Polymer Characterization II: Glass and Melting Transitions, Crystallinity, Orientation, Spectroscopic, Thermal and Microscopic Analysis
2. The melting transition and the factors that set Tm
Melting of the crystalline regions is a first-order thermodynamic transition with latent heat ΔH_m and, because crystals are of different sizes and perfection, a broad melting range, quoted as the peak of the DSC endotherm. At equilibrium Tm⁰ = ΔH_m/ΔS_m: a high Tm needs a large enthalpy of fusion (strong intermolecular forces such as hydrogen bonding) or a small entropy of fusion (a stiff chain that loses little conformational freedom on melting). Thus nylon 66 (hydrogen bonds, about 265 °C) and polyethylene terephthalate (stiff aromatic ring, about 260 °C) melt high, while polyethylene (flexible, no polar groups, about 130–140 °C) and polypropylene (about 165 °C for the isotactic grade) melt lower.
Further factors on Tm: symmetry and regularity raise it; lamellar thickness, since thin or imperfect crystals melt lower (Gibbs–Thomson: Tm = Tm⁰[1 − 2σ_e/(ΔH_u ρ_c l)], where l is the lamellar thickness and σ_e the fold-surface energy), so annealing raises Tm and fast cooling gives a lower, broader peak; copolymerization with a second unit or branching (LDPE against HDPE) lowers Tm; plasticisers lower it; and molar mass raises Tm only at low Mn. A rough rule due to Boyer and Beaman is Tg/Tm ≈ ½ (in kelvin) for symmetric polymers and about ⅔ for unsymmetrical ones.
| Property | Glass transition | Melting |
|---|---|---|
| Phase involved | Amorphous | Crystalline |
| Thermodynamic order | Kinetic, second-order-like | First-order |
| DSC signal | Step in heat capacity | Endothermic peak, area = ΔH_m |
| Depends on cooling rate | Yes, strongly | Through crystal size and perfection |
3. Amorphous and crystalline states, and what favours crystallinity
A polymer cannot crystallise completely, because chains are long, entangled and of different length; the usual picture is a two-phase model of crystalline lamellae embedded in an amorphous matrix, the degree of crystallinity being the fraction of crystalline material (typically 10–80%). Crystalline regions give stiffness, strength, density, opacity (light scattering at crystal–amorphous boundaries) and solvent resistance; amorphous regions give toughness above Tg, transparency, and permeability. The glass transition of the amorphous part is retained in semicrystalline polymers.
Factors favouring crystallinity: (i) regular structure, with stereoregular isotactic or syndiotactic chains (atactic polystyrene, atactic polypropylene stay amorphous); (ii) symmetry, as in polyethylene or PTFE; (iii) small, regular side groups and the absence of bulky ones; (iv) strong intermolecular forces (hydrogen bonding in polyamides lock chains in register); (v) chain flexibility enough to fold and pack, but not so much as to resist ordering (stiff chains crystallise slowly); (vi) slow cooling or annealing that allows time to organise, and temperature between Tg and Tm; (vii) orientation and strain, which promote crystallisation. Hindering factors are random copolymerization, branching, crosslinking, bulky or irregular substituents, and fast quenching (quenched PET is clear and amorphous).
Morphology. From the melt or concentrated solution, polymers form spherulites: spherical aggregates, 1–100 μm across, in which radial ribbon-like lamellae, about 10–20 nm thick, grow outward from a nucleus, with chains running roughly perpendicular to the lamella plane and folding back at its faces; amorphous material and tie molecules lie between lamellae and hold the structure together. Fewer nuclei (slow cooling) give large spherulites and a brittle, hazy product; nucleating agents or fast cooling give many small spherulites with better clarity and toughness. From dilute solution, polymers grow thin single-crystal lamellae; under flow or stress, row structures and fibrillar crystals (shish-kebab) form.
