Equations of state

1.1 The perfect gas equation of state

1.2 Using the perfect gas law

Box 1.1 The gas laws and the weather

1.3 Mixtures of gases: partial pressures

The kinetic model of gases

1.4 The pressure of a gas according to the kinetic model

1.5 The average speed of gas molecules

1.6 The Maxwell distribution of speeds

1.7 Diffusion and effusion

1.8 Molecular collisions

Real gases

Box 1.2 The Sun as a ball of perfect gas

1.9 Molecular interactions

1.10 The critical temperature

1.11 The compression factor

1.12 The virial equation of state

1.13 The van der Waals equation of state

1.14 The liquefaction of gases

CHECKLIST OF KEY IDEAS

FURTHER INFORMATION 1.1

DISCUSSION QUESTIONS

EXERCISES

Although gases are simple, both to describe and interms of their internal structure, they are of immenseimportance. We spend our whole lives surroundedby gas in the form of air, and the local variation in its properties is what we call the ‘weather’. To understand the atmospheres of this and other planetswe need to understand gases. As we breathe, wepump gas in and out of our lungs, where it changescomposition and temperature. Many industrial processes involve gases, and both the outcome of the reaction and the design of the reaction vesselsdepend on a knowledge of their properties.

Equations of state

We can specify the state of any sample of substanceby giving the values of the following properties (allof which are defined in the Introduction):

V, the volume of the sample

p, the pressure of the sample

T, the temperature of the sample

n, the amount of substance in the sample

However, an astonishing experimental fact is thatthese four quantities are not independent of one another. For instance, we cannot arbitrarily chooseto have a sample of 0.555 mol H2O in a volume of100 cm3 at 100 kPa and 500 K: it is found experi-mentally that the state simply does not exist. If weselect the amount, the volume, and the temperature,then we find that we have to accept a particularpressure (in this case, close to 230 kPa). The same istrue of all substances, but the pressure in generalwill be diCerent for each one. This experimental

Chapter 1

The properties of gases

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EQUATIONS OF STATE

generalization is summarized by saying the substanceobeys an equation of state, an equation of the form

p = f(n,V,T ) (1.1)

This expression tells us that the pressure is somefunction of amount, volume, and temperature andthat if we know those three variables, then the pressure can have only one value.

The equations of state of most substances are not known, so in general we cannot write down anexplicit expression for the pressure in terms of theother variables. However, certain equations of stateare known. In particular, the equation of state of alow-pressure gas is known, and proves to be verysimple and very useful. This equation is used to describe the behaviour of gases taking part in reac-tions, the behaviour of the atmosphere, as a startingpoint for problems in chemical engineering, andeven in the description of the structures of stars.

1.1 The perfect gas equation of state

The equation of state of a low-pressure gas wasamong the first results to be established in physicalchemistry. The original experiments were carried outby Robert Boyle in the seventeenth century and therewas a resurgence in interest later in the century whenpeople began to fly in balloons. This technologicalprogress demanded more knowledge about the re-sponse of gases to changes of pressure and temper-ature and, like technological advances in other fieldstoday, that interest stimulated a lot of experiments.

The experiments of Boyle and his successors led to the formulation of the following perfect gasequation of state:

pV = nRT (1.2)

In this equation (which has the form of eqn 1.1when we rearrange it into p = nRT/V), the gas constant R is an experimentally determined quantitythat turns out to have the same value for all gases. It may be determined by evaluating R = pV/nRTas the pressure is allowed to approach zero or bymeasuring the speed of sound (which depends on R).Values of R in diCerent units are given in Table 1.1.

The perfect gas equation of state—more briefly,the ‘perfect gas law’—is so-called because it is anidealization of the equations of state that gases actu-ally obey. Specifically, it is found that all gases obeythe equation ever more closely as the pressure is

reduced towards zero. That is, eqn 1.2 is an exampleof a limiting law, a law that becomes increasinglyvalid as the pressure is reduced and is obeyed exactlyin the limit of zero pressure.

A hypothetical substance that obeys eqn 1.2 at allpressures is called a perfect gas.1 From what has justbeen said, an actual gas, which is termed a real gas,behaves more and more like a perfect gas as its pres-sure is reduced towards zero. In practice, normal atmospheric pressure at sea level (p ≈ 100 kPa) is already low enough for most real gases to behave almost perfectly, and unless stated otherwise weshall always assume in this text that the gases we encounter behave like a perfect gas. The reason whya real gas behaves diCerently from a perfect gas canbe traced to the attractions and repulsions that existbetween actual molecules and which are absent in aperfect gas (Chapter 16).

The perfect gas law summarizes three sets of experi-mental observations. One is Boyle’s law:

At constant temperature, the pressure of a fixedamount of gas is inversely proportional to its volume.

Mathematically:

Boyle’s law: at constant temperature,

We can easily verify that eqn 1.2 is consistent withBoyle’s law: by treating n and T as constants, theperfect gas law becomes pV = constant, and hence p ∝ 1/V. Boyle’s law implies that if we compress (reduce the volume of) a fixed amount of gas at constant temperature into half its original volume,then its pressure will double. Figure 1.1 shows the

pV

∝1

13

Table 1.1

The gas constant in various units

R = 8.314 47 dm3 K−1 mol−1

8.314 47 dm3 kPa K−1 mol−1

8.205 74 × 10−2 dm3 atm K−1 mol−1

62.364 dm3 Torr K−1 mol−1

1.987 21 cal K−1 mol−1

1 dm3 = 10−3 m3.

1 The term ‘ideal gas’ is also widely used.

TEOC01 11/4/04 5:08 PM Page 13

0

10

20

–10

–200 12–1–2

y

x

graph obtained by plotting experimental values of p against V for a fixed amount of gas at diCerenttemperatures and the curves predicted by Boyle’slaw. Each curve is called an isotherm because it depicts the variation of a property (in this case, thepressure) at a single constant temperature. It isdiAcult, from this graph, to judge how well Boyle’slaw is obeyed. However, when we plot p against1/V, we get straight lines, just as we would expectfrom Boyle’s law (Fig. 1.2).

A note on good practice. It is generally the case that aproposed relation is easier to verify if the experimentaldata are plotted in a form that should give a straight line.

The isotherms are hyperbolas, graphs of xy = constant, or y = constant/x. The illustrationshows the graphs of xy = 1 and xy = 2 for positiveand negative values of x and y.

CHAPTER 1: THE PROPERTIES OF GASES

The second experimental observation summarizedby eqn 1.2 is Charles’s law:

At constant pressure, the volume of a fixed amountof gas varies linearly with the temperature.

Mathematically:

Charles’s law: at constant pressure, V = A + Bθ

where θ is the temperature on the Celsius scale andA and B are constants that depend on the amount of gas and the pressure. Figure 1.3 shows typicalplots of volume against temperature for a series ofsamples of gases at diCerent pressures and confirmsthat (at low pressures, and for temperatures that are not too low) the volume varies linearly with the Celsius temperature. We also see that all the volumes extrapolate to zero as θ approaches thesame very low temperature (−273.15°C, in fact), regardless of the identity of the gas. Because a vol-ume cannot be negative, this common temperaturemust represent the absolute zero of temperature, atemperature below which it is impossible to cool an object. Indeed, the Kelvin scale ascribes the value T = 0 to this absolute zero of temperature. In termsof the Kelvin temperature, therefore, Charles’s lawtakes the simpler form

Charles’s law: at constant pressure, V ∝ T

14

Volume, V

Pre

ssu

re, p

Increasingtemperature

Fig. 1.1 The volume of a gas decreases as the pressure on it is increased. For a sample that obeys Boyle’s law andthat is kept at constant temperature, the graph showing thedependence is a hyperbola, as shown here. Each curve cor-responds to a single temperature, and hence is an isotherm.

1/Volume, 1/V

Pre

ssu

re, p

Perfectgas

Observed

Fig. 1.2 A good test of Boyle’s law is to plot the pressureagainst 1/V (at constant temperature), when a straight lineshould be obtained. This diagram shows that the observedpressures (the blue line) approach a straight line as the volume is increased and the pressure reduced. A perfectgas would follow the straight line at all pressures; real gasesobey Boyle’s law in the limit of low pressures.

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EQUATIONS OF STATE

It follows that doubling the temperature (on the Kelvinscale, such as from 300 K to 600 K, corresponding toan increase from 27°C to 327°C) doubles the volume,provided the pressure remains the same. Now we cansee that eqn 1.2 is consistent with Charles’s law.First, we rearrange it into V = nRT/p, and then notethat when the amount n and the pressure p are bothconstant, we can write V ∝ T, as required.

The third feature of gases summarized by eqn 1.2is Avogadro’s principle:

At a given temperature and pressure, equal volumesof gas contain the same numbers of molecules.

