Question

Compute diffusion coefficients for the interdiffusion of carbon in both (a) α-iron (BCC) and (b) γ-iron (FCC) at 1000˚C. Assume that D0 for the interdiffusion of carbon in α-iron and in γ-iron are 1.1 × 10-6 and 2.3 × 10-5 m2/s, respectively, and that Qd are 80 and 148 kJ/mol, respectively.

Answer 1

Explanation:

According to Ficks first law, the relation between diffusion flux and diffusion coefficient is as follows.

             J = $-D \frac{dC}{dx}$

where, D = function of temperature

         D = $D_{e} exp (\frac{-Q_{d}}{RT})$

where,   $D_{e}$ = pre-exponential independent of temperature

(a)  Substituting the given values into the above formula, we will calculate the value of D as follows.

                   D = $D_{e} exp (\frac{-Q_{d}}{RT})$

                      = $1.1 \times 10^{-6} m^{2}/s exp (\frac{-80 \times 10^{3} J/mol}{8.314 J/mol \times 1273 K})$

                      = $1.1 \times 10^{-6} m^{2}/s exp (\frac{-80 \times 10^{3} J/mol}{8.314 J/mol \times 1273 K})$

                      = $1.1 \times 10^{-6} \times 1.007$

                     = $1.108 \times 10^{-6} m^{2}/s$

Therefore, diffusion coefficient for the inter-diffusion of carbon in $\alpha$-iron is $1.108 \times 10^{-6} m^{2}/s$.

(b)   For $\gamma$-iron we will calculate the value of D as follows.

                D = $D_{e} exp (\frac{-Q_{d}}{RT})$

                      = $2.3 \times 10^{-5} m^{2}/s exp (\frac{-148 \times 10^{3} J/mol}{8.314 J/mol \times 1273 K})$  

                      = $2.3 \times 10^{-5} \times exp(-13.9)$

                      = $2.3 \times 10^{-5} \times 9.189 \times 10^{-7}$

                      = $2.113 \times 10^{-12} m^{2}/s$

Therefore, diffusion coefficient for the inter-diffusion of carbon in $\gamma$-iron is $2.113 \times 10^{-12} m^{2}/s$.