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

Transcription

1 Lecture 20 Phase ransitions Phase diagrams Latent heats Phase-transition fun Reading for this Lecture: Elements Ch 13 Lecture 20, p 1

2 Solid-gas equilibrium: vapor pressure Consider solid-gas equilibrium at constant volume and temperature. Some substances (e.g., CO 2 ) don t exist as liquids at atmospheric pressure. he solid has negligible entropy (compared to the gas), so we ll ignore it. In that case: F s = U s -S s = -N µ s = - he gas: µ g = k ln(n/n ) Binding energy of an atom in the solid Set them equal and solve for the equilibrium gas pressure: n p k ln = k ln = p pvapor = pe n p he equilibrium pressure is called the vapor pressure of the solid at temperature. If p < p vapor, atoms will leave the solid until p gas = p vapor. his is called sublimation. For liquids, it s called evaporation. Examples: Si (28 g/mol): p vapor = ( atm)(28) 3/2 e -3eV/.026eV = atm CO 2 (44 g/mol): p vapor = ( atm)(44) 3/2 e -0.26eV/.026eV = 535 atm - /k V, Some solids don t sublimate. Some do. uestion: Does water ice sublimate? Note the different s Lecture 20, p 2

3 Chemical Potential of an Ideal Gas Remember: µ = ln n p g k k ln, where p nk n = p = It s negative, because (unless highly compressed) n >> n and p >> p. 4 For helium at = 300K and p = 1 atm: ( )( ) p = = p 6 So, µ He = k ln = k ln( ) = (.026 ev)( 12.69) = 0.33 ev p All ideal gases will behave similarly, with a logarithmic pressure dependence and an approximately linear temperature dependence. (he curvature is due to the dependence of n.) atm atm Here s a graph: Increase the temperature, holding p constant:: µ decreases. µ K Increase the pressure, holding constant:: µ increases ev p > 1 atm p < 1 atm p = 1 atm Lecture 20, p 3

4 Vapor Pressure of a Solid and the p- Phase Diagram We saw that µ solid is essentially constant (- ). We can use this to determine graphically the equilibrium pressure (vapor pressure) of the solid as a function of. For a given temperature, at what pressure does the curve cross µ = -? µ p < 1 atm p > 1 atm - Equilibrium p = 1 atm From this, we can plot the equilibrium curve, p(), where the two phases can coexist. his graph is called a p- phase diagram. It tells us what regions in the (p,) plane make a gas and which make a solid. he equilibrium curve separates them. It s just a way to visualize the equilibrium equation: p Solid Gas At low or high p, equilibrium is all solid. he phases coexist on the equilibrium curve. At high or low p, equilibrium is all gas. p µ g = µ s k ln = p= pe p / k Lecture 20, p 4

5 Solids and Liquids, More Accurately We ignored the entropy of the solid, because the entropy of the gas was so much larger. When comparing solids and liquids, we can t ignore entropy. he entropy is small, but not completely negligible. Compare solids and liquids (for a given substance): S > L. Atoms in the solid are more strongly bound. S S < S L. Atoms in the liquid have more available microstates (due to motion). Compare the chemical potentials: Remember that F = U-S, and µ is the free energy per particle. At very low, the entropy is not important, so µ S < µ L. As increases, both chemical potentials decrease, but the liquid decreases faster. herefore, the solid phase is stable at low temperature, while the liquid phase is stable at high temperatures. here is an equilibrium temperature, the melting/freezing temperature. As the substance passes through freeze, there is a phase transition. L S freeze µ Liquid-solid equilibrium µ L µ S Lecture 20, p 5

6 Act 1: Solid-Liquid 1) Which point corresponds to a liquid? A B C µ A B C µ L µ S 2) he substance is in state C. What will happen? A) he substance will melt B) Free energy will minimize itself C) Entropy will maximize Lecture 20, p 6

7 Solution 1) Which point corresponds to a liquid? A B C µ A B C Both points lie on the liquid s µ() curve. Point A does not lie on the solid s curve, so it must be liquid µ L µ S (but not one in equilibrium [lowest possible µ]. At point B the liquid and solid are in equilibrium. We don t actually know how much of each there is. FYI: Global equilibrium condition: Minimum G = N g µ g + N s µ s + N l µ l G = Gibbs Free Energy Except at a coexistence temperature, c, two of the N s will be zero in equilibrium. 2) he substance is in state C. What will happen? A) he substance will melt B) Free energy will minimize itself C) Entropy will maximize Lecture 20, p 7

8 Solution 1) Which point corresponds to a liquid? A B C µ A B C Both points lie on the liquid s µ() curve. Point A does not lie on the solid s curve, so it must be liquid µ L µ S (but not one in equilibrium [lowest possible µ]. At point B the liquid and solid are in equilibrium. We don t actually know how much of each there is. FYI: Global equilibrium condition: Minimum G = N g µ g + N s µ s + N l µ l G = Gibbs Free Energy Except at a coexistence temperature, c, two of the N s will be zero in equilibrium. 2) he substance is in state C. What will happen? A) he substance will melt B) Free energy will minimize itself C) Entropy will maximize B and C are equivalent. hat s what systems do unless we interfere. A is also correct, because µ L < µ S at that temperature. Lecture 20, p 8

