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C H A P T E R 2Properties of Gases

The study of the behavior of gases has given rise to numerous chemical and physicaltheories. In many ways, the gaseous state is the easiest to investigate. In this chapter,we examine several gas laws based on experimental observations, introduce the con-cept of temperature, and discuss the kinetic theory of gases.

2.1 Some Basic Definitions

Before we discuss the gas laws, it is useful to define a few basic terms that will be usedthroughout the book. We often speak of the system in reference to a particular part ofthe universe in which we are interested. Thus, a system could be a collection of oxy-gen molecules in a container, a NaCl solution, a tennis ball, or a Siamese cat. Havingdefined a system, we call the rest of the universe the surroundings. There are threetypes of systems (Figure 2.1). An open system is one that can exchange both mass andenergy with its surroundings. A closed system is one that does not exchange masswith its surroundings but can exchange energy. An isolated system is one that canexchange neither mass nor energy with its surroundings. To completely define a sys-tem, we need to understand certain experimental variables, such as pressure, volume,temperature, and composition, that collectively describe the state of the system.

7

A system is separated fromthe surroundings by definiteboundaries such as walls orsurfaces.

Water vapor

Heat Heat

(a) (b) (c)

Figure 2.1(a) An open system allows the exchange of both mass and energy; (b) a closed system allowsthe exchange of energy but not mass; and (c) an isolated system allows exchange of neithermass nor energy.

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Most of the properties of matter may be divided into two classes: extensive

properties and intensive properties. Consider, for example, two beakers containing thesame amounts of water at the same temperature. If we combine these two systems bypouring the water from one beaker to the other, we find that the volume of the wateris doubled and so is its mass. On the other hand, the temperature and the density ofthe water do not change. Properties whose values are directly proportional to theamount of the material present in the system are called extensive properties; thosethat do not depend on the amount are called intensive properties. Extensive prop-erties include mass, area, volume, energy, and electrical charge. As mentioned, tem-perature and density are both intensive properties; so are pressure and electricalpotential.

2.2 An Operational Definition of Temperature

Temperature is a very important quantity in many branches of science, and not sur-prisingly, it can be defined in di¤erent ways. Daily experience tells us that tempera-ture is a measure of coldness and hotness, but for our purposes we need a moreprecise operational definition. Consider the following system of a container of gas A.The walls of the container are flexible so that its volume can expand and contract.This is a closed system that allows heat but not mass to flow into and out of thecontainer. The initial pressure ðPÞ and volume ðVÞ are PA and VA. Now we bring thecontainer in contact with a similar container of gas B at PB and VB. Heat exchangewill take place until thermal equilibrium is reached. At equilibrium, the pressure andvolume of A and B will be altered to P 0A;V

0A and P 0B;V

0B. It is possible to remove

container A temporarily, readjust its pressure and volume to P 00A and V 00A, and stillhave A in thermal equilibrium with B at P 0B and V 0B. In fact, an infinite set of suchvalues ðP 0A;V 0AÞ; ðP 00A;V 00A Þ; ðP 000A ;V

000A Þ; . . . can be obtained that will satisfy the equilib-

rium conditions. Figure 2.2 shows a plot of these points.For all these states of A to be in thermal equilibrium with B, they must have the

same value of the variable we call temperature. It follows that if two systems are inthermal equilibrium with a third system, they must also be in thermal equilibriumwith each other. This statement is generally known as the zeroth law of thermody-

namics. The curve in Figure 2.2 is the locus of all the points that represent the statesthat can be in thermal equilibrium with system B. Such a curve is called an isotherm,or ‘‘same temperature.’’ At another temperature, a di¤erent isotherm is obtained.

2.3 Ideal Gases

In this section we shall briefly examine the various gas laws formulated for thebehavior of an ideal gas.

Boyle’s Law

In a 1662 study of the physical behavior of gases, the English chemist RobertBoyle (1627–1691) found that the volume ðVÞ of a given amount of gas at constanttemperature is inversely proportional to its pressure ðPÞ:

V m1

P

or

PV ¼ constant ð2:1Þ

P

V

PA',VA'

PA'',VA''

PA''',VA'''

PA'''',VA''''

Figure 2.2Plot of pressure versus volume atconstant temperature for a givenamount of a gas. Such a graph iscalled an isotherm.

8 Chapter 2: Properties of Gases

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Equation 2.1 is known as Boyle’s law. A plot of P versus V at a given temperaturegives a hyperbola, which is the isotherm illustrated in Figure 2.2.

Boyle’s law is used to predict the pressure of a gas when its volume changesand vice versa. Letting the initial values of pressure and volume be P1 and V1 and thefinal values of pressure and volume be P2 and V2, we have

P1V1 ¼ P2V2 ðconstant n and temperatureÞ ð2:2Þ

where n is the number of moles of the gas present.

Charles’ and Gay-Lussac’s Law

Boyle’s law depends on the amount of gas and the temperature of the systemremaining constant. But suppose the temperature changes. How does a change intemperature a¤ect the volume and pressure of a gas? Let us first look at the e¤ect oftemperature on the volume of a gas. The earliest investigators of this relationshipwere the French physicists Jacques Alexandre Charles (1746–1823) and Joseph LouisGay-Lussac (1778–1850). Their studies showed that, at constant pressure, the volumeof a gas sample expands when heated and contracts when cooled. The quantitativerelations involved in changes in gas temperature and volume turn out to be remark-ably consistent. For example, we observe an interesting phenomenon when we studythe temperature–volume relationship at various pressures. At any given pressure,the plot of volume versus temperature yields a straight line. By extending the line tozero volume, we find the intercept on the temperature axis to be �273:15�C. At anyother pressure, we obtain a di¤erent straight line for the volume–temperature plot,but we get the same zero-volume temperature intercept at �273:15�C (Figure 2.3).

(In practice, we can measure the volume of a gas over only a limited temperaturerange, because all gases condense at low temperatures to form liquids.)

In 1848, the Scottish mathematician and physicist William Thomson (LordKelvin, 1824–1907) realized the significance of this phenomenon. He identified�273:15�C as absolute zero, which is theoretically the lowest attainable temperature.Then he set up an absolute temperature scale, now called the kelvin temperaturescale, with absolute zero as the starting point. On the kelvin scale, one kelvin (K) isequal in magnitude to one degree Celsius. The only di¤erence between the absolutetemperature scale and the Celsius scale is that the zero position is shifted. The rela-tion between the two scales is

T=K ¼ t=�Cþ 273:15

Dividing the symbol by theunit gives us a pure number.Thus, if T ¼ 298 K, thenT/K ¼ 298.

V

273.15°C

P1

P3

P4

P2

t

Increasingpressure Figure 2.3

Plots of the volume of a given amount of gas versustemperature ðtÞ at di¤erent pressures. All gases ultimatelycondense if they are cooled to low enough temperatures.When these lines are extrapolated, they all converge atthe point representing zero volume and a temperatureof �273.15�C.

2.3 Ideal Gases 9

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Important points on the two scales match up as follows:

Kelvin Scale Celsius Scale

Absolute zero 0 K �273.15�C

Freezing point of water 273.15 K 0�C

Boiling point of water 373.15 K 100�C

In most cases, we shall use 273 instead of 273.15 as the term relating the two scales.By convention, we use T to denote absolute (kelvin) temperature and t to indicatetemperature on the Celsius scale. As we shall soon see, the absolute zero of temper-ature has major theoretical significance; absolute temperatures must be used in gaslaw problems and thermodynamic calculations.

At constant pressure, the volume of a given amount of gas is directly propor-tional to the absolute temperature:

V mT

or

V

T¼ constant ð2:3Þ

Equation 2.3 is known as Charles’ law, or the law of Charles and Gay-Lussac. Analternative form of Charles’ law relates the pressure of a given amount of gas to itstemperature at constant volume:

PmT

or

P

T¼ constant ð2:4Þ

Equations 2.3 and 2.4 permit us to relate the volume–temperature and pressure–temperature values of a gas in states 1 and 2 as follows:

V1

T1¼ V2

T2ðconstant n and PÞ ð2:5Þ

P1

T1¼ P2

T2ðconstant n and VÞ ð2:6Þ

A practical consequence of Equation 2.6 is that automobile tire pressure should bechecked only when the car has been idle for a while. After a long drive, tires becomequite hot and the air pressure in them rises.

Avogadro’s Law

Another important gas law was formulated by Amedeo Avogadro in 1811. Heproposed that equal volumes of gases at the same temperature and pressure containthe same number of molecules. This concept means that

V m n

The normal freezing point andnormal boiling point are mea-sured at 1 atm pressure.


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or

V

n¼ constant ðconstant T and PÞ ð2:7Þ

Equation 2.7 is known as Avogadro’s law.

The Ideal-Gas Equation

According to Equations 2.1, 2.3, and 2.7, the volume of a gas depends on thepressure, temperature, and number of moles as follows:

V m1

Pðconstant T and nÞ ðBoyle’s lawÞ

V mT ðconstant P and nÞ ðCharles’ lawÞ

V m n ðconstant T and PÞ ðAvogadro’s lawÞ

Therefore, V must be proportional to the product of these three terms, that is,

V mnT

P

V ¼ RnT

P

or

PV ¼ nRT (2.8)

where R, a proportionality constant, is the gas constant. Equation 2.8 is called theideal-gas equation. The ideal-gas equation is an example of an equation of state,which provides the mathematical relationships among the properties that define thestate of the system, such as P;T , and V .

The value of R can be obtained as follows. Experimentally, it is found that 1mole of an ideal gas occupies 22.414 L at 1 atm and 273.15 K (a condition known asstandard temperature and pressure, or STP). Thus,

R ¼ ð1 atmÞð22:414 LÞð1 molÞð273:15 KÞ ¼ 0:08206 L atm K�1 mol�1

To express R in units of J K�1 mol�1, we use the conversion factors

1 atm ¼ 1:01325� 105 Pa

1 L ¼ 1� 10�3 m3

and obtain

R ¼ ð1:01325� 105 N m�2Þð22:414� 10�3 m3Þð1 molÞð273:15 KÞ

¼ 8:314 N m K�1 mol�1

¼ 8:314 J K�1 mol�1 ð1 J ¼ 1 N mÞ

2.3 Ideal Gases 11

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From the two values of R, we can write

0:08206 L atm K�1 mol�1 ¼ 8:314 J K�1 mol�1

or

1 L atm ¼ 101:3 J

and

1 J ¼ 9:87� 10�3 L atm

Example 2.1

Air entering the lungs ends up in tiny sacs called alveoli, from which oxygen di¤usesinto the blood. The average radius of the alveoli is 0.0050 cm, and the air inside contains14 mole percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and thetemperature is 37�C, calculate the number of oxygen molecules in one of the alveoli.