That is, 1.00 dm3 of oxygen at 100 kPa and 300 Kcontains the same number of molecules as 1.00 dm3

of carbon dioxide, or any other gas, at the same tem-perature and pressure. The principle implies that ifwe double the number of molecules, but keep thetemperature and pressure constant, then the volumeof the sample will double. We can therefore write:

Avogadro’s principle: at constant temperatureand pressure, V ∝ n

This result follows easily from eqn 1.2 if we treat p and T as constants. Avogadro’s suggestion is aprinciple rather than a law (a direct summary of experience), because it is based on a model of a gas,in this case as a collection of molecules.

The molar volume, Vm, of any substance (not just a gas) is the volume it occupies per mole ofmolecules. It is calculated by dividing the volume ofthe sample by the amount of molecules it contains:

(1.3)

With volume in cubic decimetres and amount inmoles, the units of molar volume are cubic deci-metres per mole (dm3 mol−1). Avogadro’s principle implies that the molar volume of a gas should be the same for all gases at the same temperature andpressure. The data in Table 1.2 show that this con-clusion is approximately true for most gases undernormal conditions (normal atmospheric pressure ofabout 100 kPa and room temperature).

1.2 Using the perfect gas law

Here we review three elementary applications of theperfect gas equation of state. The first is the predic-tion of the pressure of a gas given its temperature, itschemical amount, and the volume it occupies. Thesecond is the prediction of the change in pressurearising from changes in the conditions. The third isthe calculation of the molar volume of a perfect gasunder any conditions. Calculations like these under-lie more advanced considerations, including the waythat meteorologists understand the changes in theatmosphere that we call the weather (Box 1.1).

VVnm =

15

Perfectgas

Observed

Temperature, q/°C

Volu

me,

V

–273.15

Increasingpressure

Fig. 1.3 This diagram illustrates the content and implicationsof Charles’s law, which asserts that the volume occupied by a gas (at constant pressure) varies linearly with the tem-perature. When plotted against Celsius temperatures (ashere), all gases give straight lines that extrapolate to V = 0 at−273.15°C. This extrapolation suggests that −273.15°C isthe lowest attainable temperature.

Table 1.2

The molar volumes of gases at standard ambienttemperature and pressure (298.15 K and 1 bar)

Gas Vm/(dm3 mol−1)

Perfect gas 24.7896*Ammonia 24.8Argon 24.4Carbon dioxide 24.6Nitrogen 24.8Oxygen 24.8Hydrogen 24.8Helium 24.8

* At STP (0°C, 1 atm), Vm = 24.4140 dm3 mol−1.

Volume

Amount

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CHAPTER 1: THE PROPERTIES OF GASES16

Box 1.1 The gas laws and the weather

The composition of the Earth’s atmosphere

Substance PercentageBy volume By mass

Nitrogen, N2 78.08 75.53Oxygen, O2 20.95 23.14Argon, Ar 0.93 1.28Carbon dioxide, CO2 0.031 0.047Hydrogen, H2 5.0 × 10−3 2.0 × 10−4

Neon, Ne 1.8 × 10−3 1.3 × 10−3

Helium, He 5.2 × 10−4 7.2 × 10−5

Methane, CH4 2.0 × 10−4 1.1 × 10−4

Krypton, Kr 1.1 × 10−4 3.2 × 10−4

Nitric oxide, NO 5.0 × 10−5 1.7 × 10−6

Xenon, Xe 8.7 × 10−6 3.9 × 10−5

Ozone, O3 summer: 7.0 × 10−6 1.2 × 10−5

winter: 2.0 × 10−6 3.3 × 10−6

Pressure, p/p0

0A

ltit

ud

e, h

/km

6

0.5 1

12

18

24

0

The variation of atmospheric pressure with altitude aspredicted by the barometric formula.

troposphere, then the pressure varies with altitude, h, according to the barometric formula:

p = p0 e−h/H

where p0 is the pressure at sea level and H is a constantapproximately equal to 8 km. More specifically, H = RT/Mg,where M is the average molar mass of air and T is thetemperature. The barometric formula fits the observedpressure distribution quite well even for regions wellabove the troposphere (see the first illustration). It impliesthat the pressure of the air and its density fall to half theirsea-level value at h = H ln 2, or 6 km.

Local variations of pressure, temperature, and com-position in the troposphere are manifest as ‘weather’. Asmall region of air is termed a parcel. First, we note that a parcel of warm air is less dense than the same parcel of cool air. As a parcel rises, it expands without transfer of heat from its surroundings and so it cools (see Section 1.14 for an explanation). Cool air can absorb lower concentrations of water vapour than warm air, sothe moisture forms clouds. Cloudy skies can therefore be associated with rising air and clear skies are often associated with descending air.

The motion of air in the upper altitudes may lead to anaccumulation in some regions and a loss of moleculesfrom other regions. The former result in the formation of

The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases with the compositionsummarized in the table. The composition is maintainedmoderately constant by diffusion and convection (winds,particularly the local turbulence called eddies) but thepressure and temperature vary with altitude and with thelocal conditions, particularly in the troposphere (the ‘sphereof change’), the layer extending up to about 11 km.

One of the most variable constituents of air is watervapour, and the humidity it causes. The presence of watervapour results in a lower density of air at a given temper-ature and pressure, as we may conclude from Avogadro’sprinciple. The numbers of molecules in 1 m3 of moist airand dry air are the same (at the same temperature andpressure), but the mass of an H2O molecule is less thanthat of all the other major constituents of air (the molarmass of H2O is 18 g mol−1, the average molar mass of airmolecules is 29 g mol−1), so the density of the moist sample is less than that of the dry sample.

The pressure and temperature vary with altitude. In the troposphere the average temperature is 15°C at sealevel, falling to −57°C at the bottom of the tropopause at 11 km. This variation is much less pronounced whenexpressed on the Kelvin scale, ranging from 288 K to 216 K,an average of 268 K. If we suppose that the temperaturehas its average value all the way up to the edge of the

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EQUATIONS OF STATE 17

H

H

LL

L

L

L

L

A typical weather map; this one for the continental UnitedStates on 14 July 1999. Regions of high pressure are denoted H and those of low pressure L.

L

L

Wind

Rotation

N

S

The horizontal flow of air relative to an area of low pressure in the northern and southern hemispheres.

regions of high pressure (‘highs’ or anticyclones) and thelatter result regions of low pressure (‘lows’, depressions,or cyclones). These regions are shown as H and L on the accompanying weather map. The lines of constantpressure—differing by 4 mbar (400 Pa, about 3 Torr)—marked on it are called isobars. The elongated regions ofhigh and low pressure are known, respectively, as ridgesand troughs.

In meteorology, large-scale vertical movement is calledconvection. Horizontal pressure differentials result in theflow of air that we call wind. Because the Earth is rotatingfrom west to east, winds are deflected towards the rightin the northern hemisphere and towards the left in thesouthern hemisphere. Winds travel nearly parallel to theisobars, with low pressure to their left in the northernhemisphere and to the right in the southern hemisphere.At the surface, where wind speeds are lower, the windstend to travel perpendicular to the isobars from high tolow pressure. This differential motion results in a spiraloutward flow of air clockwise in the northern hemispherearound a high and an inward counterclockwise flowaround a low.

The air lost from regions of high pressure is restored as an influx of air converges into the region and descends.As we have seen, descending air is associated with clearskies. It also becomes warmer by compression as it descends, so regions of high pressure are associated with high surface temperatures. In winter, the cold surfaceair may prevent the complete fall of air, and result in atemperature inversion, with a layer of warm air over a layer

of cold air. Geographical conditions may also trap cool air,as in Los Angeles, and the photochemical pollutants weknow as smog may be trapped under the warm layer. Aless dramatic manifestation of an inversion layer is thepresence of hazy skies, particularly in industrial areas. Hazyskies also form over vegetation that generate aerosols ofterpenes or other plant transpiration products. These hazesgive rise to the various ‘Blue Mountains’ of the world,such as the Great Dividing Range in New South Wales,the range in Jamaica, and the range stretching from central Oregon into southeastern Washington, which are dense with eucalyptus, tree ferns, and pine and fir, respectively. The Blue Ridge section of the Appalachiansis another example.

Exercise 1 Balloons were used to obtain much of theearly information about the atmosphere and continue tobe used today to obtain weather information. In 1782,Jacques Charles used a hydrogen-filled balloon to fly from Paris 25 km into the French countryside. What is the mass density of hydrogen relative to air at the sametemperature and presure? What mass of payload can be lifted by 10 kg of hydrogen, neglecting the mass of the balloon?

Exercise 2 Atmospheric pollution is a problem that hasreceived much attention. Not all pollution, however, is fromindustrial sources. Volcanic eruptions can be a significantsource of air pollution. The Kilauea volcano in Hawaii emits200–300 t of SO2 per day. If this gas is emitted at 800°Cand 1.0 atm, what volume of gas is emitted?

