The density and associated percent crystallinity ( for two poly(ethylene terephthalate) materials are as follows: \begin{tabular}{lc} \hline\(\rho\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) & crystallinity \((\%)\) \\\ \hline \(1.408\) & \(74.3\) \\ \hline \(1.343\) & \(31.2\) \\ \hline \end{tabular} (a) Compute the densities of totally crystalline and totally amorphous poly(ethylene terephthalate). (b) Determine the percent crystallinity of a specimen having a density of \(1.382 \mathrm{~g} / \mathrm{cm}^{3}\).

Short Answer

Expert verified
Answer: The densities of completely amorphous and completely crystalline poly(ethylene terephthalate) (PET) are approximately 0.3656 g/cm³ and 0.5098 g/cm³, respectively. The percent crystallinity of the specimen with a density of 1.382 g/cm³ is approximately 64.98%.

Step by step solution

01

Set up equations for densities of completely amorphous and completely crystalline PET

We are given two sets of values for density and percent crystallinity. For both sets, we can set up the equations as follows: 1.408 = ρ_a * (1 - 0.743) + ρ_c * 0.743 1.343 = ρ_a * (1 - 0.312) + ρ_c * 0.312
02

Solve the equations to find the densities of completely amorphous and completely crystalline PET

Solve the system of equations to find ρ_a and ρ_c values: ρ_a = ρ_c * 0.743 - 0.408 ρ_a = ρ_c * 0.312 - 0.188 By comparing the two expressions, we get 0.743 * ρ_c - 0.408 = 0.312 * ρ_c - 0.188. Solving for ρ_c: ρ_c = (0.408 - 0.188) / (0.743 - 0.312) ρ_c = 0.220 / 0.431 ρ_c ≈ 0.5098 \text{g/cm³} Now, we can find the value of ρ_a using one of the expressions: ρ_a = 0.312 * 0.5098 - 0.188 ρ_a ≈ 0.3656 \text{g/cm³} So, the densities of completely amorphous and completely crystalline poly(ethylene terephthalate) are approximately 0.3656 g/cm³ and 0.5098 g/cm³, respectively.
03

Determine the percent crystallinity of a specimen with density 1.382 g/cm³

We are now to find the percent crystallinity of a specimen with a density of 1.382 g/cm³. We can use the rule of mixtures formula again: 1.382 = 0.3656 * (1 - X_c) + 0.5098 * X_c Solve for X_c: X_c = (1.382 - 0.3656) / (0.5098 - 0.3656) X_c ≈ 0.6498 Therefore, the percent crystallinity of the specimen is approximately 64.98%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Computation
Understanding the computation of density is vital in materials science and engineering. Density, expressed as mass per unit volume (\rho), is a fundamental physical property of a material. Materials with complex structures, like polymers, can have varying densities based on their degree of crystallinity.

In the context of poly(ethylene terephthalate) (PET), which is often used in textiles and packaging, calculating density involves considering both crystalline and amorphous forms. The crystalline regions are tightly packed and regularly ordered, which makes them denser. Contrarily, the amorphous regions are more randomly ordered and typically less dense.

When it comes to computation, the rule of mixtures is applied to estimate the overall density of a partially crystalline polymer. This method assumes linear mixing between the amorphous and crystalline regions, calculated by the formula: \[ \rho = \rho_a \times (1 - X_c) + \rho_c \times X_c \] where \rho_a represents the density of the amorphous phase, \rho_c is the density of the crystalline phase, and X_c is the fraction of the crystalline phase.
Percent Crystallinity
Percent crystallinity is a measure of how much of a polymer's volume is taken up by crystalline regions. High levels of crystallinity can affect a material's mechanical properties and transparency. For instance, higher crystallinity in PET can lead to increased strength and rigidity, making it suitable for beverage bottles and containers.

To calculate the percent crystallinity, we use the density values of the fully amorphous and fully crystalline states as reference points. Then we express the crystallinity of a specimen as a percentage of the difference between these reference states and the specimen's density.

Using the aforementioned rule of mixtures, the fraction crystallinity (X_c) can be found from a known density, \( \rho \), of the specimen. It's given by: \[ X_c = \frac{\rho - \rho_a}{\rho_c - \rho_a} \] Multiplying this fraction by 100 gives us the percent crystallinity. Solving exercises involving percent crystallinity is not only about performing calculations but also about understanding the material's structure-properties relationship.
Amorphous and Crystalline Densities
In the study of polymers, particularly PET, it's crucial to differentiate between the amorphous and crystalline densities, as they play a significant role in determining the material's overall properties. The density of the amorphous phase (\( \rho_a \)) characterizes the less ordered, more flexible regions of the polymer, while the crystalline density (\( \rho_c \)) reflects the tightly packed, highly ordered structures.

These densities are not just theoretical numbers; they have practical implications in the application of polymers. For example, they influence thermal resistance, strength, and the diffusion rate of gases.

To find these values, we typically rely on laboratory data or direct measurement. However, in the absence of direct data, we can use mathematical relationships between known densities and percent crystallinities of given samples: \[ \rho = \rho_a \times (1 - X_c) + \rho_c \times X_c \] By working with systems of equations derived from these formulas, one can deduce both \( \rho_a \) and \( \rho_c \). This enables material scientists and engineers to tailor polymers for specific uses by varying their processing conditions to adjust their crystallinity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An alternating copolymer is known to have a ( number-average molecular weight of \(100,000 \mathrm{~g} / \mathrm{mol}\) and a degree of polymerization of 2210 . If one of the repeat units is ethylene, which of styrene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? Why?

Molecular weight data for some polymer are tabulated here. Compute (a) the number average molecular weight, and (b) the weight-average molecular weight. (c) If it is known that this material's degree of polymerization is \(477,\) which one of the polymers listed in Table 14.3 is this polymer? Why? $$\begin{array}{ccc} \hline \text {Molecular Weight} & & \\ \text {Range }(\mathrm{g} / \mathrm{mol}) & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{w}_{\boldsymbol{i}} \\ \hline 8,000-20,000 & 0.05 & 0.02 \\ 20,000-32,000 & 0.15 & 0.08 \\ 32,000-44,000 & 0.21 & 0.17 \\ 44,000-56,000 & 0.28 & 0.29 \\ 56,000-68,000 & 0.18 & 0.23 \\ 68,000-80,000 & 0.10 & 0.16 \\ 80,000-92,000 & 0.03 & 0.05 \\ \hline \end{array}$$

Explain briefly why the tendency of a polymer to crystallize decreases with increasing molecular weight.

(a) Is it possible to grind up and reuse phenol- ( formaldehyde? Why or why not? (b) Is it possible to grind up and reuse polypropylene? Why or why not?

A random poly(styrene-butadiene) copoly\(\Theta\) mer has a number-average molecular weight of \(350,000 \mathrm{~g} /\) mol and a degree of polymerization of 5000. Compute the fraction of styrene and butadiene repeat units in this copolymer.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free