4. Measuring crystallinity; orientation in polymers
Density method. With ρ the sample density, ρ_c that of the 100% crystalline phase and ρ_a that of the amorphous phase (additive volumes), the volume fraction crystallinity is X_v = (ρ − ρ_a)/(ρ_c − ρ_a) and the mass fraction is X_w = ρ_c(ρ − ρ_a)/[ρ(ρ_c − ρ_a)]. Example: ρ = 0.950 g/cm³ with ρ_c = 1.000 and ρ_a = 0.855 g/cm³ gives X_v = 0.095/0.145 = 0.655 and X_w = 0.655 × 1.000/0.950 = 0.690. Always read which of the two fractions is asked: mass and volume fractions differ because the crystal is denser.
DSC method. The mass-fraction crystallinity is X_c = (ΔH_m − ΔH_cc)/(ΔH_m⁰ · w_p), with ΔH_m the measured enthalpy of fusion (J/g, from the peak area), ΔH_cc any cold-crystallisation exotherm during heating, ΔH_m⁰ the enthalpy of fusion of the 100% crystalline polymer (a literature value, to be given), and w_p the polymer mass fraction in a filled sample. Example: ΔH_m = 100 J/g and ΔH_m⁰ = 290 J/g, with no cold crystallisation, give X_c = 0.345. XRD gives crystallinity from the ratio of crystalline peak area to total scattered area (peaks plus amorphous halo), and the different methods agree only roughly since each defines crystallinity differently.
Orientation. Drawing a polymer above Tg (cold drawing, fibre spinning, film stretching) aligns chains along the draw direction. The sample becomes anisotropic: modulus and strength rise strongly along the axis (a gel-spun high-molar-mass polyethylene fibre approaches the theoretical chain stiffness) and fall transversely, with optical birefringence Δn that measures orientation. The Hermans orientation function f = (3⟨cos²θ⟩ − 1)/2 equals 1 for perfect alignment, 0 for random and −½ for perpendicular alignment. Orientation, via strain-induced crystallisation, also raises crystallinity, and the oriented, crystallised chains shrink back on heating above Tg if not heat-set, the principle of heat-shrink films.
5. Analysis by IR, XRD and thermal methods (DSC, DMTA, TGA)
| Method | Measures | Information for polymers |
|---|---|---|
| IR (FTIR, ATR) | Absorption of IR by vibrations that change the dipole moment | Functional groups, composition, end groups, oxidation (carbonyl growth), tacticity and crystallinity-sensitive bands |
| XRD (WAXD, SAXS) | Diffraction of X-rays by ordered planes | Crystal structure, d-spacing, crystallinity, crystallite size, orientation; SAXS gives the lamellar long period |
| DSC | Heat flow against temperature at a fixed rate | Tg (step), Tc (exotherm), Tm (endotherm), ΔH_m and crystallinity, cure exotherm and degree of cure |
| DMTA (DMA) | Storage modulus E′, loss modulus E″, tan δ = E″/E′ under oscillating strain against T | Tg (tan δ or E″ peak, more sensitive than DSC), secondary relaxations, crosslink density from rubbery plateau, blend miscibility |
| TGA | Mass against temperature or time in N₂ or air | Thermal stability, onset of degradation, moisture and volatiles, filler and ash content, composition of blends |
IR practice. A broad band near 3200–3600 cm⁻¹ is O–H or N–H stretching, a strong band near 1700–1740 cm⁻¹ is the carbonyl stretch (an ester near 1735 cm⁻¹; growth of a carbonyl band in polyethylene signals oxidation), and a sharp band near 2240 cm⁻¹ is the nitrile C≡N of polyacrylonitrile. Beer–Lambert behaviour, A = εcl, allows quantitation. Symmetric vibrations without a dipole change, such as the C=C of an ethylene-like backbone, are weak in IR and strong in Raman.
