Flux Calculation Using Fick's First Law

The problem asks for the magnitude of the diffusion flux ($J$) at a specific location ($x = 2$ mm) within a solid. We are given the concentration ($C$) profile and the diffusion coefficient ($D$).

Concentration Gradient Derivation

The concentration ($C$) of the solute is given by the expression:

$C = a_1x^2 + a_2x$

where $x$ is the position in mm.

Fick's First Law relates flux to the concentration gradient:

$J = -D \frac{dC}{dx}$

First, we need to find the concentration gradient, $\frac{dC}{dx}$. Differentiating the concentration expression with respect to $x$:

$\frac{dC}{dx} = \frac{d}{dx}(a_1x^2 + a_2x)$

$\frac{dC}{dx} = 2a_1x + a_2$

Evaluating Flux Magnitude

We are given the following values:

• $a_1 = 1 \text{ atoms}\cdot\text{mm}^{-5}$
• $a_2 = 1 \text{ atoms}\cdot\text{mm}^{-4}$
• $x = 2 \text{ mm}$
• $D = 3 \times 10^{-3} \text{ mm}^2 \cdot \text{s}^{-1}$

Substitute the values of $a_1$, $a_2$, and $x$ into the concentration gradient equation:

$\frac{dC}{dx} \bigg|_{x=2\text{ mm}} = 2(1 \text{ atoms}\cdot\text{mm}^{-5})(2 \text{ mm}) + (1 \text{ atoms}\cdot\text{mm}^{-4})$

$\frac{dC}{dx} \bigg|_{x=2\text{ mm}} = 4 \text{ atoms}\cdot\text{mm}^{-4} + 1 \text{ atoms}\cdot\text{mm}^{-4}$

$\frac{dC}{dx} \bigg|_{x=2\text{ mm}} = 5 \text{ atoms}\cdot\text{mm}^{-4}$

Now, calculate the flux ($J$) using Fick's First Law:

$J = -D \frac{dC}{dx}$

$J = -(3 \times 10^{-3} \text{ mm}^2 \cdot \text{s}^{-1}) \times (5 \text{ atoms}\cdot\text{mm}^{-4})$

$J = -15 \times 10^{-3} \text{ atoms} \cdot \text{mm}^{-2} \cdot \text{s}^{-1}$

The question asks for the *magnitude* of the flux. The magnitude is the absolute value:

$\text{Magnitude of } J = |-15 \times 10^{-3}| \text{ atoms} \cdot \text{mm}^{-2} \cdot \text{s}^{-1}$

$\text{Magnitude of } J = 15 \times 10^{-3} \text{ atoms} \cdot \text{mm}^{-2} \cdot \text{s}^{-1}$

Rounding to the nearest integer for the value preceding $\times 10^{-3}$, the magnitude is 15.