Question

The diffusion coefficients for silver in copper are given at two temperatures: VMSE: D0 and Qd from Experimental Data T (°C) D (m2/s) 650 5.5 × 10−16 900 1.3 × 10−13 (a) Determine the values of D0 and Qd. (b) What is the magnitude of D at 875°C?

Answer 1

Answer:

Explanation:

From the formula;

• RT = Qd / (lnD0 - lnD)

• R = universal gas constant,

• Qd = activation energy

When T = 923 K and D = 5.5 x 10-16 m2/s

923 x R = Qd / [lnDo - ln(5.5 x 10-16 m2/s)] ----- (1)

When T = 1173 K and D = 1.3 x 10-13 m2/s

Plugging the values ; and dividing equation 1 by 2

1173 x R = Qd / [lnDo - ln(1.3 x 10-13 m2/s)] ----- (2)

• By dividing equation(1) and (2) we get

• 923 / 1173 = [lnDo - ln(5.5 x 10-16 m2/s)] / [lnDo - ln(1.3 x 10-13 m2/s)]

• Do = 0.0000754 m2/s

• Putting the value of Do in equation (1) we get

• 923 x 8.314 JK-1mol-1 = Qd / [ln(7.54x10-5 m2/s) - ln(5.5 x 10-16 m2/s)]

• Qd = (923 x 8.314 JK-1mol-1) x [ln(7.54x10-5 m2/s) - ln(5.5 x 10-16 m2/s)] = 196783.83 J/atom

b) T = 875°C = 875+273 = 1148K

Hence 1148 x 8.314 JK-1mol-1 = 196783.83 J / [ln(7.54x10-5 m2/s) - lnD]

D = 8.38x10-14 m2/s