Question

The diffusion coefficients for copper diffusing in gold are given at two temperatures (hint: logarithm):(a) Determine the values of D0and the activation energy Qd?(b) What is the magnitude of D at 1100°C?

Answer 1

Answer:

Explanation:

From the formula;

• RT = Qd / (lnD0 - lnD)

• R = universal gas constant,

• Qd = activation energy

• When T = 1273 K and D = 9.4 x 10-16 m2/s

1273 x R = Qd / [lnD0 - ln(9.4 x 10-16 m2/s)] ----- (1)

When T = 1473 K and D = 2.4 x 10-14 m2/s

• Plugging the values ; and dividing equation 1 by 2

• 1473 x R = Qd / [lnD0 - ln(2.4 x 10-14 m2/s)] ----- (2)

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

1273 / 1473 = [lnD0 - ln(2.4 x 10-14 m2/s)] / [lnD0 - ln(9.4 x 10-16 m2/s)]

D0 = 0.0000216919 m2/s

• Putting the value of D0 in equation (1) we get

• 1273 x 8.314 JK-1mol-1 = Qd / [ln(2.17x10-5 m²/s) - ln(9.4 x 10-16 m²/s)]

• Qd = (1273Kx8.314 JK-1mol-1) x [ln(2.17x10-5 m²/s) - ln(9.4 x 10-16 m²/s)] = 252.56 KJ or 252562 J/atom

b) b) T = 1100 C = 1100+273 = 1373 K

• Hence 1373 x 8.314 JK-1mol-1 = 252562J / [ln(2.17x10-5 m2/s) - lnD]

• D = 5.344x10-15 m²/s

Answer 2

Explanation:

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