Lect17 - Misce ous Note llane s The end is near dont get...

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Lecture 17, p 1 Miscellaneous Notes The end is near – don’t get behind. All Excuses must be taken to 233 Loomis before noon, Tuesday, Nov. 30. The PHYS 213 final exam times are * 7-9 PM, Monday, Dec. 13 and * 8-10 AM, Tuesday, Dec. 14 . The deadline for changing your final exam time is 10pm, Tuesday, Nov. 30. Homework 6 is due Saturday , Dec. 4 at 8 am. Course Survey = 2 bonus points (accessible at the top of HW6)
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Lecture 17, p 2 Law of Atmospheres, revisited Semiconductors Reference for this Lecture: Elements Ch 12 Reference for Lecture 19: Elements Ch 13 Lecture 17 Applications of Free Energy Minimum
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Lecture 17, p 3 Thumbnail review of free energy: Equilibrium corresponds to maximum S tot = S reservoir + S small system . When we calculate S, we only need to know the temperature of the reservoir. In minimizing F (equivalent to maximizing S tot ) we don’t have to deal explicitly with S reservoir . Consider exchange of material (particles) between two containers. These are two small systems in equilibrium with a reservoir (not shown) at temperature T. In equilibrium, dF/dN 1 = 0: The derivative of free energy with respect to particle number is so important that we define a special name and symbol for it: For two subsystems exchanging Last Time Maximum Total Entropy Minimum Free Energy Equal chemical potentials 1 2 1 2 1 1 1 1 2 1 2 1 2 0 dF dF dF dF dF dN dN dN dN dN dF dF dN dN = + = - = = 1 N F = F 1 +F 2 N 1 N 2 The chemical potential of subsystem “i” i i i dF dN μ 1 2 =
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Lecture 17, p 4 Chemical Potential with Potential Energy The potential energy per particle, PE, makes an additional contribution N·PE to the free energy: F = U - TS. So, the chemical potential ( μ = dF/dN) gains an additional contribution: d(N·PE)/dN = PE. It’s the energy that one particle adds to the system. Examples: A ln T n kT PE n μ= + n = N/V particle density ln T n kT mgh n + ln T n kT n - ∆
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Lecture 17, p 5 Question: How do the gas density n and pressure p of the gas vary with height? Compare region 2 at height h with region 1 at height 0. (equal volumes) For every state in region 1 there’s a corresponding state in region 2 with mgh more energy. Assume that the temperature is in equilibrium (T 1 =T 2 ). Write the ideal gas law like this: p = nkT. (n N / V) The Law of Atmospheres 0 h p 2 , n 2 p 1 , n 1
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Lecture 17, p 6 The two regions can exchange particles (molecules), so imagine that they are connected by a narrow tube. The rest of the atmosphere is the thermal reservoir. Equilibrium: μ 1 = 2 Chemical potential (ideal gas): Solution: Ideal gas law: p/n = kT (T 1 = T 2 in equilibrium) 1 2 2 1 2 1 / ln ln mgh ln T T mgh kT n n kT kT n n n kT mgh n n n e - = + = - = 1 1 2 2 ln ln mgh T T n kT n n kT n = = + The Law of Atmospheres (2) p 2 N 2 h p 1 N 1 2 1 / mgh kT p p e - =
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Lecture 17, p 7 Electrons in Semiconductors In many materials, electrons cannot have every conceivable energy. There valence band ”) and a high energy range (the A “gap” of disallowed energies separates them. (The reason for the gap is a Physics 214 topic.) At T = 0, every valence band state is occupied. (S=0. Why? ) At T 0,
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This note was uploaded on 02/21/2011 for the course PHYS 213 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Lect17 - Misce ous Note llane s The end is near dont get...

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