A N S W E R

The volume of one alveolus is

V ¼ 4

3pr3 ¼ 4

3pð0:0050 cmÞ3

¼ 5:2� 10�7 cm3 ¼ 5:2� 10�10 L ð1 L ¼ 103 cm3Þ

The number of moles of air in one alveolus is given by

n ¼ PV

RT¼ ð1:0 atmÞð5:2� 10�10 LÞð0:08206 L atm K�1 mol�1Þð310 KÞ

¼ 2:0� 10�11 mol

Because the air inside the alveolus is 14% oxygen, the number of oxygen molecules is

2:0� 10�11 mol air� 14% O2

100% air� 6:022� 1023 O2 molecules

1 mol O2

¼ 1:7� 1012 O2 molecules

Dalton’s Law of Partial Pressures

So far, we have discussed the pressure–volume–temperature behavior of a puregas. Frequently, however, we work with mixtures of gases. For example, a chemistresearching the depletion of ozone in the atmosphere must deal with several gaseouscomponents. For a system containing two or more di¤erent gases, the total pressureðPTÞ is the sum of the individual pressures that each gas would exert if it were aloneand occupied the same volume. Thus,

PT ¼ P1 þ P2 þ � � �

PT ¼X

i

Pi (2.9)


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where P1;P2; . . . are the individual or partial pressures of components 1; 2; . . . andP

is the summation sign. Equation 2.9 is known as Dalton’s law of partial pressures

(after the English chemist and mathematician John Dalton, 1766–1844).Consider a system containing two gases (1 and 2) at temperature T and volume

V . The partial pressures of the gases are P1 and P2, respectively. From Equation 2.8,

P1V ¼ n1RT or P1 ¼n1RT

V

P2V ¼ n2RT or P2 ¼n2RT

V

where n1 and n2 are the numbers of moles of the two gases. According to Dalton’slaw,

PT ¼ P1 þ P2

¼ n1RT

Vþ n2

RT

V

¼ ðn1 þ n2ÞRT

V

Dividing the partial pressures by the total pressure and rearranging, we get

P1 ¼n1

n1 þ n2PT ¼ x1PT

and

P2 ¼n2

n1 þ n2PT ¼ x2PT

where x1 and x2 are the mole fractions of gases 1 and 2. A mole fraction, defined asthe ratio of the number of moles of one gas to the total number of moles of all gasespresent, is a dimensionless quantity. Furthermore, by definition, the sum of all themole fractions in a mixture must be unity, that is

Xi

xi ¼ 1 ð2:10Þ

In general, in a mixture of gases the partial pressure of the ith component, Pi, isgiven by

Pi ¼ xiPT (2.11)

How are partial pressures determined? A manometer can measure only the totalpressure of a gaseous mixture. To obtain partial pressures, we need to know the molefractions of the components. The most direct method of measuring partial pressuresis using a mass spectrometer. The relative intensities of the peaks in a mass spectrumare directly proportional to the amounts, and hence to the mole fractions, of thegases present.

The gas laws played a key role in the development of atomic theory, and manypractical illustrations of them appear in everyday life. The two brief examples below

2.3 Ideal Gases 13

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are particularly important to scuba divers. Seawater has a slightly higher density thanfresh water—approximately 1.03 g mL�1 compared with 1.00 g mL�1. The pressureexerted by a 33-ft (10-m) column of seawater is equivalent to 1 atm pressure. Whatwould happen if a diver were to rise to the surface rather quickly, holding his breath?If the ascent started at 40 ft under water, the decrease in pressure from this depth tothe surface would be (40 ft/33 ft)� 1 atm, or 1.2 atm. Assuming constant tempera-ture, when the diver reached the surface, the volume of air trapped in his lungs wouldhave increased by a factor of ð1þ 1:2Þ atm/1 atm, or 2.2 times! This sudden expan-sion of air could damage or rupture the membranes of his lungs, seriously injuring orkilling the diver.

Dalton’s law has a direct application to scuba diving. The partial pressure ofoxygen in air is approximately 0.2 atm. Because oxygen is essential for our sur-vival, we may have a hard time believing that it could be harmful to breathe in morethan normal. In fact, the toxicity of oxygen is well documented.* Physiologically, ourbodies function best when the partial pressure of oxygen is 0.2 atm. For this reason,the composition of the air in a scuba tank is adjusted when the diver is submerged.For example, at a depth where the total pressure (hydrostatic plus atmospheric) is4 atm, the oxygen content in the air supply should be reduced to 5% by volumeto maintain the optimal partial pressure ð0:05� 4 atm ¼ 0:2 atmÞ. At a greaterdepth, the oxygen content must be even lower. Although nitrogen would seem to bethe obvious choice for mixing with oxygen in a scuba tank because it is the majorcomponent of air, it is not the best choice. When the partial pressure of nitrogenexceeds 1 atm, a su‰cient amount will dissolve in the blood to cause nitrogen narco-

sis. Symptoms of this condition, which resembles alcohol intoxication, include light-headedness and impaired judgment. Divers su¤ering from nitrogen narcosis havebeen known to do strange things, such as dancing on the sea floor and chasingsharks. For this reason, helium is usually employed to dilute oxygen in diving tanks.Helium, an inert gas, is much less soluble in blood than nitrogen, and it does notproduce narcotic e¤ects.

2.4 Real Gases

The ideal-gas equation holds only for gases that have the following properties: (1) thegas molecules possess negligible volume, and (2) there is no interaction, attractive orrepulsive, among the molecules. Obviously, no such gases exist. Nevertheless, Equa-tion 2.8 is quite useful for many gases at high temperatures or moderately low pres-sures (a10 atm).

When a gas is being compressed, the molecules are brought closer to one an-other, and the gas will deviate appreciably from ideal behavior. One way to measurethe deviation from ideality is to plot the compressibility factor ðZÞ of a gas versuspressure. Starting with the ideal-gas equation, we write

PV ¼ nRT

or

* At partial pressures above 2 atm, oxygen becomes toxic enough to produce convulsions andcoma. Years ago, newborn infants placed in oxygen tents often developed retrolental fibroplasia,damage of the retinal tissues by excess oxygen. This damage usually resulted in partial or totalblindness.

The pressure exerted by sea-water is called the hydrostaticpressure.


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Z ¼ PV

nRT¼ PV

RT(2.12)

where V is the molar volume of the gas ðV=nÞ or the volume of 1 mole of the gas atthe specified temperature and pressure. For an ideal gas, Z ¼ 1 for any value of P ata given T . However, as Figure 2.4 shows, the compressibility factors for real gasesexhibit fairly divergent dependence on pressure. At low pressures, the compressibilityfactors of most gases are close to unity. In fact, in the limit of P approaching zero, wehave Z ¼ 1 for all gases. This finding is expected because all real gases behave ideallyat low pressures. As pressure increases, some gases have Z < 1, which means thatthey are easier to compress than an ideal gas. Then, as pressure increases further, allgases have Z > 1. Over this region, the gases are harder to compress than an idealgas. These behaviors are consistent with our understanding of intermolecular forces.In general, attractive forces are long-range forces, whereas repulsive forces operateonly within a short range (more on this topic in Chapter 13). When molecules are farapart (for example, at low pressures), the predominant intermolecular interaction isattraction. As the distance of separation between molecules decreases, the repulsiveinteraction among molecules becomes more significant.

Over the years, considerable e¤ort has gone into modifying the ideal-gas equa-tion for real gases. Of the numerous such equations proposed, we shall consider two:the van der Waals equation and the virial equation of state.

The van der Waals Equation

The van der Waals equation of state (after the Dutch physicist Johannes Diderikvan der Waals, 1837–1923) attempts to account for the finite volume of individualmolecules in a nonideal gas and the attractive forces between them.

Pþ an2

V 2

� �ðV � nbÞ ¼ nRT (2.13)

The pressure exerted by the individual molecules on the walls of the container de-pends on both the frequency of molecular collisions with the walls and the momen-tum imparted by the molecules to the walls. Both contributions are diminished by theattractive intermolecular forces (Figure 2.5). In each case, the reduction in pressure

Z

Ideal gas

0 200 400 600 800 1000

2.0

1.5

1.0

0.5

N2

CH4

P /atm

He

Figure 2.4Plot of the compressibility factor versus pressure for real gases and an ideal gas at 273 K.Note that for an ideal gas Z ¼ 1, no matter how great the pressure.

Figure 2.5E¤ect of intermolecular forceson the pressure exerted by a gas.The speed of a molecule that ismoving toward the containerwall (red sphere) is reduced bythe attractive forces exerted byits neighbors (gray spheres).Consequently, the impact thismolecule makes with the wall isnot as great as it would be if nointermolecular forces were pre-sent. In general, the measuredgas pressure is lower than thepressure the gas would exert ifit behaved ideally.

2.4 Real Gases 15

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depends on the number of molecules present or the density of the gas, n=V , so that

reduction in pressure due to attractive forcesmn

V

� �n

V

� �

¼ an2

V 2

where a is a proportionality constant.Note that in Equation 2.13, P is the measured pressure of the gas and

ðPþ an2=V 2Þ would be the pressure of the gas if there were no intermolecular forcespresent. Because an2=V 2 must have units of pressure, a is expressed as atm L2 mol�2.To allow for the finite volume of molecules, we replace V in the ideal-gas equationwith ðV � nbÞ, where nb represents the total e¤ective volume of n moles of the gas.Thus, nb must have the unit of volume, and b has the units L mol�1. Both a and b areconstants characteristic of the gas under study. Table 2.1 lists the values of a and b

for several gases. The value of a is related to the magnitude of attractive forces. Usingthe boiling point as a measure of the strength of intermolecular forces (the higher theboiling point, the stronger the intermolecular forces), we see that there is a roughcorrelation between the values of a and the boiling points of these substances. Thequantity b is more di‰cult to interpret. Although b is proportional to the size of themolecule, the correlation is not always straightforward. For example, the value of b

for helium is 0.0237 L mol�1 and that for neon is 0.0174 L mol�1. Based on thesevalues, we might expect that helium is larger than neon, which we know is not true.

The van der Waals equation is valid over a wider range of pressure and temper-ature than is the ideal-gas equation. Furthermore, it provides a molecular interpre-tation for the equation of state. At very high pressures and low temperatures, the vander Waals equation also becomes unreliable.

The Virial Equation of State

Another way of representing gas nonideality is the virial equation of state. In thisrelationship, the compressibility factor is expressed as a series expansion in inversepowers of molar volume V :

Z ¼ 1þ B

Vþ C

V 2þ D

V 3þ � � � (2.14)

Table 2.1van der Waals Constants and Boiling Points of Some Substances

Substance a/atm � L2 �mol�2 b/L �mol�1 Boiling Point/K

He 0.0341 0.0237 4.2

Ne 0.214 0.0174 27.2

Ar 1.34 0.0322 87.3

H2 0.240 0.0264 20.3

N2 1.35 0.0386 77.4

O2 1.34 0.0312 90.2

CO 1.45 0.0395 83.2

CO2 3.60 0.0427 195.2

CH4 2.26 0.0430 109.2

H2O 5.47 0.0305 373.15

NH3 4.25 0.0379 239.8


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where B;C;D, are called the second, third, fourth, . . . virial coe‰cients. The firstvirial coe‰cient is 1. The second and higher virial coe‰cients are all temperaturedependent. For a given gas, they are evaluated from the P–V –T data of the gas bya curve-fitting procedure using a computer. For an ideal gas, the second and highervirial coe‰cients are zero and Equation 2.14 becomes Equation 2.8.

