Quiz 13 10 points The difison ceficien lfinty foranc crpe ms

Quiz 13 1(0 points) The difison ceficien lfinty foranc crpe m/s at 1200°C Determine the difusnity for Fick\'s first law of difusion is difusion fm cin copper at 800 C ick\'s second law of diffusion is Cn :D The Arthenius expression for the difusion coeficientis

Solution

Step 1: Given data:

D_x@1000C=4.94x10^-13m²/s

D_x@1200C=5.73x10^-12m₂/s

Step 2: Find out the Dx@800C:

From the Arrhenius expression

D=D₀exp[-E/RT]

Step 3: Find out the energy

T1=1000+273=1273K

T2=1200+273=1473K

E=-R(ln(D1)-ln(D2))/(1/T1-1/T2)

E=-8.314(ln(4.94x10^-13)-ln(5.73x10^-12)/(1/1273-1/1473)

E=191747.0695J/mol

Step 4: Find out the value of D₀:

D₀=D1exp[E/RT1]

D₀=4.94x10^-13 exp[191747/(8.314x1273)]

D₀=3.64x10^-5m²/sec

Step 5: Find out the diffusivity for zinc in copper at 800C:

D=D₀exp[-E/RT]

D=3.64x10^-5exp[-191747.06/((8.314x(800+273))]

D=1.66x10^-14m²/sec