Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]C_o = 0.20 \ wt\% \\ \\ C_s = 1 \ wt\% \\ \\ t = 51 \ h \\ \\ x = 3.9 \ mm \\ \\ C_x = 0.35 \ wt\%[/tex]
[tex]\text{Using Fick's 2{nd} \ law \ of \ diffusion;} \\ \\ \dfrac{C_x- C_o}{C_s-C_o}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]\dfrac{0.35-0.20}{1-0.20}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]0.1875 = 1 - erf ( \dfrac{x}{2\sqrt{DT}}) \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 1 - 0.1875 \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 0.8125[/tex]
[tex]\text{To find the value of Z by Obtaining Data from Tabulation of Error Function}[/tex] [tex]\text{Table Values:}[/tex]
Z erf(z)
0.90 → 0.7970
0.95 → 0.8209
? → 0.8225
∴
[tex]\dfrac{z-0.90}{0.95-0.90}= \dfrac{0.8125-0.7970}{0.8209-0.7970}[/tex]
[tex]\dfrac{z-0.90}{0.05}= \dfrac{0.0155}{0.0239}[/tex]
[tex]z = 0.9324[/tex]
[tex]\text{To determine the diffusion coefficient;}[/tex]
[tex]erf (0.9324) = 0.8125 = erf (\dfrac{x}{2\sqrt{Dt}}) \\ \\[/tex]
[tex]\dfrac{x}{2 \sqrt{Dt}}= 0.9324 \\ \\ \dfrac{3.9 \times 10^{-3}}{2 \times \sqrt{D\times 51 \times 3600}} = 0.92324 \\ \\ \sqrt{D} = 4.88 \times 10^{-6} \\ \\ D = \sqrt{4.88 \times 10^{-6}} \\ \\ D = 2.38 \times 10^{-11} \ m^2 /s[/tex]