An alternate form of the virial equation is given by a series expansion of thecompressibility factor in terms of the pressure, P:

Z ¼ 1þ B 0Pþ C 0P2 þD 0P3 þ � � � (2.15)

Because P and V are related, it is not surprising that relationships exist between B

and B 0, C and C 0, and so on. In each equation, the values of the coe‰cients decreaserapidly. For example, in Equation 2.15, the magnitude of the coe‰cients is such thatB 0gC 0gD 0 so that at pressures between zero and 10 atm, say, we need to includeonly the second term provided the temperature is not very low:

Z ¼ 1þ B 0P ð2:16Þ

Equations 2.13 and 2.14 or 2.15 exemplify two rather di¤erent approaches. Thevan der Waals equation accounts for the nonideality of gases by correcting for thefinite molecular volume and intermolecular forces. Although these corrections doresult in a definite improvement over the ideal gas equation, Equation 2.13 is stillan approximate equation. The reason is that our present knowledge of intermolec-ular forces is insu‰cient to quantitatively explain macroscopic behavior. Of course,we could further improve this equation by adding more corrective terms; indeed,numerous other equations of state have been proposed since van der Waals first pre-sented his analysis. On the other hand, Equation 2.14 is accurate for real gases, but itdoes not provide us with any direct molecular interpretation. The nonideality of thegas is accounted for mathematically by a series expansion in which the coe‰cientsB;C; . . . can be determined experimentally. These coe‰cients do not have any phys-ical meaning, although they can be related to intermolecular forces in an indirectway. Thus, our choice in this case is between an approximate equation that gives ussome physical insight or an equation that describes the gas behavior accurately (if thecoe‰cients are known), but tells us nothing about molecular behavior.

Example 2.2

Calculate the molar volume of methane at 300 K and 100 atm, given that the secondvirial coe‰cient ðBÞ of methane is �0.042 L mol�1. Compare your result with thatobtained using the ideal-gas equation.

A N S W E R

From Equation 2.14, neglecting terms containing C;D; . . . ;

Z ¼ 1þ B

V

¼ 1þ BP

RT

¼ 1þ ð�0:042 L mol�1Þð100 atmÞð0:08206 L atm K�1 mol�1Þð300 KÞ

¼ 1� 0:17 ¼ 0:83

2.4 Real Gases 17

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V ¼ ZRT

P

¼ ð0:83Þð0:08206 L atm K�1 mol�1Þð300 KÞ100 atm

¼ 0:20 L mol�1

For an ideal gas,

V ¼ RT

P

¼ ð0:08206 L atm K�1 mol�1Þð300 KÞ100 atm

¼ 0:25 L mol�1

C O M M E N T

At 100 atm and 300 K, methane is more compressible than an ideal gas (Z ¼ 0:83compared with Z ¼ 1) due to the attractive intermolecular forces between the CH4

molecules.

2.5 Condensation of Gases and the Critical State

The condensation of gas to liquid is a familiar phenomenon. The first quantitativestudy of the pressure–volume relationship of this process was made in 1869 by theIrish chemist Thomas Andrews (1813–1885). He measured the volume of a givenamount of carbon dioxide as a function of pressure at various temperatures andobtained a series of isotherms like those shown in Figure 2.6. At high temperatures,the curves are roughly hyperbolic, indicating that the gas obeys Boyle’s law. As thetemperature is lowered, deviations become evident, and a drastically di¤erent be-havior is observed at T4. Moving along the isotherm from right to left, we see that

P

V

Critical point

Liquid and vaporat equilibrium

T1

T2

T3

T4

T7

T6

T5

Figure 2.6Isotherms of carbon dioxide at various temperatures(temperature increases from T1 to T7). The criticaltemperature is T5. Above this temperature carbondioxide cannot be liquefied no matter how great thepressure.


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although the volume of the gas decreases with pressure, the product PV is no longera constant (because the curve is no longer a hyperbola). Increasing the pressure fur-ther, we reach a point that is the intersection between the isotherm and the dashedcurve on the right. If we could observe this process, we would note the formation ofliquid carbon dioxide at this pressure. With the pressure held constant, the volumecontinues to decrease (as more vapor is converted to liquid) until all the vapor hasbeen condensed. Beyond this point (the intersection between the horizontal line andthe dashed curve on the left), the system is entirely liquid, and any further increase inpressure will result in only a very small decrease in volume, because liquids are muchless compressible than gases (Figure 2.7).

The pressure corresponding to the horizontal line (the region in which vapor andliquid coexist) is called the equilibrium vapor pressure or simply the vapor pressure ofthe liquid at the temperature of the experiment. The length of the horizontal linedecreases with increasing temperature. At a particular temperature (T5 in this case),the isotherm is tangential to the dashed curve and only one phase (the gas phase) ispresent. The horizontal line is now at a point known as the critical point. The corre-sponding temperature, pressure, and volume at this point are called the critical tem-perature ðTcÞ, critical pressure ðPcÞ, and critical volume ðVcÞ, respectively. The criti-

cal temperature is the temperature above which no condensation can occur no matterhow great the pressure. The critical constants of several gases are listed in Table 2.2.*Note that the critical volume is usually expressed as a molar quantity, called themolar critical volume ðVcÞ, given by Vc=n, where n is the number of moles of thesubstance present.

The phenomenon of condensation and the existence of a critical temperature aredirect consequences of the nonideal behavior of gases. After all, if molecules did notattract one another, no condensation would occur and if molecules had no volume,then we would not be able to observe liquids. As mentioned earlier, the nature ofmolecular interaction is such that the force among molecules is attractive when they

* The van der Waals constants of a gas (see Equation 2.13) can be obtained from its critical con-stants. For mathematical details see the physical chemistry texts listed in Chapter 1.

P

V

T1

Gas

Gas

LiquidLiquid

(a) (b) (c) (d)

(a)(b)(c)

Figure 2.7The liquefaction of carbon dioxide at T1 (see Figure 2.6). At (a), the first drop of the liquidappears. From (b) to (c) the gas is gradually and completely converted to liquid at constantpressure. Beyond (c) the volume decreases only slightly with increasing pressure becauseliquids are highly incompressible. As temperature increases, the horizontal line becomesshorter until it becomes a point at T5, the critical temperature.

2.5 Condensation of Gases and the Critical State 19

Black plate (20,1)

are relatively far apart, but as they get closer to one another (for example, a liquidunder pressure) this force becomes repulsive, because of electrostatic repulsions be-tween nuclei and between electrons. In general, the attractive force reaches a maxi-mum at a certain finite intermolecular distance. At temperatures below Tc, it is pos-sible to compress the gas and bring the molecules within this attractive range, wherecondensation will occur. Above Tc, the kinetic energy of the gas molecules is suchthat they will always be able to break away from this attraction and no condensa-tion can take place. Figure 2.8 shows the critical phenomenon of sulfur hexafluoride(SF6).

Table 2.2Critical Constants of Some Substances

Substance Pc/atm Vc=L �mol�1 Tc=K

He 2.25 0.0578 5.2

Ne 26.2 0.0417 44.4

Ar 49.3 0.0753 151.0

H2 12.8 0.0650 32.9

N2 33.6 0.0901 126.1

O2 50.8 0.0764 154.6

CO 34.5 0.0931 132.9

CO2 73.0 0.0957 304.2

CH4 45.4 0.0990 190.2

H2O 217.7 0.0560 647.6

NH3 109.8 0.0724 405.3

SF6 37.6 0.2052 318.7

(a) (b) (c) (d)

Figure 2.8The critical phenomenon of sulfur hexafluoride (Tc ¼ 45:5�C; Pc ¼ 37:6 atm). (a) Belowthe critical temperature, a clear liquid phase is visible. (b) Above the critical temperature,the liquid phase disappears. (c) The substance is cooled just below its critical temperature.(d) Finally, the liquid phase reappears.


Black plate (21,1)

In recent years, there has been much interest in the practical applications ofsupercritical fluids (SCF), that is, of the state of matters above the critical tempera-ture. One of the most studied SCFs is carbon dioxide. Under appropriate conditionsof temperature and pressure, SCF CO2 can be used as a solvent for removing ca¤einefrom raw co¤ee beans and cooking oil from potato chips to produce crisp, oil-freechips. It is also being used in environmental cleanups because it dissolves chlorinatedhydrocarbons. SCFs of CO2, NH3, and certain hydrocarbons such as hexane andheptane are used in chromatography. SCF CO2 has been shown to be an e¤ectivecarrier medium for substances such as antibiotics and hormones, which are unstableat the high temperatures required for normal chromatographic separations.

2.6 Kinetic Theory of Gases

The study of gas laws exemplifies the phenomenological, macroscopic approach tophysical chemistry. Equations describing the gas laws are relatively simple, and ex-perimental data are readily accessible. Yet studying gas laws gives us no real physicalinsight into processes that occur at the molecular level. Although the van der Waalsequation attempts to account for nonideal behavior in terms of intermolecular inter-actions, it does so in a rather vague manner. It does not answer such questions as:How is the pressure of a gas related to the motion of individual molecules, and whydo gases expand when heated at constant pressure? The next logical step, then, isto try to explain the behavior of gases in terms of the dynamics of molecular motion.To interpret the properties of gas molecules in a more quantitative manner, we turnto the kinetic theory of gases.

The Model

Any time we try to develop a theory to account for experimental observations,we must first define our system. If we do not understand all the properties of a sys-tem, as is usually the case, we must make a number of assumptions. Our model for

the kinetic theory of gases is based on the following assumptions:

1. A gas is made up of a great number of atoms or molecules, separated bydistances that are large compared to their size.

2. The molecules have mass, but their volume is negligibly small.

3. The molecules are constantly in random motion.

4. Collisions among molecules and between molecules and the walls of thecontainer are elastic; that is, kinetic energy may be transferred from onemolecule to another, but it is not converted to other forms of energy.

5. There is no interaction, attractive or repulsive, between the molecules.

Assumptions 2 and 5 should be familiar from our discussion of ideal gases. The dif-ference between the ideal-gas laws and the kinetic theory of gases is that for the latter,we shall use the foregoing assumptions in an explicit manner to derive expressions formacroscopic properties, such as pressure and temperature, in terms of the motion ofindividual molecules.

