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2 but ( Intro Experimental Results&Discussion or
3 kinetics
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5 Prof. Juan Manuel Garcia Ruiz Gypsum (36 feet ~ 11 m) CaSO 4 2H 2 O Cueva de los Cristales (Cave of Crystals)
6 he Cullinan Diamond is the largest rough gem-quality diamond ever found, at 3, carats ( g). It was found by Frederick Wells, surface manager of the Premier Diamond Mining Company in Cullinan, Gauteng, South Africa, on January 25, he stone was named after Sir homas Cullinan, the owner of the diamond mine.
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11 Polystyrene particles d = 200 nm (Duke Scientific Corporation)
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13 Grain X µm µm
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15 Si wafer, d = 300 mm, (100) oriented
16 Watson, Crick double helix DNA
17 1914 (Max von Laue) X 1915 (Sir William Henry Bragg) (William Lawrence Bragg) X 1926 (Jean Baptiste Perrin) 1937 (Clinton Joseph Davisson) (George Paget homson)
18 1956 (William Bradford Shockley) (John Bardeen) (Walter Houser Brattain) 1973 (Ivar Giaever) 2000 Z. I. (Zhores I. Alferov) H. (Herbert Kroemer) (Jack S. Kilby) (IC)
19 1946 James Batcheller Sumner John Howard Northrop Wendell Meredith Stanley 1962 Max Ferdinand Perutz John Cowdery Kendrew
20 1964 Dorothy Crowfoot Hodgkin X 1985 Herbert Aaron Hauptman Jerome Karle
21 1988 Johann Deisenhofer Robert Huber Hartmut Michel
22 Extraction Purification Crystal Growth *124 *89 *63 *19 PDB holdings: X-ray: NMR: 7259 (30-Mar-2010) Bottle Neck!!! Structure Analysis *15 *A summary of progress in the Human Proteome Structural Genomics Pilot Project (
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25 1962 Cavendish Laboratory of Cambridge University, Sir Lawrence Bragg, Max Perutz, John Kendrew (James Dewey Watson) (Francis Harry Compton Crick) (Maurice Hugh Frederick Wilkins) (Rosalind Elsie Franklin)
26 2 Silicon IC, LSI
27 Silicon jpg
28 Ca 5 (PO 4 ) 3 OH CAl 2 O 3
29
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31 MOCVD GaN
32 µ = k ln (C / C e ), µ = l / e
33
34 Solid Liquid Gas
35 S = k N, V SB V ln( = N! N k ) ln Nk W B B ln( V N )
36 G = U + PV S = A + PV = H S G Liquid Solid e
37 µ dn dv N, V, S, E P e δq = ds
38 de = ds PdV + µ dn dg = Sd + VdP + µ dn...(1) G(, P, ln ) = lg(, P, N ) G = N µ (, P)...(2) (1), (2) dg = Nd µ + µ dn Nd µ = Sd + VdP
39 P N L N L -δn N S Crystal N s +δn Melting Freezing riple point G(, P, N = G G N S δ S = G (, P, N Liquid = ) µδ Gas S S N ) + G Critical Lpoint = ( µ µ ) δ µ = µ L µ S L (, P, N G µ S, = µ L, δn S = δn L = δn N L N L )
40 P µ = S µ L II Melting Liquid Critical point Freezing I Crystal riple point µ = = µ µ S L G Gas
41 3
42 P = const. M Crystal δq Liquid Q
43 s s s N Q M S L M = = ) ( δ G e Liquid Solid M S L S L S L M S S M S L M L L h P P h h h s s s h s h = = = = = = ), ( ), ( ) ( µ µ µ µ
44 µ M = h P P P P P P P P M M S M L M S M L M S M L S L L = + = = ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( µ µ µ µ µ µ µ µ µ
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3 3.1 [ ]< 85, 86 > ( ) ds > 0. (1) dt ds dt =0, S = S max. (2) ( δq 1 = TdS 1 =0) (δw 1 < 0) (du 1 < 0) (δq 2 > 0) (ds = ds 2 = TδQ 2 > 0) 39 3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3)
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1 Physical Chemistry I (Basic Chemical Thermodynamics) [I] [II] [III] [IV] Introduction Energy(The First Law of Thermodynamics) Work Heat Capacity C p and C v Adiabatic Change Exact(=Perfect) Differential
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10 15 01 02 1 2 3 03 04 4 05 06 5 153,476 232,822 6,962 19,799 133,362 276,221 344,360 440,112 412,477 846,445 164,935 422,265 1,433,645 26,694 336,206 935,497 352,675 451,321 1,739,493 30,593 48,894 153,612
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