Phase Transitions A homogeneous equilibrium state of matter is the most natural one, given the fact that the interparticle interactions are translationally invariant. Nevertheless there is no contradiction with the fundamentals of the Statistical Mechanics if under some conditions we have|at a fixed volume| different phases of one and the same substance: liquid and vapor, solid and liquid, liquid and vapor, etc. What are the conditions for such a situation to occur, and why do different phases exist in principle? First, let us take the existence of different phases for granted, and consider the conditions for their coexistence. Each phase is assumed to be homogeneous and characterized by its equation of state in the form µ = µ(P; T ) ; (1) in which all the three variables are intensive. Suppose we have two coexisting phases, a and b. Then, each of them, with respect to the other one, simultaneously plays the following three roles: (i) a piston, (ii) a heat bath, and (iii) a particle reservoir. This means that (i) the pressures, (ii) the temperatures, and (iii) the chemical potentials of the two phases coincide, leading to the following condition of the phase coexistence in terms of the equations of state: µa(P; T ) = µb(P; T ) : (2) Eq. (2) implies that for a given temperature, T , there can exist only one special critical pressure, Pc(T ), at which the coexistence of the two phases is possible. Equivalently, for a given pressure, P , there can exist only one critical temperature, Tc(P ). The functions Pc(T ) and Tc(P ) can be found from (2) in an implicit form: µa(Pc(T );T ) ≡ µb(Pc(T );T ) ; (3) µa(P; Tc(P )) ≡ µb(P; Tc(P )) : (4) Hence, the relation (2) defines a critical line in the (P; T )-plane on which|and only on which|the coexistence of two phases is possible. Clearly, coexistence of three different phases is possible only at an isolated point|so-called triple point|on the (P; T )-plane, as in this case we have one more independent constraint, say, µb(P; T ) = µc(P; T ) : (5) Eqs. (2) and (5) can be simultaneously satisfied only at an isolated point: two independent equations for two unknown variables. In a general case, the coexistence of more than three different phases is impossible because we will get three or more independent equations for just two variables. We see that at a fixed pressure there is only one special temperature, Tc(P ), at which two (or three) phases can coexist. Hence, if we pass this special temperature the system will change its phase in a jump-like way. This phenomenon is known as the first-order phase transition. Now we are in a position to understand the reason for a phase transition to occur, and, in particular, the reason for existing different phases, since there is no phase transitions without phases. Fixed pressure and varying temperature are very convenient variables, because the state of the system remains homogeneous up to a single special point Tc. The reason for different phases to exist is thermodynamic favorability. In the space of states of a macroscopic system there can be two or more different regions corresponding to qualitatively different properties of the system (liquid, vapor, solid, ferromagnetic/paramagnetic, etc.). In accordance with Gibbs distribution, the system can be found in each of these regions with the probability proportional 1 to the contribution to the total partition function, associated with the corresponding region. Given two phases, a and b, we thus write the corresponding probabilities as Za Zb Pa = ; Pb = : (6) Za + Zb Za + Zb Here we face a technical problem: We know how to calculate the partition function at a fixed volume, as the volume is a natural parameter fixing the spectrum of the energy eigenvalues. But now we want to fix pressure, not volume. A trick is to treat the pressure as resulting from a constant force (equal to pressure times piston area) applied to our system. We thus get a situation of an external potential equal to −force × coordinate = P × surface area × coordinate = PV: (7) That is for a system of the volume V we get the extra contribution of PV to the energy (and Helmholtz free energy), or, equivalently, the factor of exp(P V=T ) to the partition function. Recalling that F + PV = G; (8) where G is the Gibbs free energy, we conclude that at a fixed pressure the effective partition function for the phase a reads −Ga=T Za = e ; (9) and the same for the phase b. Hence, Pa Za = = e(Gb−Ga)=T : (10) Pb Zb The probabilities become equal at the point where Ga(P; T ) = Gb(P; T ) ; (11) and are radically different away from this point, since the difference Gb − Ga is an extensive quantity, proportional to the total number of particles.