Homework_3_solution - Introduction To Materials Science,...

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Polyethylene Spring 2005 Homework #3 (Solution) Problem 1 Diffusion is a thermally activated process. Parameters of a thermally activated process often obey an exponential equation. For diffusion coefficients it has the form: D = D 0 e Q D RT D is a function of temperature. More, precisely, log(D) is a linear function of 1/T, as shown in the figure in Homework #3. In this equation you have two unknown values, D o and Q. You just need to read out two points (D 1 ,T 1 ) and (D 2 ,T 2 ) on the curve and then you can determine the values of D o and Q. Pay attention to the axes of the figure: logD and 1/T. Taking natural logarithmic of this equation yields: ln D = ln D 0 Q D R 1 T From the graph (i) when D = 10 -13 m 2 /s; T=1183˚K (ii) when D=10 -15 m 2 /s, T=968˚K Substitute the values of D and T above into the equation gives QD = 204 kJ/mol D0 = 1.011 10 -4 m 2 /s Problem 2 It is important to recognize that this is a situation conforming to: D 1 t 1 = D
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Homework_3_solution - Introduction To Materials Science,...

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