( ) ( ) F G µ = = (16) N,V =const N,=const We now write F and G as functions of a new arameter - N. Out of these two functions, the Gibbs free energy, G = G(,,N) is an extensive variable: namely, it is linearly roortional to the article number, N. G(,,N) = NG(,,1) Ng(,), (17) where g is the Gibbs free energy er article. From Equations 16 and 17 it follows immediately that µ = g(,) (18) hus, the chemical otential is the Gibbs free energy er article, and the condition for hase equilibrium is µ 1 = µ 2 g 1 (,) = g 2 (,) (19) 3. Equilibrium conditions (II) Exerimentally, it is difficult to deal with isolated systems. It is easier to handle systems at constant ressure and temerature. Let us determine the condition for hase equilibrium under these conditions. We know that for a system at fixed temerature and ressure, the condition for equilibrium is that the Gibbs free energy is minimized. hus, and and using we get dg = [ N 1 g 1 +N 2 G = G 1 +G 2 = N 1 g 1 (,)+N 2 g 2 (,) (20) dg = N 1 dg 1 +g 1 dn 1 +N 2 dg 2 +g 2 dn 2 = 0, (21) dg 1,2 = g 1,2 d + g 1,2, (22) ] [ g 2 g 1 d + N 1 +N 2 ] g 2 +g 1 (,)dn 1 +g 2 (,)dn 2 (23)
5 5 InequilibriumGisaminimum, anddg = 0, undertheconditionsthat anareconstants, thus = d = 0. his leaves us with dg = g 1 (,)dn 1 +g 2 (,)dn 2 = 0. (24) Furthermore, thetotalnumberofarticlesisconserved, N 1 +N 2 = N, andthusdn 1 = dn 2. his imlies that dg = [g 1 (,) g 2 (,)]dn 1 = 0. (25) We thus retrieve again the Equilibrium condition in Equation 19, g 1 = g 2. We thus find that the hase equilibrium condition for a system held at constant temerature and ressure is the same as that of an isolated system. his should not be surrising: for a system to be in equilibrium between two hases, they must be at the same ressure, temerature and chemical otential, irresective of the alied (external) constraints. 4. Imlications of the equilibrium conditions he equilibrium condition in Equation 19, g 1 = g 2 defines a curve in lane. Recall that the condition for equilibrium is a minimization of Gibbs free energy. hus, if the system is in a oint that does not lie on this curve, it means that the minimum of Gibbs free energy is achieved if all the substance molecules are in hase 1 (namely, N 1 = N, N 2 = 0, G = N 1 g(,)), or hase 2. he curve g 1 = g 2 thus divides the (,) lane into regions where one or the other hase reresents a stable equilibrium state. In is only on the curve that the two hases can coexist in equilibrium (see Figure 2). his curve is called a hase equilibrium curve. Let us now consider equilibrium of 3 different hases (solid, liquid and vaor) of a onecomonent system. Reeating the same calculation as we have done before, we obtain the equilibrium condition g 1 (,) = g 2 (,) = g 3 (,) (26) Equations 26 reresent the intersection of two curves: g 1 = g 2 and g 2 = g 3 in the (,) diagram. his is known as the trile oint. It is shown in Figure 3, which is known as the hase diagram of the system Clearly, at the trile oint, all three hases are in equilibrium with each other. Pure substances may be caable of existing in more than one allotroic form (e.g., diamond and coal), in which case they will have several trile oints. his is illustrated in Figure 3 (right).
6 6 Fig. 2. emerature - Pressure diagram of a hase equilibrium curve (g 1 = g 2 ). he three hase equilibrium curves divide the (, ) lane into three regions in which the solid, liquid and gaseous hases resectively are the stable state. here are in addition meta-stable states (e.g., suercooled liquids), but these are not stable. 5. he Clausius-Claeyron Equation he Clausius-Claeyron equation is an equation describing the hase equilibrium curve, namely the sloe /d at any oint along the curve. hink of two nearby oints along the hase equilibrium curve. We know that g 1 (,) = g 2 (,) g 1 ( +d,+) = g 2 ( +d,+). (27)
7 7 Fig. 3. Left: A hase diagram of a one-comonent system ossessing one trile oint. Right: Schematic hase diagram for a sulhur. A sulhur can exist in two different crystalline forms, rhombic and monoclinic, and has three trile oints. By aylor exanding the second Equation and subtracting the first one, ( ) ( ) ( ) ( ) g1 g1 g2 g2 d + = d + (28) or [ ( g1 ) and thus the sloe of the curve is d = ( ( ) ] g2 d = ( g2 g 2 ) ) ( g 1 ( g 1 [( ) g2 ) ) = ( ) g ( ) ( ) ] g1 (29) In order to roceed, we recall that the change in Gibbs free energy is given by Equation 15 for each hase searately: where µ i = g i. Also, G = Ng, from which g (30) dg i = S i d +V i +µ i dn i. (31) dg = Ndg +gdn = Sd +V+µdN dg = S N d + V N. (32)
8 8 his result enables us to write: ( ) gi = S i N i ; ( ) gi = V i N i (33) Using the result of Equation 33 in Equation 30 enables us to write S2 d = N 2 S 1 N 1 V 2, (34) N 2 V 1 N 1 and if we refer to the same amount of substance in each hase, then N 1 = N 2, and we have d = S 2 S 1 = S V 2 V 1 V Equation 35 is known as the Clausius-Claeyron equation. For every hase change which is accomanied by a change in entroy S, there is emission or absortion of heat - known as the latent heat, L. he entroy change in hase transition at temerature is S = S 2 S 1 = L Using this in Equation 35, we find that in a hase transition d = L V where recall again that L and V refer to the same amount of substance. he difference in entroy between the two hases imly that hase change involves a latent heat. his tye of hase change is called first order transition. Many known examles are like that - e.g., solid-liquid-vaor hase change, or allotroic transitions (e.g., grey to white tin). here are hase changes in which the entroy continuously changes, so there is no latent heat involved. hese do involve higher order derivatives of g i and other thermodynamic quantities, such as the heat caacity. We will not consider those here. For the rocesses of melting, evaoration and sublimation, S > 0. his is easily understood as the change is from an ordered hase to a less ordered one. In vaorization or sublimation, the density decreases, and so V > 0. hus, from the Clausius-Claeyron equation, for vaorization and sublimation, (35) (36) (37) d = S V > 0 (38) Most substances exand in melting; however, there are excetions - the most notable one is, of course, water, which contract when melting. For these, /d < 0. he hase diagrams thus looks like resented in Figure 4.
