Thermodynamics of phase transitions

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Thermodynamics of phase transitions Katarzyna Sznajd-Weron Department of Theoretical of Physics Wroc law University of Science and Technology, Poland March 12, 2017 Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 1 / 37

The energy of the system can be changed by external forces that perform work Work dw performed by tension f that extends a metal rod by the length dx : dw = fdx. (1) External magnetic field h does the work (an increase of magnetization): dw = hdm. (2) Pressure p is the force that changes the volume by dv, and a corresponding work: dw = pdv (3) Chemical potential is the force that changes the number of particles N: dw = µdn (4) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 2 / 37

Thermodynamic (macroscopic) parameters Generalized coordinates defining the state of the system: distance X, magnetization M, volume V, the number of particles N Generalized external forces: tension f, magnetic field h, pressure p, chemical potential µ Can you find some common property of generalized coordinates? Can you find some common property of generalized forces? Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 3 / 37

Extensive and intensive parameters 1 Extensive (additive) the value of such a parameter for the whole system is equal to the sum of the parameters for subsystems making up the system. Examples of such parameters are distance X, magnetization M, volume V, the number of particles N 2 Intensive the value for the whole system is equal to the value of that parameter for each of the identical subsystems making up a given system. Examples of such parameters are temperature T, pressure p, chemical potential µ and... Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 4 / 37

Is the change of energy possible without work (change in the macroscopic coordinates)? Heat - the result of changes in microscopic motion Do we need all microscopic coordinates (coordinates of all particles) to describe these changes? We introduce a new generalized coordinate to describe the microscopic motion in a collective way Entropy S is a new generalized coordinate and corresponding generalized force? dq = TdS (5) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 5 / 37

Internal energy - conservation of energy State function (extensive): U = U(S, V, N): U = W + Q, (6) W - work over the system W > 0, U increases Q - heat added to the system p z = F n A F n = p z A L p L k A V p = AL p V k = AL k W = F n L k L p = p z A L k L p = p z V k V p < 0 Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 6 / 37

Four Postulates of Thermodynamics - traditional The Zero-th Law of Thermodynamics: If C is initially in thermal equilibrium with both A and B, then A and B are also in thermal equilibrium with each other. Conclusion: Two systems are in thermal equilibrium if and only if they have the same temperature. The First Law of Thermodynamics: U = W + Q The Second Law of Thermodynamics: It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body. The Third Law of Thermodynamics: Entropy S = 0 at T = 0 Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 7 / 37

Problems with thermodynamics Thermodynamics is an axiomatic theory. Is it a problem? Thermodynamics can be expressed in several equivalent formulations. It is not easy to see that they are equivalent! Problem - what is the assumption and what is the conclusion? Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 8 / 37

Four Postulates in Thermodynamics - after Callen Simple system: macroscopically homogeneous, isotropic and uncharged and large enough to neglect surface effects; not acted by electric, magnetic or gravitational fields Postulate I: There exist particular states (called equilibrium) of simple systems that, macroscopically, are characterized completely by the internal energy U, the volume V and the mole numbers N 1, N 2,..., N r of chemical components. Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 9 / 37

Four Postulates in Thermodynamics - after Callen Postulate II: There exists a function (called entropy S) of the extensive parameters of the extensive variables of any closed system, defined for all equilibrium states and having the following property: the values assumed by the extensive parameters in the absence of an internal constraint are those which maximize the entropy over the manifold of constrained equilibrium states. The key postulate equivalent to part of the Second Law of Thermodynamics. We say nothing about nonequilibrium states Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 10 / 37

Fundamental equations Internal energy as a function of the extensive parameters contains all the thermodynamic information about the system: U = U(S, V, N 1,..., N r ) (7) Entropy as a function of the extensive parameters contains all the thermodynamic information about the system: S = S(U, V, N 1,..., N r ) (8) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 11 / 37

Four Postulates in Thermodynamics - after Callen Postulate III: The entropy of a composite system is additive over the constituent subsystems. The entropy is continuous and differentiable and is a monotonically increasing function of the energy. S(U 1 + U 2, V 1 + V 2, N 1 + N 2 ) = S 1 (U 1, V 1, N 1 ) + S 2 (U 2, V 2, N 2 ) ( S/ U) > 0 U = U(S, V, N 1,..., N r ) Postulate IV: The entropy of any systems vanishes in the state ( ) U = 0 S V,N 1,...,N r that is at zero temperature. Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 12 / 37

