Concepts in Thermal Physics ( PDFDrive )

324 Phase transitions 28.2 Chemical potential and phase changes We have seen in Section 16.5 that the Gibbs function is the quantity that must be minimized when systems are held at constant pressure and temperature. In Section 22.5 we found that the chemical potential is the Gibbs function per particle. We were also able to write in eqn 22.52 that dG = V dp − SdT + μidNi. (28.8) i Now consider the situation in Fig. 28.2, in which N1 particles of phase 1 are in equilibrium with N2 particles of phase 2. Then the total Gibbs free energy is Gtot = N1μ1 + N2μ2, (28.9) Fig. 28.2 Two phases in equilibrium and since we are in equilibrium we must have at constant pressure (the constraint of constant pressure is maintained by the dGtot = 0, (28.10) piston). and hence dGtot = dN1μ1 + dN2μ2 = 0. (28.11) But if we increase the number of particles in phase 1, the number of particles in phase 2 must decrease by the same amount, so that dN1 = −dN2. Hence we have that μ1 = μ2. (28.12) Thus in phase equilibrium, each coexisting phase has the same chemical potential. The lowest μ phase is the stable phase. Along a line of coexistence, μ1 = μ2. 28.3 The Clausius–Clapeyron equation Fig. 28.3 Two phases in the p–T plane We now want to find the equation that describes the phase boundary in coexist at the phase boundary, shown the p–T plane (see Fig. 28.3). This line of coexistence of the two phases by the solid line. is determined by the equation μ1(p, T ) = μ2(p, T ). (28.13) If we move along this phase boundary, we must also have μ1(p + dp, T + dT ) = μ2(p + dp, T + dT ), (28.14) so that when we change p to p + dp and T to T + dT we must have dμ1 = dμ2. (28.15) This implies that (using eqns 16.22 and 22.48) −s1dT + v1dp = −s2dT + v2dp, (28.16)

28.3 The Clausius–Clapeyron equation 325 where s1 and s2 are the entropy per particle in phases 1 and 2, and v1 and v2 are the volume per particle in phases 1 and 2. Rearranging this equation therefore gives that dp = s2 − s1 . (28.17) dT v2 − v1 If we define the latent heat per particle as l = T Δs, we then have that dp l (28.18) =, dT T (v2 − v1) or equivalently dp L (28.19) =, dT T (V2 − V1) which is known as the Clausius–Clapeyron equation. This shows 6This can be obtained from the differ- ence in densities. that the gradient of the phase boundary of the p–T plane is purely determined by the latent heat, the temperature at the phase boundary, and the difference in volume between the two phases.6 Example 28.2 Derive an equation for the phase boundary of the liquid and gas phases under the assumptions that the latent heat L is temperature inde- pendent, that the vapour can be treated as an ideal gas, and that Vvapour = V Vliquid. Solution: Assuming that Vvapour = V Vliquid and that pV = RT for one mole, the Clausius–Clapeyron equation becomes dp Lp (28.20) dT = RT 2 . This can be rearranged to give dp LdT (28.21) p = RT 2 , and hence integrating we obtain ln p = − L + constant. (28.22) RT Hence the equation of the phase boundary is L (28.23) In this case, the constant p0 is given p(T ) = p0 exp − RT , by p0 = p(∞), the pressure at infinite temperature. where the exponential looks like a Boltzmann factor e−βl with l = L/NA, the latent heat per particle. Remember again that R = NAkB.