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CHAPTER 1: THE PROPERTIES OF GASES

In some cases, we are given the pressure under one set of conditions and are asked to predict thepressure of the same sample under a diCerent set ofconditions. We use the perfect gas law as follows.Suppose the initial pressure is p1, the initial tem-perature is T1, and the initial volume is V1. Then by dividing both sides of eqn 1.2 by the temperature wecan write

Suppose now that the conditions are changed to T2and V2, and the pressure changes to p2 as a result.Then under the new conditions eqn 1.2 tells us that

The nR on the right of these two equations is thesame in each case, because R is a constant and theamount of gas molecules has not changed. It followsthat we can combine the two equations into a singleequation:

(1.4)

This expression is known as the combined gas equation. We can rearrange it to calculate any oneunknown (such as p2, for instance) in terms of theother variables.

Self-test 1.2

What is the final volume of a sample of gas that hasbeen heated from 25°C to 1000°C and its pressure increased from 10.0 kPa to 150.0 kPa, given that its initial volume was 15 cm3?

[Answer: 4.3 cm3]

Finally, we see how to use the perfect gas law tocalculate the molar volume of a perfect gas at anytemperature and pressure. Equation 1.3 expressesthe molar volume in terms of the volume of a sample;eqn 1.2 in the form V = nRT/p expresses the volumein terms of the pressure. When we combine the two,we get

(1.5)VVn

nRT pn

RTpm

/ = = =

p VT

p VT

1 1

1

2 2

2

=

p VT

nR2 2

2

=

p VT

nR1 1

1

=

18

Example 1.1

Predicting the pressure of a sample of gas

A chemist is investigating the conversion of atmo-spheric nitrogen to usable form by the bacteria that inhabit the root systems of certain legumes, and needsto know the pressure in kilopascals exerted by 1.25 g of nitrogen gas in a flask of volume 250 cm3 at 20°C.

Strategy For this calculation we need to arrange eqn 1.2 (pV = nRT ) into a form that gives the unknown(the pressure, p) in terms of the information supplied:

To use this expression, we need to know the amount of molecules (in moles) in the sample, which we can obtain from the mass and the molar mass (by using n = m /M ) and to convert the temperature to the Kelvinscale (by adding 273.15 to the Celsius temperature).Select the value of R from Table 1.1 using the units thatmatch the data and the information required (pressurein kilopascals and volume in cubic decimetres).

Solution The amount of N2 molecules (of molar mass28.02 g mol−1) present is

The temperature of the sample is

T/K = 20 + 273.15

Therefore, from p = nRT/V,

We have used the relations 1 J = 1 Pa m3 and 1 kPa =103 Pa. Note how the units cancel like ordinary numbers.

A note on good practice. It is best to postpone the actual numerical calculation to the last possible stage,and carry it out in a single step. This procedure avoidsrounding errors.

Self-test 1.1

Calculate the pressure exerted by 1.22 g of carbon dioxide confined to a flask of volume 500 dm3 at 37°C.

[Answer: 143 kPa]

p ( . / . ) ( . ) ( . )

.

( . / . ) ( . )

.

.

=× × +

×

=× ×

×

= × =

− −

−

1 25 28 02 8 314 47 20 273 15

2 50 10

1 25 28 02 8 314 47 293

2 50 10

4 35 10 435

1 1

3 3

3 3

5

mol J K mol K

m

J

m

Pa kPa

nm

MNN

122

g

g molmol= = =−

.

.

..

1 25

28 02

1 2528 02

pnRTV

=

n R T = 293 K

V = 250 cm3

V = nRT/p

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EQUATIONS OF STATE

This expression lets us calculate the molar volume ofany gas (provided it is behaving perfectly) from itspressure and its temperature. It also shows that, fora given temperature and pressure, provided they arebehaving perfectly, all gases have the same molarvolume.

Chemists have found it convenient to report muchof their data at a particular set of ‘standard’ condi-tions. By standard ambient temperature and pres-sure (SATP) they mean a temperature of 25°C (moreprecisely, 298.15 K) and a pressure of exactly 1 bar(100 kPa). The standard pressure is denoted pa, sopa = 1 bar exactly. The molar volume of a perfectgas at SATP is 24.79 dm3 mol−1, as can be verified by substituting the values of the temperature andpressure into eqn 1.5. This value implies that atSATP, 1 mol of perfect gas molecules occupies about25 dm3 (a cube of about 30 cm on a side). An earlierset of standard conditions, which is still encoun-tered, is standard temperature and pressure (STP),namely 0°C and 1 atm. The molar volume of a perfect gas at STP is 22.41 dm3 mol−1.

1.3 Mixtures of gases: partial pressures

We are often concerned with mixtures of gases, such as when we are considering the properties ofthe atmosphere in meteorology, the composition of exhaled air in medicine, or the mixtures of hydro-gen and nitrogen used in the industrial synthesis of ammonia. We need to be able to assess the con-tribution that each component of a gaseous mixturemakes to the total pressure.

In the early nineteenth century, John Dalton carriedout a series of experiments that led him to formulatewhat has become known as Dalton’s law:

The pressure exerted by a mixture of perfect gases is the sum of the pressures that each gas would exert if it were alone in the container at the sametemperature:

p = pA + pB + . . . (1.6)

In this expression, pJ is the pressure that the gas J (J = A, B, . . . ) would exert if it were alone in the container at the same temperature. Dalton’s law isstrictly valid only for mixtures of perfect gases (orfor real gases at such low pressures that they are

behaving perfectly), but it can be treated as validunder most conditions we encounter.

For any type of gas (real or perfect) in a mixture,the partial pressure, pJ, of the gas J is defined as

pJ = xJp (1.7)

where xJ is the mole fraction of the gas J in the mixture. The mole fraction of J is the amount of J molecules expressed as a fraction of the totalamount of molecules in the mixture. In a mixturethat consists of nA A molecules, nB B molecules, and

19

5

2025

kPa kPa kPa

pA pB p pA B +

A BA

andB

Fig. 1.4 The partial pressure pA of a perfect gas A is the pressure that the gas would exert if it occupied acontainer alone; similarly, the partial pressure pB of aperfect gas B is the pressure that the gas would exert if it occupied the same container alone. The total pres-sure p when both perfect gases simultaneously occupythe container is the sum of their partial pressures.

Illustration 1.1

Dalton’s law

Suppose we were interested in the composition of inhaled and exhaled air, and we knew that a certainmass of carbon dioxide exerts a pressure of 5 kPa whenpresent alone in a container, and that a certain mass ofoxygen exerts 20 kPa when present alone in the samecontainer at the same temperature. Then, when bothgases are present in the container, the carbon dioxide inthe mixture contributes 5 kPa to the total pressure andoxygen contributes 20 kPa; according to Dalton’s law,the total pressure of the mixture is the sum of thesetwo pressures, or 25 kPa (Fig. 1.4).

TEOC01 11/4/04 5:08 PM Page 19

For a mixture of perfect gases, we can identify thepartial pressure of J with the contribution that Jmakes to the total pressure. Thus, if we introduce p = nRT/V into eqn 1.7, we get

The value of nJRT/V is the pressure that an amountnJ of J would exert in the otherwise empty container.That is, the partial pressure of J as defined by eqn 1.7 is the pressure of J used in Dalton’s law, provided all the gases in the mixture behave perfectly.If the gases are real, their partial pressures are stillgiven by eqn 1.7, as that definition applies to allgases, and the sum of these partial pressures is the total pressure (because the sum of all the molefractions is 1); however, each partial pressure is nolonger the pressure that the gas would exert whenalone in the container.

Illustration 1.2

Calculating partial pressures

From Self-test 1.3, we have xN2= 0.780, xO2

= 0.210,and xAr = 0.009 for dry air at sea level. It then followsfrom eqn 1.7 that when the total atmospheric pressureis 100 kPa, the partial pressure of nitrogen is

pN2= xN2

p = 0.780 × (100 kPa) = 78.0 kPa

Similarly, for the other two components we find pO2=

21.0 kPa and pAr = 0.9 kPa. Provided the gases are perfect, these partial pressures are the pressures thateach gas would exert if it were separated from the mixture and put in the same container on its own.

Self-test 1.4

The partial pressure of oxygen in air plays an importantrole in the aeration of water, to enable aquatic life tothrive, and in the absorption of oxygen by blood in ourlungs (see Box 6.1). Calculate the partial pressures of asample of gas consisting of 2.50 g of oxygen and 6.43 gof carbon dioxide with a total pressure of 88 kPa.

[Answer: 31 kPa, 57 kPa]

p x p xnRT

V

n RT

VJ J JJ = = × =

so on (where the nJ are amounts in moles), the molefraction of J (where J = A, B, . . . ) is

(1.8a)

where n = nA + nB + . . . . Mole fractions are unitlessbecause the unit mole in numerator and denomin-ator cancels. For a binary mixture, one that consistsof two species, this general expression becomes

(1.8b)

When only A is present, xA = 1 and xB = 0. Whenonly B is present, xB = 1 and xA = 0. When both A and B are present in the same amounts, xA = andxB = (Fig. 1.5).

Self-test 1.3

Calculate the mole fractions of N2, O2, and Ar in dry airat sea level, given that 100.0 g of air consists of 75.5 gof N2, 23.2 g of O2, and 1.3 g of Ar. (Hint. Begin by con-verting each mass to an amount in moles.)

