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Chapter 16: Problem 23

Cite the distinction between carbon and graphite.

Short Answer

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Answer: The key differences between carbon and graphite are: 1. Molecular Structure: Carbon exists in various allotropes with different molecular structures. Graphite has a hexagonal lattice, forming layers of two-dimensional, planar structures held together by weak van der Waals forces. 2. Physical Properties: Diamond, an allotrope of carbon, is the hardest naturally occurring substance, while graphite is soft and slippery. Graphite has high electrical conductivity, while carbon in other forms may not. Additionally, graphite appears opaque, while diamond is transparent. 3. Common Uses: Graphite is used in pencil leads, dry lubricants, lithium-ion battery anodes, and industrial applications such as crucibles, electrodes, and refractory materials. Carbon has various applications depending on its allotrope.

Step by step solution

01

Introduction to Carbon and Graphite

Carbon is a chemical element with the symbol C and atomic number 6. It exists in various forms, known as allotropes, which have different molecular structures and physical properties. Graphite is one of these allotropes of carbon, along with diamond, amorphous carbon, and other lesser-known forms.
02

Molecular Structure

The primary distinction between carbon and graphite lies in their molecular structures. Carbon atoms bond together in various ways to form different allotropes. In the case of graphite, carbon atoms are arranged in a hexagonal lattice, forming layers of two-dimensional, planar structures. These layers are held together by weak van der Waals forces, allowing them to slide over each other easily.
03

Physical Properties

The differences in molecular structure lead to different physical properties for carbon and graphite. 1. Hardness: Carbon, in its diamond allotrope, is known as the hardest naturally occuring substance. On the other hand, graphite is quite soft and has a slippery texture due to its layered structure. 2. Conductivity: Graphite has high electrical conductivity because its molecular structure allows electrons to move easily through the layers. Carbon, in other forms, may not possess this property. 3. Opacity: Graphite appears dark and opaque, while diamond form of carbon has extraordinary transparency.
04

Common Uses

Carbon is used in various forms and has numerous applications. Some examples for graphite include: 1. Pencil leads: Due to its slippery nature, graphite is used as the main component in pencil leads, as it can easily slide over paper to leave a mark. 2. Lubricants: Graphite is used as a dry lubricant, especially in high-temperature and high-pressure environments where conventional lubricants are not suitable. 3. Batteries: Graphite is used as a component in the anode of lithium-ion batteries. 4. Industrial applications: Graphite is used in making crucibles, electrodes, and other refractory materials due to its ability to withstand high temperatures and good electrical conductivity. In summary, carbon and graphite are chemically the same, but their structural differences lead to contrasting physical properties and applications. While carbon exists in various allotropes, graphite is one such form with a unique hexagonal lattice structure resulting in its soft nature, high electric conductivity, and use in various practical applications.

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Most popular questions from this chapter

16.28 Briefly describe pultrusion, filament winding, and prepreg production fabrication processes; cite the advantages and disadvantages of each.

16.7 (a) For a fiber-reinforced composite, the efficiency of reinforcement \(\eta\) is dependent on fiber length \(l\) according to $$ \eta=\frac{l-2 x}{l} $$ where \(x\) represents the length of the fiber at each end that does not contribute to the load transfer. Make a plot of \(\eta\) versus \(l\) to \(l=40\) \(\mathrm{mm}\) (1.6 in.), assuming that \(x=0.75 \mathrm{~mm}\) \((0.03 \mathrm{in} .)\) (b) What length is required for a \(0.80\) efficiency of reinforcement?

16.9 Is it possible to produce a continuous and oriented aramid fiber-epoxy matrix composite having longitudinal and transverse moduli of elasticity of \(57.1 \mathrm{GPa}\left(8.28 \times 10^{6} \mathrm{psi}\right)\) and \(4.12 \mathrm{GPa}\left(6 \times 10^{5} \mathrm{psi}\right)\), respectively? Why or why not? Assume that the modulus of elasticity of the epoxy is \(2.4 \mathrm{GPa}\left(3.50 \times 10^{5} \mathrm{psi}\right)\).

\(16.14\) A continuous and aligned fiber-reinforced composite having a cross- sectional area of \(1130 \mathrm{~mm}^{2}\left(1.75 \mathrm{in} .^{2}\right)\) is subjected to an \(\mathrm{ex}\) ternal tensile load. If the stresses sustained by the fiber and matrix phases are \(156 \mathrm{MPa}\) \((22,600 \mathrm{psi})\) and \(2.75 \mathrm{MPa}(400 \mathrm{psi})\), respectively; the force sustained by the fiber phase is \(74,000 \mathrm{~N}\left(16,600 \mathrm{lb}_{6}\right)\); and the total longitudinal strain is \(1.25 \times 10^{-3}\), determine (a) the force sustained by the matrix phase, (b) the modulus of elasticity of the composite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.

The mechanical properties of aluminum may be improved by incorporating fine particles of aluminum oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\). Given that the moduli of elasticity of these materials are, respectively, \(69 \mathrm{GPa}\left(10 \times 10^{6} \mathrm{psi}\right)\) and 393 GPa ( \(\left.57 \times 10^{6} \mathrm{psi}\right)\), plot modulus of elasticity versus the volume percent of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in \(\mathrm{Al}\) from 0 to 100 vol\%, using both upper- and lower-bound expressions.
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