hw8sol - Homework #8 1. nv/N = 3.091 10 at 1234C = 1507 K 5...

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3.091 Fall Term 2004 Homework #8 Solutions 1. n v /N = 3.091 × 10 –5 at 1234 ° C = 1507 K = 5.26 × 10 –3 at mp = 2716 K n v = A exp H v N R T v 3.091 × 10 –5 = A exp – H (1) 1507 R v 5.26 × 10 –3 = A exp – H (2) 2716 R H v + H v (1)/(2) = 5.876 × 10 –3 = exp 1507 R 2716 R Taking the logarithm of both sides gives H v 1 1 v – 5.137 = R 1507 + 2716 = –2.954 × 10 –4 H ⇒ ∆ H v = 1.497 × 10 5 J/mol R 2 θ sin 2 θ normalized clear fractions (hkl)? sin 2 θ h 2 + k 2 + l 2 38.40 0.108 1.00 3 111 0.0360 44.50 0.143 1.32 4 200 0.0358 64.85 0.288 2.67 8 220 0.0359 77.90 0.395 3.66 11 311 0.0358 81.85 0.429 3.97 12 222 0.0358 98.40 0.573 5.31 16 400 0.0358 111.20 0.681 6.31 19 331 0.0358
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3.091 Fall Term 2004 Homework #7 Solutions page 2 2. All we need to know is the temperature dependence of the vacancy density: H v n v = Ae RT where T is in Kelvins and the m.p. of Al is 660ºC N 0.08 v v Ae = −∆ H/RT 1 , where T 1 = 923K; 0.01 = Ae 2 , where T 2 = 757K 100 100 Taking the ratio: 81 0 4 A e 1 H v 1 1 × v R T 1 T 2 H 1 1 v = = e ln 8 = 11 4 2 × A v R
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This note was uploaded on 11/02/2009 for the course CHEMISTRY 3.091 taught by Professor Donsadoway during the Fall '04 term at MIT.

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hw8sol - Homework #8 1. nv/N = 3.091 10 at 1234C = 1507 K 5...

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