Question

The diffusion coefficients for carbon in nickel are given at two temperatures: T (°C) 600 celcius 700 celcius (a) Determine the values of D0 and Qd. (b) What is the magnitude of D at 850°C? D (m2/s)= 5.5 × 10–14@ 600 and 3.9 × 10–13@ 700

Answer 1

The diffusion coeffecients at different temperatures are

@ 600°C / 873 K = 5.5 × 10–14

@ 700°C ./ 973K =3.9 × 10–13

we know that

lnD1 = lnD0 - Qd / R (1/T1).....(1)

lnD2 = lnD0 - Qd / R (1/T2).....(2)

substract (2) from (1)

lnD1-lnD2 = -Qd / R [1/T2 - 1/T1]

So Qd = -R X ln(D1/D2) / [1/T1 - 1/T2]

Qd = - (8.314) X (-1.959) / [0.00114- 0.001027)

Qd = 16.287 / 0.000113

Qd = 144132.7J/mole

Now

lnD1 = lnD0 - Qd / R (1/T1).....(1)

ln 5.5 × 10–14 = ln D0 - 144132.7/ 8.314 X 873

ln 5.5 × 10–14 = ln D0 - 19.858

Taking antilog

5.5 × 10–14 = D0 + (3.136 X 10^-9)

D0 = 1.0 X 10^-5 m^2 /s

b) we can calculate D at 850 C or 1123 K

D = D0 exp ( Qd / RT)

D = 1.0 X 10^-5 exp ( 144132.7 / 8.314 X 1123)

D = 3.69 X 10^-12 m^2 /sec