Lecture 22 Freezing/boiling Point Elevation/depression Supercooling/superheating Midterm review

1. Phase Diagram: p Solid Liquid p3 p2 Gas p1 T Lecture 22Freezing/boiling Point Elevation/depressionSupercooling/superheatingMidterm review

2. Phase Transitions and Volume Change m Tfreeze T S L p Solid Liquid Gas T So far, our liquid and solid m(T) curves have beenindependent of pressure. We justified this with the fact that solids and liquids are nearly incompressible (which is true). This means that Tfreeze should not depend on pressure. So: Why is the line that separates the liquid and solid phases not vertical? The reason is that, although they are incompressible, the volumes of the liquid and solid are not the same.Most substances expand when they melt. Suppose the volume change per mole is DV (whichmeans DV/NA per particle). Thus, (during melting) when a particle moves from solid to liquid, work is done by it (it’s expanding):W = pDV/NA. This is a positive contribution to the liquid’s chemical potential, and it increases with pressure. If you look at the chemical potential graph above, you’ll see that this causes Tfreeze to increase with pressure.

3. Water is Unusual The typical phase diagram The H2O phase diagram p Solid Liquid T = 0.01° Cp = 0.006 atm Gas T Most materials freeze if you push hard enough, but water melts!This results from the fact that water expands when it freezes (ice floats). Consequences:Fish can live in wintry lakes, because they don’t freeze solid.Ice is slippery when you skate on it … NOT!It’s true that ice is slippery, but not for this reason - the pressure required is too large. Here’s an article about it: http://people.virginia.edu/~lz2n/mse305/ice-skating-PhysicsToday05.pdf

4. m T G L S Variation of Freezing/boiling Temperatures Lowering the density of the gas lowers its chemical potential curve mG. This will move the boiling transition to lower temperature. This is the reason water boils at lower T at high altitude. We can similarly lower the mL curve by lowering the density of the molecules in the liquid. This will lower the freezing temperature of the liquid. Example: Putting salt on ice lowers the melting point. Why? The salt dissolves in the liquid, lowering the density (and thus m) of the remaining liquid H2O. The salt does not easily penetrate the solid ice, so has little effect on its m. This illustrates a general principle: Dissolving substance A in substance B lowers the chemical potential. If the chemical potential weren’t lowered, it wouldn’t dissolve! Note: The density argument we made above isn’t always the important point. Sometimes, the energy change is also important (even dominant). http://antoine.frostburg.edu/chem/senese/101/solutions/faq/thermo-explanation-of-freezingpoint-depression.shtml

5. Freezing point depression/Boiling point elevation Lowering the density of the gas lowers the chemical potential curve mg . We can similarly lower the mL curve by lowering the density of the molecules in the liquid. Tfreezing T Tboiling mg ms mL Gasmg LiquidmL Solid ms http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml

6. Freezing point depression/Boiling point elevation Gasmg LiquidmL Solid ms Liquid with solvent Lowering the density of the gas lowers the chemical potential curve mg . We can similarly lower the mL curve by lowering the density of the molecules in the liquid. How do we lower the density of molecules? Replace them with other sorts of molecules (It is only the molecules of the same species as the solid that count toward equilibrium). There are very many solutes which: • Dissolve easily in the liquid phase • Don’t fit well at all in the crystal • Have very low vapor pressure (absent from gas) These have almost no effect on H2O’s mg or mS. • anti-freeze • salt on roads • sugar would work too, but • doesn’t have as many molecules/kg • costs more • would leave a sticky mess! Tfreezing T’boiling T Tboiling T’freezing mg ms mL mL,solvent They always lower ml (stabilize liquid in each case). So they lower Tf and they raise Tb

