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5 Evaporation, Vapor Pressure, and Boiling Point

5 Evaporation, Vapor Pressure, and Boiling Point

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10.5 EVAPORATION, VAPOR PRESSURE, AND BOILING POINT



Evaporation and vapor pressure are both explained on a molecular level by the

kinetic–molecular theory, developed in Section 9.6 to account for the behavior of



Number of molecules



gases. The molecules in a liquid are in constant motion but at a variety of speeds

depending on the amount of kinetic energy they have. In considering a large sample,

molecular kinetic energies follow a distribution curve like that shown in Figure 10.12,

with the exact shape of the curve dependent on the temperature. The higher the temperature and the lower the boiling point of the substance, the greater the fraction of

molecules in the sample that have sufficient kinetic energy to break free from the surface of the liquid and escape into the vapor.



Lower

temperature



363



Remember...

The kinetic–molecular theory is a group

of five postulates that can be used to

account for the behavior of gases and to

derive the ideal gas law. Temperature and

kinetic energy are related according to the

equation Ek = (3/2)RT. (Section 9.6)



Only the faster-moving molecules have

sufficient kinetic energy to escape from

the liquid and enter the vapor.

The higher the temperature,

the larger the number of

molecules that have enough

energy to escape.



Energy needed

to escape liquid



Higher

temperature



Figure 10.12

Kinetic energy



The distribution of molecular kinetic

energies in a liquid.



Molecules that enter the vapor phase in an open container can escape from the

liquid and drift away until the liquid evaporates entirely, but molecules in a closed

container are trapped. As more and more molecules pass from the liquid to the

vapor, the chances increase that random motion will cause some of them to return

occasionally to the liquid. Ultimately, the number of molecules returning to the

liquid and the number escaping become equal, at which point a dynamic equilibrium

exists. Although individual molecules are constantly passing back and forth from one

phase to the other, the total numbers of molecules in both liquid and vapor phases

remain constant.

The numerical value of a liquid’s vapor pressure depends on the magnitude of

the intermolecular forces present and on the temperature. The smaller the intermolecular forces, the higher the vapor pressure, because loosely held molecules escape

more easily. The higher the temperature, the higher the vapor pressure, because a

larger fraction of molecules have sufficient kinetic energy to escape.



The Clausius–Clapeyron Equation

As indicated in Figure 10.13, the vapor pressure of a liquid rises with temperature in a

nonlinear way. A linear relationship is found, however, when the natural logarithm

of the vapor pressure, ln Pvap, is plotted against the inverse of the Kelvin temperature,

1/T. Table 10.8 gives the appropriate data for water, and Figure 10.13 shows the plot.

As noted in Section 9.2, a linear graph is characteristic of mathematical equations of

the form y = mx + b. In the present instance, y = ln Pvap, x = 1/T, m is the slope of

the line (- ¢Hvap/R), and b is the y-intercept (a constant, C). Thus, the data fit an

expression known as the Clausius–Clapeyron equation:

natural

logarithm



Clausius–Clapeyron equation



ln Pvap =

y



=







ΔHvap 1

+ C

R

T

m



x



+



b



where ¢Hvap is the heat of vaporization of the liquid, R is the gas constant (Section 9.3),

and C is a constant characteristic of each specific substance.



Chapter 10 LIQUIDS, SOLIDS, AND PHASE CHANGES



TABLE 10.8

Temp

(K)



Vapor Pressure of Water at Various Temperatures

Pvap

(mm Hg)



ln Pvap



1/T



Temp

(K)



Pvap

(mm Hg)



ln Pvap



1/T



273

283

293

303

313



4.58

9.21

17.5

31.8

55.3



1.522

2.220

2.862

3.459

4.013



0.003 66

0.003 53

0.003 41

0.003 30

0.003 19



333

343

353

363

373



149.4

233.7

355.1

525.9

760.0



5.007

5.454

5.872

6.265

6.633



0.003 00

0.002 92

0.002 83

0.002 75

0.002 68



323



92.5



4.527



0.003 10



378



906.0



6.809



0.002 65



Vapor pressure (mm Hg)



1000



Ether



7



Ethanol Water



6



800



5



600



ln Pvap



364



1 atm



400



4

3



200



2



0

260



280



300



320



340



360



380



Temperature (K)

The vapor pressures of ether, ethanol, and

water show a nonlinear rise when plotted

as a function of temperature.