Intensive parameters of thermodynamics In energy representation the fundamental equation of a pure fluid: du = U = U(S, V, N). ( ) ( ) U U ds + dv + S V,N V S,N ( ) U dn (9) N S,V From comparison with usual expression for the conservation energy: ( ) U T temperature (10) S V,N 1,...,N ( ) r U P pressure (11) V S,N 1,...,N ( ) r U µ j chemical potential (12) N j S,V,...N k Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 13 / 37

Intensive parameters of thermodynamics magnetic system U = U(S, M, N) ( ) U du = S where M,N ds + ( ) U dm + M S,N ( ) U M S,N ( ) U dn (13) N S,M h. (14) For magnetic systems N = const dn = 0. Moreover we will usually use variables per particle u = U/S, m = M/N, s = S/N: u = u(s, m) du = Tds + hdm (15) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 14 / 37

Response functions mechanical response The coefficient of thermal expansion: α 1 ( ) V V T measures the relative thermal expansion of the volume at constant pressure. The isothermal compressibility κ T 1 ( ) V V P T,N measures the relative change of the volume with increasing pressure at fixed temperature. The adiabatic compressibility κ S 1 V ( ) V P p,n (16) (17). (18) S,N Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 15 / 37

Response functions thermal response The specific heat at constant pressure c P T ( ) S = 1 N T p,n N ( ) dq. (19) dt p,n We measure the ratio between the injected heat into a system under constant pressure and the consequent change of temperature. The specific heat at constant volume: c V T ( ) S = 1 N T V,N N ( ) dq. (20) dt V,N Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 16 / 37

Specific heat Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 17 / 37

Response functions magnetic response Isothermal susceptibility: Adiabatic susceptibility: ξ T = ξ S = ( ) m. (21) h T ( ) m. (22) h S Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 18 / 37

Stability of the system Thermal stability: c p, c V 0 Mechanical stability κ T, κ S 0 c p, c V, κ T, κ S, α are functions of T, p, N Maybe S, V, N are not the most useful as independent variables? Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 19 / 37

Thermodynamic potentials In the energy representation U = U(S, V, N) extensive parameters S, V, N are the independent variables In the entropy representation S = S(U, V, N) extensive parameters U, V, N are the independent variables It is more convenient to work with independent variables of easier experimental access give me examples... Thermodynamic potentials - Legendre transformations of the fundamental equation in the energy representation U = U(S, V, N) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 20 / 37

Legendre transformation Let us consider a function instead of X k want Y = Y (X 0, X 1,..., X r ) (23) P k Y X k (24) to be independent. It is possible due to Legendre transformation: Ψ = Y k P k X k. (25) Legendre transformation is an operation that reverses the convexity of the function: convex concave and vice versa Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 21 / 37

Example 1: Legendre transformation of a quadratic function Let us calculate Legendre transformation of the function Legendre transformation: y = y(x) = ax 2 + bx + c (26) ψ(p) = ax 2 + bx + c px, (27) where Finally: p = dy dx = 2ax + b x = p b 2a ψ(p) = 1 4a p2 + b 2a p b2 4a (28) (29) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 22 / 37

Example 2: Hamiltonian formulation of classical mechanics Lagrange function L = L(q, q, t) - configuration space The generalized momentum is defined as: p = L q (30) The function of Hamilton H = H(q, p, t) - phase space H(q, p, t) = L p q (31) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 23 / 37

The Helmholtz free energy (S T ) U = U(S, V, N 1, N 2,...) S T( ) U T = S V,N 1,...,N r (32) F = U TS (33) F = F (T, V, N 1, N 2,...) (34) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 24 / 37

The enthalpy (V p) U = U(S, V, N 1, N 2,...) V p p = U V H = U + Vp (35) H = H(S, p, N 1, N 2 ) (36) (37) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 25 / 37

