Lect12 - Misc. Notes The end is near don't get behind. All...

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Physics 213: Lecture 12, Pg 1 Misc. Notes The end is near – don’t get behind. All Excuses must be taken to 233 Loomis before noon, Friday, Dec. 4. The PHYS 213 final exam times are * 8:00 – 10:00 am Wed., Dec. 16 and * 1:30 - 3:30 pm Wed., Dec. 16. The deadline for changing your final exam time is 10pm, Tuesday, Dec. 1. Homework 6 is due Saturday , Dec. 5 at 8 am. Course Survey = 2 bonus points (accessible at the top of HW6) BEWARE!!! The midterm average (85%) was significantly higher than usual. Don’t become complacent!!
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Physics 213: Lecture 12, Pg 2 Lecture 12 Lecture 12 Applications of Free-Energy Minimum Applications of Free-Energy Minimum Agenda for today Agenda for today Law of Atmospheres, revisited Semiconductors Reference for this Lecture: Elements Ch 12 Reference for Lecture 13: Elements Ch 13
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Physics 213: Lecture 12, Pg 3 Last time: Free Energy, F = U - TS Last time: Free Energy, F = U - TS Answer: Equilibrium corresponds to maximum S tot = S reservoir + S small system . The free energy incorporates the effect of the reservoir by its temperature T. In minimizing F (equivalent to maximizing S tot ) we don’t have to deal explicitly with S reservoir . Consider exchange of gas between two containers: The derivative of free energy with respect to particle number is so important in determining an equilibrium condition that we define a special name and symbol for it: For these 2 subsystems exchanging particles , we see that the condition for “chemical equilibrium” is: V 1 V 2 2 1 μ = μ chemical potential of subsystem “i” = i i i dN dF = μ Maximum Total Entropy Minimum Free Energy Equal chemical potentials 0 dN dF dN dF dN dF dN dF dN dF 2 2 1 1 1 2 1 1 1 = - = + = Equilibrium condition: (dF/dN 1 = 0) dN dF dN dF 2 2 1 1 = 1 N equilibrium value F = F 1 +F 2 Small system at temperature T
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Physics 213: Lecture 12, Pg 4 Chemical potential of particles with Chemical potential of particles with potential energy potential energy (e.g., (e.g., atoms in a gravitational field or atoms bound into molecules) atoms in a gravitational field or atoms bound into molecules) The potential energy gets added to the translational energy: + = PE kT 2 3 N U ln T n kT PE n μ = + n = N/V particle density Atom in gravity: ln mgh Molecule with binding energy : ln T T n kT n n kT n = + = - ∆ So, the chemical potential ( μ = dF/dN) gains an additional contribution: d(N·PE)/dN = PE. (For now we are ignoring the rotation and vibration of molecules.) PE = potential energy per particle The potential energy makes an additional contribution (N·PE) to the free energy: F = U - TS. The resulting chemical potential in this case is therefore: Examples Remember: Chemical potential is the change in free energy when we add one particle (at constant V,T).
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Physics 213: Lecture 12, Pg 5 The Law of Atmospheres: Revisited The Law of Atmospheres: Revisited Problem: Find how the gas density n and pressure
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Lect12 - Misc. Notes The end is near don't get behind. All...

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