7. FYI: Quantitative Freezing Point Depression, H2O How it works (exact for dilute solutions): A solvent molecule entering from another phase gives extra V for each solute molecule to roam in. That increases the solute’s S, reduces G= U+pV-TS. Smaller G on adding a solvent particle means reduced of solvent. • At low solute concentration, n= N/V, the only effect of the solute on  is via getting more S when the solvent volume increases. • (too few solutes to interact, solute-H2O interaction doesn’t change when you get more H2O) • We can calculate the effect exactly via state counting! This result does not depend on the type of solute particle! • At higher solute concentrations other terms (solute-solute interactions) also matter • Adding 1 molecule of water increases the solution volume V by dV=3*10-29 m3. • So V available to each solute molecule increases by a factor (1+dV/V). • So its S increases by k ln(1+dV/V)= k dV/V • This happens to nV molecules. • So S = knV dV/V = nkdV. • At 1 Molar solute, n=6*1026/m3: DS = nkdV = 2.5*10-25 J/K • DS of freezing = Latent Heat/T= (6000 J/Mole) /273K ~ 22 J/K-Mole, ~3.7 10-23 J/K-molecule • So a 1M solute increases the melting DS by ~ 0.7% (2.5*10-25 / 3.7 10-23 ) • To keep the TDS term in DG balancing DH at TF,TF must drop by about 0.7%/M or 1.9 K/M (works well experimentally) • Obtained via plain old state counting, and minimizing G in equilibrium.

8. m T G L S ACT 1: Boiling Temperature What does the addition of salt to liquid water do to its boiling temperature? • A) Raises the boiling temperature • B) Lowers the boiling temperature • C) Does not change the boiling temperature

9. m T G L S Solution What does the addition of salt to liquid water do to its boiling temperature? • A) Raises the boiling temperature • B) Lowers the boiling temperature • C) Does not change the boiling temperature Adding salt to the liquid lowers the liquid’s m, (because it lowers the H2O density) raising the boiling temperature. However, we must be careful. What does the salt do to the gas’s chemical potential? In this case, salt does not evaporate easily (it has a very low vapor pressure), so it has a negligible effect on mG. If we were to dissolve a more volatile substance (e.g., alcohol) in the liquid, the answer might be different.

10. Supercooling & Superheating Gasmg LiquidmL Solid ms If we cool a liquid to Tfreezing, normally it will freeze to lower its free energy. However, if there are no nucleation sites, it is possible to “supercool” the liquid well below the freezing temperature. This is an unstable equilibrium. What happens when the system suddenly freezes? Tfreezing Tsc T Tboiling mg Supercooled liquid ms mL

11. Solution Gasmg LiquidmL Solid ms If we cool a liquid to Tfreezing, normally it will freeze to lower its free energy. However, if there are no nucleation sites, it is possible to “supercool” the liquid well below the freezing temperature. This is an unstable equilibrium. What happens when the system suddenly freezes? The free energy drops. This “latent heat” is suddenly released. It is similarly possible to superheat a liquid above its boiling temperature; adding nucleation sites leads to rapid (and often dangerous!) boiling. Tfreezing Tsc T Tboiling mg Supercooled liquid ms mL Superheated liquid

12. Supercooling/superheating Demos Liquid water can be • heated above 100°C • http://www.animations.physics.unsw.edu.au/jw/superheating.htm • Cooled well below 0°C • http://www.youtube.com/watch?v=13unrtlvfrw Why does the water only partly boil or freeze (slush)? What is T of the leftover water?: Boiling: 100°C Freezing: 0°C

13. Supercooling/superheating Demos (2) Freezing: warmed to 0°C. The molecules U* dropped as they joined the ice, releasing U* that warmed things up to 0°C At which point freezing stopped leaving slush. *To be precise, it’s U+pV (enthalpy, H) here at constant p Boiling: cooled to 100°C. U+pV (enthalpy, H) increased as molecules joined the gas, soaking up H from surroundings, cooling them to 100°C At which point boiling stopped leaving some liquid.

