## Step-by Solution ###Given Data** **Surface concentration \( C_s \):** .1 wt% N - **Initial concentration in iron, \( C_ \):** wt% N (pure iron) -Distance from surface, \( x \):** 1 mm = .001 m - **Time, \( t \):** 10 h = 10 × 360 s = 36,000 s - **Diffusion coefficient, \( D \):** \( 2.5 \times 10^{-11} \) m²/s - **Temperature:** 700°C (already used in D) --- ### **Step 1: Identify the Diffusion Equation** For this case (semi-infinite solid, constant surface concentration), use the error function solution to Fick’s Second Law: \[ \frac{C_x - C_}{C_s - C_} = 1 - \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) \] Where: - \( C_x \): Concentration at depth \( x \) and time \( t \) - \( \text{erf} \): Error function --- ### **Step 2: Plug in the Values** - \( x = .001 \) m - \( D = 2.5 \times 10^{-11} \) m²/s - \( t = 36,000 \) s Calculate the argument of the error function: \[ \frac{x}{2\sqrt{Dt}} = \frac{.001}{2\sqrt{2.5 \times 10^{-11} \times 36,000}} \] First, calculate \( Dt \): \[ Dt = 2.5 \times 10^{-11} \times 36,000 = 9 \times 10^{-7} \] Now, \( \sqrt{Dt} \): \[ \sqrt{9 \times 10^{-7}} = 3 \times 10^{-3.5} = 3 \times 10^{-3} \times 10^{-.5} \approx 3 \times .316 \times 10^{-3} = .9487 \times 10^{-3} \] So, \[ 2\sqrt{Dt} = 2 \times .9487 \times 10^{-3} = 1.8974 \times 10^{-3} \] Now, the argument: \[ \frac{x}{2\sqrt{Dt}} = \frac{.001}{.0018974} \approx .527 \] --- ### **Step 3: Find the Error Function Value** \[ \text{erf}(.527) \approx .545 \] --- ### **Step 4: Substitute into the Solution** \[ \frac{C_x - }{.1 - } = 1 - .545 = .455 \] \[ C_x = .1 \times .455 = .0455 \text{ wt% N} \] --- ## **Final Answer** **The concentration of nitrogen at 1 mm from the surface after 10 hours is:** \[ \boxed{.0455 \text{ wt% N}} \] --- ### **Summary of Steps** 1. Used the error function solution for diffusion into a semi-infinite solid. 2. Converted all units and plugged in values. 3. Calculated the argument for the error function. 4. Used the error function value to find the final concentration.