Pressure of a Gas

Using the model for the kinetic theory of gases, we can derive an expressionfor the pressure of a gas in terms of its molecular properties. Consider an ideal gas

2.6 Kinetic Theory of Gases 21

Black plate (22,1)

made up of N molecules, each of mass m, confined in a cubic box of length l. Atany instant, the molecular motion inside the container is completely random. Letus analyze the motion of a particular molecule with velocity v. Because velocity is avector quantity—it has both magnitude and direction—v can be resolved into threemutually perpendicular components vx; vy, and vz. These three components give therates at which the molecule is moving along the x; y, and z directions, respectively; v

is simply the resultant velocity (Figure 2.9). The projection of the velocity vector onthe xy plane is OA, which, according to Pythagoras’ theorem, is given by

OA2 ¼ v2x þ v2

y

Similarly,

v2 ¼ OA2 þ v2z

¼ v2x þ v2

y þ v2z ð2:17Þ

Let us for the moment consider the motion of a molecule only along the x di-rection. Figure 2.10 shows the changes that take place when the molecule collideswith the wall of the container (the yz plane) with velocity component vx. Because thecollision is elastic, the velocity after collision is the same as before but opposite indirection. The momentum of the molecule is mvx, where m is its mass, so that thechange in momentum is given by

mvx �mð�vxÞ ¼ 2mvx

The sign of vx is positive when the molecule moves from left to right and negativewhen it moves in the opposite direction. Immediately after the collision, the moleculewill take time l=vx to collide with the other wall, and in time 2l=vx the molecule willstrike the same wall again.* Thus, the frequency of collision between the moleculeand a given wall (that is, the number of collisions per unit time) is vx=2l, and thechange in momentum per unit time is ð2mvxÞðvx=2lÞ, or mv2

x=l. According to New-ton’s second law of motion,

force ¼ mass� acceleration

¼ mass� distance� time�2

¼ momentum time�1

* We assume that the molecule does not collide with other molecules along the way. A more rig-orous treatment including molecular collision gives exactly the same result.

A

O y

z

vx

vz

vy

v 2vx v 2vy

Velocity vector v

x

Figure 2.9Velocity vector v and its componentsalong the x; y, and z directions.

After Before

vxvx

Figure 2.10Change in velocity upon colli-sion of a molecule moving withvx with the wall of the container.


Black plate (23,1)

Therefore, the force exerted by one molecule on one wall as a result of the collision ismv2

x=l, and the total force due to N molecules is Nmv2x=l. Because pressure is force/

area and area is l2, we can now express the total pressure exerted on one wall as

P ¼ F

A

¼ Nmv2x

lðl2Þ ¼Nmv2

x

V

or

PV ¼ Nmv2x ð2:18Þ

where V is the volume of the cube (equal to l 3). When we are dealing with a largecollection of molecules (for example, when N is on the order of 6� 1023), there is atremendous spread of molecular velocities. It is more appropriate, therefore, to re-place v2

x in Equation 2.18 with the mean or average quantity, v2x. Referring to Equa-

tion 2.17, we see that the relation between the average of the square of the velocitycomponents and the average of the square of the velocity, v2, is still

v2 ¼ v2x þ v2

y þ v2z

The quantity v2 is called the mean-square velocity, defined as

v2 ¼ v21 þ v2

2 þ � � � þ v2N

Nð2:19Þ

When N is a large number, it is correct to assume that molecular motions along thex; y, and z directions are equally probable. This means that

v2x ¼ v2

y ¼ v2z ¼

v2

3

and Equation 2.18 can now be written as

P ¼ Nmv2

3V

Multiplying the top and bottom by 2 and recalling that the kinetic energy of themolecule Etrans is given by 1

2 mv2 (where the subscript trans denotes translational mo-tion; that is, motion through space of the whole molecule), we obtain

P ¼ 2N

3V

1

2mv2

� �¼ 2N

3VEtrans ð2:20Þ

This is the pressure exerted by N molecules on one wall. The same result can beobtained regardless of the direction (x; y, or z) we describe for the molecular motion.We see that the pressure is directly proportional to the average kinetic energy or,more explicitly, to the mean-square velocity of the molecule. The physical meaningof this dependence is that the larger the velocity, the more frequent the collisions andthe greater the change in momentum. Thus, these two independent terms give us thequantity v2 in the kinetic theory expression for the pressure.

2.6 Kinetic Theory of Gases 23

Black plate (24,1)

Kinetic Energy and Temperature

Let us compare Equation 2.20 with the ideal-gas equation (Equation 2.8):

PV ¼ nRT

¼ N

NART

or

P ¼ NRT

NAVð2:21Þ

where NA is the Avogadro constant. Combining the pressures in Equations 2.20 and2.21, we get

2

3

N

VEtrans ¼

N

NA

RT

V

or

Etrans ¼3

2

RT

NA¼ 3

2kBT (2.22)

where R ¼ kBNA and kB is the Boltzmann constant, equal to 1:380658� 10�23 J K�1

[after the Austrian physicist Ludwig Eduard Boltzmann (1844–1906)]. (In most cal-culations, we shall round kB to 1:381� 10�23 J K�1.) We see that the mean kineticenergy of one molecule is proportional to absolute temperature.

The significance of Equation 2.22 is that it provides an explanation for the tem-perature of a gas in terms of molecular motion. For this reason, random molecularmotion is sometimes referred to as thermal motion. It is important to keep in mindthat the kinetic theory is a statistical treatment of our model; hence, it is meaninglessto associate temperature with the kinetic energy of just a few molecules. Equation2.22 also tells us that whenever two ideal gases are at the same temperature T , theymust have the same average kinetic energy. The reason is that Etrans in Equation 2.22is independent of molecular properties such as size or molar mass or amount of thegas present, as long as N is a large number.

It is easy to see that v2 would be a very di‰cult quantity to measure, if indeedit could be measured at all. To do so, we would need to measure each individualvelocity, square it, and then take the average (see Equation 2.19). Fortunately, v2 canbe obtained quite directly from other quantities. From Equation 2.22, we write

1

2mv2 ¼ 3

2

RT

NA¼ 3

2kBT

so that

v2 ¼ 3RT

mNA¼ 3kBT

m

or

ffiffiffiffiffiv2

p¼ vrms ¼

ffiffiffiffiffiffiffiffiffiffi3RT

M

r¼

ffiffiffiffiffiffiffiffiffiffiffiffi3kBT

m

rðM ¼ mNAÞ ð2:23Þ

The kinetic energy of 1 moleof the gas is given byð3=2ÞNAkBT ¼ ð3=2ÞRT .

Note that kB refers to onemolecule and R refers to onemole of such molecules;kB ¼ R=NA.


Black plate (25,1)

where vrms is the root-mean-square velocity* and m is the mass (in kg) of one mole-cule; M is the molar mass (in kg mol�1). Note that vrms is directly proportional to thesquare root of temperature and inversely proportional to the square root of molarmass of the molecule. Therefore, the heavier the molecule, the slower its motion.

2.7 The Maxwell Distribution Laws

The root-mean-square velocity gives us an average measure that is very useful in thestudy of a large number of molecules. When we are studying, say, one mole of a gas,it is impossible to know the velocity of each individual molecule for two reasons.First, the number of molecules is so huge that there is no way we can follow all theirmotions. Second, although molecular motion is a well-defined quantity, we cannotmeasure its velocity exactly. Therefore, rather than concerning ourselves with indi-vidual molecular velocities, we ask this question: For a given system at some knowntemperature, how many molecules are moving at velocities between v and vþ Dv atany moment? Or, how many molecules in a macroscopic gas sample have velocities,say, between 306.5 m s�1 and 306.6 m s�1 at any moment?

Because the total number of molecules is very large, there is a continuous spread,or distribution, of velocities as a result of collisions. We can therefore make the ve-locity range Dv smaller and smaller, and in the limit it becomes dv. This fact has greatsignificance, because it enables us to replace the summation sign with the integral signin calculating the number of molecules whose velocities fall between v and vþ dv.Mathematically speaking, it is easier to integrate than to sum a large series. Thisdistribution-of-velocities approach was first employed by the Scottish physicist JamesClerk Maxwell (1831–1879) in 1860 and later refined by Boltzmann. They showedthat for a system containing N ideal gas molecules at thermal equilibrium with itssurroundings, the fraction of molecules dN=N moving at velocities between vx andvx þ dvx along the x direction is given by

dN

N¼ m

2pkBT

� �1=2

e�mv2x=2kBT dvx ð2:24Þ

where m is the mass of the molecule, kB the Boltzmann constant, and T the absolutetemperature.

As mentioned earlier, velocity is a vector quantity. In many cases, we need todeal only with the speed of molecules (c), which is a scalar quantity; that is, it hasmagnitude but no directional properties. The fraction of molecules dN=N movingbetween speeds c and cþ dc is given by

dN

N¼ 4pc2 m

2pkBT

� �3=2

e�mc2=2kBT dc

¼ f ðcÞdc ð2:25Þ

where f ðcÞ, the Maxwell speed distribution function, is given by

f ðcÞ ¼ 4pc2 m

2pkBT

� �3=2

e�mc2=2kBT (2.26)

* Because velocity is a vector quantity, the average molecular velocity, v, must be zero; there arejust as many molecules moving in the positive direction as there are in the negative direction. On theother hand, vrms is a scalar quantity; that is, it has magnitude but no direction.

2.7 The Maxwell Distribution Laws 25

Black plate (26,1)

Figure 2.11 shows the dependence of the speed distribution curve on tempera-ture and molar mass. At any given temperature, the general shape of a distributioncurve can be explained as follows. Initially, at small c values, the c2 term in Equation2.25 dominates so f ðcÞ increases with increasing c. At larger values of c, the terme�mc2=2kBT becomes more important. These two opposing terms cause the curve toreach a maximum beyond which it decreases roughly exponentially with increasing c.The speed corresponding to the maximum value of f ðcÞ is called the most probable

speed, cmp, because it is the speed of the largest number of molecules.Figure 2.11a shows how the shape of the distribution curve is influenced by

temperature. At low temperatures the distribution has a rather narrow range. As thetemperature increases, the curve becomes flatter, meaning that there are now morefast-moving molecules. This temperature dependence of the distribution curve hasimportant implications in chemical reaction rates. As we shall see in Chapter 9, inorder to react, a molecule must possess a minimum amount of energy, called activa-

tion energy. At low temperatures the number of fast-moving molecules is small; hencemost reactions proceed at a slow rate. Raising the temperature increases the numberof energetic molecules and causes the reaction rate to increase. In Figure 2.11b we seethat heavier gases have a narrower range of speed distribution than lighter gases atthe same temperature. This is to be expected considering that heavier gases moveslower, on the average, than lighter gases. The validity of the Maxwell speed distri-bution has been verified experimentally.

The usefulness of the Maxwell speed distribution function is that it enables us tocalculate average quantities. In fact, we can obtain three related expressions for speedcalled the most probable speed (cmp), as defined above, average speed (c), which isdefined as the sum of the speeds of all the molecules divided by the number of mol-ecules, and root-mean-square speed (crms):*

* For the derivation of cmp, see Problem 2.66; for the derivation of c, see the physical chemistrytexts listed in Chapter 1. Because the square of the average velocity is a scalar quantity, it followsthat v2 ¼ c2; hence vrms ¼ crms.

Molecular speed (m s 1)

(a)

0 500

100 K

300 K

700 K

1000 1500 2500

Nu

mb

er

of

mo

lecu

les

Molecular speed (m s 1)

(b)

0 500 1000 1500 2000

Nu

mb

er

of

mo

lecu

lesN2 (28.02 g mol 1)

Cl2 (70.90 g mol 1)

N2 (28.02 g mol 1)

He (4.003 g mol 1)

T 300 K

Figure 2.11(a) The distribution of speeds for nitrogen gas at three di¤erent temperatures. At highertemperatures, more molecules are moving at faster speeds. (b) The distribution of speeds forthree gases at 300 K. At a given temperature, the light molecules are moving faster, on theaverage.