14. Thank you,Erika !!

15. Physics 213 Midterm Most Missed Problems

16. [30%] On a planet not entirely unlike earth, the ratio of the partial pressure of N2 to that of O2 equals 1 at an altitude of 1 km: pN2/pO2 = 1. Assuming that T = 200K, and the gravitational constant is 5 m/s2, what is the ratio pN2/pO2 at an altitude of 0 km. (Hint: You can use the Boltzmann distribution to calculate the law of atmospheres.) a. pN2/pO2 = 0.90 b. pN2/pO2 = 0.99 c. pN2/pO2 = 1.00 d. pN2/pO2 = 1.01 • pN2/pO2 = 1.10 9 [55%]. Assuming that the total number of gas particles is constant, what will happen to the density of the N2 at zero altitude as the temperature drops from 200 K  190K. a. the density will decrease b. the density will stay the same c. the density will increase

17. A Cu sphere of mass 0.5 kg is thermally linked by three wires to a large thermal reservoir at 0 °C. The length of each wire is 1m and the radius is 1 mm. The initial temperature of the sphere is 100 °C. The specific heat of Cu is 386 J/kg-K. The heat conductivity of copper is κ=401 W/m-K. 2. [45%] How much time will it take for the temperature of the Cu sphere to decrease from 100°C to 99.9°C ? a. 10 s b. 50 s c. 100 s d. 10 μs • 50,000 s 3. [80%] Assume now that we cut two out of three wires connected the Cu sphere to the Cu block. How will the time required to cool the sphere from 100 °C to 99.9 °C change? a. The time becomes longer by a factor of 9. b. The time becomes shorter by a factor of 3. c. The time becomes longer by a factor of 3.

18. excited state, degeneracy = 2 ground state, no degeneracy E = ε + μB, “μ opposite B” E = ε – μB, “μ along B” ground state Consider a collection of atoms, all with a single ground state, and a doubly degenerate first excited state. 5. [65%] Now we apply a magnetic field which happens to split the first excited level into two (non-degenerate) states, one of which has the energy ε + μB (“μ opposite B”), the other with energy ε – μB (“μ along B”), as shown. At T = 200 K we observe that the relative fraction of atoms with spin up to those with spin down is: F = N(“μ opposite B”)/N(“μ along B”) = 0.1. What is μB? a. 0.02 eV b. 0.46 eVc. 0.8 eV 6. [70%] Now we increase the temperature T (a lot). What will happen to F = N(“μ opposite B”)/N(“μ along B”)? a. F → 1b. F → ∞c. F → 0 d. F stays the same e. Cannot tell from the information given 7. [47%] As the temperature T becomes very large, what is the dimensionless entropy for a system with N of these atoms (still with small magnetic field on)? a. N ln 2 b. N ln 3 c. cannot be determined from the information given

19. Good Luck !!

20. Supplement: Semi-quantitative Analysisof Freezing Point Depression We can figure out how big the effect is if we know how much the solute reduces water’s m. Let’s estimate by assuming the solute is ideal, and counting states. Adding 1 molecule of water increases the solution volume V by dV=3*10-29 m3. (You can check.) So V available to each solute molecule increases by a factor (1+dV/V). So its entropy (s) is changed by ln(1+dV/V)= dV/V If the solute concentration is n=N/V, this happens to nV molecules. So the net effect on their total s is ds = nV dV/V = ndV. At 1 Molar, n=6*1026/m3: ndV=6*1026/m3 *3*10-29 m3= 1.8*10-2 So when a water molecule is added there’s an extra ds = 1.8*10-2 Per water molecule: dS = 2.5*10-25 J/K. Per mole: extra DS= 0.15 J/K DS of freezing = Latent Heat/T= (6000 J/Mole) /273K ~ 22 J/K-Mole. So a 1M solute increases the melting DS by ~ 0.7%. To keep the TDS term in DG balancing DH at TF,TF must drop by about 0.7%/M or 1.9 K/M The real number from a table is about 1.86 K/M. We got that by going back to plain old state counting, and remembering to minimize G in equilibrium.