[Answer: 0.780, 0.210, 0.009]

12

12

xn

n nx

nn n

x xAA

A BB

B

A BA B=

+=

++ =

1

xn

nJJ =

CHAPTER 1: THE PROPERTIES OF GASES20

xA = 0.167

xA = 0.452

xB = 0.833

xB = 0.548

xB = 0.167xA = 0.833

AB

Fig. 1.5 A representation of the meaning of mole fraction. In each case, a small square represents one molecule of A(coloured squares) or B (white squares). There are 84 squaresin each sample.

Amount of J

Total amount of moleculesp = nRT/V

xJ = nJ/n

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THE KINETIC MODEL OF GASES

The kinetic model of gases

We saw in the Introduction that a gas may be pic-tured as a collection of particles in ceaseless, randommotion (Fig. 1.6). Now we develop this model of the gaseous state of matter to see how it accounts for the perfect gas law. One of the most importantfunctions of physical chemistry is to convert qualit-ative notions into quantitative statements that can be tested experimentally by making measurementsand comparing the results with predictions. Indeed,an important component of science as a whole is itstechnique of proposing a qualitative model and thenexpressing that model mathematically. The ‘kineticmodel’ (or the ‘kinetic molecular theory’, KMT) ofgases is an excellent example of this procedure: themodel is very simple, and the quantitative prediction(the perfect gas law) is experimentally verifiable.

The kinetic model of gases is based on three assumptions:

1 A gas consists of molecules in ceaseless randommotion.

2 The size of the molecules is negligible in the sensethat their diameters are much smaller than theaverage distance travelled between collisions.

3 The molecules do not interact, except during collisions.

The assumption that the molecules do not interactunless they are in contact implies that the potentialenergy of the molecules (their energy due to their position) is independent of their separation and maybe set equal to zero. The total energy of a sample ofgas is therefore the sum of the kinetic energies (theenergy due to motion) of all the molecules present init. It follows that the faster the molecules travel (andhence the greater their kinetic energy), the greaterthe total energy of the gas.

1.4 The pressure of a gas according to thekinetic model

The kinetic model accounts for the steady pressureexerted by a gas in terms of the collisions themolecules make with the walls of the container.Each collision gives rise to a brief force on the wall,but as billions of collisions take place every second,the walls experience an almost constant force, andhence the gas exerts a steady pressure. On the basisof this model, the pressure exerted by a gas of molarmass M in a volume V is2

(1.9)

Here c is the root-mean-square speed (rms speed) ofthe molecules. This quantity is defined as the squareroot of the mean value of the squares of the speeds,v, of the molecules. That is, for a sample consistingof N molecules with speeds v1, v2, . . . , vN, we squareeach speed, add the squares together, divide by thetotal number of molecules (to get the mean, denotedby ⟨ . . . ⟩), and finally take the square root of the result:

(1.10)

The rms speed might at first encounter seem to be a rather peculiar measure of the mean speeds of the molecules, but its significance becomes clearwhen we make use of the fact that the kinetic energyof a molecule of mass m travelling at a speed v is EK = mv2, which implies that the mean kinetic 1

2

c vv v v

NN

. . . /

/

= =+ + +

⟨ ⟩2 1 2 12

22 2 1 2

pnMc

V =

2

3

21

Fig. 1.6 The model used for discussing the molecular basisof the physical properties of a perfect gas. The point-likemolecules move randomly with a wide range of speeds and in random directions, both of which change when theycollide with the walls or with other molecules.

2 See Further information 1.1 for a derivation of eqn 1.9.

TEOC01 11/4/04 5:08 PM Page 21

energy, ⟨EK⟩, is the average of this quantity, ormc2. It follows that

(1.11)

Therefore, wherever c appears, we can think of it as a measure of the mean kinetic energy of themolecules of the gas. The rms speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, C, ofthe molecules:

(1.12)

For samples consisting of large numbers of mole-cules, the mean speed is slightly smaller than the rms speed. The precise relation is

c ≈ 0.921c (1.13)

For elementary purposes, and for qualitative argu-ments, we do not need to distinguish between thetwo measures of average speed, but for precise workthe distinction is important.

Self-test 1.5

Cars pass a point travelling at 45.00 (5), 47.00 (7), 50.00 (9), 53.00 (4), 57.00 (1) km h−1, where the num-ber of cars is given in parentheses. Calculate (a) the rms speed and (b) the mean speed of the cars. (Hint.Use the definitions directly; the relation in eqn 1.13 isunreliable for such small samples.)

[Answer: (a) 49.06 km h−1, (b) 48.96 km h−1]

Equation 1.9 already resembles the perfect gasequation of state, as we can rearrange it into

pV = nMc2 (1.14)

and compare it to pV = nRT. This conclusion is a major success of the kinetic model, as the modelimplies an experimentally verified result.

1.5 The average speed of gas molecules

We now suppose that the expression for pV derivedfrom the kinetic model is indeed the equation of state

13

C /

=

83

1 2

π

C . . .

=+ + +v v v

NN1 2

cEm

/

=

21 2⟨ ⟩K

12

CHAPTER 1: THE PROPERTIES OF GASES

of a perfect gas. That being so, we can equate the ex-pression on the right of eqn 1.14 to nRT, which gives

nMc2 = nRT

The ns now cancel. The great usefulness of this expression is that we can rearrange it into a for-mula for the rms speed of the gas molecules at anytemperature:

(1.15)

Substitution of the molar mass of O2 (32.0 g mol−1)and a temperature corresponding to 25°C (that is,298 K) gives an rms speed for these molecules of 482 m s−1. The same calculation for nitrogen mole-cules gives 515 m s−1. Both these values are not far oC the speed of sound in air (346 m s−1 at 25°C).That similarity is reasonable, because sound is a waveof pressure variation transmitted by the movementof molecules, so the speed of propagation of a waveshould be approximately the same as the speed atwhich molecules can adjust their locations.

The important conclusion to draw from eqn 1.15is that the rms speed of molecules in a gas is propor-tional to the square root of the temperature. Becausethe mean speed is proportional to the rms speed, thesame is true of the mean speed too. Therefore, doub-ling the temperature (on the Kelvin scale) increasesthe mean and the rms speed of molecules by a factorof 21/2 = 1.414 . . . .

Illustration 1.3

The effect of temperature on mean speeds

Cooling a sample of air from 25°C (298 K) to 0°C (273 K)reduces the original rms speed of the molecules by afactor of

So, on a cold day, the average speed of air molecules(which is changed by the same factor) is about 4 percent less than on a warm day.

1.6 The Maxwell distribution of speeds

So far, we have dealt only with the average speed ofmolecules in a gas. Not all molecules, however,

273298

273298

0 9571 2 1 2

./ /

KK

=

=

cRTM

/

=

31 2

13

22

TEOC01 11/4/04 5:08 PM Page 22

THE KINETIC MODEL OF GASES

travel at the same speed: some move more slowlythan the average (until they collide, and get acceler-ated to a high speed, like the impact of a bat on aball), and others may briefly move at much higherspeeds than the average, but be brought to a suddenstop when they collide. There is a ceaseless redistribu-tion of speeds among molecules as they undergo collisions. Each molecule collides once every nano-second (1 ns = 10−9 s) or so in a gas under normalconditions.

The mathematical expression that tells us thefraction of molecules that have a particular speed atany instant is called the distribution of molecularspeeds. Thus, the distribution might tell us that at20°C, 19 out of 1000 O2 molecules have a speed inthe range between 300 and 310 m s−1, that 21 out of 1000 have a speed in the range 400 to 410 m s−1,and so on. The precise form of the distribution wasworked out by James Clerk Maxwell towards theend of the nineteenth century, and his expression is known as the Maxwell distribution of speeds.According to Maxwell, the fraction f of moleculesthat have a speed in a narrow range between s and s + ∆s (for example, between 300 m s−1 and 310 m s−1,corresponding to s = 300 m s−1 and ∆s = 10 m s−1) is

f = F(s)∆s with

(1.16)

This formula was used to calculate the valuesquoted above.

Although eqn 1.16 looks complicated, its featurescan be picked out quite readily. One of the skills todevelop in physical chemistry is the ability to inter-pret the message carried by equations. Equationsconvey information, and it is far more important to be able to read that information than simply toremember the equation. Let’s read the informationin eqn 1.16 piece by piece.

Before we begin, and in preparation for a largenumber of occurrences of exponential functionsthroughout the text, it will be useful to know theshape of exponential functions. Here we deal withtwo types, e−ax and e−ax2

. An exponential function ofthe form e−ax starts oC at 1 when x = 0 and decaystowards zero, which it reaches as x approachesinfinity (Fig. 1.7). This function approaches zeromore rapidly as a increases. The function e−ax2

is

F sMRT

s Ms RT( ) /

/=

−42

3 22 22π

πe

called a Gaussian function. It also starts oC at 1when x = 0 and decays to zero as x increases; how-ever, its decay is initially slower but then plungesdown more rapidly than e−ax. The illustration alsoshows the behaviour of the two functions for negat-ive values of x. The exponential function e−ax risesrapidly to infinity, but the Gaussian function fallsback to zero and traces out a bell-shaped curve.

