VIPSolutionsAct like a helpful tutor and exlain me step-by-step solution with explanation and final answer:2. Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.1 wt% N, what will be the concentration 1 mm from the surface after 10 h? The diffusion coefficient for nitrogen in iron at 700°C is 2.5 x 10" ms. (5 Marks)
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Act like a helpful tutor and exlain me step-by-step solution with explanation and final answer:
2. Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the
surface concentration is maintained at 0.1 wt% N, what will be the concentration
1 mm from the surface after 10 h? The diffusion coefficient for nitrogen in iron at
700°C is 2.5 x 10" ms. (5 Marks)
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Asked by: francis
Created at: 2026-02-19 07:49:30
Asked by: francis
Created at: 2026-02-19 07:49:30
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# Step-by-Step Solution
Let's break down the solution for this diffusion problem:
---
## **Given Data**
- **Surface concentration, \( C_s \):** .1 wt% N
- **Initial concentration in iron, \( C_ \):** wt% N (pure iron)
- **Distance from surface, \( x \):** 1 mm = \( 1 \times 10^{-3} \) m
- **Time, \( t \):** 10 h = 10 × 360 = 36,000 s
- **Diffusion coefficient, \( D \):** \( 2.5 \times 10^{-11} \) m²/s
---
## **Step 1: Recognize the Appropriate Equation**
This is a **semi-infinite solid diffusion** problem (constant surface concentration).
We use the 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 distance \( x \) and time \( t \)
- \( \text{erf} \) = error function
---
## **Step 2: Plug in Known Values**
### Calculate the denominator inside the error function:
\[
2\sqrt{Dt} = 2\sqrt{2.5 \times 10^{-11} \times 36,000}
\]
First, calculate \( Dt \):
\[
Dt = 2.5 \times 10^{-11} \times 36,000 = 9 \times 10^{-7}
\]
Now, calculate \( \sqrt{Dt} \):
\[
\sqrt{Dt} = \sqrt{9 \times 10^{-7}} = 3 \times 10^{-3.5}
\]
\[
10^{-3.5} = 10^{-3} \times 10^{-.5} \approx .001 \times .3162 = .0003162
\]
\[
3 \times .0003162 = .0009486
\]
So,
\[
2\sqrt{Dt} = 2 \times .0009486 = .0018972
\]
---
## **Step 3: Calculate the Argument for the Error Function**
\[
\frac{x}{2\sqrt{Dt}} = \frac{.001}{.0018972} \approx .527
\]
---
## **Step 4: Find the Error Function Value**
From error function tables or using a calculator:
\[
\text{erf}(.527) \approx .537
\]
---
## **Step 5: Substitute Back into the Fick's Law Solution**
\[
\frac{C_x - }{.1 - } = 1 - .537 = .463
\]
\[
C_x = .1 \times .463 = .0463 \text{ wt% N}
\]
---
## **Final Answer**
### **The concentration of nitrogen 1 mm from the surface after 10 hours is:**
\[
\boxed{.046 \ \text{wt}\% \ \text{N}}
\]
---
### **Summary Table**
| Parameter | Value |
|-------------------|-------------------------|
| \( C_s \) | .1 wt% N |
| \( C_ \) | wt% N |
| \( x \) | 1 mm = \( 1 \times 10^{-3} \) m |
| \( t \) | 10 h = 36,000 s |
| \( D \) | \( 2.5 \times 10^{-11} \) m²/s |
| \( C_x \) | \( .046 \) wt% N |
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