1

0.0025



0.0030



0.0035



0.0040



1/T

A plot of ln Pvap versus 1/T (Kelvin) for

water, prepared from the data in Table

10.8, shows a linear relationship.



Figure 10.13



Vapor pressure of liquids at different temperatures.



᭡ What is the vapor pressure of

the liquid at its boiling point?



The Clausius–Clapeyron equation makes it possible to calculate the heat of

vaporization of a liquid by measuring its vapor pressure at several temperatures and

then plotting the results to obtain the slope of the line. Alternatively, once the heat of

vaporization and the vapor pressure at one temperature are known, the vapor pressure of the liquid at any other temperature can be calculated, as shown in Worked

Example 10.5.

When the vapor pressure of a liquid rises to the point where it becomes equal to

the external pressure, the liquid boils and changes into vapor. On a molecular level,

you might picture boiling in the following way: Imagine that a few molecules in the

interior of the liquid momentarily break free from their neighbors and form a microscopic bubble. If the external pressure from the atmosphere is greater than the vapor

pressure inside the bubble, the bubble is immediately crushed. At the temperature

where the external pressure and the vapor pressure in the bubble are the same, however, the bubble is not crushed. Instead, it rises through the denser liquid, grows

larger as more molecules join it, and appears as part of the vigorous action we associate with boiling.

The temperature at which a liquid boils when the external pressure is exactly

1 atm is called the normal boiling point. On the plots in Figure 10.13, the normal

boiling points of the three liquids are reached when the curves cross the dashed line

representing 760 mm Hg—for ether, 34.6 °C (307.8 K); for ethanol, 78.3 °C (351.5 K);

and for water, 100.0 °C (373.15 K).



10.5 EVAPORATION, VAPOR PRESSURE, AND BOILING POINT



If the external pressure is less than 1 atm, then the vapor pressure necessary for

boiling is reached earlier than 1 atm and the liquid boils at a lower than normal temperature. On top of Mt. Everest, for example, where the atmospheric pressure is only

about 260 mm Hg, water boils at approximately 71 °C rather than 100 °C. Conversely,

if the external pressure on a liquid is greater than 1 atm, the vapor pressure necessary

for boiling is reached later and the liquid boils at a greater than normal temperature.

Pressure cookers take advantage of this effect by causing water to boil at a higher

temperature, thereby allowing food to cook more rapidly.

WORKED EXAMPLE 10.5



CALCULATING A VAPOR PRESSURE USING

THE CLAUSIUS–CLAPEYRON EQUATION

The vapor pressure of ethanol at 34.7 °C is 100.0 mm Hg, and the heat of vaporization

of ethanol is 38.6 kJ/mol. What is the vapor pressure of ethanol in millimeters of

mercury at 65.0 °C?

STRATEGY



There are several ways to do this problem. One way is to use the vapor pressure at

T = 307.9 K (34.7 °C) to find a value for C in the Clausius–Clapeyron equation. You

could then use that value to solve for ln Pvap at T = 338.2 K (65.0 °C).

Alternatively, because C is a constant, its value is the same at any two pressures and

temperatures. That is:

C = ln P1 +



¢Hvap

RT1



= ln P2 +



¢Hvap

RT2



This equation can be rearranged to solve for the desired quantity, ln P2:

ln P2 = ln P1 + a



¢Hvap

R



ba



1

1

b

T1

T2



where P1 = 100.0 mm Hg and ln P1 = 4.6052, ¢Hvap = 38.6 kJ/mol, R = 8.3145 J>

(K # mol), T2 = 338.2 K (65.0 °C), and T1 = 307.9 K (34.7 °C).

SOLUTION



J

1

1

mol

≤a

ln P2 = 4.6052 + ±

b

J

307.9 K

338.2 K

8.3145 #

K mol

38,600



ln P2 = 4.6052 + 1.3509 = 5.9561

P2 = antiln (5.9561) = 386.1 mm Hg

Antilogarithms are reviewed in Appendix A.2.



WORKED EXAMPLE 10.6



CALCULATING A HEAT OF VAPORIZATION USING

THE CLAUSIUS–CLAPEYRON EQUATION

Ether has Pvap = 400 mm Hg at 17.9 °C and a normal boiling point of 34.6 °C. What is

the heat of vaporization, ¢Hvap, for ether in kJ/mol?