Now let’s consider the content of eqn 1.16.

1 Because f is proportional to the range ofspeeds ∆s, we see that the fraction in the range ∆sincreases in proportion to the width of the range. If at a given speed we double the range of interest(but still ensure that it is narrow), then the fractionof molecules in that range doubles too.

2 Equation 1.16 includes a decaying exponen-tial function, the term e−Ms2/ 2RT. Its presence impliesthat the fraction of molecules with very high speedswill be very small because e−x2

becomes very smallwhen x2 is large.

3 The factor M/2RT multiplying s2 in the expo-nent is large when the molar mass, M, is large, so theexponential factor goes most rapidly towards zerowhen M is large. That tells us that heavy moleculesare unlikely to be found with very high speeds.

4 The opposite is true when the temperature, T, ishigh: then the factor M/2RT in the exponent is small,so the exponential factor falls towards zero relativelyslowly as s increases. This tells us that at high temper-atures, a greater fraction of the molecules can be ex-pected to have high speeds than at low temperatures.

23

0

e–x2

e–x

x x0

1

e–x

e–x2

1

Fig. 1.7 The exponential function, e−x, and the bell-shapedGaussian function, e−x 2

. Note that both are equal to 1 at x = 0 but the exponential function rises to infinity as x → −∞.The enlargement in the right shows the behaviour for x > 0in more detail.

TEOC01 11/4/04 5:08 PM Page 23

5 A factor s2 (the term before the e) multipliesthe exponential. This factor goes to zero as s goes tozero, so the fraction of molecules with very lowspeeds will also be very small.

The remaining factors (the term in parentheses ineqn 1.16 and the 4π) simply ensure that when weadd together the fractions over the entire range ofspeeds from zero to infinity, we get 1.

Figure 1.8 is a plot of the Maxwell distribution,and shows these features pictorially for the same gas(the same value of M) but diCerent temperatures. Aswe deduced from the equation, we see that onlysmall fractions of molecules in the sample have verylow or very high speeds. However, the fraction withvery high speeds increases sharply as the temper-ature is raised, as the tail of the distribution reachesup to higher speeds. This feature plays an importantrole in the rates of gas-phase chemical reactions because (as we shall see in Section 10.10) the rate of a reaction in the gas phase depends on the energywith which two molecules crash together, which inturn depends on their speeds.

Figure 1.9 is a plot of the Maxwell distributionfor molecules with diCerent molar masses at thesame temperature. As can be seen, not only do heavymolecules have lower average speeds than light

CHAPTER 1: THE PROPERTIES OF GASES

molecules at a given temperature, but they also havea significantly narrower spread of speeds. That nar-row spread means that most molecules will be foundwith speeds close to the average. By contrast, lightmolecules (such as H2) have high average speeds and a wide spread of speeds: many molecules will befound travelling either much more slowly or muchmore quickly than the average. This feature plays an important role in determining the composition of planetary atmospheres, because it means that a significant fraction of light molecules travel atsuAciently high speeds to escape from the planet’sgravitational attraction. The ability of light moleculesto escape is one reason why hydrogen (molar mass2.02 g mol−1 ) and helium (4.00 g mol−1 ) are veryrare in the Earth’s atmosphere.

The Maxwell distribution has been verified experi-mentally by passing a beam of molecules from anoven at a given temperature through a series ofcoaxial slotted discs. The speed of rotation of thediscs brings the slots into line for molecules travel-ling at a particular speed, so only molecules withthat speed pass through and are detected. By varyingthe rotation speed, the shape of the speed distribution

24

Nu

mb

er o

f m

ole

cule

s

Speed

Lowtemperature

Intermediatetemperature

Hightemperature

Fig. 1.8 The Maxwell distribution of speeds and its variationwith the temperature. Note the broadening of the distribu-tion and the shift of the rms speed (denoted by the locationsof the vertical lines) to higher values as the temperature is increased.

Nu

mb

er o

f m

ole

cule

s

Speed

Lowmolar mass

Intermediatemolar mass

Highmolar mass

Fig. 1.9 The Maxwell distribution of speeds also dependson the molar mass of the molecules. Molecules of low molar mass have a broad spread of speeds, and a significantfraction may be found travelling much faster than the rmsspeed. The distribution is much narrower for heavy molecules,and most of them travel with speeds close to the rms value(denoted by the locations of the vertical lines).

TEOC01 11/4/04 5:08 PM Page 24

THE KINETIC MODEL OF GASES

can be explored and is found to match that pre-dicted by eqn 1.16.

1.7 Diffusion and effusion

DiIusion is the process by which the molecules ofdiCerent substances mingle with each other. Theatoms of two solids diCuse into each other when thetwo solids are in contact, but the process is veryslow. The diCusion of a solid through a liquid solvent is much faster but mixing normally needs tobe encouraged by stirring or shaking the solid in theliquid (the process is then no longer pure diCusion).Gaseous diCusion is much faster. It accounts for the largely uniform composition of the atmospherebecause, if a gas is produced by a localized source(such as carbon dioxide from the respiration of ani-mals, oxygen from photosynthesis by green plants,and pollutants from vehicles and industrial sources),then the molecules of gas will diCuse from thesource and in due course be distributed throughoutthe atmosphere. In practice, the process of mixing is accelerated by winds: such bulk motion of matteris called convection. The process of eIusion is theescape of a gas through a small hole, as in a puncturein an inflated balloon or tyre (Fig. 1.10).

The rates of diCusion and eCusion of gases increase with increasing temperature, as both processes depend on the motion of molecules, andmolecular speeds increase with temperature. Therates also decrease with increasing molar mass, asmolecular speeds decrease with increasing molarmass. The dependence on molar mass, however, issimple only in the case of eCusion. In eCusion, onlya single substance is in motion, not the two or moreintermingling gases involved in diCusion.

The experimental observations on the depend-ence of the rate of eCusion of a gas on its molar mass are summarized by Graham’s law of eIusion,proposed by Thomas Graham in 1833:

At a given pressure and temperature, the rate of eLusion of a gas is inversely proportional to thesquare root of its molar mass:

(1.17)

Rate in this context means the number (or numberof moles) of molecules that escape per second.

Rate of effusion /∝11 2M

Illustration 1.4

Relative rates of effusion

The rates (in terms of amounts of molecules) at whichhydrogen (molar mass 2.016 g mol−1) and carbon diox-ide (44.01 g mol−1) effuse under the same conditions ofpressure and temperature are in the ratio

The mass of carbon dioxide that escapes in a given interval is greater than the mass of hydrogen, becausealthough nearly 5 times as many hydrogen mole-cules escape, each carbon dioxide molecule has over20 times the mass of a molecule of hydrogen.

A note on good practice. Always make it clear whatterms mean: in this instance ‘rate’ alone is ambiguous;you need to specify that it is the rate in terms of amountof molecules.

Rate of effusion of H

Rate of effusion of CO

g molg mol

CO

H

2

2

1 2

1

1

1 2

1 2

2

2

44 012 016

44 012 016

4 672

. .

.

. .

/

/

/

=

=

=

=

−

−

M

M

25

(a)

(b)

Fig. 1.10 (a) Diffusion is the spreading of the molecules ofone substance into the region initially occupied by anotherspecies. Note that molecules of both substances move, andeach substance diffuses into the other. (b) Effusion is the escape of molecules through a small hole in a confining wall.

TEOC01 11/4/04 5:08 PM Page 25

The high rate of eCusion of hydrogen and heliumis one reason why these two gases leak from con-tainers and through rubber diaphragms so readily.The diCerent rates of eCusion through a porous bar-rier are used in the separation of uranium-235 fromthe more abundant and less useful uranium-238 inthe processing of nuclear fuel. The process dependson the formation of uranium hexafluoride, a volatilesolid. However, because the ratio of the molarmasses of 238UF6 and 235UF6 is only 1.008, the ratioof the rates of eCusion is only (1.008)1/2 = 1.004.Thousands of successive eCusion stages are there-fore required to achieve a significant separation. Therate of eCusion of gases was once used to determinemolar mass by comparison of the rate of eCusion ofa gas or vapour with that of a gas of known molarmass. However, there are now much more precisemethods available, such as mass spectrometry.

Graham’s law is explained by noting that the rmsspeed of molecules of a gas is inversely proportionalto the square root of the molar mass (eqn 1.15).Because the rate of eCusion through a hole in a con-tainer is proportional to the rate at which moleculespass through the hole, it follows that the rate shouldbe inversely proportional to M1/2, which is in accordwith Graham’s law.

1.8 Molecular collisions

The average distance that a molecule travels betweencollisions is called its mean free path, λ (lambda).The mean free path in a liquid is less than the diameter of the molecules, because a molecule in a liquid meets a neighbour even if it moves only afraction of a diameter. However, in gases, the meanfree paths of molecules can be several hundredmolecular diameters. If we think of a molecule as thesize of a tennis ball, then the mean free path in a typ-ical gas would be about the length of a tennis court.

