Lecture 14. Phases of Pure Substances (Ch.5)

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1 ecture 4. hases of ure Substances (Ch.5) Up to now we have dealt almost exclusively with systems consisting of a single phase. In this lecture, we will learn how more complicated, multiphase systems can be eated by the methods of thermodynamics. he guiding principle will be the minimization of the ibbs free energy in equilibrium for all systems, including the multiphase ones. he generic phase diagram of a substance in the - coordinates is shown above. Every point of this diagram is an equilibrium state; different states of the system in equilibrium are called phases. he lines dividing different phases are called the coexistence curves. Along these curves, the phases coexist in equilibrium, and the system is macroscopically inhomogeneous. All three coexistence curves can meet at the iple point in this case, all three phases coexist at (, ).

2 he Coexistence Curves Any system in contact with the thermal bath is governed by the minimum free energy principle. he shape of coexistence curves on the - phase diagram is determined by the condition: and, since ( ) ( ),, Nμ ( ) (, ) μ, μ - otherwise, the system would be able to decrease its ibbs free energy by ansforming the phase with a higher μ into the phase with lower μ. wo phases are in a state of diffusive equilibrium: there are as many molecules migrating from to as the molecules migrating from to. Also for equilibrium between the phases: Along the coexistence curves, two different phases and coexist in equilibrium (e.g., ice and water coexist at 0 0 C and bar). he system undergoes phase separation each time we cross the equilibrium curve (the system is spatially inhomogeneous along the equilibrium curves). - as for any two sub-systems in equilibrium - the phase boundary does not move hough is continuous across the ansition, H demonsates a step-like behavior: Nμ U + V S H S Δ H ΔS (different phases have different values of the enopy)

3 Example: the as-iquid ransformation μ N, F N, V [ F U S] U N, V S N, V as: U/ N term is small and positive (kin. energy of a single molecule), ( S/ N) term is large and positive μ is negative, and rapidly decreases with increasing. iquid: U/ N term is negative (ataction between molecules), ( S/ N) term is smaller than that in gas and positive μ is also negative, and slowly increases with decreasing. able on page 404 (a very useful source of information) provides the values of H and for different phases of many substances. he data are provided per mole, at 98 K and bar. For example, let s check that at the boiling point, the values of for liquid water and water vapor are equal to each other:, N S S(water) 70 J/K S(vapor) 89 J/K ( ) ( ) ( ) S, N,, N, 0 0 μ 0 liquid phase ansformation gas ( liq) 7 0 J / mol 75K 70J /( K mol) 4 0 J / mol 7K ( ) J / mol 75K 89J /( K mol) 4 0 J / mol vap 7K

hases of Carbon he phase equilibrium on the,-plane is determined by ( ) (, ) or μ (, ) ( ),, μ At normal conditions, graphite is more stable than diamond: (graphite) 0, (diamond).

4 hases of Carbon he phase equilibrium on the,-plane is determined by ( ) (, ) or μ (, ) ( ),, μ At normal conditions, graphite is more stable than diamond: (graphite) 0, (diamond).9 kj (diamonds are not forever...). What happens at higher pressures?, N V ( ) ( ) + ( )V, N, N, since the molar volume of graphite is greater than the molar volume of diamond, (graphite) will grow with pressure faster than (diamond) [we neglected V V()] D. becomes more stable than. only at >.5 Ma.9 kj diamond graphite 00K With increasing, the pressure range of graphite stability becomes wider:, N S ( ) ( ) ( ) S, N,, N, 0 0 S(graphite) 5.74 J/K, S(diamond).8 J/K,.9 kj diamond graphite (Ma) bar (K)

5 S gas const gas+liquid liquid solid liquid+solid he evaporation is generally greater than the melting (the disorder inoduced by evaporation is greater than that inoduced by melting). he First-Order ransitions Because molecules aggregate differently in different phases, we have to provide (or remove) energy when crossing the coexistence curves. he energy difference is called the latent heat ; crossing the coexistence curve, the system releases (absorbs) a latent heat. he enopy of the system changes abruptly: δ Q ΔS S mixed phase he ansitions which displays a jump in enopy and a latent heat associated with this jump are called the first-order phase ansitions. critical point S gas S liquid C Q: Can the critical point exist along the melting coexistence curve? beyond critical point, gas is indistinguishable from liquid temperature

6 solid liquid (r. 5.9).,N const gas On the graph () at,n const, the slope d/d is always negative: S, N he First-Order ransitions (cont.) Note that in the first-order ansitions, the () curves have a real meaning even beyond the intersection point, this results in metastability and hysteresis. here is usually an energy barrier that prevents a ansition occurring from the higher μ to the lower μ phase (e.g., gas, being cooled below does not immediately condense, since surface energy makes the formation of very small droplets energetically unfavorable).. water can exist at far lower than the freezing temperature: water in organic cells can avoid freezing down to 0 0 C in insects and down to 47 0 C in plants. S C C S ΔS /, N

roblem he enopy of water at atmospheric pressure and 00 0 C is. J/g K, and the enopy of water vapor at the same and is 7.4 J/g K. (a) What is the heat of vaporization at this temperature?