STRATEGY



The heat of vaporization, ¢Hvap, of a liquid can be obtained either graphically from the

slope of a plot of ln Pvap versus 1/T or algebraically from the Clausius–Clapeyron

equation. As derived in Worked Example 10.5,

ln P2 = ln P1 + a



¢Hvap

R



ba



1

1

b

T1

T2

continued on next page



365



366



Chapter 10 LIQUIDS, SOLIDS, AND PHASE CHANGES



which can be solved for ¢Hvap:

¢Hvap =



(ln P2 - ln P1)(R)

a



1

1

b

T1

T2



where P1 = 400 mm Hg and ln P1 = 5.991, P2 = 760 mm Hg at the normal boiling

point and ln P2 = 6.633, R = 8.3145 J/(K # mol), T1 = 291.1 K (17.9 °C), and

T2 = 307.8 K (34.6 °C).

SOLUTION



(6.633 - 5.991)a8.3145

¢Hvap =



J



K # mol



1

1

291.1 K

307.8 K



b

= 28,600 J/mol = 28.6 kJ/mol



Ī PROBLEM 10.9 The normal boiling point of benzene is 80.1 °C, and the heat of

vaporization is ¢Hvap = 30.7 kJ/mol. What is the boiling point of benzene in °C on top of

Mt. Everest, where P = 260 mm Hg?



Bromine has Pvap = 400 mm Hg at 41.0 °C and a normal boiling

point of 331.9 K. What is the heat of vaporization, ¢Hvap, of bromine in kJ/mol?



Ī PROBLEM 10.10



10.6 KINDS OF SOLIDS

It’s clear from a brief look around that most substances are solids rather than liquids

or gases at room temperature. It’s also clear that there are many different kinds of

solids. Some solids, such as iron and aluminum, are hard and metallic. Others, such

as sugar and table salt, are crystalline and easily broken. And still others, such as rubber and many plastics, are soft and amorphous.

The most fundamental distinction between kinds of solids is that some are crystalline and others are amorphous. Crystalline solids are those whose constituent

particles—atoms, ions, or molecules—have an ordered arrangement extending

over a long range. This order on the atomic level is also seen on the visible level because

crystalline solids usually have flat faces and distinct angles (Figure 10.14a). Amorphous

solids, by contrast, are those whose constituent particles are randomly arranged and

have no ordered long-range structure (Figure 10.14b). Rubber is an example.



(a) A crystalline solid, such



Figure 10.14



Some different kinds of solids.



as this amethyst, has flat

faces and distinct angles.

These regular macroscopic

features reflect a similarly

ordered arrangement of

particles at the atomic level.



(b) An amorphous solid like rubber has a

disordered arrangement of its constituent particles.



10.6 KINDS OF SOLIDS



Crystalline solids can be further categorized as ionic, molecular, covalent network,

or metallic.

Ionic solids are those like sodium chloride, whose constituent particles are ions.

A crystal of sodium chloride is composed of alternating Na + and Cl - ions ordered in

a regular three-dimensional arrangement and held together by ionic bonds, as discussed in Sections 2.11 and 6.7.

Molecular solids are those like sucrose or ice, whose constituent particles are

molecules held together by the intermolecular forces discussed in Section 10.2. A

crystal of ice, for example, is composed of H2O molecules held together in a regular

way by hydrogen bonding (Figure 10.15a).

Covalent network solids are those like quartz (Figure 10.15b) or diamond, whose

atoms are linked together by covalent bonds into a giant three-dimensional array. In

effect, a covalent network solid is one very large molecule.

Metallic solids, such as silver or iron, also consist of large arrays of atoms, but

their crystals have metallic properties such as electrical conductivity. We’ll discuss

metals in Chapter 21.



(a) Ice consists of individual H2O molecules held



(b) Quartz (SiO2) is essentially one very large



together in a regular manner by hydrogen bonds.



molecule with Si– O covalent bonds. Each silicon

atom has tetrahedral geometry and is bonded to

four oxygens; each oxygen has approximately

linear geometry and is bonded to two silicons.



(c) This shorthand

representation shows

how SiO4 tetrahedra

join at their corners to

share oxygen atoms.



Figure 10.15



Crystal structures of ice, a molecular solid, and quartz, a covalent network solid.



A summary of the different types of crystalline solids and their characteristics is

given in Table 10.9.