The collision frequency, z, is the average rate ofcollisions made by one molecule. Specifically, z is theaverage number of collisions one molecule makes in a given time interval divided by the length of theinterval. It follows that the inverse of the collisionfrequency, 1/z, is the time of flight, the average timethat a molecule spends in flight between two colli-sions (for instance, if there are 10 collisions per second, so the collision frequency is 10 s−1, then theaverage time between collisions is of a second and1

10

CHAPTER 1: THE PROPERTIES OF GASES

the time of flight is s). As we shall see, the colli-sion frequency in a typical gas is about 109 s−1 at 1 atm and room temperature, so the time of flight in a gas is typically 1 ns.

Because speed is distance travelled divided by thetime taken for the journey, the rms speed c, whichwe can think of loosely as the average speed, is theaverage length of the flight of a molecule betweencollisions (that is, the mean free path, λ) divided bythe time of flight (1/z). It follows that the mean freepath and the collision frequency are related by

(1.18)

Therefore, if we can calculate either λ or z, then wecan find the other from this equation and the valueof c given in eqn 1.15.

To find expressions for λ and z we need a slightlymore elaborate version of the kinetic model. Thebasic kinetic model supposes that the molecules areeCectively point-like; however, to obtain collisions,we need to assume that two ‘points’ score a hitwhenever they come within a certain range d of eachother, where d can be thought of as the diameter ofthe molecules (Fig. 1.11). The collision cross-section,σ (sigma), the target area presented by one moleculeto another, is therefore the area of a circle of radiusd, so σ = πd2. When this quantity is built into the kinetic model, we find that

(1.19)λσ

σ /

/

= =RTN p

zN cpRT2

21 2

1 2

A

A

cz

z /

= = =mean free pathtime of flight

λ1

λ

110

26

Diameter, dRadius, d

Fig. 1.11 To calculate features of a perfect gas that are related to collisions, a point is regarded as being surroundedby a sphere of diameter d. A molecule will hit anothermolecule if the centre of the former lies within a circle of radius d. The collision cross-section is the target area, pd 2.

TEOC01 11/4/04 5:08 PM Page 26

REAL GASES

Table 1.3 lists the collision cross-sections of somecommon atoms and molecules.

Illustration 1.5

Calculating a mean free path

From the information in Table 1.3 we can calculate thatthe mean free path of O2 molecules in a sample of oxygen at SATP (25°C, 1 bar) is

We have used R in one of its SI unit forms: this form is usually appropriate in calculations based on the kinetic model; we have also used 1 J = 1 Pa m3 and 1 nm = 10−9 m. Under the same conditions, the collisionfrequency is 6.2 × 109 s−1, so each molecule makes 6.2 billion collisions each second.

Once again, we should interpret the essence of the two expressions in eqn 1.19 rather than trying to remember them.

l ( . ) ( )

( . ) ( . ) ( . )

.

. . .

.

/

/

=×

× × × × × ×

=×

× × × ×

= × =

− −

− −

−

8 314 47 298

2 6 022 10 0 40 10 1 00 10

8 314 47 298

2 6 022 0 40 1 00 10

7 3 10 73

1 1

1 2 23 1 18 2 5

1 2 10 2

8

J K mol K

mol m Pa

J

Pa m

m nm

1 Because λ ∝ 1/p, we see that the mean free path decreases as the pressure increases. Thisdecrease is a result of the increase in the number ofmolecules present in a given volume as the pressure is increased, so each molecule travels a shorter dis-tance before it collides with a neighbour.

For example, the mean free path of an O2molecule decreases from 73 nm to 36 nm when the pressure is increased from 1.0 bar to 2.0 bar at 25°C.

2 Because λ ∝ 1/σ, the mean free path is shorterfor molecules with large collision cross-sections.

For example, the collision cross-section of a benzene molecule (0.88 nm2) is about four timesgreater than that of a helium atom (0.21 nm2), andat the same pressure and temperature its mean freepath is four times shorter.

3 Because z ∝ p, the collision frequency increases with the pressure of the gas. This dependence follows from the fact that, provided the temperature is the same, the molecules take less time to travel to its neighbour in a denser,higher-pressure gas.

For example, although the collision frequency for an O2 molecule in oxygen gas at SATP is 6.2 × 109 s−1, at 2.0 bar and the same temperaturethe collision frequency is doubled, to 1.2 × 1010 s−1.

4 Because eqn 1.19 shows that z ∝ c, and we know that c ∝ 1/M1/2, heavy molecules have lower collision frequencies than light molecules,providing their collision cross-sections are the same. Heavy molecules travel more slowly on average than light molecules do (at the same temperature), so they collide with other moleculesless frequently.

Real gases

So far, everything we have said applies to perfectgases, in which the average separation of themolecules is so great that they move independentlyof one another. In terms of the quantities introducedin the previous section, a perfect gas is a gas forwhich the mean free path, λ , of the molecules in thesample is much greater than d, the separation atwhich they are regarded as being in contact:

Condition for perfect gas behaviour: λ � d

27

Table 1.3

Collision cross-sections of atoms andmolecules

Species σ/nm2

Argon, Ar 0.36Benzene, C6H6 0.88Carbon dioxide, CO2 0.52Chlorine, Cl2 0.93Ethene, C2H4 0.64Helium, He 0.21Hydrogen, H2 0.27Methane, CH4 0.46Nitrogen, N2 0.43Oxygen, O2 0.40Sulfur dioxide, SO2 0.58

1 nm2 = 10−18 m2.

R T

NA s p

TEOC01 11/4/04 5:08 PM Page 27

As a result of this large average separation, a perfectgas is a gas in which the only contribution to the energy comes from the kinetic energy of the motionof the molecules and there is no contribution to thetotal energy from the potential energy arising fromthe interaction of the molecules with one another.However, all molecules do in fact interact with oneanother provided they are close enough together, sothe ‘kinetic energy only’ model is only an approx-imation. Nevertheless, under most conditions the

CHAPTER 1: THE PROPERTIES OF GASES

criterion λ � d is satisfied and the gas can be treated as though it is perfect; the criterion is alsosatisfied under some other rather surprising con-ditions (Box 1.2).

1.9 Molecular interactions

There are two types of contribution to the inter-action between molecules. At relatively large separa-tions (a few molecular diameters), molecules attract

28

Box 1.2 The Sun as a ball of perfect gas

The kinetic theory of gases is valid when the size of theparticles is negligible compared with their mean free path.It may seem absurd, therefore, to expect the kinetic theoryand, as a consequence, the perfect gas law, to be applicableto the dense matter of stellar interiors. In the Sun, for instance, the density is 1.50 times that of liquid water atits centre and comparable to that of water about halfway toits surface. However, we have to realize that the state ofmatter is that of a plasma, in which the electrons have beenstripped from the atoms of hydrogen and helium that makeup the bulk of the matter of stars. As a result, the particlesmaking up the plasma have diameters comparable to thoseof nuclei, or about 10 fm. Therefore, a mean free path of only0.1 pm satisfies the criterion for the validity of the kinetictheory and the perfect gas law. We can therefore use pV = nRT as the equation of state for the stellar interior.

As for any perfect gas, the pressure in the interior of theSun is related to the mass density, r = m /V, by

The problem is to know the molar mass to use. Atoms arestripped of their electrons in the interior of stars, so if we suppose that the interior consists of ionized hydrogenatoms, the mean molar mass is one-half the molar massof hydrogen, or 0.5 g mol−1 (the mean of the molar massof H+ and e−, the latter being almost 0). Halfway to thecentre of the Sun, the temperature is 3.6 MK and themass density is 1.20 g cm−3 (slightly denser than water);so the pressure there is

or 720 Mbar (about 720 million atmospheres).

p ( . ) ( . ) ( . )

.

.

=× × × ×

×

= ×

− − −

− −

1 20 10 8 3145 3 6 10

0 50 10

7 2 10

3 3 1 1 6

3 1

13

kg m J K mol K

kg mol

Pa

pnRTV

mRTMV

RTM

= = =r

We can combine this result with the expression for thepressure from kinetic theory (p = NMc2/V ). Because thetotal kinetic energy of the particles is EK = Nmc2, we canwrite p = EK/V. That is, the pressure of the plasma is related to the kinetic energy density, rK = EK/V, the kineticenergy of the molecules in a region divided by the volumeof the region, by

p = rK

It follows that the kinetic energy density halfway to thecentre of the Sun is

rK = p = × (7.2 × 1013 Pa) = 1.1 × 1014 J m−3

or 0.11 GJ cm−3. By contrast, on a warm day (25°C) onEarth, the (translational) kinetic energy density of our atmosphere is only 1.5 × 105 J m−3 (corresponding to 0.15 J cm−3).

Exercise 1 A star eventually depletes some of the hydro-gen in its core, which contracts and results in higher temperatures. The increased temperature results in an increase in the rates of nuclear reaction, some of whichresult in the formation of heavier nuclei, such as carbon.The outer part of the star expands and cools to produce a red giant. Assume that halfway to the centre a red giant has a temperature of 3500 K, is composed primarilyof fully ionized carbon atoms and electrons, and has amass density of 1200 kg m−3. What is the pressure at this point?