7 roblem he enopy of water at atmospheric pressure and 00 0 C is. J/g K, and the enopy of water vapor at the same and is 7.4 J/g K. (a) What is the heat of vaporization at this temperature? (b) he enthalpy of vapor under these conditions is 680 J/g. Calculate the enthalpy of water under these conditions. (c) Compute the ibbs free energies of water and steam under these conditions. (a) (b) he heat of vaporization: ΔS 7K 6. J/g K75 J/g he differential of enthalpy dh ds+vd. Hence, H water H vapor ds H vapor (680-75)J/g 405 J/g (c) Since H-S, water H water S water 405J/g - 7K.J/g K -80J/g vapor H vapor S vapor 680J/g - 7K 7.4J/g K -80J/g

8 he Second Order ransitions he vaporization coexistence curve ends at a point called the critical point ( C, C ). As one moves along the coexistence curve toward the critical point, the distinction between the liquid phase on one side and the gas phase on the other gradually decreases and finally disappears at ( C, C ). he -driven phase ansition that occurs exactly at the critical point is called a second-order phase ansition. Unlike the st - order ansitions, the nd -order ansition does not require any latent heat (0). In the higher-order ansitions (order-disorder ansitions or critical phenomena) the enopy is continuous across the ansition. he specific heat C (δs/δ) diverges at the ansition (a cusp-like λ singularity). Whereas in the st -order ansitions the () curves have a real meaning even beyond the intersection point, nothing of the sort can occur for a nd -order ansition the ibbs free energy is a continuous function around the critical temperature. Second-order ansition S C C S, N ΔS0

9 d d he Clausius-Clapeyron Relation (, ) (, ) Along the phase boundary: ( ) ( ),, For the slope of the boundary we have: S V (, ) S(, ) (, ) V (, ) ( ) (, ), μ μ Consider two distinct displacements along the coexistence curve, one immediately above the curve (in phase ), the other immediately below the curve, in phase. Because the chemical potentials remain equal along the curve, dμ dμ Since S - S / ( is the latent heat), we arrive at the Clausius-Clapeyron Relation : - the slope is determined by the enopies and volumes of the two phases. he larger the difference in enopy between the phases the steeper the coexistence curve, the larger the difference in molar volumes the shallower the curve. (compare the slopes of melting and vaporization curves) d d Δ ( ) V ( ) Δ Δ phase boundary S d + Vd Sd + Vd (applies to all coexistence curves) d d Example: [ Vgas ( ) Vliq ( )] ( ) - since V gas > V liq, and > 0 for the liquid gas ansformation, the boiling temperature increases with pressure. he freezing temperature with increasing pressure either increases or decreases, depending on the sign V liq V solid (exception He).

10 roblem kg of water at 0 0 C is converted into ice at -0 0 C (all this happens at bar). he latent heat of ice melting melt 4 kj/kg, the heat capacity of water at constant pressure is 4. kj/(kg K) and that of ice. kj/(kg K). (a) What is the total change in enopy of the water-ice system? (b) If the density of water at 0 0 C is taken as 0% greater than that of ice, what is the slope of the melting curve of ice at this temperature? ive both sign and size. (a). From 0 0 C to 0 0 C:. Melting of ice. From 0 0 C to -0 0 C: ΔS m (b) d ( S ) d ( v ) Q mcwaterd 9K ds Q mcd ΔS mcwaterln 7K 9K melt melt 6K 7K mciced 7K ΔS mcice ln 6K 7K 9 4kJ / kg ΔS kg 4.kJ / 7 7K J J J J K K K K Sice J /( K kg) water 7. 0 a K v 0. 0 m / water ice ( kg K ) ln.kj /( kg K ) 7 ln 6

11 he Vapor ressure Equation he differential Clausius-Clapeyron equation can be solved to find the shape of the entire coexistence curve (r. 5.5). d d vap S V gas gas S V liq liq ( ) V ( ) Δ For the liquid-gas phase ansition, we can make the following reasonable assumptions: the molar volume of the liquid is much smaller than that of the gas (neglect V liquid ) the molar volume of gas is given by the ideal gas law V R/ the latent heat is almost -independent, () d d vap V gas R d R d vap exp R (mbar) 4 He (mbar) H O (K) ( 0 C)

