Miscellaneous Notes

1. Miscellaneous Notes • The end is near – don’t get behind. • All Excuses must be taken to 233 Loomis before 4:15, Monday, April 30. • The PHYS 213 final exam times are * 8-10 AM, Monday, May 7 * 8-10 AM, Tuesday, May 8 and* 1:30-3:30 PM, Friday, May 11. The deadline for changing your final exam time is 10pm, Monday, April 30. • Homework 6 is due Tuesday, May 1 at 8 am. (NO late turn-in). • Course Survey = 2 bonus points (now on SmartPhysics – due Wed. May 2)

2. f 1 Phase Diagram: p T1 T2 Solid Liquid T3 0.5 p3 T3 > T2 > T1 p2 0 Gas p p0(T2) p1 T Lecture 21Review & Examples

3. Phases: Solid, liquid, gas LIQUID: Particles interact strongly, But no regular pattern Medium U, S, small V GAS: Particles Scarcely Interact Highest U, S, V H2O ? SOLID: Particles interact strongly, Form regular pattern Lowest U, S Small V The stable phase at some (p, T) has the lowest . (G/particle)

4. The p-T Diagram with Three Phases p1 p2 p3 m m m T T T Phase Diagram: p Solid Liquid p3 p2 Gas p1 T T1 Gas Solid Liquid At low pressure, the liquid phase is not stable for any T.The substance sublimes at temperature T1. If we increase the pressure, the gas curve moves to the right. There is a critical pressure for which all three curves pass through a point. The three phases can coexist only at (T2,p2). At high pressure, there are two phase transitions:solid-liquid at T3, and liquid-gas at T4. T1 T2 T3 T4

5. Phase Diagrams On special transition lines, two phases are stable. At very special “triple points”, those line cross and three phases are stable. p T solid liquid gas • Notice that in almost all the (p.T) plane, only ONE phase is stable. Why is that?

6. Why usually ONE stable phase? • Ordinary chemical reaction equilibria: • At low concentration ni, mi drops without limit (e.g. kTln(ni/nT)) until it matches the other m’s, at some particular ratio. • Therefore have positive concentrations of every possible molecular type. • true at fixed p or at fixed V • Phase transitions: • Each phase has adefinite density at a given (T,p) and hence a definite m, regardless of how much is present! • Adding material changes the volume of the phase, not the density, sodoesn’t change. • Therefore at most (T, p) only ONE phase, with lowest , is stable. • At special (T, p), two can have the same , so they can coexist in any ratio • Under constant V conditions, there’s some dependence of p, and thus , on the amounts of the phases, and thus some coexistence ranges.

7. Phases: Solid, liquid, gas and ? LIQUID: Particles interact strongly, But no regular pattern Medium U, S, small V Liquid is much less compressible than gas. As p goes up, at the boiling point the volume difference between the gas and liquid decreases. When they cross, there is no longer any transition, just a miscellaneous “supercritical” fluid. Critical point GAS: Particles scarcely interact Highest U, S, V H2O

8.  vs. T at fixed p Special lines where Two phases coexist H2O Tsc T Tboiling Tfreezing mg Gasmg LiquidmL ms Solid ms mL To minimize G  state with the lowest m At special values of (p,T), mg= mL or ms= mg or ms= mL so different phases coexist. Along the p= 1 atm line

9. Why are phases separate? G/100kT Nmolecules /kT nucleus Could liquids solidify gradually via little ordered clusters? Those 11 molecules lost just about as much Sas if they were in the middle of a big crystal. But they gained only 21 nice crystalline contacts (bonds),instead of the 33 they would gain in the middle of a crystal. So their G is much bigger than it would if they were part of a big crystal. • The surfaces between phases add G • Extra G/area is called the surface tension • Phases separate to lower G Even below the freezing point, • a small solid cluster is most likely to melt. • So it’s hard for something to freeze even if S < L • Requires a rare “critical nucleus”

