503hw2 - Solution to homework#2 1 Since C t C s 0.95Cs C s...

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Solution to homework #2 1. Since C t C C C C C C s s s s s ( ) . . . - - = - - = < 0 0 95 0 0 05 08 , the first term of the solution can be used: C t C C C s s ( ) - - 0 =0.05= 8 2 2 2 p p exp - ° L ± ² l ³ Dt L therefore, L Dt 2 2 2 0 05 8 = - ° L ± ² l ³ p p ln . L= - ° L ± ² l ³ - p p 2 5 2 10 1 0 05 8 x x x ln . = 0.006cm 2. (a) For the detector to lose its sensitivity due to diffusion at different temperature, the diffusion distance is constant: D T t D T t ( ) ( ) 1 1 2 2 = or D D D Q RT D Q RT t t 1 2 0 1 0 2 2 1 = - - = exp[ / ( )] exp[ / ( )] therefore, Q R T T t t 1 1 2 1 2 1 - ° L ± ² l ³ = ° L ± ² l ³ ln 2(a) and ( ) ( ) Q t t T T R x kJ mol = - = + - + = ln / / / ln / / ( ) / ( ) . . ( / ) 2 1 2 1 1 1 250 55 1 273 110 1 273 162 8 314 40 3 . According to Eqn. 2(a) and using t 1 =55hr, T 1 =435K and T 3 =298K: ln( / ) . . t x 3 3 55 40 3 10 8 314 1 298 1 435 = - ° L ± ² l ³ t 3 =9300 (hr) (b) Assume that the detector loses its sensitivity when the diffusion distance exceeds the spacing of the interlayer, i.e. Dt nm = 13 Therefore, D = (13x10 -7 cm) 2 /(55hr x 3600 s/hr)=8.5x10 -18 cm 2 s -1 3. (a) Since Dt cm s x s cm cm r = = << = - - 10 1 0 01 0 5 4 2 1 . . , the short time solution can be applied: C(x,t)=A+Berf x Dt 2 ° L ± ² l ³ From C(0,t)=0 and C(x,0) = C 0 , we have A=0 and B=C 0 .

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Therefore, C(x,t)=C 0 erf x Dt 2 ° L ± ² l ³ where C 0 =25x10 -6 x7.8gcm -3 =1.95x10 -4 gcm -3 . The flux of hydrogen from the surface is J D C x DC x erf x Dt DC Dt x Dt x x x x | | | exp | = = = = = - = - ° L ± ² l ³ Ø º OE ø û oe = - - ° L ± ² l ³ 0 0 0 0 0 2 0 2 2 1 2 4 p J C D t x x gcm s x | . . ( ) = - - - - - = - = - ´ = - 0 0 4 4 6 2 1 195 10 10 1 11 10 p p
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