Statistical Physics

124 8 First-Order Phase TransitionsUsually, the entropy and volume of the liquid phase are larger than thoseof the solid phase. Therefore, the slope of the coexistence line is usually alsopositive in the case of solid–liquid phase transitions. However, there are exceptions.One such exception is the case of water. As we know, ice floats onwater: the volume of a sample of ice is larger than that of a sample of water ofthe same mass. On the other hand, ice has a lower entropy, which is evidentbecause we need heat to melt ice. Therefore, the slope of the phase boundarybetween ice and liquid water is negative. Ice melts when pressure is applied. 4A rough value for the slope can be estimated from our daily experience. Tomake iced coffee, we first almost fill the glass with ice at 0 ◦ C. Since thereis space between the blocks of ice, we can pour nearly the same amount ofboiling coffee into the glass. Then most of the ice melts, and we get cold coffee.This tells us that the latent heat, the quantity of heat required to melta sample of ice, is nearly the same as the quantity of heat required to heatthe same amount of water from 0 ◦ C to 100 ◦ C. The latter quantity of heat isabout 100 cal g −1 , as this can be related to the definition of the calorie, whichis equal to 4.18 J in SI units. The volume change can be guessed from the wayice floats on water. About 10% of the ice is above the surface of the water,and so the volume change is about 10%, or 10 −7 m 3 g −1 . Using these values,we can obtain the estimatedPdT = − 418 J273 K × 10 −7 m 3 =1.5 × 107 Pa K −1 . (8.14)Thus the freezing temperature decreases by 0.007 K when a pressure of 1 atmis applied. 5Another exception is the coexistence line between liquid and solid 3 He,the isotope of helium with mass number 3. Liquid helium-3 remains liquideven at zero temperature when the pressure is below about 3.4 MPa. Abovethat pressure, it solidifies. The phase diagram is as shown in Fig. 8.5. Inthe low-temperature portion of the coexistence line, between T ≃ 1mK andT ≃ 0.316 K, the slope is negative. How is such a behavior possible? The solidphase has a smaller volume. This is evident because the solid phase is on thehigh-pressure side. This is the normal behavior. Therefore, the entropy mustbehave abnormally. Namely, the entropy of the solid must be larger than thatof the liquid. The larger entropy of the solid cannot originate from the motion4 This melting of ice under pressure was once considered to be the main reasonwhy we can enjoy skiing or ice-skating. However, this is not correct. The loweringof the melting temperature under a skater’s blade is estimated to be only about3 K, which is too small to explain the fact that we can skate or ski even whenthe temperature of the ice or snow is −30 ◦ C. The true reason is the existence ofa thin liquid layer covering the ice, although the reason for the existence of thisliquid is not fully understood [4].5 The correct values for the latent heat and the molar-volume difference at ambientpressure are Q L =6.01×10 3 Jmol −1 and ∆V =1.62×10 −6 m 3 mol −1 . Therefore,dP/dT =1.36 × 10 7 Pa K −1 .