12 roblem (he pressure cooker) he boiling point of water at atmospheric pressure ( bar) is 7 K, and the latent heat of vaporization at this is.5 MJ/kg. Assuming the latent heat to be constant over the relevant temperature range, that water vapor is an ideal gas, and that the volume of the liquid is negligible relative to that of the vapor, find the temperature inside a pressure cooker operating at an absolute pressure of bar. d R d ln R R 8.J / mol ln 0.7 K J / kg 8 0 kg / mol 94 6

13 roblem For Hydrogen (H ) near its iple point ( 4K), the latent heat of vaporization vap.0 kj/mol. he liquid density is 7 kg m -, the solid density is 8 kg m -, and the melting temperature is given by m.99+/., where m and measured in K and Ma respectively. Compute the latent heat of sublimation solid liquid Near the iple point: vap ( S S ) ( S S ) ( S S ) melt S sub S sub melt vap gas V melt ( S S ) + ( ) J/mol 7 J/mol sub S vap melt ( V V ) d / d molar mass ( ) ( kg/mol) m /mol density( kg/m ) S 0 kg/mol 0 kg/mol 4-7 kg/m 8 kg/m 6 /. 0 6 J

14 roblem (cont.) solid liquid he vapor pressure equation for H : where 0 90 Ma. vap 0 exp R sub gas Compute the slope of the vapor pressure curve (d/d) for the solid H near the iple point, assuming that the H vapor can be eated as an ideal gas. At the solid-gas phase boundary: d d sub sub ( V VS ) V Assuming that the H vapor can be eated as an ideal gas V R V R exp - m 7 ( / R ) 9 0 a exp -00 J/ ( 8. J/K mol 4 K) R vap / mol 8. J/K mol 4 K [ ] d d V 7 J/mol a/k 4 K m / mol sub

Ice I hase Diagram of H O For most normal substances, the slope of the melting curve is positive (S >S S, V >V S ).

he ice is less dense than water (V <V S ): the hydrogen bonds determine the teahedral coordination and openness of the sucture of

15 Ice I hase Diagram of H O For most normal substances, the slope of the melting curve is positive (S >S S, V >V S ). he phase diagram for water shows the characteristic negative slope of the solid-liquid equilibrium curve. he ice is less dense than water (V <V S ): the hydrogen bonds determine the teahedral coordination and openness of the sucture of ice. As ice melts into water the change in enopy (or the latent heat) is positive, while the change in volume is negative, hence the negative slope. d d ( V V ) liq < 0 he negative slope of the solid-liquid coexistence curve makes ice skating possible: ice melts under the pressure exerted by the skate blade. he Clausius-Clapeyron equation provides the connection between ice skating and the observation that ice floats on water. sol

16 roblem Ice skating becomes unpleasant if the weather is too cold so that the ice becomes too hard. Estimate the lowest temperature for which ice skating is still enjoyable for a person of normal weight. he given data are that the latent heat of fusion of water is J/g, that the density of liquid water is g/cm, and that ice cubes float with ~9/0 of their volume submerged. he lowest temperature for enjoyable skating is the temperature at which the pressure exerted by the skater on ice is equal to the pressure on the coexistence curve at this. At 0 bar, ice melts at 0 7.5K (0 0 C). d d pressure: ( V V ) liq mg 00kg 0m / s kg s m 0 a area 0. 0 m 6 ( Vliq Vsol ) ( ) 7 K ( 0. 0 m / g) 8 0 a 8 K sol 0 Δ 0 he lowest temperature: C, about right. J / g et s verify that from two points on the melting curve, (0.006 bar 7.6K) and ( bar 7.5K) we can get a reasonable estimate for : ( 7K ) ( Vliq Vsol )( ) ( ) 7K m /g K 5 ( atm) a 7J/g

17 Why Is Ice Slippery? (R. Rosenberg, hysics today, Dec. 005) ressure melting does not explain why skating is possible in climates as cold as 0 o C. his popular explanation of the thermodynamics of ice skating is not the whole story (the experiments by Robert Wood and other researchers). he mechanism(s) is much more complicated. he physicists interested in the problem: Faraday, ibbs, etc. wo other important factors: Frictional heating. S. Colbeck in his experiments ( ) attached a thermocouple to a skate blade (and to the bottom of skis) and showed that the increase in temperature with velocity was consistent with frictional heating. iquid layer on ice surface below zero. here is a disordered (liquid-like) layer on the surface of ice (its thickness - ~ 0 nm) down to ~ -0 o C.

the λ ansition 4 He hase Diagrams of 4 He He is the only element that remains a liquid at 0 and bar because (a) the zero-point oscillations of light atoms are large, and (b) the binding forces