Black plate (27,1)

cmp ¼ffiffiffiffiffiffiffiffiffiffi2RT

M

r(2.27)

c ¼ffiffiffiffiffiffiffiffiffiffi8RT

pM

r(2.28)

crms ¼ffiffiffiffiffiffiffiffiffiffi3RT

M

r(2.29)

Example 2.3

Calculate the values of cmp; c, and crms for O2 at 300 K.

A N S W E R

The constants are

R ¼ 8:314 J K�1 mol�1 T ¼ 300 K

M ¼ 0:03200 kg mol�1

The most probable speed is given by

cmp ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 8:314 J K�1 mol�1 � 300 K

0:03200 kg mol�1

s

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:56� 105 J kg�1

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:56� 105 m2 s�2p

¼ 395 m s�1

Similarly, we can show that

c ¼ffiffiffiffiffiffiffiffiffiffi8RT

pM

r¼ 446 m s�1

and

crms ¼ffiffiffiffiffiffiffiffiffiffi3RT

M

r¼ 484 m s�1

C O M M E N T

The calculation shows, and indeed it is generally true, that crms > c > cmp. That cmp isthe smallest of the three speeds is due to the asymmetry of the curve (see Figure 2.11).The reason crms is greater than c is that the squaring process in Equation 2.19 isweighted toward larger values of c.

2.7 The Maxwell Distribution Laws 27

Black plate (28,1)

Finally, note that both the c and crms values of N2 and O2 are close to the speedof sound in air. Sound waves are pressure waves. The propagation of these waves isdirectly related to the movement of molecules and hence to their speeds.

2.8 Molecular Collisions and the Mean Free Path

Now that we have an explicit expression for the average speed, c, we can use it tostudy some dynamic processes involving gases. We know that the speed of a moleculeis not constant but changes frequently as a result of collisions. Therefore, the questionwe ask is: How often do molecules collide with one another? The collision frequencydepends on the density of the gas and the molecular speed, and therefore on thetemperature of the system. In the kinetic theory model, we assume each molecule tobe a hard sphere of diameter d. A molecular collision is one in which the separationbetween the two spheres (measured from each center) is d.

Let us consider the motion of a particular molecule. A simple approach is to as-sume that at a given instant, all molecules except this one are standing still. In time t,this molecule moves a distance ct (where c is the average speed) and sweeps out acollision tube that has a cross-sectional area pd 2 (Figure 2.12). The volume of the

cylinder is ðpd 2ÞðctÞ. Any molecule whose center lies within this cylinder will collidewith the moving molecule. If there are altogether N molecules in volume V , then thenumber density of the gas is N=V , the number of collisions in time t is pd 2ctðN=VÞ,and the number of collisions per unit time, or the collision frequency, Z1, ispd 2cðN=VÞ. The expression for the collision frequency needs a correction. If we as-sume that the rest of the molecules are not frozen in position, we should replace c

with the average relative speed. Figure 2.13 shows three di¤erent collisions for twomolecules. The relative speed for the case shown in Figure 2.13c is

ffiffiffi2p

c, so that

Moleculein motion

Collision tube

Miss

Justmiss

Justhit

Hit

Area πd 2

ct

Figure 2.12The collision cross section and the collision tube.Any molecule whose center lies within or touches thetube will collide with the moving molecule (red sphere).

(a) (b) (c)

90°

Figure 2.13Three di¤erent approaches for two colliding molecules. The situations shown in (a) and (b)represent the two extreme cases, while that shown in (c) may be taken as the ‘‘average’’ casefor molecular encounter.


Black plate (29,1)

Z1 ¼ffiffiffi2p

pd 2cN

V

� �collisions s�1 ð2:30Þ

This is the number of collisions a single molecule makes in one second. Because thereare N molecules in volume V and each makes Z1 collisions per second, the totalnumber of binary collisions, or collisions between two molecules, per unit volume perunit time, Z11, is given by

Z11 ¼1

2Z1

N

V

� �

¼ffiffiffi2p

2pd 2c

N

V

� �2

collisions m�3 s�1ð2:31Þ

The factor 1=2 is introduced in Equation 2.31 to ensure that we are counting eachcollision between two molecules only once. The probability of three or more mole-cules colliding at once is very small except at high pressures. Because the rate of achemical reaction generally depends on how often reacting molecules come in contactwith one another, Equation 2.31 is quite important in gas-phase chemical kinetics.We shall return to this equation in Chapter 9.

A quantity closely related to the collision number is the average distance traveledby a molecule between successive collisions. This distance, called the mean free path,l (Figure 2.14), is defined as

l ¼ ðaverage speedÞ � ðaverage time between collisionsÞ

Because the average time between collisions is the reciprocal of the collision fre-quency, we have

l ¼ c

Z1¼ cffiffiffi

2p

pd 2cðN=VÞ¼ 1ffiffiffi

2p

pd 2ðN=VÞð2:32Þ

Notice that the mean free path is inversely proportional to the number density of thegas ðN=VÞ. This behavior is reasonable because in a dense gas, a molecule makesmore collisions per unit time and hence travels a shorter distance between successivecollisions. The mean free path can also be expressed in terms of the gas pressure.Assuming ideal behavior,

P ¼ nRT

V

¼ ðN=NAÞRT

V

N

V¼ PNA

RT

Equation 2.32 can now be written as

l ¼ RTffiffiffi2p

pd 2PNA

(2.33)

Figure 2.14The distances traveled by amolecule between successivecollisions. The average of thesedistances is called the meanfree path.

2.8 Molecular Collisions and the Mean Free Path 29

Black plate (30,1)

Example 2.4

The concentration of dry air at 1.00 atm and 298 K is about 2:5� 1019 molecules cm�3.Assuming that air contains only nitrogen molecules, calculate the collision frequency, thebinary collision number, and the mean free path of nitrogen molecules under theseconditions. The collision diameter of nitrogen is 3.75 A. (1 A ¼ 10�8 cm.)

A N S W E R

Our first step is to calculate the average speed of nitrogen. From Equation 2.28, we findc ¼ 4:8� 102 m s�1. The collision frequency is given by

Z1 ¼ffiffiffi2p

pð3:75� 10�8 cmÞ2ð4:8� 104 cm s�1Þð2:5� 1019 molecules cm�3Þ¼ 7:5� 109 collisions s�1

Note that we have replaced the unit ‘‘molecules’’ with ‘‘collisions’’ because, in thederivation of Z1, every molecule in the collision volume represents a collision. Thebinary collision number is

Z11 ¼Z1

2

N

V

� �

¼ ð7:5� 109 collisions s�1Þ2

� 2:5� 1019 molecules cm�3

¼ 9:4� 1028 collisions cm�3 s�1

Again, we converted molecules to collisions in calculating the total number of binarycollisions. Finally, the mean free path is given by

l ¼ c

Z1¼ 4:8� 104 cm s�1

7:5� 109 collisions s�1

¼ 6:4� 10�6 cm collision�1

¼ 640 A collision�1

C O M M E N T

It is usually su‰cient to express mean free path in terms of distance alone rather thandistance per collision. Thus, in this example, the mean free path of nitrogen is 640 A, or6:4� 10�6 cm.

2.9 Graham’s Laws of Diffusion and Effusion

Perhaps without thinking about it, we witness molecular motion on a daily basis. Thescent of perfume and the shrinking of an inflated helium rubber balloon are examplesof di¤usion and e¤usion, respectively. We can apply the kinetic theory of gases toboth processes.

The phenomenon of gas di¤usion o¤ers direct evidence of molecular motion.Were it not for di¤usion, there would be no perfume industry, and skunks would bejust another cute, furry species. Removing a partition separating two di¤erent gasesin a container quickly leads to a complete mixing of molecules. These are spontane-ous processes for which we shall discuss the thermodynamic basis in Chapter 4.During e¤usion, a gas travels from a high-pressure region to a low-pressure one


Black plate (31,1)

through a pinhole or orifice (Figure 2.15). For e¤usion to occur, the mean free pathof the molecules must be large compared with the diameter of the orifice. Thisensures that a molecule is unlikely to collide with another molecule when it reachesthe opening but will pass right through it. It follows then that the number of mole-cules passing through the orifice is equal to the number that would normally strike anarea of wall equal to the area of the hole.

Although the basic molecular mechanisms for di¤usion and e¤usion are quitedi¤erent (the former involves bulk flow, whereas the latter involves molecular flow),these two phenomena obey laws of the same form. Both laws were discovered by theScottish chemist Thomas Graham (1805–1869), the law of di¤usion in 1831 and thelaw of e¤usion in 1864. These laws state that under the same conditions of tempera-ture and pressure, the rates of di¤usion (or e¤usion) of gases are inversely propor-tional to the square roots of their molar masses. Thus, for two gases 1 and 2, we have

r1

r2¼

ffiffiffiffiffiffiffiffiM2

M1

r(2.34)

where r1 and r2 are the rates of di¤usion (or e¤usion) of the two gases.

Suggestions for Further Reading

Books

Hirschfelder, J. O., C. F. Curtiss, and R. B. Bird, The

Molecular Theory of Gases and Liquids, John Wiley &Sons, New York, 1954.

Hildebrand, J. H. An Introduction to Molecular Kinetic

Theory, Chapman & Hall, London, 1963 (VanNostrand Reinhold Company, New York).

Walton, A. J. The Three Phases of Matter, 2nd ed., OxfordUniversity Press, New York, 1983.

Tabor, D. Gases, Liquids, and Solids, 3rd ed., CambridgeUniversity Press, New York, 1996.

Articles

Gas Laws and Equations of State

‘‘The van der Waals Gas Equation,’’ F. S. Swinbourne, J.

Chem. Educ. 32, 366 (1955).‘‘A Simple Model for van der Waals,’’ S. S. Winter, J.

Chem. Educ. 33, 459 (1959).‘‘Comparisons of Equations of State in E¤ectively

Describing PVT Relations,’’ J. B. Ott, J. R. Goales,and H. T. Hall, J. Chem. Educ. 48, 515 (1971).

‘‘Scuba Diving and the Gas Laws,’’ E. D. Cooke, J. Chem.

Educ. 50, 425 (1973).

OrificeVacuum

Figure 2.15An e¤usion process in which molecules move through an opening (orifice) into an evacuatedregion. The conditions for e¤usion are that the mean free path of the molecules is largecompared to the size of the opening and the wall containing the opening is thin so that nomolecular collisions occur during the exit. Also, the pressure in the right chamber must below enough so as not to obstruct the molecular movement through the hole.

2.9 Graham’s Laws of Diffusion and Effusion 31

Black plate (32,1)

‘‘Derivation of the Ideal Gas Law,’’ S. Levine, J. Chem.

Educ. 62, 399 (1985).‘‘The Ideal Gas Law at the Center of the Sun,’’ D. B.

Clark, J. Chem. Educ. 66, 826 (1989).‘‘The Many Faces of van der Waals’s Equation of State,’’

J. G. Eberhart, J. Chem. Educ. 66, 906 (1989).‘‘Does a One-Molecule Gas Obey Boyle’s Law?’’ G.

Rhodes, J. Chem. Educ. 69, 16 (1992).‘‘Equations of State,’’ M. Ross in Encyclopedia of Applied

Physics, G. L. Trigg, Ed., VCH Publishers, NewYork, 1993, Vol. 6, p. 291.