XRD practice. The Bragg condition is nλ = 2d sinθ, where θ is half the measured scattering angle 2θ. With Cu Kα radiation (λ = 0.1542 nm), a peak at 2θ = 20° corresponds to d = 0.1542/(2 sin 10°) = 0.444 nm. Crystalline polymers give sharp peaks on a broad amorphous halo; the width gives the crystallite size by the Scherrer equation L = Kλ/(β cosθ) (K ≈ 0.9, β the peak width in radians). Wide-angle scattering reflects atomic spacings of about 0.1–1 nm, while small-angle scattering probes the lamellar repeat of 10–30 nm.
6. Microscopy: optical, SEM, TEM and AFM
| Technique | Probe and principle | Resolution and use | Sample preparation |
|---|---|---|---|
| Optical (polarised, phase contrast, hot stage) | Visible light, transmitted or reflected | About 0.2 μm limit; spherulites, Maltese cross, crystallisation in situ | Thin film or microtomed section |
| SEM | Scanned electron beam; secondary electrons form image | nm scale, large depth of field; surface topography, fracture surfaces, blend and filler dispersion | Conductive coating (gold or carbon) of insulating polymers, vacuum |
| TEM | Electron beam transmitted through an ultrathin sample | Highest resolution (sub-nm to nm); internal structure, lamellae, block copolymer domains, particle size in latex | Ultramicrotomy to 50–100 nm; staining of one phase (OsO₄ for unsaturated rubber, RuO₄) |
| AFM | Sharp tip on a cantilever scans the surface; deflection from forces | nm lateral, sub-nm vertical; 3D topography, phase imaging of stiffness, lamellae, nanomechanics | Little preparation; no coating or vacuum needed |
Contrast is the central problem in polymers because they are made of light elements. TEM obtains it by staining (heavy-metal oxides bind to C=C or to one phase), SEM by differences in topography or by etching one phase, optical microscopy by birefringence under crossed polarisers or phase contrast, and AFM by tip–sample stiffness and adhesion in tapping-mode phase images. Electron beams damage soft polymers, so doses are kept low or the sample is cooled.
Choosing a tool. Spherulite size and growth rate on a hot stage: polarised optical microscopy. Dispersion of rubber particles in a toughened plastic, or the domain structure of a block copolymer: TEM with staining. Fracture surface and filler distribution: SEM. Surface roughness and lamellar structure on an uncoated film in air: AFM. Each of these gives morphology (how the phases and crystals are arranged in space), unlike DSC or XRD, which give averages over the sample.
Key takeaways
- Tg is a kinetic, second-order-like step in heat capacity of the amorphous phase, raised by stiffness, bulky or polar groups and crosslinks, lowered by plasticiser; Fox: 1/Tg = Σw_i/Tg,i (kelvin).
- Tm is first-order, Tm⁰ = ΔH_m/ΔS_m; stiff chains, hydrogen bonds and regularity raise it, while thin lamellae, copolymer units and branches lower it; Tg/Tm ≈ ½ to ⅔.
- Crystallinity is favoured by regularity, symmetry, small side groups, hydrogen bonding, slow cooling and orientation; X_v = (ρ − ρ_a)/(ρ_c − ρ_a), X_w = (ρ_c/ρ)X_v, X_c = ΔH_m/ΔH_m⁰ from DSC.
- Orientation aligns chains, raises axial modulus and strength and gives birefringence; the Hermans function is 1 (aligned), 0 (random), −½ (perpendicular).
- IR identifies functional groups, XRD (Bragg nλ = 2d sinθ) gives crystal structure and crystallinity, DSC gives Tg, Tm, ΔH, DMTA gives E′, E″, tan δ and sensitive Tg, TGA gives stability and composition.
- Spherulites (1–100 μm, radial lamellae of 10–20 nm with chain folding) are seen by polarised optical microscopy; SEM images surfaces, TEM internal structure with staining, AFM topography and stiffness without coating.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which statement correctly describes the glass transition of an amorphous polymer on heating?
