Question

At one atmosphere pressure, \(\alpha\)-Fe transforms to \(\gamma\)-Fe above 912 \(^{\circ}\)C. Density of \(\gamma\)-Fe is more than that of \(\alpha\)-Fe. Choose the correct statement.

• A. Increasing the pressure above one atmosphere lowers the \(\alpha\)-Fe to \(\gamma\)-Fe transformation temperature.
• B. Increasing the pressure above one atmosphere raises the \(\alpha\)-Fe to \(\gamma\)-Fe transformation temperature.
• C. Molar volume of \(\gamma\)-Fe is higher than the molar volume of \(\alpha\)-Fe.
• D. Pressure change will not have any effect on the \(\alpha\)-Fe to \(\gamma\)-Fe transformation temperature.

Hint:

Remember the simple rule from Le Chatelier's principle: "Pressure favors density." Increasing pressure will always shift the equilibrium towards the denser phase. For water freezing, ice is less dense, so increasing pressure lowers the freezing point. For the \(\alpha\)-\(\gamma\) iron transformation, \(\gamma\)-Fe is denser, so increasing pressure also lowers the transformation temperature.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question relates the effect of pressure on a phase transformation temperature. This can be understood using Le Chatelier's principle or the Clapeyron equation, which connects changes in pressure, temperature, and volume during a phase change.
Step 2: Key Formula or Approach:
Le Chatelier's principle states that if a change of condition (like pressure) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In this case, increasing pressure will favor the phase with the smaller volume.
The Clapeyron equation is given by: \[ \frac{dP}{dT} = \frac{\Delta H}{T \Delta V} \] where \(\Delta H\) is the enthalpy of transformation and \(\Delta V\) is the change in volume.
Step 3: Detailed Explanation:
The transformation is \(\alpha\)-Fe \(⇒\) \(\gamma\)-Fe. We are given that Density(\(\gamma\)-Fe) > Density(\(\alpha\)-Fe).
Since Density = Mass/Volume, for a given mass (e.g., one mole), a higher density implies a lower volume.
Therefore, Molar Volume(\(\gamma\)-Fe) < Molar Volume(\(\alpha\)-Fe).
This means the change in molar volume during the transformation, \(\Delta V = V_{\gamma} - V_{\alpha}\), is negative (\(\Delta V<0\)).
According to Le Chatelier's principle, if we increase the pressure, the equilibrium will shift to favor the phase with the lower volume, which is \(\gamma\)-Fe.
Favoring the formation of \(\gamma\)-Fe means that the transformation from \(\alpha\)-Fe to \(\gamma\)-Fe can occur at a lower temperature. Thus, increasing the pressure lowers the transformation temperature.
Let's check this with the Clapeyron equation. The transformation occurs with heating, so it is endothermic (\(\Delta H>0\)). We found \(\Delta V<0\). Therefore: \[ \frac{dP}{dT} = \frac{(+)}{T(-)}<0 \] A negative slope \(dP/dT\) means that as pressure (P) increases, the equilibrium temperature (T) must decrease to maintain the phase boundary. This confirms that increasing pressure lowers the transformation temperature.
Step 4: Final Answer:
Increasing the pressure above one atmosphere lowers the \(\alpha\)-Fe to \(\gamma\)-Fe transformation temperature. Option (C) is incorrect because higher density means lower molar volume. Option (D) is incorrect because there is a volume change, so pressure will have an effect.
Step 5: Why This is Correct:
Based on Le Chatelier's principle and the Clapeyron equation, an increase in pressure favors the denser phase (\(\gamma\)-Fe), causing the transformation to occur at a lower temperature.