‘‘Interpretation of the Second Virial Coe‰cient,’’ J.Wisniak, J. Chem. Educ. 76, 671 (1999).

The Critical State

‘‘The Critical Temperature: A Necessary Consequence ofGas Nonideality,’’ F. L. Pilar, J. Chem. Educ. 44, 284(1967).

‘‘Supercritical Fluids: Liquid, Gas, Both, or Neither? ADi¤erent Approach,’’ E. F. Meyer and T. P. Meyer,J. Chem. Educ. 63, 463 (1986).

‘‘Past, Present, and Possible Future Applications ofSupercritical Fluid Extraction Technology,’’ C. L.Phelps, N. G. Smart, and C. M. Wai, J. Chem. Educ.

73, 1163 (1996).

Kinetic Theory of Gases

‘‘Kinetic Energies of Gas Molecules,’’ J. C. Aherne, J.

Chem. Educ. 42, 655 (1965).

‘‘Kinetic Theory, Temperature, and Equilibrium,’’ D. K.Carpenter, J. Chem. Educ. 43, 332 (1966).

‘‘Graham’s Laws of Di¤usion and E¤usion,’’ E. A. Masonand B. Kronstadt, J. Chem. Educ. 44, 740 (1967).

‘‘The Cabin Atmosphere in Manned Space Vehicles,’’W. H. Bowman and R. M. Lawrence, J. Chem. Educ.

48, 152 (1971).‘‘The Assumption of Elastic Collisions in Elementary Gas

Kinetic Theory,’’ B. Rice and C. J. G. Raw, J. Chem.

Educ. 51, 139 (1974).‘‘Velocity and Energy Distribution in Gases,’’ B. A.

Morrow and D. F. Tessier, J. Chem. Educ. 59, 193(1982).

‘‘Applications of Maxwell-Boltzmann DistributionDiagrams,’’ G. D. Peckham and I. J. McNaught, J.

Chem. Educ. 69, 554 (1992).‘‘Misuse of Graham’s Laws,’’ S. J. Hawkes, J. Chem.

Educ. 70, 836 (1993).‘‘Graham’s Law and Perpetuation of Error,’’ S. J. Hawkes,

J. Chem. Educ. 74, 1069 (1997).

General

‘‘The Lung,’’ J. H. Comroe, Sci. Am. February 1966.‘‘The Invention of the Balloon and the Birth of Modern

Chemistry,’’ A. F. Scott, Sci. Am. January 1984.‘‘Temperature, Cool but Quick,’’ S. M. Cohen, J. Chem.

Educ. 63, 1038 (1986).‘‘Mountain Sickness,’’ C. S. Houston, Sci. Am. October

1992.

Problems

Ideal Gases

2.1 Classify each of the following properties as inten-sive or extensive: force, pressure (P), volume (V ), tem-perature (T), mass, density, molar mass, molar volume(V ).

2.2 Some gases, such as NO2 and NF2, do not obeyBoyle’s law at any pressure. Explain.

2.3 An ideal gas originally at 0.85 atm and 66�C wasallowed to expand until its final volume, pressure, andtemperature were 94 mL, 0.60 atm, and 45�C, respec-tively. What was its initial volume?

2.4 Some ballpoint pens have a small hole in the mainbody of the pen. What is the purpose of this hole?

2.5 Starting with the ideal-gas equation, show how youcan calculate the molar mass of a gas from a knowledgeof its density.

2.6 At STP (standard temperature and pressure),0.280 L of a gas weighs 0.400 g. Calculate the molarmass of the gas.

2.7 Ozone molecules in the stratosphere absorb muchof the harmful radiation from the sun. Typically, thetemperature and partial pressure of ozone in the strato-

sphere are 250 K and 1:0� 10�3 atm, respectively. Howmany ozone molecules are present in 1.0 L of air underthese conditions? Assume ideal-gas behavior.

2.8 Calculate the density of HBr in g L�1 at 733 mmHgand 46�C. Assume ideal-gas behavior.

2.9 Dissolving 3.00 g of an impure sample of CaCO3 inan excess of HCl acid produced 0.656 L of CO2 (mea-sured at 20�C and 792 mmHg). Calculate the percent bymass of CaCO3 in the sample.

2.10 The saturated vapor pressure of mercury is0.0020 mmHg at 300 K and the density of air at 300 Kis 1.18 g L�1. (a) Calculate the concentration of mer-cury vapor in air in mol L�1. (b) What is the number ofparts per million (ppm) by mass of mercury in air?

2.11 A very flexible balloon with a volume of 1.2 L at1.0 atm and 300 K is allowed to rise to the stratosphere,where the temperature and pressure are 250 K and3:0� 10�3 atm, respectively. What is the final volumeof the balloon? Assume ideal-gas behavior.

2.12 Sodium bicarbonate (NaHCO3) is called bakingsoda because when heated, it releases carbon dioxide


Black plate (33,1)

gas, which causes cookies, doughnuts, and bread to riseduring baking. (a) Calculate the volume (in liters) ofCO2 produced by heating 5.0 g of NaHCO3 at 180�Cand 1.3 atm. (b) Ammonium bicarbonate (NH4HCO3)has also been used as a leavening agent. Suggest oneadvantage and one disadvantage of using NH4HCO3

instead of NaHCO3 for baking.

2.13 A common, non-SI unit for pressure is pounds persquare inch (psi). Show that 1 atm ¼ 14:7 psi. An auto-mobile tire is inflated to 28.0 psi gauge pressure whencold, at 18�C. (a) What will the pressure be if the tire isheated to 32�C by driving the car? (b) What percentageof the air in the tire would have to be let out to reducethe pressure to the original 28.0 psi? Assume that thevolume of the tire remains constant with temperature.(A tire gauge measures not the pressure of the air insidebut its excess over the external pressure, which is 14.7psi.)

2.14 (a) What volume of air at 1.0 atm and 22�C isneeded to fill a 0.98-L bicycle tire to a pressure of 5.0atm at the same temperature? (Note that 5.0 atm is thegauge pressure, which is the di¤erence between the pres-sure in the tire and atmospheric pressure. Initially, thegauge pressure in the tire was 0 atm.) (b) What is thetotal pressure in the tire when the gauge reads 5.0 atm?(c) The tire is pumped with a hand pump full of air at1.0 atm; compressing the gas in the cylinder adds all theair in the pump to the air in the tire. If the volume ofthe pump is 33% of the tire’s volume, what is the gaugepressure in the tire after 3 full strokes of the pump?

2.15 A student breaks a thermometer and spills most ofthe mercury (Hg) onto the floor of a laboratory thatmeasures 15.2 m long, 6.6 m wide, and 2.4 m high.(a) Calculate the mass of mercury vapor (in grams) inthe room at 20�C. (b) Does the concentration of mer-cury vapor exceed the air quality regulation of 0.050mg Hg m�3 of air? (c) One way to treat small quantitiesof spilled mercury is to spray powdered sulfur over themetal. Suggest a physical and a chemical reason for thistreatment. The vapor pressure of mercury at 20�C is1:7� 1:0�6 atm.

2.16 Nitrogen forms several gaseous oxides. One ofthem has a density of 1.27 g L�1 measured at 764mmHg and 150�C. Write the formula of the compound.

2.17 Nitrogen dioxide (NO2) cannot be obtained in apure form in the gas phase because it exists as a mixtureof NO2 and N2O4. At 25�C and 0.98 atm, the densityof this gas mixture is 2.7 g L�1. What is the partialpressure of each gas?

2.18 An ultra-high-vacuum pump can reduce the pres-sure of air from 1.0 atm to 1:0� 10�12 mmHg. Calcu-late the number of air molecules in a liter at this pres-sure and 298 K. Compare your results with the numberof molecules in 1.0 L at 1.0 atm and 298 K. Assumeideal-gas behavior.

2.19 An air bubble with a radius of 1.5 cm at the bot-tom of a lake where the temperature is 8.4�C and the

pressure is 2.8 atm rises to the surface, where the tem-perature is 25.0�C and the pressure is 1.0 atm. Calculatethe radius of the bubble when it reaches the surface. As-sume ideal-gas behavior. [Hint: The volume of a sphereis given by ð4=3Þpr3, where r is the radius.]

2.20 The density of dry air at 1.00 atm and 34.4�C is1.15 g L�1. Calculate the composition of air (percent bymass) assuming that it contains only nitrogen and oxy-gen and behaves like an ideal gas. (Hint: First calculatethe ‘‘molar mass’’ of air, then the mole fractions, andthen the mass fractions of O2 and N2.)

2.21 A gas that evolved during the fermentation of glu-cose has a volume of 0.78 L when measured at 20.1�Cand 1.0 atm. What was the volume of this gas at thefermentation temperature of 36.5�C? Assume ideal-gasbehavior.

2.22 Two bulbs of volumes VA and VB are connectedby a stopcock. The number of moles of gases in thebulbs are nA and nB, and initially the gases are at thesame pressure, P, and temperature, T . Show that thefinal pressure of the system, after the stopcock has beenopened, equals P. Assume ideal-gas behavior.

2.23 The composition of dry air at sea level is 78.03%N2, 20.99% O2, and 0.033% CO2 by volume. (a) Calcu-late the average molar mass of this air sample. (b) Cal-culate the partial pressures of N2, O2, and CO2 in atm.(At constant temperature and pressure, the volume of agas is directly proportional to the number of moles ofthe gas.)

2.24 A mixture containing nitrogen and hydrogenweighs 3.50 g and occupies a volume of 7.46 L at 300 Kand 1.00 atm. Calculate the mass percent of these twogases. Assume ideal-gas behavior.

2.25 The relative humidity in a closed room with a vol-ume of 645.2 m3 is 87.6% at 300 K, and the vaporpressure of water at 300 K is 0.0313 atm. Calculate themass of water in the air. [Hint: The relative humidity isdefined as ðP=PsÞ � 100%, where P and Ps are the par-tial pressure and saturated partial pressure of watervapor, respectively.]

2.26 Death by su¤ocation in a sealed container is nor-mally caused not by oxygen deficiency but by CO2 poi-soning, which occurs at about 7% CO2 by volume. Forwhat length of time would it be safe to be in a sealedroom 10� 10� 20 ft? [Source: ‘‘Eco-Chem,’’ J. A.Campbell, J. Chem. Educ. 49, 538 (1972).]

2.27 A flask contains a mixture of two ideal gases, Aand B. Show graphically how the total pressure of thesystem depends on the amount of A present; that is,plot the total pressure versus the mole fraction of A. Dothe same for B on the same graph. The total number ofmoles of A and B is constant.

2.28 A mixture of helium and neon gases is collectedover water at 28.0�C and 745 mmHg. If the partialpressure of helium is 368 mmHg, what is the partialpressure of neon? (Note: The vapor pressure of water at28�C is 28.3 mmHg.)

