Question

The crystallization of a polymer can only proceed in a temperature range limited to glass transition temperature (\(T_g\)) on the lower side, and the equilibrium melting point (\(T_m^0\)) on the higher side. Around \(T_g\), the mobility of the polymer chains is lower, while in the proximity of \(T_m^0\), crystal nucleation is inhibited.
In a miscible polymer blend with only one component being crystalline, which option(s) correctly match(es) the Temperature conditions with Events? 

• A. P-3; Q-2; R-4
• B. P-3; Q-1; R-2
• C. P-4; Q-1; R-3
• D. P-4; Q-1; R-2

Hint:

In polymer blends, the relationship between the glass transition temperature (\(T_g\)) and the equilibrium melting point (\(T_m^0\)) is crucial for determining the crystallization behavior. Higher \(T_g\) leads to slower crystallization, and lower \(T_g\) favors crystallization.

Answer 1

The Correct Option is B, D

In this question, we need to match the temperature conditions with the corresponding events related to crystallization. The events depend on the glass transition temperature (\(T_g\)) and the equilibrium melting point (\(T_m^0\)) of the polymer. Let’s analyze each case based on the given conditions.
- Condition P: "The \(T_g\) of the amorphous component is lower than the crystallizable one." This means that the amorphous component will have a lower glass transition temperature, which could affect the crystallization process. As the amorphous component has a lower \(T_g\), crystallization is more favorable, and the temperature range over which crystallization can occur becomes narrower. This corresponds to Event 3: "The temperature range over which crystallization can occur becomes smaller." Hence, P-3 is correct.
- Condition Q: "The \(T_g\) of the amorphous component is higher than the crystallizable one." When the \(T_g\) of the amorphous component is higher, it results in inhibited crystallization, as the component will not easily transition into the crystalline state. Therefore, Q corresponds to Event 1: "Crystallization is inhibited." Hence, Q-1 is correct.
- Condition R: "The blend \(T_g\) is higher than the \(T_m^0\) of the crystallizable one." This condition indicates that the crystallization temperature range is widened, as the \(T_g\) is significantly higher than the \(T_m^0\), allowing for more favorable conditions for crystallization. Therefore, R corresponds to Event 2: "Crystallization is favored." Hence, R-2 is correct.
Thus, the correct answers are (B) P-3; Q-1; R-2 and (D) P-4; Q-1; R-2.