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Answer: B — It is a step in heat capacity with no latent heat and shifts with rate
Tg is a kinetic, second-order-like change: volume and enthalpy are continuous but heat capacity and expansion coefficient step up, and faster rates give higher Tg. Latent heat and a fixed value belong to a first-order transition such as melting.Which change is expected to raise the glass transition temperature of a given amorphous polymer?
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Answer: C — Light crosslinking that restricts the motion of chain segments
Crosslinks limit segmental motion and so raise Tg. Plasticisers and chain ends add free volume, and flexible links ease rotation, all of which lower Tg.Nylon 66 melts near 265 °C while polyethylene melts near 135 °C. The main reason is that nylon 66 has
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Answer: D — hydrogen bonds between amide groups, which raise the enthalpy of fusion
Tm⁰ = ΔH_m/ΔS_m: hydrogen bonding raises the enthalpy that has to be supplied to melt. A flexible backbone would raise ΔS_m and lower Tm, and molar mass above a modest value barely changes Tm.A sample of isotactic polypropylene is annealed for several hours at a temperature between its Tg and Tm. Compared with the quenched sample, the annealed one will show
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Answer: B — a higher melting peak and higher crystallinity as lamellae thicken and perfect
Annealing gives chains time to organise: lamellae thicken and defects heal, so by the Gibbs–Thomson relation Tm rises, and crystallinity increases. Holding above Tg but below Tm promotes crystallisation; it does not make the polymer amorphous.Which of the following polymers, made by conventional free-radical or stereo-irregular routes, is expected to be essentially amorphous?
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Answer: A — Atactic polystyrene, with random phenyl placement along the chain
Random stereochemistry of the bulky phenyl groups prevents regular packing, so atactic polystyrene is amorphous. Linear symmetric polyethylene and PTFE and hydrogen-bonded nylon 66 pack into crystals readily.A quenched PET sample is heated in a DSC. It shows a melting endotherm of ΔH_m = 40 J/g and a cold-crystallisation exotherm of ΔH_cc = 36 J/g during the same scan. With ΔH_m⁰ = 140 J/g for 100% crystalline PET, the degree of crystallinity of the sample before heating was about
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Answer: D — 3%, from (ΔH_m − ΔH_cc)/ΔH_m⁰
Crystals formed on heating (the cold-crystallisation exotherm) were not present originally, so X_c = (40 − 36)/140 = 0.029, about 3%. Using ΔH_m alone, 29%, would credit the sample with crystals that formed during the scan.The X-ray diffraction pattern of a polymer film shows only a single broad halo and no sharp peaks. This indicates that the film is
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Answer: C — amorphous, having only short-range order
Sharp Bragg peaks need long-range periodic order. A broad halo arises from the average neighbour spacing in a disordered state. Very large crystallites would give narrower, not broader, peaks, and a single crystal gives spots or sharp reflections.A thin section of a styrene–butadiene–styrene triblock copolymer is exposed to osmium tetroxide vapour and imaged by TEM. Which domains appear dark?
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Answer: B — The polybutadiene domains, because OsO₄ adds to the C=C bonds and deposits heavy osmium
OsO₄ stains unsaturated rubber through reaction at C=C, putting heavy atoms in the polybutadiene phase, which then scatters electrons more and appears dark; saturated polystyrene is unstained and appears light. The staining is selective, not equal for both phases.Thermogravimetric analysis (TGA) of a filled polymer in air up to 700 °C leaves a residue. This residue is usually interpreted as
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Answer: B — the inorganic filler content (ash) that does not volatilise
TGA records mass against temperature; the organic polymer burns off in air and an inorganic filler such as glass or silica remains as residue. Moisture leaves at low temperature, and neither crystallinity nor molar mass is read from mass loss.Which of the following changes raise the glass transition temperature of an amorphous polymer?