TABLE 10.9



Types of Crystalline Solids and Their Characteristics



Type of Solid



Intermolecular Forces



Properties



Examples



Ionic



Ion–ion forces



Brittle, hard, high-melting



NaCl, KBr, MgCl2



Molecular



Dispersion forces, dipole–dipole

forces, hydrogen bonds



Soft, low-melting, nonconducting



H2O, Br2, CO2, CH4



Covalent network



Covalent bonds



Hard, high-melting



C (diamond), SiO2



Metallic



Metallic bonds



Variable hardness and melting

point, conducting



Na, Zn, Cu, Fe



367



368



Chapter 10 LIQUIDS, SOLIDS, AND PHASE CHANGES



10.7 PROBING THE STRUCTURE OF SOLIDS:

X-RAY CRYSTALLOGRAPHY

How can the structure of a solid be found experimentally? According to a principle

of optics, the wavelength of light used to observe an object must be less than twice

the length of the object itself. Since atoms have diameters of around 2 * 10-10 m and

the visible light detected by our eyes has wavelengths of 4–7 * 10-7 m, it’s impossible to see atoms using even the finest optical microscope. To “see” atoms, we must

use “light” with a wavelength of approximately 10-10 m, which is in the X-ray region

of the electromagnetic spectrum (Section 5.1).

The origins of X-ray crystallography go back to the work of Max von Laue in

1912. On passing X rays through a crystal of sodium chloride and letting them strike

a photographic plate, Laue noticed that a pattern of spots was produced on the plate,

indicating that the X rays were being diffracted by the atoms in the crystal. A typical

diffraction pattern is shown in Figure 10.16.



A beam of X rays impinges

on a crystal and strikes a

photographic film.

Crystal



X-ray tube

X-ray beam



Photographic

film

The rays are diffracted by atoms in

the crystal, giving rise to a regular

pattern of spots on the film.



Lead

screen

Electron

beam



Figure 10.16



An X-ray diffraction experiment.



Diffraction of electromagnetic radiation occurs when a beam is scattered by an

object containing regularly spaced lines (such as those in a diffraction grating) or

points (such as the atoms in a crystal). This scattering can happen only if the spacing

between the lines or points is comparable to the wavelength of the radiation.

As shown schematically in Figure 10.17, diffraction is due to interference between

two waves passing through the same region of space at the same time. If the waves

are in-phase, peak to peak and trough to trough, the interference is constructive and

the combined wave is increased in intensity. If the waves are out-of-phase, however,

the interference is destructive and the wave is canceled. Constructive interference

gives rise to the intense spots observed on Laue’s photographic plate, while destructive interference causes the surrounding light areas.



10.7 PROBING THE STRUCTURE OF SOLIDS: X-RAY CRYSTALLOGRAPHY



Sum



Sum



Constructive interference occurs if

the waves are in-phase, producing a

wave with increased intensity.



Destructive interference occurs if

the waves are out-of phase, resulting

in cancellation.



Figure 10.17



Interference of electromagnetic waves.



How does the diffraction of X rays by atoms in a crystal give rise to the

observed pattern of spots on a photographic plate? According to an explanation

advanced in 1913 by the English physicist William H. Bragg and his 22-year-old

son, William L. Bragg, the X rays are diffracted by different layers of atoms in the

crystal, leading to constructive interference in some instances but destructive interference in others.

To understand the Bragg analysis, imagine that incoming X rays with wavelength

l strike a crystal face at an angle u and then bounce off at the same angle, just as light

bounces off a mirror (Figure 10.18). Those rays that strike an atom in the top layer are

all reflected at the same angle u, and those rays that strike an atom in the second layer

are also reflected at the angle u. But because the second layer of atoms is farther from

the X-ray source, the distance that the X rays have to travel to reach the second layer

is farther than the distance they have to travel to reach the first layer by an amount

indicated as BC in Figure 10.18. Using trigonometry, you can show that the extra distance BC is equal to the distance between atomic layers d (= AC) times the sine of the

angle u:

BC

d



sin u =



so



BC = d sin u



The extra distance BC = CB’ must also be traveled again by the reflected rays as

they exit the crystal, making the total extra distance traveled equal to 2d sin u.

BC + CB œ = 2d sin u



Incident

rays



Reflected

rays



λ



A



θ

θ



θ



If the extra distance is a

whole number of wavelengths,

the reflected rays are in-phase

and interfere constructively,

making it possible to calculate

the distance d between the

layers.

