Question

In Fe-Fe₃C phase diagram, the eutectoid composition is 0.8 weight % of carbon at 725 °C. The maximum solubility of carbon in $\alpha$-ferrite phase is 0.025 weight % of carbon. A steel sample, having no other alloying element except 0.5 weight % of carbon, is slowly cooled from 1000 °C to room temperature. The fraction of pro-eutectoid $\alpha$-ferrite in the above steel sample at room temperature is:

• A. 0.387
• B. 0.864
• C. 0.475
• D. 0.775

Hint:

When applying the lever rule in phase diagrams, remember to use the compositions of the phases and the overall composition. The fraction of each phase will be determined by how the overall composition "leans" towards the phase boundaries.

Answer 1

The Correct Option is A

Step 1: Use the lever rule to determine the fraction of phases.
The lever rule is used to determine the fraction of the phases in a two-phase region. The formula for the lever rule is: \[ \text{Fraction of Phase 1} = \frac{C_2 - C_0}{C_2 - C_1} \] Where:
- \( C_0 \) is the overall composition (given as 0.5 wt% C),
- \( C_1 \) is the composition of the phase at the phase boundary (in this case, the eutectoid composition \( C_1 = 0.8 \) wt% C),
- \( C_2 \) is the composition of the other phase (in this case, the maximum solubility of carbon in \( \alpha \)-ferrite at eutectoid, \( C_2 = 0.025 \) wt% C).
Step 2: Substitute the values into the lever rule formula.
For the fraction of pro-eutectoid \( \alpha \)-ferrite: \[ \text{Fraction of } \alpha = \frac{C_1 - C_0}{C_1 - C_2} = \frac{0.8 - 0.5}{0.8 - 0.025} = \frac{0.3}{0.775} \approx 0.387 \] Therefore, the fraction of pro-eutectoid \( \alpha \)-ferrite at room temperature is approximately \(\boxed{0.387}\). Final Answer: \[ \boxed{0.387} \]