Question 5 of 5: Boltzmann–Matano Interdiffusion Coefficient

Question 5: Boltzmann–Matano Interdiffusion Coefficient (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. $D_{AB}$ at $C_B=0.02\ \text{mol/cm}^3$ by the Boltzmann–Matano graphical method.

[Figure not reproduced. See the official exam paper.]

Approach. The Boltzmann substitution $\eta=(x-x_M)/\sqrt{t}$ reduces Fick’s second law with a concentration-dependent diffusivity, $\partial C/\partial t=\partial/\partial x[D(C)\,\partial C/\partial x]$, to an ordinary differential equation in $\eta$ alone. Integrating it from the far field down to a chosen composition $C_1$ gives the Matano relation $$D(C_1)=-\frac{1}{2t}\left(\frac{dx}{dC}\right)_{C_1}\int_{0}^{C_1}\left(x-x_M\right)dC$$ in which distances must be measured from the Matano interface $x_M$ — the plane that balances the two areas between the profile and its terminal compositions, $\int_0^{C_L}(x-x_M)\,dC=0$. The printed graph’s own left-hand edge is not that plane: the drawn curve leaves its $C_B=0.10$ plateau only at $x\approx0.078$ cm, so $x_M$ has to be located from the curve before anything is integrated.

The profile was read from the printed figure and calibrated against the figure’s own scale bar: its two arrow tips touch adjacent vertical grid lines and the dimension is labelled 0.01 cm, so one grid square is 0.01 cm. On that scale the drawn curve runs from $x=0.078$ cm at $C_B\approx0.10$ to $x=0.236$ cm at $C_B\approx0.005$.

1. Locate the Matano interface. Writing $x=x_M+s\,\mathrm{erfc}^{-1}(2C_B/C_L)$ with the left-hand terminal composition $C_L=0.10$, the 44 digitized points fall on a straight line ($R^2=0.987$) — that is, this couple is an ordinary error-function profile. The least-squares line gives $$x_M=0.159\ \text{cm},\qquad s\equiv2\sqrt{Dt}=0.0581\ \text{cm}$$ The equal-area condition $\int_0^{0.10}(x-x_M)\,dC=0$, integrated numerically over the digitized curve with the fitted tails, returns the same plane ($x_M=0.1588$ cm), and the half-height point $C_B=C_L/2=0.05$ sits at $x=0.155$ cm, as it must for a symmetric couple.
2. Read $x$ at $C_1=0.02$. From the digitized curve, $$x(C_B=0.02)=0.191\ \text{cm}\qquad\Rightarrow\qquad x-x_M=0.032\ \text{cm}$$
3. Tangent slope at $C_1=0.02$. Differentiating the fitted profile at that composition — equivalently, the tangent drawn on Fig. 4 — gives $$\left(\frac{dx}{dC}\right)_{0.02}=-1.47\ \text{cm}\,/\,(\text{mol/cm}^3)$$ A local straight-line fit through the printed-curve points bracketing $C_B=0.02$ returns $-1.95$ instead; the gap between the two is the honest precision of a tangent read from a printed graph, and it is carried through below.
4. Matano integral (area). Trapezoidal integration of $(x-x_M)$ over $0\le C_B\le0.02$, with the $C_B\to0$ tail supplied by the fitted profile: $$\int_0^{0.02}\left(x-x_M\right)dC=1.20\times10^{-3}\ \text{cm}\cdot(\text{mol/cm}^3)$$
5. Assemble $D_{AB}$. With $t=28\times3600=1.008\times10^5\ \text{s}$: $$D_{AB}=\frac{1}{2t}\left|\frac{dx}{dC}\right|\int_0^{0.02}(x-x_M)\,dC=\frac{(1.47)(1.20\times10^{-3})}{2(1.008\times10^{5})}$$ $$\boxed{D_{AB}(C_B=0.02)\approx8.7\times10^{-9}\ \text{cm}^2/\text{s}=8.7\times10^{-13}\ \text{m}^2/\text{s}}$$ Using the steeper raw tangent instead would give $1.2\times10^{-8}\ \text{cm}^2/\text{s}$, so to the precision a graphical read can support the answer is $D_{AB}\sim1\times10^{-8}\ \text{cm}^2/\text{s}=1\times10^{-12}\ \text{m}^2/\text{s}$.
6. Independent check. Because the profile is an error function, its fitted width alone fixes the diffusivity with no tangent and no area at all: $$D=\frac{(s/2)^2}{t}=\frac{(0.0291)^2}{1.008\times10^{5}}=8.4\times10^{-9}\ \text{cm}^2/\text{s}$$ within 4% of the Matano result. Both sit in the published band for substitutional interdiffusion in FCC metals at 1200 °C (e.g. Cu–Ni couples report $\tilde D\sim10^{-13}$–$10^{-12}\ \text{m}^2/\text{s}$ near this temperature).