Exercise 2 If the red giant in Exercise 1 consisted of neutral carbon atoms, what would be the pressure at thesame point under the same conditions?

32

32

23

23

12

13

TEOC01 11/4/04 5:08 PM Page 28

REAL GASES

each other. This attraction is responsible for thecondensation of gases into liquids at low tempera-tures. At low-enough temperatures the molecules of a gas have insuAcient kinetic energy to escapefrom each other’s attraction and they stick together.Second, although molecules attract each other whenthey are a few diameters apart, as soon as they comeinto contact they repel each other. This repulsion isresponsible for the fact that liquids and solids have adefinite bulk and do not collapse to an infinitesimalpoint.

Molecular interactions—the attractions and repul-sions between molecules—give rise to a potential energy that contributes to the total energy of a gas.Because attractions correspond to a lowering oftotal energy as molecules get closer together, theymake a negative contribution to the potential energy.Repulsions, however, make a positive contributionto the total energy as the molecules squash together.Figure 1.12 illustrates the general form of the varia-tion of the intermolecular potential energy. At largeseparations, the energy-lowering interactions aredominant, but at short distances the energy-raisingrepulsions dominate.

Molecular interactions aCect the bulk propertiesof a gas and, in particular, their equations of state.

For example, the isotherms of real gases have shapesthat diCer from those implied by Boyle’s law, par-ticularly at high pressures and low temperatureswhen the interactions are most important. Figure 1.13shows a set of experimental isotherms for carbondioxide. They should be compared with the per-fect gas isotherms shown in Fig. 1.1. Although the experimental isotherms resemble the perfect gasisotherms at high temperatures (and at low pres-sures, oC the scale on the right of the graph), thereare very striking diCerences between the two at temperatures below about 50°C and at pressuresabove about 1 bar.

1.10 The critical temperature

To understand the significance of the isotherms inFig. 1.13, let’s begin with the isotherm at 20°C. Atpoint A the sample is a gas. As the sample is com-pressed to B by pressing in a piston, the pressure increases broadly in agreement with Boyle’s law, andthe increase continues until the sample reaches pointC. Beyond this point, we find that the piston can bepushed in without any further increase in pressure,through D to E. The reduction in volume from E to F requires a very large increase in pressure. Thisvariation of pressure with volume is exactly whatwe expect if the gas at C condenses to a compact liquid at E. Indeed, if we could observe the sample

29

0

Po

ten

tial

en

erg

y

Attractions dominant

Repulsions dominant

Separation

Fig. 1.12 The variation of the potential energy of two mole-cules with their separation. High positive potential energy (at very small separations) indicates that the interactions between them are strongly repulsive at these distances. At intermediate separations, where the potential energy isnegative, the attractive interactions dominate. At large separations (on the right) the potential energy is zero andthere is no interaction between the molecules.

140

120

100

80

60

40

20

00 0.2 0.4 0.6

50 C°

40 C°

31.04 C ( )° Tc

p/a

tm

Vm /(dm mol )3 –1

E DB

*

20°C

Critical point

C

F

0°C

A

Fig. 1.13 The experimental isotherms of carbon dioxide atseveral temperatures. The critical isotherm is at 31.04°C.

TEOC01 11/4/04 5:08 PM Page 29

we would see it begin to condense to a liquid at C,and the condensation would be complete when thepiston was pushed in to E. At E, the piston is restingon the surface of the liquid. The subsequent reduc-tion in volume, from E to F, corresponds to the veryhigh pressure needed to compress a liquid into asmaller volume. In terms of intermolecular inter-actions, the step from C to E corresponds to themolecules being so close on average that they attracteach other and cohere into a liquid. The step from E to F represents the eCect of trying to force themolecules even closer together when they are alreadyin contact, and hence trying to overcome the strongrepulsive interactions between them.

If we could look inside the container at point D,we would see a liquid separated from the remain-ing gas by a sharp surface (Fig. 1.14). At a slightlyhigher temperature (at 30°C, for instance), a liquidforms, but a higher pressure is needed to produce it.It might be diAcult to make out the surface becausethe remaining gas is at such a high pressure that itsdensity is similar to that of the liquid. At the specialtemperature of 31.04°C (304.19 K) the gaseousstate appears to transform continuously into thecondensed state and at no stage is there a visible

CHAPTER 1: THE PROPERTIES OF GASES

surface between the two states of matter. At thistemperature, which is called the critical temper-ature, Tc, and at all higher temperatures, a singleform of matter fills the container at all stages of thecompression and there is no separation of a liquidfrom the gas. We have to conclude that a gas cannotbe condensed to a liquid by the application of pres-sure unless the temperature is below the critical temperature.

Table 1.4 lists the critical temperatures of somecommon gases. The data there imply, for example,that liquid nitrogen cannot be formed by the applica-tion of pressure unless the temperature is below 126 K (−147°C). The critical temperature is some-times used to distinguish between the terms ‘vapour’and ‘gas’: a vapour is the gaseous phase of a sub-stance below its critical temperature (and which can therefore be liquefied by compression); a gas is

30

Increasing temperature

Fig. 1.14 When a liquid is heated in a sealed container, thedensity of the vapour phase increases and that of the liquidphase decreases, as depicted here by the changing densityof shading. There comes a stage at which the two densitiesare equal and the interface between the two fluids dis-appears. This disappearance occurs at the critical temperature.The container needs to be strong: the critical temperature ofwater is at 373°C and the vapour pressure is then 218 atm.

Table 1.4

The critical temperatures of gases

Critical temperature/°C

Noble gases

Helium, He −268 (5.2 K)Neon, Ne −229Argon, Ar −123Krypton, Kr −64Xenon, Xe 17

Halogens

Chlorine, Cl2 144Bromine, Br2 311

Small inorganic molecules

Ammonia, NH3 132Carbon dioxide, CO2 31Hydrogen, H2 −240Nitrogen, N2 −147Oxygen, O2 −118Water, H2O 374

Organic compounds

Benzene, C6H6 289Methane, CH4 −83Tetrachloromethane, CCl4 283

TEOC01 11/4/04 5:08 PM Page 30

REAL GASES

the gaseous phase of a substance above its criticaltemperature (and which cannot therefore be liquefiedby compression alone). Oxygen at room temper-ature is therefore a true gas; the gaseous phase ofwater at room temperature is a vapour.

The text’s web site contains links to online databases of properties of gases.

The dense fluid obtained by compressing a gaswhen its temperature is higher than its critical tem-perature is not a true liquid, but it behaves like a liq-uid in many respects—it has a density similar to thatof a liquid, for instance, and can act as a solvent.However, despite its density, the fluid is not strictlya liquid because it never possesses a surface that separates it from a vapour phase. Nor is it much likea gas, because it is so dense. It is an example of a supercritical fluid. Supercritical fluids (SCFs) arecurrently being used as solvents. For example, super-critical carbon dioxide is used to extract caCeine in the manufacture of decaCeinated coCee where,unlike organic solvents, it does not result in the forma-tion of an unpleasant and possibly toxic residue.Supercritical fluids are also currently of great interestin industrial processes, as they can be used instead of chlorofluorocarbons (CFCs) and hence avoid the environmental damage that CFCs are known to cause. Because supercritical carbon dioxide is obtained either from the atmosphere or from renewable organic sources (by fermentation), its use does not increase the net load of atmosphericcarbon dioxide.

1.11 The compression factor

A useful quantity for discussing the properties of realgases is the compression factor, Z, which is the ratioof the actual molar volume of a gas to the molar volume of a perfect gas under the same conditions:

(1.20a)

The molar volume of a perfect gas is RT/p (recalleqn 1.3), so we can rewrite this definition as

ZV

V = m

mperfect

(1.20b)

where Vm is the molar volume of the gas we arestudying. For a perfect gas, Z = 1, so deviations of Z from 1 are a measure of how far a real gas departsfrom behaving perfectly.

When Z is measured for real gases, it is found tovary with pressure as shown in Fig. 1.15. At lowpressures, some gases (methane, ethane, and ammo-nia, for instance) have Z < 1. That is, their molarvolumes are smaller than that of a perfect gas, sug-gesting that the molecules are pulled together slightly.We can conclude that for these molecules and theseconditions, the attractive interactions are dominant.The compression factor rises above 1 at high pres-sures whatever the identity of the gas, and for somegases (hydrogen in Fig. 1.15) Z > 1 at all pressures.3

The observation that Z > 1 tells us that the molarvolume of the gas is now greater than that expected

ZV

RT ppVRT

/

= =m m

31

Fig. 1.15 The variation of the compression factor, Z, withpressure for several gases at 0°C. A perfect gas has Z = 1 atall pressures. Of the gases shown, hydrogen shows positivedeviations at all pressures (at this temperature); all the other gases show negative deviations initially but positivedeviations at high pressures. The negative deviations are a result of the attractive interactions between moleculesand the positive deviations are a result of the repulsive interactions.

