A04_DIFFUSION_Report_Template_2023

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Rensselaer Polytechnic Institute *

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Material Science

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Apr 3, 2024

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pdf

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1 Aidan McVeigh Section: 3 LaGraff ENGR1600: Materials Science for Engineers Interactive Activity #04 DIFFUS ION SIMUL ATION REPOR T TEMPLATE Objective : Run simulations of random walk model of diffusion to, (1) understand diffusion at the atomic level, (2) calculate diffusion coefficient, D, by simulating diffusion distance versus time at four different temperatures, (3) create an Arrhenius plot to determine the activation energy, Q. Procedure : Download CDF App of Diffusion-Simulation from LMS. Follow instructions on the CDF app. Step 1 (Part B: Obtain diffusion coefficient, D, from RMS displacement versus time at four different temperatures) : a) On a single graph, insert a figure with the RMS displacement versus square root of time (obtained in Part B), at four different temperatures, showing the best linear curve fit for each (y = mx + b) and the R^2 value. The slopes of each line is the square root of 6D, or (6D) 1/2 , from which the diffusion coefficient at each temperature can be calculated. 1K: y = 0.2862x + 0.0479 ; R^2 = 0.9955 1.5K: y = 0.4197x + 0.047 ; R^2 = 0.9957 2K: y = 0.5144x + 0.015 ; R^2 = 0.9978 0 1 2 3 4 5 6 7 0 2 4 6 8 10 12 RMS Displacement (nm) Sqrt(t) RMS Displacement vs. Sqrt(t) of 500 Particles 1K 1.5K 2K 2.5K IMPORTANT: Discussion is encouraged, but each student should do their own simulations, calculations, and graph preparation.
2 2.5K: y = 0.5646x + 0.152 ; R^2 = 0.9937 b) Insert a table of the Diffusion Coefficient at the 4 different temperatures you selected from part (a). Include correct units Temp. (K) Slope Diffusion Coefficient (nm^2 / t) D (m^2 / s) 1000 0.2862 0.01366 1.366 * 10^-20 1500 0.4197 0.02935 2.935 * 10^-20 2000 0.5144 0.04411 4.411 * 10^-20 2500 0.5646 0.05313 5.313 * 10^-20 Step 2 (Obtain activation energy from Arrhenius plot) : Plot the natural log of D versus inverse temperature (K -1 ). From the slope of the curve determine the activation energy, Q. Q = 19.044 * 10^3 J/mol Discussions : 1. What did you notice as you changed the number of particles in the simulation (Part A)? What were the similarities and differences? The first thing I noticed is that as the number of particles increased, so did the chance of my computer crashing.
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