1st layer

of atoms



d



B



B′

C



2nd layer

of atoms



Rays striking atoms in the second

layer travel a distance equal to

BC + CB′ farther than rays striking

atoms in the first layer.



Figure 10.18



Diffraction of X rays of wavelength l from atoms in the top two layers of a crystal.



369



370



Chapter 10 LIQUIDS, SOLIDS, AND PHASE CHANGES



The key to the Bragg analysis is the realization that the different rays striking the

two layers of atoms are in-phase initially but can be in-phase after reflection only if

the extra distance BC + CB’ is equal to a whole number of wavelengths nl, where n

is an integer (1, 2, 3, ...). If the extra distance is not a whole number of wavelengths,

then the reflected rays will be out-of-phase and will cancel. Setting the extra distance

2d sin u = nl and rearranging to solve for d gives the Bragg equation:

BC + CB œ = 2d sin u = nl

Bragg equation d =



nl

2 sin u



Of the variables in the Bragg equation, the value of the wavelength l is known,

the value of sin u can be measured, and the value of n is a small integer, usually 1.

Thus, the distance d between layers of atoms in a crystal can be calculated. For their

work, the Braggs shared the 1915 Nobel Prize in Physics. The younger Bragg was

25 years old at the time.

Computer-controlled X-ray diffractometers are now available that automatically

rotate a crystal and measure the diffraction from all angles. Analysis of the X-ray

diffraction pattern then makes it possible to measure the interatomic distance

between any two nearby atoms in a crystal. For molecular substances, this knowledge of interatomic distances indicates which atoms are close enough to form a

bond. X-ray analysis thus provides a means for determining the structures of molecules (Figure 10.19).



Figure 10.19



A computer-generated structure of

adenosine triphosphate (ATP) as

determined by X-ray crystallography.

ATP has been called “the energy currency

of the living cell” because it fuels many

metabolic processes.



10.8 UNIT CELLS AND THE PACKING

OF SPHERES IN CRYSTALLINE SOLIDS

How do particles—whether atoms, ions, or molecules—pack together in crystals?

Let’s look at metals, which are the simplest examples of crystal packing because the

individual atoms are spheres. Not surprisingly, metal atoms (and other kinds of particles as well) generally pack together in crystals so that they can be as close as

possible and maximize intermolecular attractions.

If you were to take a large number of uniformly sized marbles and arrange them

in a box in some orderly way, there are four possibilities you might come up with.

One way to arrange the marbles is in orderly rows and stacks, with the spheres in

one layer sitting directly on top of those in the previous layer so that all layers are

identical (Figure 10.20a). In this arrangement, called simple cubic packing, each

sphere is touched by six neighbors—four in its own layer, one above, and one

below—and is thus said to have a coordination number of 6. Only 52% of the available volume is occupied by the spheres in simple cubic packing, making inefficient



10.8 UNIT CELLS AND THE PACKING OF SPHERES IN CRYSTALLINE SOLIDS



371



use of space and minimizing attractive forces. Of all the metals in the periodic table,

only polonium crystallizes in this way.

Alternatively, space could be used more efficiently if, instead of stacking the

spheres directly on top of one another, you slightly separate the spheres in a given

layer and offset alternating layers in an a-b-a-b arrangement so that the spheres in

the b layers fit into the depressions between spheres in the a layers, and vice versa

(Figure 10.20b). In this arrangement, called body-centered cubic packing, each sphere

has a coordination number of 8—four neighbors above and four below—and space is

used quite efficiently: 68% of the available volume is occupied. Iron, sodium, and 14

other metals crystallize in this way.

Figure 10.20



(a) In simple cubic packing,

all layers are identical

and all atoms are lined

up in stacks and rows.



Simple cubic packing and body-centered

cubic packing.



(b) In body-centered cubic packing, the



spheres in layer a are separated slightly and

the spheres in layer b are offset so

that they fit into the depressions

between atoms in layer a. The third layer is a

repeat of the first.

Layer a



Layer b



Layer a

Simple cubic

Coordination number 6:

Each sphere is touched by six

neighbors, four in the same

layer, one directly above,

and one directly below.



Body-centered cubic

Coordination number 8:

Each sphere is touched by

eight neighbors, four in the

layer below, and four in the

layer above.