Problems 33

Black plate (34,1)

2.29 If the barometric pressure falls in one part of theworld, it must rise somewhere else. Explain why.

2.30 A piece of sodium metal reacts completely withwater as follows:

2NaðsÞ þ 2H2OðlÞ ! 2NaOHðaqÞ þH2ðgÞ

The hydrogen gas generated is collected over water at25.0�C. The volume of the gas is 246 mL measured at1.00 atm. Calculate the number of grams of sodiumused in the reaction. (Note: The vapor pressure of waterat 25�C is 0.0313 atm.)

2.31 A sample of zinc metal reacts completely with anexcess of hydrochloric acid:

ZnðsÞ þ 2HClðaqÞ ! ZnCl2ðaqÞ þH2ðgÞ

The hydrogen gas produced is collected over water at25.0�C. The volume of the gas is 7.80 L, and the pres-sure is 0.980 atm. Calculate the amount of zinc metal ingrams consumed in the reaction. (Note: The vapor pres-sure of water at 25�C is 23.8 mmHg.)

2.32 Helium is mixed with oxygen gas for deep seadivers. Calculate the percent by volume of oxygen gasin the mixture if the diver has to submerge to a depthwhere the total pressure is 4.2 atm. The partial pressureof oxygen is maintained at 0.20 atm at this depth.

2.33 A sample of ammonia (NH3) gas is completelydecomposed to nitrogen and hydrogen gases overheated iron wool. If the total pressure is 866 mmHg,calculate the partial pressures of N2 and H2.

2.34 The partial pressure of carbon dioxide in air varieswith the seasons. Would you expect the partial pressurein the Northern Hemisphere to be higher in the summeror winter? Explain.

2.35 A healthy adult exhales about 5:0� 102 mL of agaseous mixture with each breath. Calculate the num-ber of molecules present in this volume at 37�C and1.1 atm. List the major components of this gaseousmixture.

2.36 Describe how you would measure, by either chem-ical or physical means (other than mass spectrometry),the partial pressures of a mixture of gases: (a) CO2 andH2, (b) He and N2.

2.37 The gas laws are vitally important to scuba divers.The pressure exerted by 33 ft of seawater is equivalentto 1 atm pressure. (a) A diver ascends quickly to thesurface of the water from a depth of 36 ft withoutexhaling gas from his lungs. By what factor would thevolume of his lungs increase by the time he reaches thesurface? Assume that the temperature is constant.(b) The partial pressure of oxygen in air is about 0.20atm. (Air is 20% oxygen by volume.) In deep-sea diving,the composition of air the diver breathes must bechanged to maintain this partial pressure. What mustthe oxygen content (in percent by volume) be when thetotal pressure exerted on the diver is 4.0 atm?

2.38 A 1.00-L bulb and a 1.50-L bulb, connected by astopcock, are filled, respectively, with argon at 0.75 atmand helium at 1.20 atm at the same temperature. Calcu-late the total pressure and the partial pressures of eachgas after the stopcock has been opened and the molefraction of each gas. Assume ideal-gas behavior.

2.39 A mixture of helium and neon weighing 5.50 goccupies a volume of 6.80 L at 300 K and 1.00 atm.Calculate the composition of the mixture in masspercent.

Nonideal Gases

2.40 Suggest two demonstrations to show that gases donot behave ideally.

2.41 Which of the following combinations of conditionsmost influences a gas to behave ideally: (a) low pressureand low temperature, (b) low pressure and high temper-ature, (c) high pressure and high temperature, and(d) high pressure and low temperature.

2.42 The van der Waals constants of a gas can beobtained from its critical constants, where a ¼ð27R2Tc

2=64PcÞ and b ¼ ðRTc=8PcÞ. Given that Tc ¼562 K and Pc ¼ 48:0 atm for benzene, calculate its aand b values.

2.43 Using the data shown in Table 2.1, calculate thepressure exerted by 2.500 moles of carbon dioxide con-fined in a volume of 1.000 L at 450 K. Compare thepressure with that calculated assuming ideal behavior.

2.44 Without referring to a table, select from the fol-lowing list the gas that has the largest value of b in thevan der Waals equation: CH4, O2, H2O, CCl4, Ne.

2.45 Referring to Figure 2.4, we see that for He theplot has a positive slope even at low pressures. Explainthis behavior.

2.46 At 300 K, the virial coe‰cients (B) of N2 andCH4 are �4.2 cm3 mol�1 and �15 cm3 mol�1, re-spectively. Which gas behaves more ideally at thistemperature?

2.47 Calculate the molar volume of carbon dioxide at400 K and 30 atm, given that the second virial coe‰-cient (B) of CO2 is �0.0605 L mol�1. Compare your re-sult with that obtained using the ideal-gas equation.

2.48 Consider the virial equation Z ¼ 1þ B 0Pþ C 0P2,which describes the behavior of a gas at a certain tem-perature. From the following plot of Z versus P, deducethe signs of B 0 and C 0 (< 0;¼ 0; > 0).

Z

1.0

1.1

1.2

1.3

1.4

0.8

0.9200 400 600 800 1000

P /atm


Black plate (35,1)

Kinetic Theory of Gases

2.49 Apply the kinetic theory of gases to explainBoyle’s law, Charles’ law, and Dalton’s law.

2.50 Is temperature a microscopic or macroscopic con-cept? Explain.

2.51 In applying the kinetic molecular theory to gases,we have assumed that the walls of the container areelastic for molecular collisions. Actually, whether thesecollisions are elastic or inelastic makes no di¤erence aslong as the walls are at the same temperature as the gas.Explain.

2.52 If 2:0� 1023 argon (Ar) atoms strike 4.0 cm2 ofwall per second at a 90� angle to the wall when movingwith a speed of 45,000 cm s�1, what pressure (in atm)do they exert on the wall?

2.53 A square box contains He at 25�C. If the atomsare colliding with the walls perpendicularly (at 90�) atthe rate of 4:0� 1022 times per second, calculate theforce and the pressure exerted on the wall given that thearea of the wall is 100 cm2 and the speed of the atomsis 600 m s�1.

2.54 Calculate the average translational kinetic energyfor a N2 molecule and for 1 mole of N2 at 20�C.

2.55 To what temperature must He atoms be cooled sothat they have the same vrms as O2 at 25�C?

2.56 The crms of CH4 is 846 m s�1. What is the temper-ature of the gas?

2.57 Calculate the value of the crms of ozone moleculesin the stratosphere, where the temperature is 250 K.

2.58 At what temperature will He atoms have the samecrms value as N2 molecules at 25�C? Solve this problemwithout calculating the value of crms for N2.

Maxwell Speed Distribution

2.59 List the conditions used for deriving the Maxwellspeed distribution.

2.60 Plot the speed distribution function for (a) He, O2,and UF6 at the same temperature, and (b) CO2 at300 K and 1000 K.

2.61 Account for the maximum in the Maxwell speeddistribution curve (Figure 2.11) by plotting the follow-ing two curves on the same graph: (1) c2 versus c and(2) e�mc2=2kBT versus c. Use neon (Ne) at 300 K for theplot in (2).

2.62 A N2 molecule at 20�C is released at sea level totravel upward. Assuming that the temperature is con-stant and that the molecule does not collide with othermolecules, how far would it travel (in meters) beforecoming to rest? Do the same calculation for a He atom.[Hint: To calculate the altitude, h, the molecule willtravel, equate its kinetic energy with the potential en-ergy, mgh, where m is the mass and g the accelerationdue to gravity (9.81 m s�2).]

2.63 The speeds of 12 particles (in cm s�1) are 0.5, 1.5,1.8, 1.8, 1.8, 1.8, 2.0, 2.5, 2.5, 3.0, 3.5, and 4.0. Find(a) the average speed, (b) the root-mean-square speed,and (c) the most probable speed of these particles.Explain your results.

2.64 At a certain temperature, the speeds of six gaseousmolecules in a container are 2.0 m s�1, 2.2 m s�1, 2.6m s�1, 2.7 m s�1, 3.3 m s�1, and 3.5 m s�1. Calculatethe root-mean-square speed and the average speed ofthe molecules. These two average values are close toeach other, but the root-mean-square value is alwaysthe larger of the two. Why?

2.65 The following diagram shows the Maxwell speeddistribution curves for a certain ideal gas at two di¤er-ent temperatures (T1 and T2). Calculate the value of T2.

500 1000 1500 20000

f (c )

T2 ?

T1 300 K

c /m s 1

2.66 Derive an expression for cmp. [Hint: Di¤erentiatef ðcÞ with respect to c in Equation 2.26 and set the resultto zero.]

2.67 Calculate the values of crms; cmp, and c for argonat 298 K.

2.68 Calculate the value of cmp for C2H6 at 25�C.What is the ratio of the number of molecules with aspeed of 989 m s�1 to the number of molecules with thisvalue of cmp?

Molecular Collisions and the Mean Free Path

2.69 Considering the magnitude of molecular speeds,explain why it takes so long (on the order of minutes) todetect the odor of ammonia when someone opens a bot-tle of concentrated ammonia at the other end of a labo-ratory bench.

2.70 How does the mean free path of a gas depend on(a) the temperature at constant volume, (b) the density,(c) the pressure at constant temperature, (d) the volumeat constant temperature, and (e) the size of molecules?

2.71 A bag containing 20 marbles is being shaken vigo-rously. Calculate the mean free path of the marbles ifthe volume of the bag is 850 cm3. The diameter of eachmarble is 1.0 cm.

2.72 Calculate the mean free path and the binary num-ber of collisions per liter per second between HI mole-cules at 300 K and 1.00 atm. The collision diameter of

Problems 35

Black plate (36,1)

the HI molecules may be taken to be 5.10 A. Assumeideal-gas behavior.

2.73 Ultra-high-vacuum experiments are routinely per-formed at a total pressure of 1:0� 10�10 torr. Calculatethe mean free path of N2 molecules at 350 K underthese conditions. The collision diameter of N2 is 3.75 A.

2.74 Suppose that helium atoms in a sealed containerall start with the same speed, 2:74� 104 cm s�1. Theatoms are then allowed to collide with one another untilthe Maxwell distribution is established. What is thetemperature of the gas at equilibrium? Assume thatthere is no heat exchange between the gas and itssurroundings.

2.75 Compare the collision number and the mean freepath for air molecules at (a) sea level (T ¼ 300 K anddensity ¼ 1:2 g L�1) and (b) in the stratosphere (T ¼250 K and density ¼ 5:0� 10�3 g L�1). The molarmass of air may be taken as 29.0 g, and the collisiondiameter is 3.72 A.

2.76 Calculate the values of Z1 and Z11 for mercury(Hg) vapor at 40�C, both at P ¼ 1:0 atm and atP ¼ 0:10 atm. How do these two quantities depend onpressure? The collision diameter of Hg is 4.26 A.

Gas Diffusion and Effusion

2.77 Derive Equation 2.34 from Equation 2.23.

2.78 An inflammable gas is generated in marsh landsand sewage by a certain anaerobic bacterium. A puresample of this gas was found to e¤use through an ori-fice in 12.6 min. Under identical conditions of tempera-ture and pressure, oxygen takes 17.8 min to e¤usethrough the same orifice. Calculate the molar mass ofthe gas, and suggest what this gas might be.