|If it is non-zero, then it is macroscopically large. Hence, at the critical point the systems \jumps" from one phase into the other one. In accordance with Eq. (10), the most favorable phase is the phase with the lowest Gibbs free energy. Is there any connection between the condition (11) and the condition (2) of phase coexistence? Yes, and quite a trivial one, because G ≡ µN : (12) Problem 47. Establish the identity (12). Rewriting (10) as Pa Za = = eN(µb−µa)=T ; (13) Pb Zb we see that at the fixed pressure the most favorable phase is the phase with a smaller chemical potential. The Gibbs free energies of the two phases are equal at the phase transition point.|What about the rest of quantities? Taking into account that G = E − TS + PV; (14) and that, by definition, the pressure and temperature remain the same at the phase transition point, from (11) we get Ea − TSa + PVa = Eb − TSb + PVb : (15) 2 Note that neither energy, nor entropy, nor volume is supposed to be the same in the two phases. Eq. (16) can be rewritten as Eb − Ea = T (Sb − Sa) + P (Va − Vb) : (16) In this form it describes the energy balance. The left-hand side is the change of energy. The second term in the right-hand side is the work performed over the system. Correspondingly the term Lab = T (Sb − Sa) (17) is the heat transferred to the system from the heat bath during the transition. It is called the latent heat. During the I-order phase transition the heat transfer does not result in any temperature change. This means that the heat capacity behaves like a δ-function of (T − Tc). Problem 48. Derive the Clapeyron equation: dP L c = ab : (18) dT T (Vb − Va) Hint. Differentiate the identity (3), remembering (12). Consider some point on the phase transition line T = Tc(P ). At the given temperature and pressure the system can have two different volumes, Va, and Vb. What happens if starting from, say, volume Va < Vb, we will be keeping temperature constant while increasing volume, V ? The system cannot be totally in the phase b, until V reaches the value Vb. It cannot be totally in the phase a either, since V > Va. The only possibility is the phase separation, when a part of the system volume corresponds to the phase b, while the rest of the volume is in the phase a. And this is precisely the situation of the phase coexistence, described by the relation (2). From (2) it follows that if T is fixed, then the pressures of the two coexisting phases correspond to the critical pressure Pc(T ) and thus are independent of volume. The same is true for the densities, since the temperature and pressure unambiguously define the density of each of the two phases. Hence, the only quantities that are sensitive to V in the region V 2 [Va;Vb] are the volumes V1 and V2 of the components a and b, respectively. These volumes are easily found from the relation V V 1 + 2 = 1 ; (19) Va Vb and an obvious equation V1 + V2 = V: (20) Problem 49. Show that Eq. (19) follows from the requirement that the total number of particles be fixed. The Critical Point. Second-Order Phase Transitions Suppose we take a fixed volume with coexisting liquid and vapor, and start to increase the temperature. At high enough T , there should be no phase separation: in the limit of T ! 1 the system behaves like an ideal gas, because the kinetic energy of particles is much larger than the potential energy of their interaction. Hence, with increasing temperature we always cross a phase transition line. There are two generic scenarios of what will be happening: (i) the liquid component expands until the vapor component vanishes; (ii) the liquid component evaporates and vanishes. By continuity argument, we conclude that for the given number of particles, N, there should exist such a critical volume, V∗(N)|it is a bit more physical to talk of a critical density, n∗, because V∗ ≡ N=n∗, that the volumes of both coexisting phases remain finite up to the very phase transition point. The phase transition point at n = n∗ turns out to be different from the generic (first-order) liquid-vapor phase transition point. At this point there is no latent heat. All the thermodynamic potentials remain continuous. When approaching this point, the densities of liquid and vapor approach each other, so that beyond this point it becomes fundamentally impossible to tell one from another. 3 The critical point in the liquid-vapor system corresponds to what is known as the second-order phase transition.