The remaining two packing arrangements of spheres are both said to be

closest-packed. The hexagonal closest-packed arrangement (Figure 10.21a) has two

alternating layers, a-b-a-b. Each layer has a hexagonal arrangement of touching

spheres, which are offset so that spheres in a b layer fit into the small triangular

depressions between spheres in an a layer. Zinc, magnesium, and 19 other metals

crystallize in this way.

The cubic closest-packed arrangement (Figure 10.21b) has three alternating layers,

a-b-c-a-b-c. The a-b layers are identical to those in the hexagonal closest-packed

arrangement, but the third layer is offset from both a and b layers. Silver, copper, and

16 other metals crystallize with this arrangement.

In both kinds of closest-packed arrangements, each sphere has a coordination

number of 12—six neighbors in the same layer, three above, and three below—and

74% of the available volume is filled. The next time you’re in a grocery store, look

to see how the oranges or apples are stacked in their display box. They’ll almost

certainly have a closest-packed arrangement.



᭣ What kind of packing arrangement

do these oranges have?



372



Chapter 10 LIQUIDS, SOLIDS, AND PHASE CHANGES



Figure 10.21



Hexagonal closest-packing and cubic

closest-packing. In both kinds of packing,

each sphere is touched by 12 neighbors, 6

in the same layer, 3 in the layer above, and

3 in the layer below.



(a) Hexagonal closest-packing:

Two alternating hexagonal layers

a and b are offset so that the spheres

in one layer sit in the small triangular

depressionsof neighboring layers.

Layer a



Layer b



Layer a

Top view



(b) Cubic closest-packing: Three



alternating layers a, b, and c are

offset so that the spheres in one

layer sit in the small triangular

depressions of neighboring layers.



Both arragements have a

coordination number of 12.



Layer a



Layer c



Layer b



Top view



Layer a



Unit Cells



᭡ Just as these bricks are stacked together

in a regular way on the pallet, a crystal is

made of many small repeating units

called unit cells that stack together in a

regular way.



Having taken a bulk view of how spheres can pack in a crystal, let’s now take a closeup view. Just as a large brick wall is made up of many identical bricks stacked

together in a repeating pattern, a crystal is made up of many small repeat units called

unit cells stacked together in three dimensions.

Fourteen different unit-cell geometries occur in crystalline solids. All are

parallelepipeds—six-sided geometric solids whose faces are parallelograms. We’ll be

concerned here only with those unit cells that have cubic symmetry; that is, cells

whose edges are equal in length and whose angles are 90°.

There are three kinds of cubic unit cells: primitive-cubic, body-centered cubic, and

face-centered cubic. As shown in Figure 10.22a, a primitive-cubic unit cell for a metal

has an atom at each of its eight corners, where it is shared with seven neighboring

cubes that come together at the same point. As a result, only 1/8 of each corner atom

“belongs to” a given cubic unit. This primitive-cubic unit cell, with all atoms arranged

in orderly rows and stacks, is the repeat unit found in simple cubic packing.

A body-centered cubic unit cell has eight corner atoms plus an additional atom

in the center of the cube (Figure 10.22b). This body-centered cubic unit cell, with two

repeating offset layers and with the spheres in a given layer slightly separated, is the

repeat unit found in body-centered cubic packing.



10.8 UNIT CELLS AND THE PACKING OF SPHERES IN CRYSTALLINE SOLIDS



(a)



(b)



Eight primitive-cubic unit

cells stack together to share

a common corner.



Figure 10.22



Geometries of (a) primitive-cubic and (b) body-centered cubic unit cells. Both skeletal (top)

and space-filling views (bottom) are shown.



A face-centered cubic unit cell has eight corner atoms plus an additional atom

in the center of each of its six faces, where it is shared with one other neighboring

cube (Figure 10.23a). Thus, 1/2 of each face atom belongs to a given unit cell. This

face-centered cubic unit cell is the repeat unit found in cubic closest-packing, as can

be seen by looking down the body diagonal of a unit cell (Figure 10.23b). The faces of

the unit-cell cube are at 54.7° angles to the layers of the atoms.



(a)



(b)



A view down a body

diagonal shows how this

unit cell is found in cubic

closest-packing.



The faces are tilted

at 54.7° angles to the

three repeating

atomic layers.



Figure 10.23



Geometry of a face-centered cubic

unit cell.



A summary of stacking patterns, coordination numbers, amount of space used,

and unit cells for the four kinds of packing of spheres is given in Table 10.10. Hexagonal

closest-packing is the only one of the four that has a noncubic unit cell.



373



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