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Chapter 15: Problem 23

List the two molecular characteristics that are essential for elastomers.

Short Answer

Expert verified
Answer: The two essential molecular characteristics for elastomers are cross-linking and low glass transition temperature.

Step by step solution

01

1. Understanding Elastomers

Elastomers are a type of polymer with unique properties which enable them to be stretched and return to their original shape. These properties make them widely used in various applications such as rubber bands, tires, and seals.
02

2. Molecular Characteristic 1: Cross-linking

Cross-linking refers to the process where individual polymer chains are connected through covalent bonds or weaker intermolecular forces. This connection between chains allows for the formation of a three-dimensional network, giving the elastomer its characteristic elasticity. When the elastomer is stretched, the network deforms, but the bonds can return to their original configuration once the force is removed.
03

3. Molecular Characteristic 2: Low Glass Transition Temperature

The glass transition temperature (T_g) is the temperature at which a polymer transitions from a hard, glassy state into a more flexible, rubbery state. For elastomers, this temperature is typically below room temperature. In this rubbery state, the polymer chains gain mobility, allowing for the elastomer鈥檚 elastic nature. This low glass transition temperature is crucial for elastomers to maintain their elastic properties at room temperature and above. To summarize, the two essential molecular characteristics for elastomers are: 1. Cross-linking 2. Low glass transition temperature

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Most popular questions from this chapter

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Nylon 6,6 may be formed by means of a condensation polymerization reaction in which hexamethylene diamine \(\left[\mathrm{NH}_{2}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2}\right]\) and adipic acid react with one another with the formation of water as a by- product. What masses of hexamethylene diamine and adipic acid are necessary to yield \(20 \mathrm{~kg}\) of completely linear nylon \(6,6 ?\) (Note: The chemical equation for this reaction is the answer to Concept Check 15.12.)
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