9 he p- Diagram with hree Phases µ p 1 p 2 µ µ p 3 Gas Solid Liquid p p 3 p 2 p 1 Phase Diagram: Solid Liquid Gas At a given p,, the system will reside in the lowest µ. Phase transitions occur where the µ() curves cross. If we raise at constant p, the sequence of phase transitions depends on the value of p. he solid and liquid curves don t change with pressure, because solids and liquids are nearly incompressible. Lecture 20, p 9

10 he p- Diagram with hree Phases µ p 1 p 2 1 µ µ p 3 Gas Solid Liquid p p 3 p 2 p 1 Phase Diagram: Solid Liquid Gas At low pressure, the liquid phase is not stable for any. he substance sublimes at temperature 1. If we increase the pressure, the gas curve moves to the right. here is a critical pressure for which all three curves pass through a point. he three phases can coexist only at ( 2,p 2 ). At high pressure, there are two phase transitions: solid-liquid at 3, and liquid-gas at 4. Lecture 20, p 10

11 Phase ransitions: OOPS! he photos below show the results of steam cleaning a railroad tank car, then sealing it before the steam had cooled Lecture 20, p 11

12 Phase Diagrams p solid liquid On special transition lines, two phases are stable. gas At very special triple points, those line cross and three phases are stable. Notice that in almost all the (p.) plane, only ONE phase is stable. Lecture 20, p 12

13 AC 2: Phase ransitions Let s think a bit about energy flow during phase transitions. Suppose heat flows slowly into 0 0 C water containing some ice cubes (maintaining equilibrium). What happens to the energy added to the system? A) It warms the ice. B) It warms the water. C) It breaks ice bonds. ice Water at 0 0 C Lecture 20, p 13

14 Solution Let s think a bit about energy flow during phase transitions. Suppose heat flows slowly into 0 0 C water containing some ice cubes (maintaining equilibrium). What happens to the energy added to the system? A) It warms the ice. B) It warms the water. C) It breaks ice bonds. ice Water at 0 0 C he temperature will remain constant until all the ice is melted. Lecture 20, p 14

15 Latent Heat Consider a liguid-gas phase transition at constant pressure. he phase diagram doesn t tell the whole story. Look at how the system changes as we add heat: p Solid Liquid Gas Heater: his all happens at the white dot. Plot temperature vs the amount of heat added: Remember the 1 st law: = U + p V. Some of the energy goes into breaking the bonds ( U), and some goes into raising the weight (p V). he p- diagram doesn t show either of these changes. liquid liquid +gas H lg Heat of transformation Latent Heat gas he heat we need to add to effect the phase change is called latent heat. Lecture 20, p 15

16 Latent Heat (2) he heat required to convert the liquid entirely into gas is directly related to the entropy change of the material: 12 = 1 S 12 = U + p V H lg H = U + pv is defined as the Enthalpy ( heat content ) of a material. For phase transitions: Heat of vaporization = enthalpy change = latent heat : H lg = H gas - H liquid = Latent Heat = L Latent heats are typically given in units of J/mol or J/gram Latent Heats of Evaporation Gas Boiling temp (K) H (J/mol) S (J/mol. K) Helium Nitrogen 77 5, Oxygen 90 6, H 2 O , here is a similar latent heat for the solid-liquid transition. For H 2 O at 1 atm: L (ice-liquid) = 80 cal/g = 333 kj/kg L (liquid-steam) = 540 cal/g = 2256 kj/kg L (solid-liquid) L (liquid-gas) Lecture 20, p 16

17 Exercise What is the entropy change per mole of water when water freezes (at p = 1 atm)? ice Water at 0 0 C Lecture 20, p 17

18 Solution What is the entropy change per mole of water when water freezes (at p = 1 atm)? L = S S = - L / = -333 J/g / 273 K * 18 g/mol = J/K mol ice Water at 0 0 C hat s a loss of σ of about 2.73 per molecule. hat is, there are about e 2.7, or 15, as many microstates available to each molecule in the liquid. Lecture 20, p 18

19 Act 3: Entropy in freezing We just saw that a mole of water loses entropy when it freezes. Consider a lake that freezes in winter. If the lake loses 10 6 J/K of entropy in the process, what is the change of entropy of the environment during this process? A) S = 0 J/K mol B) S = J/K mol C) S > J/K mol Lecture 20, p 19

20 Solution We just saw that a mole of water loses entropy when it freezes. Consider a lake that freezes in winter. If the lake loses 10 6 J/K of entropy in the process, what is the change of entropy of the environment during this process? A) S = 0 J/K mol B) S = J/K mol C) S > J/K mol he entropy increase of the enviroment must at least balance the entropy loss of the water. Otherwise, the total entropy of the waterenvironment system would decrease, in violation of the 2 nd law. Since this is an open system, some of the energy extracted from the water will spread irreversibly into the environment, i.e., the net entropy must increase. Lecture 20, p 20