The Gibbs free energy (S T, V p) U = U(S, V, N 1, N 2,...) S T V p G = U TS + Vp (38) G = G(T, p, N 1, N 2,...) (39) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 26 / 37

Latent heat Latent heat is the heat released or absorbed by a body during the phase transition It allows to distinguish experimentally between continuous and discontinuous phase transition Latent heat q - energy needed to transfer a particle from one phase to another Let N i denotes number of particles in i-th phase and: u i = U i N i v i = V i N i s i = S i N i (40) denote energy, volume and entropy per particle in i-th phase. Because of the conservation of energy: du = dq dw dq = du + dw = du + pdv (41) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 27 / 37

Latent heat Because du = dq dw dq = du + dw = du + pdv (42) q = dq N = du + pdv = (u 1 u 2 ) + p(v 1 v 2 ) (43) From the equilibrium condition: Because therefore: µ 1 = µ 2 T 1 = T 2 = T p 1 = p 2 = p. (44) µ i = G i N i = E i TS i + pv i N i = u i Ts i + pv i. (45) µ 1 µ 2 = (u 1 u 2 ) T (s 1 s 2 ) + p(v 1 v 2 ) = 0. (46) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 28 / 37

Latent heat µ 1 µ 2 = (u 1 u 2 ) T (s 1 s 2 ) + p(v 1 v 2 ) = 0 (47) Finally: q T (s 1 s 2 ) = 0 q = T (s 1 s 2 ). (48) Continuous phase transitions q = 0 s 1 = s 2 (no jump in entropy) Discontinuous phase transitions q 0 s 1 s 2 (jump in entropy) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 29 / 37

Entropy in discontinuous phase transitions Figure: A Generalized Plot of Entropy versus Temperature for a Single Substance, source: Principles of General Chemistry Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 30 / 37

Chemical potential Euler equation - how to obtain it? U = TS pv + µn. (49) Legendre transformation U = U(S, V, N) G(T, p, N): G = U TS + pv = TS pv + µn TS + pv = µn. (50) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 31 / 37

Euler equation Internal energy is extensive, i.e.: U(λS, λx 1,...) = λu(s, X 1,...) (51) Differentiate above equation with respect to λ: U(λS, λx 1,...) (λs) (λs) λ + U(λS, λx 1,...) (λx 1 ) (λx 1 ) λ +... = U(S, X 1,...). (52) Above equality is valid for any value of λ, in particular for λ = 1: For a simple system we obtain: U S S + U X 1 +... = U. (53) X 1 U = TS pv + µn (54) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 32 / 37

The slope of coexistence curve Two phases: ice (A, B) and water A, B States A, A and B, B are on the coexistence curve Let us denote: then dp dt dp = P B P A = P B P A (55) dt = T B T A = T B T A (56) is a slop of the coexistence curve. p State A State B Ice Water State A State B T Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 33 / 37

Clapeyrona equation Equilibrium condition: But: µ A = µ A, (57) µ B = µ B, (58) µ B µ A = µ B µ A. (59) µ B µ A = sdt + vdp (60) µ B µ A = s dt + v dp (61) sdt + vdp = s dt + v dp (62) dp dt = s s v v (63) dp dt = q T v (64) Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 34 / 37

Regelation of ice A light rigid metallic bar of rectangular cross section (width 2mm) lies on a block of ice The length of the bar in contact with ice is 25 cm Two equal masses M are hung from the end of the bar The system is at atmospheric pressure and T = 2 0 C What is minimum M for regelation? Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 35 / 37

Regelation of ice - solution Clapeyron equation permits us to find pressure at which solid-liquid transition occurs at T = 2 0 C We must first use ice cube data to obtain the difference in molar volumes v = v liquid v solid, v liquid 18cm 3 /mol, v solid 22.5 10 6 m 3 /mol. Density of ice 0.8g/cm 3. dp dt = q T v = 5 106 Pa/K (65) So the pressure difference P = 10 7 Pa and M = 2.6Kg. Try to do it at home! Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 36 / 37

Regelation of ice - see movies by Veritasium Does Pressure Melt Ice? Ice Cutting Experiment Katarzyna Sznajd-Weron (WUST) Thermodynamics of phase transitions March 12, 2017 37 / 37