10. T QL (liquid-gas) QL (solid-liquid) Q Latent Heat The heat required to convert the liquid entirely into gas is directly related to the entropy change of the material: Q12 = T1DS12 = DU + pDV  DHlg H = U + pV is defined as the Enthalpy (‘heat content’) of a material.For phase transitions: “Heat of vaporization” = “enthalpy change” = “latent heat”:DHlg = Hgas - Hliquid = Latent Heat = QL Latent heats are typically given in units of J/mol or J/gram There is a similar latent heat for the solid-liquid transition. For H2O at 1 atm: QL(ice-liquid) = 80 cal/g = 333 kJ/kg QL(liquid-steam) = 540 cal/g = 2256 kJ/kg Latent Heats of EvaporationGas Boiling temp (K) DH (J/mol) DS (J/mol.K)Helium 4.2 92 22Nitrogen 77 5,600 72Oxygen 90 6,800 76H2O 370 40,000 120

11. Exercise What is the entropy change per mole of water when ice melts (at p = 1 atm)? ice Waterat 0 0C Q

12. Solution What is the entropy change per mole of water when ice freezes (at p = 1 atm)? ice Waterat 0 0C Q QL = TDSDS = -QL / T = -333 J/g / 273 K * 18 g/mol = -22.0 J/K mol That’s a loss of s of about 2.73 per molecule.That is, there are about e2.7, or 15, as many microstates available to each molecule in the liquid.

13. Act 1: 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 106 J/K of entropy in the process, what is the change of entropy of the environment during this process? • A) DS = 0 J/K mol • B) DS = +106 J/K mol • C) DS > +106 J/K mol

14. 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 106 J/K of entropy in the process, what is the change of entropy of the environment during this process? • A) DS = 0 J/K mol • B) DS = +106 J/K mol • C) DS > +106 J/K mol The entropy increase of the enviroment must at least balance the entropy loss of the water. Otherwise, the total entropy of the water-environment system would decrease, in violation of the 2nd 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.

15. Chemical Equilibrium Internal energy per molecule Reactions like: aA +bB  cC,where A, B, and C are the particle types and a, b, and c are integers. In equilibrium the total free energy, F, is a minimum.We must have DF = 0 when the reaction is in equilibrium, for any reaction that takes us away from equilibrium: Therefore: amA + bmB = cmC Treating the components as ideal gases or solutes: Plug these chemical potentials into the equilibrium condition, amA + bmB = cmC, and solve for the density ratios: K(T) is called the “equilibrium constant”.

16. Formation of H2 from hydrogen atoms:H2 2H, so mH2 = 2mH.Equilibrium condition:We can use nQH, because it’s monatomic. We don’t know how to calculate nQH2, because it is diatomic and has extra U and S.However, we can estimate the effect that H2 being diatomic has on mH2. D = H2 binding energy = 4 eV T = 1000 K Non-monatomic Gases

17. ACT 2: Chemical Potential of Diatomic Gas How does including the rotational energy and entropy of H2 change its m, compared to what you’d have if it were monatomic? In other words, what is the effect on F per molecule? A) No change B) Decreases mC) Increases mD) Not enough info

18. Solution How does including the rotational energy and entropy of H2 change its m, compared to what you’d have if it were monatomic? In other words, what is the effect on F per molecule? A) No change B) Decreases mC) Increases mD) Not enough info Molecular rotation increases U.  Increases F.It also increases S (more microstates).  Decreases F. The addition of rotation modes must decrease F !!! (proof on next slide)If F were not lowered when the molecules rotated, they would not rotate! What does this do to equilibrium? mH2 is now less than 2mH, so the reaction will proceedin the direction of increasing nH2, until equality is restored. Which effect is larger?

19. Contribution of Internal Modes to m Proof that internal modes lower F: For ideal gases, the internal contributions to F are independent of the other contributions. Fint/N is their contribution to m=dF/dN Fint= Uint - TSint We can calculate Fint by remembering that theinternal modes contribute to Cv(T): This is always true, because CV is always positive.So the thermal excitation of the internal modes of a moleculealways lowers its m, and thus increases its equilibrium concentration. Cv Vibration Rotation Translation 7/2 Nk 5/2 Nk 3/2 Nk T 10K 100K 1000K

20. Act 3: Formation of H2 We can also write it like this: We have: • 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease • 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease

21. Solution We have: • 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease • 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease Since nH √nH2, decreasing nH2 decreases nH. Makes sense: The overall density is reduced.

