Question

A lab developed a new biodegradable polymer. The polymer sample was tested with Differential scanning calorimetry (DSC). It was found that the heats of melting DH_m and crystallization DH_c of the polymer are 61 J/g and 5 J/g respectively. X-ray diffraction was used to measure the crystallinity( X_c) of the polymer sample. X_c was determined to be 55%. The polymer was then mixed with drug molecules to fabricate drug-encapsulated thin film. DSC analysis of the film showed that the heats of melting DH_m and crystallization DH_c of the polymer became 60 J/g and 11 J/g. What is the polymer crystallinity in the film?

Answer 1

According to the given data:

DHm - DHc = 61 - 5 = 56 J/g

The above value indicates that part of the polymer sample which was already in the crystalline state before we heated the polymer above the T_c

Therefore, the above fraction divided by DHm = 56/61 = 0.918

But the fraction of crystallization of drug = 0.55

i.e. The uncrystallized fraction = 0.918 - 0.55 = 0.368

After mixing the above polymer with drug, DHm - DHc = 60 - 11 = 49 J/g

Now, (49 J/g) / (60 J/g) = 0.817

Therefore, the actual fraction crystallized = 0.817 - 0.368 = 0.449

Therefore, the polymer crystallinity in the film = 44.9% ~ 45%