18 the λ ansition 4 He hase Diagrams of 4 He He is the only element that remains a liquid at 0 and bar because (a) the zero-point oscillations of light atoms are large, and (b) the binding forces between the atoms are very weak. he zero-point energy of He is larger than the latent heat of evaporation of liquid helium; the zero-point vibration amplitude is ~/ of the mean separation of atoms in the liquid state. As a result, the molar volume of 4 He ( He) is more than a factor of two (three) larger than one would calculate for a corresponding classical liquid. Also, the latent heat of evaporation is unusually small - ~/4 of its value for the corresponding classical liquid. According to Nernst s theorem, for any processes that bring a system at 0 from one equilibrium state to another, ΔS 0. If, at the same time, ΔV 0, then d/dv 0, and the slope of the coexistence curve solid-liquid must approach zero as 0. he slope of the phase boundary solid helium superfluid liquid helium is essentially 0 at < K: the enopy change must be zero, and the liquid must be as ordered as the solid! While the phase diagram shows that the solid and liquid II are equally ordered, x- rays reveal that only the solid has a long-range order in real space. herefore, we arrive at a conclusion that liquid II must be more ordered in the momentum space! d d ΔS ΔV

hase Diagram of He He Solid iquid linear scale Below 0.K the slope of the He solid-liquid phase boundary is negative. his negative slope tells us that ΔS and ΔV have opposite signs.

19 hase Diagram of He He Solid iquid linear scale Below 0.K the slope of the He solid-liquid phase boundary is negative. his negative slope tells us that ΔS and ΔV have opposite signs. he denser phase is always the one that is stable at high its molar volume is smaller, and at sufficiently high, its is smaller. When we move from liquid He to solid He, V decreases - thus, S must increase!! as V In other words, the liquid is more ordered than the solid, and therefore it takes heat to change the liquid to a solid! he omeranchuk effect: as the solidliquid mixture is compressed, heat is removed from the liquid phase as crystals form. he latent heat associated with converting mole of He liquid into solid is 0.4J. Cooling: from ~ 0 mk to mk. S S iquid 0.K S solid

20 roblem At the atmospheric pressure, He remains liquid even at 0. he minimum pressure of He solidification is min 8.9 bar. At low temperatures, the enopy of mole of liquid He is S R/ 0, where 0 0. K, the enopy of solid He is temperature-independent: S S R ln. he difference between the molar volumes of liquid and solid He ΔV V -V S.5 cm /mol (a) Find the temperature of solidification min at min (b) Find the temperature dependence of the latent heat of melting melt. (c) Find the pressure of solidification of He at 0. d d melt (a) he minimum on the solid-liquid coexistence curve ( min ) melt melt ( V VS ) ( ) [ S ( ) S ( )] min (b) ( ) [ S ( ) ( )] melt melt melt min corresponds to d/d 0, and, thus, melt ( min ) 0. min min S min min 0 ln 0. 5K R SS R ln - a parabola that goes through 0 at min. 0 he negative sign of melt for He is a unique phenomenon (the omeranchuk effect). Over the range of where melt < 0, the slope of the -S coexistence curve is negative. (Note that, in conast to d melt /d < 0 on the phase diagram for water, here the negative slope is observed for V -V S > 0). R (c) By integrating the Cl.-Cl. eq. d d V ln min Δ 0 R R 0 ln + 0 min 0 ln.7 0 const const ΔV 0 ΔV 5 ( ) a

21 vap d d ( ) V ( ) Δ Summary. he shape of coexistence curves on the - diagram: ( ) ( ),,. he latent heat in the st order phase ansitions: δ Q ΔS Nμ U + V S H S ΔH ΔS. he slope of the coexistence curve is given by the Clausius-Clapeyron Relation: d d vap V gas By integrating the CC relation, one can restore the shape of the coexistence curve, () 4. For the gas-liquid ansition, we can replace the CC relation with the vapor equation: R d R d 5. he iple point: (, ) (, ) (, ) liq gas sol vap exp ( S S ) ( S S ) ( S S ) ( ) + ( ) ( ) melt S sub S melt vap R sub

Lecture 20. Phase Transitions. Phase diagrams. Latent heats. Phase-transition fun. Reading for this Lecture: Elements Ch 13.

Lecture 20. Phase Transitions. Phase diagrams. Latent heats. Phase-transition fun. Reading for this Lecture: Elements Ch 13. Lecture 20 Phase ransitions Phase diagrams Latent heats Phase-transition fun Reading for this Lecture: Elements Ch 13 Lecture 20, p 1 Solid-gas equilibrium: vapor pressure Consider solid-gas equilibrium