5 7 5 7 We are asked to determine the position at which the nitrogen concentration is 0 5 kg m3 This problem is solved by using Equation 5 3 in the form C CB J D A xA xB If we take CA to be the point at which the concentration of nitrogen is 2 kg m3 then it becomes necessary to solve for xB as C C B xB xA D A J Assume xA is zero at the surface in which case 2 kg m3 0 5 kg m3 xB 0 1 2 10 10 m2 s 1 0 10 7 kg m2 s 1 8 x 10 3 m 1 8 mm Excerpts from this work may be reproduced by instructors for distribution on a not for profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5 30 5 23 This problem asks us to determine the values of Qd and D0 for the diffusion of Au in Ag from the Q Q plot of log D versus 1 T According to Equation 5 9b the slope of this plot is equal to d rather than d 2 3 R R since we are using log D rather than ln D and the intercept at 1 T 0 gives the value of log D0 The slope is equal to slope log D1 log D2 log D 1 1 1 T T T 1 2 Taking 1 T1 and 1 T2 as 1 0 x 10 3 and 0 90 x 10 3 K 1 respectively then the corresponding values of log D1 and log D2 are 14 68 and 13 57 Therefore Qd 2 3 R slope Qd 2 3 R log D1 log D2 1 1 T1 T2 14 68 13 57 2 3 8 31 J mol K 1 0 10 3 0 90 10 3 K 1 212 200 J mol Rather than trying to make a graphical extrapolation to determine D0 a more accurate value is obtained analytically using Equation 5 9b taking a specific value of both D and T from 1 T from the plot given in the problem for example D 1 0 x 10 14 m2 s at T 1064 K 1 T 0 94 x 10 3 K 1 Therefore Q D0 D exp d RT 212 200 J mol 1 0 10 14 m2 s exp 8 31 J mol K 1064 K 2 65 x 10 4 m2 s Excerpts from this work may be reproduced by instructors for distribution on a not for profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5 29 3 8 x 10 12 m2 s Note this problem may also be solved using the Diffusion module in the VMSE software Open the Diffusion module click on the D0 and Qd from Experimental Data submodule and then do the following 1 In the left hand window that appears enter the two temperatures from the table in the book converted from degrees Celsius to Kelvins viz 873 600 C and 973 700 C in the first two boxes under the column labeled T K Next enter the corresponding diffusion coefficient values viz 5 5e 14 and 3 9e 13 3 Next at the bottom of this window click the Add Curve button 4 A log D versus 1 T plot then appears with a line for the temperature dependence for this diffusion system At the top of this window are give values for D0 and Qd for this specific problem these values are 1 04 x 10 5 m2 s and 138 kJ mol respectively 5 To solve the b part of the problem we utilize the diamond shaped cursor that is located at the top of the line on this plot Click and drag this cursor down the line to the point at which the entry under the Temperature T label reads 1123 i e 850 C The value of the diffusion coefficient at this temperature is given under the label Diff Coeff D For our problem this value is 1 2 x 10 14 m2 s Excerpts from this work may be reproduced by instructors for distribution on a not for profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5 28 5 22 a Using Equation 5 9a we set up two simultaneous equations with Qd and D0 as unknowns as follows Q 1 ln D1 ln D0 d R T1 Q ln D2 ln D0 d R 1 T2 Solving for Qd in terms of temperatures T1 and T2 873 K 600 C and 973 K 700 C and D1 and D2 5 5 x 10 14 and 3 9 x 10 13 m2 s we get Qd R ln D1 ln D2 1 1 T1 T2 8 31 J mol K ln 5 5 10 14 ln 3 9 10 13 1 1 973 K 873 K 138 300 J mol Now solving for D0 from Equation 5 8 and using the 600 C value of D Q D0 D1 exp d RT1 138 300 J mol 5 5 10 14 m2 s exp 8 31 J mol K 873 K 1 05 x 10 5 m2 s b Using these values of D0 and Qd D at 1123 K 850 C is just 138 300 J mol D 1 05 10 5 m2 s exp 8 31 J mol K 1123 K Excerpts from this work may be reproduced by instructors for distribution on a not for profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5 34 5 27 a We are asked to calculate the diffusion coefficient for Mg in Al at 450 C Using the data in Table 5 2 and Equation 5 8 Q D D0 exp d RT 131 000 J mol 1 2 10 4 m2 s exp 8 31 J mol K 450 273 K 4 08 x 10 14 m2 s b This portion of the problem calls for the time required at 550 C to produce the same diffusion result as for 15 h at 450 C Equation 5 7 is employed as …