22. Solution We have: • 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease • 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease Since nH √nH2, decreasing nH2 decreases nH. Makes sense: The overall density is reduced. nH/nH2 1/√nH2, so decreasing nH2 increases the fraction of free atoms. 2H H2 requires that two atoms meet, whileH2 2H only requires a single molecule.At low density, the rate of the second process is higher, shifting equilibrium to more H. At a given T, the fraction of atoms increases at lower molecule density! There are more H atoms in outer space than H2 molecules. Why? Two particles (H + H) have more entropy than one particle (H2). Entropy maximization dominates the tendency of atoms to bind!!!

23. f 1 T1 T2 T3 0.5 T3 > T2 > T1 0 p p0(T2) Lect. 19:Adsorption of Atoms We want to know what fraction of the surface sites are occupied, for a given gas pressure p and temperature T: Using our result: We obtain a simple relation: More atoms go onto the surface at highpressure, becausemgas increases with pressure. po(T) is the characteristic pressure at which half the surface sites are occupied. It increases with temperature due to theBoltzmann factor.

24. Example: Oxygen in blood Your body needs to carry O2 from the lungs out to tissues (called T in the reactions below), using some carrier molecule. Suppose it’s myoglobin. If there are M myoglobin binding sites and NB bound oxygen molecules, (and D is the binding energy of oxygen to myoglobin), we can analyze the transport of oxygen from lungs to tissues.

25. Solution First look at the lungs. Ignore the effect of the oxygen on the myoglobin to which it binds, so it’s just like the problem of adsorption by surface sites.Equilibrium is described by: mgas = mbound. The fraction of occupied myoglobin sites is:We can’t calculate p0, but we know empirically that f  2/3 in the lungs (where pO2 = plungs  0.19 atm). That is, What happens in the tissue? p0 is the same. Empirically, f  1/3, so . About 1/3 of the myoglobin sites are used to transport oxygen to the tissues. Transport will occur as long as the O2 pressure in the tissue is less than that in the lungs.

26. ACT 4: Oxygen in Solution 1) Consider oxygen dissolving in a liquid. The reaction is O2gas O2liquid. If you double the air pressure, keeping T constant, what happens to the density of the dissolved O2? (Assume that the liquid itself is unaffected.) A) Density is halved B) Density is unchanged C) Density is doubled 2) Suppose the solution mechanism is such that oxygen molecules can only dissolve in pairs. That is, 2O2gas one dissolved “di-oxygen” object. If you double the air pressure, keeping T constant, what happens to the density of the dissolved di-oxygen? (Assume that the liquid itself is unaffected.) A) Density is doubled B) Density is tripled C) Density is quadrupled

27. Solution 1) Consider oxygen dissolving in a liquid. The reaction is O2gas O2liquid. If you double the air pressure, keeping T constant, what happens to the density of the dissolved O2? (Assume that the liquid itself is unaffected.) A) Density is halved B) Density is unchanged C) Density is doubled 2) Suppose the solution mechanism is such that oxygen molecules can only dissolve in pairs. That is, 2O2gas one dissolved “di-oxygen” object. If you double the air pressure, keeping T constant, what happens to the density of the dissolved di-oxygen? (Assume that the liquid itself is unaffected.) A) Density is doubled B) Density is tripled C) Density is quadrupled K remains unchanged, so nliquid must double.

28. Solution 1) Consider oxygen dissolving in a liquid. The reaction is O2gas O2liquid. If you double the air pressure, keeping T constant, what happens to the density of the dissolved O2? (Assume that the liquid itself is unaffected.) A) Density is halved B) Density is unchanged C) Density is doubled 2) Suppose the solution mechanism is such that oxygen molecules can only dissolve in pairs. That is, 2O2gas one dissolved “di-oxygen” object. If you double the air pressure, keeping T constant, what happens to the density of the dissolved di-oxygen? (Assume that the liquid itself is unaffected.) A) Density is doubled B) Density is tripled C) Density is quadrupled K remains unchanged, so nliquid must double. Now: 2O2gas Diox. nDiox increases by a factor of 4.