2.79 Nickel forms a gaseous compound of the formulaNi(CO)x. What is the value of x given the fact thatunder the same conditions of temperature and pres-sure, methane (CH4) e¤uses 3.3 times faster than thecompound?

2.80 In 2.00 min, 29.7 mL of He e¤use through a smallhole. Under the same conditions of temperature andpressure, 10.0 mL of a mixture of CO and CO2 e¤usethrough the hole in the same amount of time. Calculatethe percent composition by volume of the mixture.

2.81 Uranium-235 can be separated from uranium-238by the e¤usion process involving UF6. Assuming a50 : 50 mixture at the start, what is the percentage ofenrichment after a single stage of separation?

2.82 An equimolar mixture of H2 and D2 e¤usesthrough an orifice at a certain temperature. Calculatethe composition (in mole fractions) of the gas thatpasses through the orifice. The molar mass of deuteriumis 2.014 g mol�1.

2.83 The rate ðre¤ Þ at which molecules confined to avolume V e¤use through an orifice of area A is given by

ð1=4ÞnNAcA=V , where n is the number of moles of thegas. An automobile tire of volume 30.0 L and pressure1,500 torr is punctured as it runs over a sharp nail.(a) Calculate the e¤usion rate if the diameter of the holeis 1.0 mm. (b) How long would it take to lose half ofthe air in the tire through e¤usion? Assume a constante¤usion rate and constant volume. The molar mass ofair is 29.0 g, and the temperature is 32.0�C.

Additional Problems

2.84 A barometer with a cross-sectional area of 1.00cm2 at sea level measures a pressure of 76.0 cm of mer-cury. The pressure exerted by this column of mercury isequal to the pressure exerted by all the air on 1 cm2 ofEarth’s surface. Given that the density of mercury is13.6 g cm�3 and the average radius of Earth is 6371km, calculate the total mass of Earth’s atmosphere inkilograms. (Hint: The surface area of a sphere is 4pr2,where r is the radius of the sphere.)

2.85 It has been said that every breath we take, onaverage, contains molecules once exhaled by WolfgangAmadeus Mozart (1756–1791). The following calcu-lations demonstrate the validity of this statement.(a) Calculate the total number of molecules in the atmo-sphere. (Hint: Use the result from Problem 2.84 and29.0 g mol�1 as the molar mass of air.) (b) Assumingthe volume of every breath (inhale or exhale) is 500 mL,calculate the number of molecules exhaled in eachbreath at 37�C, which is the body temperature. (c) IfMozart’s life span was exactly 35 years, how many mol-ecules did he exhale in that period (given that an aver-age person breathes 12 times per minute)? (d) Calculatethe fraction of molecules in the atmosphere that wereexhaled by Mozart. How many of Mozart’s moleculesdo we inhale with each breath of air? Round your an-swer to one significant digit. (e) List three importantassumptions in these calculations.

2.86 A stockroom supervisor measured the contents ofa partially filled 25.0-gallon acetone drum on a daywhen the temperature was 18.0�C and the atmosphericpressure was 750 mmHg, and found that 15.4 gallons ofthe solvent remained. After tightly sealing the drum, anassistant dropped the drum while carrying it upstairs tothe organic laboratory. The drum was dented and itsinternal volume was decreased to 20.4 gallons. What isthe total pressure inside the drum after the accident?The vapor pressure of acetone at 18.0�C is 400 mmHg.(Hint: At the time the drum was sealed, the pressureinside the drum, which is equal to the sum of the pres-sures of air and acetone, was equal to the atmosphericpressure.)

2.87 A relation known as the barometric formula isuseful for estimating the change in atmospheric pressurewith altitude. (a) Starting with the knowledge thatatmospheric pressure decreases with altitude, we havedP ¼ �rg dh, where r is the density of air, g is the ac-celeration due to gravity (9.81 m s�2), and P and h are


Black plate (37,1)

the pressure and height, respectively. Assuming ideal-gas behavior and constant temperature, show that thepressure P at height h is related to the pressure at sealevel P0 ðh ¼ 0Þ by P ¼ P0e�gMh=RT . (Hint: For an idealgas, r ¼ PM=RT , where M is the molar mass.) (b) Cal-culate the atmospheric pressure at a height of 5.0 km,assuming the temperature is constant at 5.0�C, giventhat the average molar mass of air is 29.0 g mol�1.

2.88 In terms of the hard-sphere gas model, moleculesare assumed to possess finite volume, but there is nointeraction among the molecules. (a) Compare the P–Visotherm for an ideal gas and that for a hard-sphere gas.(b) Let b be the e¤ective volume of the gas. Write anequation of state for this gas. (c) From this equation,derive an expression for Z ¼ PV=RT for the hard-sphere gas and make a plot of Z versus P for two val-ues of T (T1 and T2, T2 > T1). Be sure to indicate thevalue of the intercepts on the Z axis. (d) Plot Z versusT for fixed P for an ideal gas and for the hard-spheregas.

2.89 One way to gain a physical understanding of b inthe van der Waals equation is to calculate the‘‘excluded volume.’’ Assume that the distance of closestapproach between two similar spherical molecules is thesum of their radii ð2rÞ. (a) Calculate the volume aroundeach molecule into which the center of another moleculecannot penetrate. (b) From your result in (a), calculatethe excluded volume for one mole of molecules, whichis the constant b. How does this compare with the sumof the volumes of 1 mole of the same molecules?

2.90 You may have witnessed a demonstration in whicha burning candle standing in water is covered by anupturned glass. The candle goes out and the water risesin the glass. The explanation usually given for this phe-nomenon is that the oxygen in the glass is consumed bycombustion, leading to a decrease in volume and hencethe rise in the water level. However, the loss of oxygenis only a minor consideration. (a) Using C12H26 as theformula for para‰n wax, write a balanced equation forthe combustion. Based on the nature of the products,show that the predicted rise in water level due to the re-moval of oxygen is far less than the observed change.(b) Devise a chemical process that would allow you tomeasure the volume of oxygen in the trapped air. (Hint:Use steel wool.) (c) What is the main reason for thewater rising in the glass after the flame is extinguished?

2.91 Express the van der Waals equation in the form ofEquation 2.14. Derive relationships between the van derWaals constants (a and b) and the virial coe‰cients(B;C, and D), given that

1

1� x¼ 1þ xþ x2 þ x3 þ � � � jxj < 1

2.92 The Boyle temperature is the temperature at whichthe coe‰cient B is zero. Therefore, a real gas behaveslike an ideal gas at this temperature. (a) Give a physical

interpretation of this behavior. (b) Using your result forB for the van der Waals equation in Problem 2.91, cal-culate the Boyle temperature for argon, given thata ¼ 1:345 atm L2 mol�2 and b ¼ 3:22� 10�2 L mol�1.

2.93 Estimate the distance (in A) between molecules ofwater vapor at 100�C and 1.0 atm. Assume ideal-gasbehavior. Repeat the calculation for liquid water at100�C, given that the density of water at 100�C is 0.96g cm�3. Comment on your results. (The diameter of aH2O molecule is approximately 3 A. 1 A ¼ 10�8 cm.)

2.94 The following apparatus can be used to measureatomic and molecular speed. A beam of metal atoms isdirected at a rotating cylinder in a vacuum. A smallopening in the cylinder allows the atoms to strike a tar-get area. Because the cylinder is rotating, atoms travel-ing at di¤erent speeds will strike the target at di¤erentpositions. In time, a layer of the metal will deposit onthe target area, and the variation in its thickness isfound to correspond to Maxwell’s speed distribution. Inone experiment, it is found that at 850�C, some bismuth(Bi) atoms struck the target at a point 2.80 cm from thespot directly opposite the slit. The diameter of the cylin-der is 15.0 cm, and it is rotating at 130 revolutions persecond. (a) Calculate the speed (m s�1) at which the tar-get is moving. (Hint: The circumference of a circle isgiven by 2pr, where r is the radius.) (b) Calculate thetime (in seconds) it takes for the target to travel 2.80cm. (c) Determine the speed of the Bi atoms. Compareyour result in (c) with the crms value for Bi at 850�C.Comment on the di¤erence.

Bi atoms

Slit

Rotating cylinder

Target

2.95 The escape velocity, v, from Earth’s gravitationalfield is given by ð2GM=rÞ1=2, where G is the universalgravitational constant (6:67� 10�11 m3 kg�1 s�2), M isthe mass of Earth (6:0� 1024 kg), and r is the distancefrom the center of Earth to the object, in meters. Com-pare the average speeds of He and N2 molecules in thethermosphere (altitude about 100 km, T ¼ 250 K).Which of the two molecules will have a greater ten-dency to escape? The radius of Earth is 6:4� 106 m.

2.96 Calculate the ratio of the number of O3 moleculeswith a speed of 1300 m s�1 at 360 K to the numberwith that speed at 293 K.

2.97 Calculate the collision frequency for 1.0 mole ofkrypton (Kr) at equilibrium at 300 K and 1.0 atm pres-sure. Which of the following alterations increases thecollision frequency more: (a) doubling the temperatureat constant pressure or (b) doubling the pressure at con-

Problems 37

Black plate (38,1)

stant temperature? (Hint: The collision diameter of Kris 4.16 A.)

2.98 Apply your knowledge of the kinetic theory ofgases to the following situations. (a) Two flasks of vol-umes V1 and V2 (where V2 > V1) contain the samenumber of helium atoms at the same temperature.(i) Compare the root-mean-square (rms) speeds andaverage kinetic energies of the helium (He) atoms in theflasks. (ii) Compare the frequency and the force withwhich the He atoms collide with the walls of their con-tainers. (b) Equal numbers of He atoms are placed intwo flasks of the same volume at temperatures T1 andT2 (where T2 > T1). (i) Compare the rms speeds of theatoms in the two flasks. (ii) Compare the frequency andthe force with which the He atoms collide with the wallsof their containers. (c) Equal numbers of He and neon(Ne) atoms are placed in two flasks of the same volume.The temperature of both gases is 74�C. Comment onthe validity of the following statements: (i) The rmsspeed of He is equal to that of Ne. (ii) The average ki-netic energies of the two gases are equal. (iii) The rmsspeed of each He atom is 1:47� 103 m s�1.

2.99 Consider 1 mole each of gaseous He and N2 at thesame temperature and pressure. State which gas (if any)has the greater value for: (a) c, (b) crms, (c) Etrans,(d) Z1, (e) Z11, (f ) density, and (g) mean free path. Thediameter of N2 is 1.7 times that of He.

2.100 The root-mean-square velocity of a certain gas-eous oxide is 493 m s�1 at 20�C. What is the molecularformula of the compound?

2.101 Calculate the mean kinetic energy ðEtransÞ injoules of the following molecules at 350 K: (a) He,(b) CO2, and (c) UF6. Explain your results.

2.102 A sample of neon gas is heated from 300 K to390 K. Calculate the percent increase in its kineticenergy.

2.103 A CO2 fire extinguisher is located on the outsideof a building in Massachusetts. During the wintermonths, one can hear a sloshing sound when the extin-guisher is gently shaken. In the summertime, there isoften no sound when it is shaken. Explain. Assume thatthe extinguisher has no leaks and that it has not beenused.


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