P: Crosslinking the chains
Q: Adding a plasticiser
R: Replacing a methylene side group by a bulky rigid phenyl group
S: Making the backbone more flexible by adding –O– linksShow answer
Answer: A — P; C — R
Crosslinks and bulky rigid side groups hinder segmental rotation and raise Tg. A plasticiser adds free volume and a more flexible backbone eases rotation, so both lower Tg.Which of the following statements about thermal analysis are correct?
P: A tan δ peak from DMTA can be used to locate Tg, often with higher sensitivity than the DSC heat-capacity step.
Q: DSC measures the storage modulus of the sample directly.
R: TGA measures the heat flow into the sample.
S: In DMTA tan δ is the ratio of the loss modulus to the storage modulus.Show answer
Answer: A — P; D — S
DMTA gives E′, E″ and tan δ = E″/E′, with a strong tan δ peak at the glass transition. DSC measures heat flow and TGA measures mass, so attributing storage modulus to DSC and heat flow to TGA reverses the techniques.Which of the following statements about the microscopy of polymers are correct?
P: Insulating polymer samples usually need a thin conductive coating for conventional SEM.
Q: AFM requires a high vacuum and a metal coating of the sample.
R: TEM requires ultrathin sections of the sample, typically below about 100 nm.
S: Spherulites show a Maltese-cross pattern between crossed polarisers in an optical microscope.Show answer
Answer: A — P; C — R; D — S
Charging of insulators in SEM is avoided by gold or carbon coating, TEM needs thin sections for electron transmission, and the radial anisotropic lamellae of spherulites give the Maltese cross. AFM works in air on uncoated samples, so the statement requiring vacuum and coating is wrong.A polyethylene sample has density ρ = 0.940 g/cm³. Take the density of the 100% crystalline phase as ρ_c = 1.000 g/cm³ and of the amorphous phase as ρ_a = 0.855 g/cm³, with additive volumes. Using X_v = (ρ − ρ_a)/(ρ_c − ρ_a) and X_w = ρ_c(ρ − ρ_a)/[ρ(ρ_c − ρ_a)], calculate the mass-fraction crystallinity X_w, to three decimal places.
Numerical answer — type the value.
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Answer: 0.624
X_v = (0.940 − 0.855)/(1.000 − 0.855) = 0.085/0.145 = 0.5862. X_w = X_v × ρ_c/ρ = 0.5862 × 1.000/0.940 = 0.6236, i.e. 0.624.A polyethylene sample shows a DSC melting endotherm of ΔH_m = 150 J/g, with no cold crystallisation and no filler. Take ΔH_m⁰ = 290 J/g for 100% crystalline polyethylene. Using X_c = ΔH_m/ΔH_m⁰, calculate the percentage crystallinity (mass basis), to one decimal place.
Numerical answer — type the value.
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Answer: 51.7
X_c = 150/290 = 0.5172, i.e. 51.7% by mass.A random copolymer contains 70% by weight of unit 1 (homopolymer Tg = 373 K) and 30% by weight of unit 2 (homopolymer Tg = 223 K). Using the Fox equation 1/Tg = w₁/Tg,₁ + w₂/Tg,₂, calculate the copolymer Tg in kelvin, to one decimal place.
Numerical answer — type the value.
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Answer: 310.4
1/Tg = 0.7/373 + 0.3/223 = 1.8767×10⁻³ + 1.3453×10⁻³ = 3.2220×10⁻³ K⁻¹, so Tg = 310.4 K.An X-ray diffraction peak of a semicrystalline polymer is observed at a scattering angle 2θ = 21.5° using Cu Kα radiation of wavelength λ = 0.1542 nm. Using the Bragg equation nλ = 2d sinθ with n = 1, calculate the lattice spacing d in nm, to three decimal places.
Numerical answer — type the value.
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Answer: 0.413
θ = 10.75°, sin θ = 0.18650, so d = λ/(2 sinθ) = 0.1542/(2 × 0.18650) = 0.4134 nm, i.e. 0.413 nm. A common slip is to use 2θ = 21.5° directly in the sine.