9 9 Fig. 4. General aearance of a hase diagram. Left: Solid exands uon melting, V > 0. Right: Solid contracts uon melting, V < 0. is 6. Alications of the Clausius-Claeyron Equation 6.1. Pressure deendence of the melting oint Consider the transition between ice and water. We know that at 0 C, the latent heat L = J/kg Furthermore, the volume (er gram) of ice and water are V ice = m 3 kg 1 V water = m 3 kg 1 (39) and therefore in melting V = V water V ice = m 3 kg 1. Using these in the Clausius-Claeyron Equation, we find d = L V = = Nm 2 K 1 = 134 atmk 1 (40) his means, that as the ressure increases, the melting oint decreases. For examle, an increase in 1000 Atmosheres, lowers the melting oint by 7.5 C. It is this effect that is resonsible for the motion of glaciers. Consider a glacier of
10 10 deth (thickness) d. he ressure at the bottom is = F A = ρgd (41) with ρ ice = 917 kgm 3 and deth of 800 m (e.g., Baring glacier in Alaska), resulting in = Pa = 71 atm, from which we find that the melting oint decreases by d 0.5 K (42) his imlies that the deeer arts of the glacier melt due to the ressure, enabling the glacier to flow. hey freeze again when the ressure decreases Pressure deendence of the boiling oint Since the volume of the gas is always larger than that of liquid, in evaoration V is always ositive. hus, increasing the ressure always increases the boiling oint. Consider again water as an examle, the latent heat of vaorization is L = J/kg. At = K, and = 1 atm, the volume (er gram) of liquid water and gas (water vaor) are V water = m 3 kg 1 V gas = m 3 kg 1 (43) (note that V gas 1000 V water!). hus, d = L V = = Nm 2 K 1 = 27 mmhgk 1 (44) At the to of the everest mountain (height 8 km) the ressure is Nm 2 (as oosed to Nm 2 at sea level). hus, the temerature difference for water evaoration at the to of mount everest is hus, water boils at 80 C at this height = 18 C (45)
11 Evaoration and sublimation We can use the following aroximations when calculating evaoration and sublimation using the Clausius-Claeyron equation: (i) Since the volume of a gas is so much larger than that of a solid or liquid, we may aroximate the change in volume as V = V 2 V 1 V 2. (ii) We can assume that the vaor behaves like a erfect gas, for which the equation of state is V = nr. or Combined into the Clausius-Claeyron equation, one finds We can write this as d = L V = L = L V 2 nr2. (46) = L d (47) nr 2 ln = L +Const, nr = C 1 e L nr = C 1 e L M (48) R where L M = L M is the latent heat er mole. hus, for small temerature change, equation 48 gives the corresonding vaor ressure. For large change in, the latent heat may change, and the aroximation no longer holds. 7. he critical oint As the ressure and temerature are increased along the transition curve between liquid and vaor (the vaor ressure curve), one reaches a critical oint. he increase of the ressure and temerature results in a decrease of the latent heat and the volume change V between the two hases, and thus as one reaches the critical oint they become zero (see Figure 5). At temeratures below the critical temerature c (corresonding to the critical oint), the fluid can co-exist in two states with different secific volumes (liquid hase and gas hase). Above c the substance exists in one fluid hase only. Inordertounderstandthat, letusconsidertheisothermsinav diagram(seefigure6). Assume that our system is comosed of a material initially at temerature 1, corresonding to oint A in Figure 6.
12 12 Fig. 5. General aearance of a hase diagram. Shown are the vaor ressure curve and the critical oint. 1. If the system is comressed isothermally, then the the ressure and the density increase where the volume decreases until the vaors reaches saturation; this occurs at oint A If the volume is continued to be reduced, condensation occurs at constant ressure. At this state, there are two coexisting hases, gas and liquid. 3. Once the material reaches oint A 1, all the vaor is condensed (became liquid). 4. Further comression (volume reduction) requires enormous ressure, due to the low comressibility of liquids. Now, if the initial temerature is higher than 1, the condensation interval becomes shorter, and eventually, for an isotherm at the critical temerature, c, it disaears. Mathematically, the critical oint is defined by ( ) ( ) 2 = = 0 (49) V = C V 2 = C
13 13 Fig. 6. Schematic m of the isotherms of a fluid. V is the secific volume (=volume er gram). C is the critical oint. Above C, the isotherms monotonically decrease in the (,V) lane, imlying that there is no distinct hase change. he roerties of the system change continuously.
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