3 The type of behaviour exhibited depends on the temperature.

Molar volume of the gas

Molar volume of perfect gas

TEOC01 11/4/04 5:08 PM Page 31

for a perfect gas of the same temperature and pressure, so the molecules are pushed apart slightly.This behaviour indicates that the repulsive forcesare dominant. For hydrogen, the attractive inter-actions are so weak that the repulsive interactionsdominate even at low pressures.

1.12 The virial equation of state

We can use the deviation of Z from its ‘perfect’ valueof 1 to construct an empirical (observation-based)equation of state. To do so, we suppose that, for realgases, the relation Z = 1 is only the first term of alengthier expression, and write instead

(1.21)

The coeAcients B, C, . . . , are called virial coeGcients;B is the second virial coeAcient, C, the third, and soon; the unwritten A = 1 is the first.4 They vary fromgas to gas and depend on the temperature. This tech-nique, of taking a limiting expression (in this case, Z = 1, which applies to gases at very large molar volumes) and supposing that it is the first term of amore complicated expression, is quite common inphysical chemistry. The limiting expression is thefirst approximation to the true expression, whateverthat may be, and the additional terms progressivelytake into account the secondary eCects that the limiting expression ignores.

The most important additional term on the rightin eqn 1.21 is the one proportional to B (more pre-cisely, under most conditions C/V2

m � B/Vm and cantherefore be neglected). From the graphs in Fig. 1.15,it follows that, for the temperature to which the data apply, B must be positive for hydrogen (so that Z > 1) but negative for methane, ethane, andammonia (so that Z < 1). However, regardless of thesign of B, the positive term C/V2

m becomes large forhighly compressed gases (when V2

m is very small)and the right-hand side of eqn 1.21 becomes greaterthan 1, just as in the curves for the other gases in Fig. 1.15. The values of the virial coeAcients formany gases are known from measurements of Zover a range of molar volumes and fitting the data to eqn 1.21 by varying the coeAcients until a goodmatch is obtained.

ZB

VC

V . . .= + + +1 2

m m

CHAPTER 1: THE PROPERTIES OF GASES

To convert eqn 1.21 into an equation of state, wecombine it with eqn 1.20b (Z = pVm/RT), which gives

We then multiply both sides by RT/Vm and replaceVm by V/n throughout to get an expression for p interms of the other variables:

(1.22)

Equation 1.22 is the virial equation of state. Whenthe molar volume is very large, the terms nB/V andn2C/V2 are both very small, and only the 1 insidethe parentheses survives. In this limit, the equationof state approaches that of a perfect gas.

1.13 The van der Waals equation of state

Although it is the most reliable equation of state, thevirial equation does not give us much immediate insight into the behaviour of gases and their con-densation to liquids. The van der Waals equation,which was proposed in 1873 by the Dutch physicistJohannes van der Waals, is only an approximateequation of state but it has the advantage of show-ing how the intermolecular interactions contributeto the deviations of a gas from the perfect gas law.We can view the van der Waals equation as anotherexample of taking a soundly based qualitative ideaand building up a mathematical expression that canbe tested quantitatively.

The repulsive interaction between two moleculesimplies that they cannot come closer than a certaindistance. Therefore, instead of being free to travelanywhere in a volume V, the actual volume in whichthe molecules can travel is reduced to an extent pro-portional to the number of molecules present andthe volume they each exclude (Fig. 1.16). We cantherefore model the eCect of the repulsive, volume-excluding forces by changing V in the perfect gasequation to V − nb, where b is the proportionalityconstant between the reduction in volume and the

pnRT

VnBV

n CV

. . .= + + +

12

2

pVRT

BV

CV

m

m m

. . .= + + +1 2

32

4 The word ‘virial’ comes from the Latin word for force, and itreflects the fact that intermolecular forces are now significant.Virial coeAcients are also denoted B2, B3, etc. in place of B, C, etc.

TEOC01 11/4/04 5:08 PM Page 32

REAL GASES

amount of molecules present in the container. Withthis modification, the perfect gas equation of statechanges from p = nRT/V to

This equation of state—it is not yet the full van derWaals equation—should describe a gas in which repulsions are important. Note that when the pres-sure is low, the volume is large compared with thevolume excluded by the molecules (which we writeV � nb). The nb can then be ignored in the denom-inator and the equation reduces to the perfect gasequation of state. It is always a good plan to verifythat an equation reduces to a known form when aplausible physical approximation is made.

The eCect of the attractive interactions betweenmolecules is to reduce the pressure that the gas exerts. We can model the eCect by supposing thatthe attraction experienced by a given molecule isproportional to the concentration, n/V, of moleculesin the container. Because the attractions slow themolecules down,5 the molecules strike the walls lessfrequently and strike it with a weaker impact. Wecan therefore expect the reduction in pressure to be proportional to the square of the molar concen-tration, one factor of n/V reflecting the reduction in frequency of collisions and the other factor the reduction in the strength of their impulse. If the constant of proportionality is written a, we can write

Reduction in pressure = ×

anV

2

pnRT

V nb

=

−

It follows that the equation of state allowing forboth repulsions and attractions is

(1.23a)

This expression is the van der Waals equation ofstate. To show the resemblance of this equation tothe perfect gas equation pV = nRT, eqn 1.23a is some-times rearranged into

(1.23b)

We have built the van der Waals equation byusing physical arguments about the volumes ofmolecules and the eCects of forces between them. It can be derived in other ways, but the presentmethod has the advantage of showing how to derivethe form of an equation out of general ideas. Thederivation also has the advantage of keeping impre-cise the significance of the van der Waals parame-ters, the constants a and b: they are much betterregarded as empirical parameters than as preciselydefined molecular properties. The van der Waals parameters depend on the gas, but are taken as independent of temperature (Table 1.5). It followsfrom the way we have constructed the equation thata (the parameter representing the role of attractions)can be expected to be large when the molecules attract each other strongly, whereas b (the parameterrepresenting the role of repulsions) can be expectedto be large when the molecules are large.

We can judge the reliability of the van der Waalsequation by comparing the isotherms it predicts,which are shown in Fig. 1.17, with the experimentalisotherms already shown in Fig. 1.13. Apart fromthe waves below the critical temperature, they do resemble experimental isotherms quite well. Thewaves, which are called van der Waals loops, are unrealistic because they suggest that under someconditions compression results in a decrease of pressure. The loops are therefore trimmed away and replaced by horizontal lines (Fig. 1.18).6 The

panV

V nb nRT ( ) +

− =2

2

pnRT

V nba

nV

=−

−

2

33

Excludedvolume

2rr

Fig. 1.16 When two molecules, each of radius r and volumevmol = pr 3 approach each other, the centre of one of themcannot penetrate into a sphere of radius 2r and therefore volume 8vmol surrounding the other molecule.

43

5 This slowing does not mean that the gas is cooler close to thewalls: the simple relation between T and mean speed in eqn 1.15is valid only in the absence of intermolecular forces.

6 Theoretical arguments show that the horizontal line shouldtrim equal areas of loop above and below where it lies.

TEOC01 11/4/04 5:08 PM Page 33

van der Waals parameters in Table 1.5 were foundby fitting the calculated curves to experimentalisotherms.

Perfect gas isotherms are obtained from the vander Waals equation at high temperatures and lowpressures. To confirm this remark, we need to note

CHAPTER 1: THE PROPERTIES OF GASES

that when the temperature is high, RT may be solarge that the first term on the right in eqn 1.23agreatly exceeds the second, so the latter may be ignored. Furthermore, at low pressures, the molarvolume is so large that V − nb can be replaced by V.Hence, under these conditions (of high temperatureand low pressure), eqn 1.23a reduces to p = nRT/V,the perfect gas equation.

34

Table 1.5 van der Waals parameters of gases

Substance a/(atm dm6 mol−2) b/(10−2 dm3 mol−1)

Air 1.4 0.039Ammonia, NH3 4.169 3.71Argon, Ar 1.338 3.20Carbon dioxide, CO2 3.610 4.29Ethane, C2H6 5.507 6.51Ethene, C2H4 4.552 5.82Helium, He 0.0341 2.38Hydrogen, H2 0.2420 2.65Nitrogen, N2 1.352 3.87Oxygen, O2 1.364 3.19Xenon, Xe 4.137 5.16

1.5

1.0

0.5

00.1 1 10

p/p

c

V/Vc

1.5

1

0.8

Fig. 1.17 Isotherms calculated by using the van der Waalsequation of state. The axes are labelled with the ‘reducedpressure’, p/pc, and ‘reduced volume’, V/Vc, where pc =a/27b2 and Vc = 3b. The individual isotherms are labelledwith the ‘reduced temperature’, T/Tc, where Tc = 8a/27Rb.The isotherm labelled 1 is the critical isotherm (the isothermat the critical temperature).

1.5

1.0

0.5

00.1 1 10

pp/

c

V V/ c

1.5

1

0.8

Fig. 1.18 The unphysical van der Waals loops are eliminatedby drawing straight lines that divide the loops into areas of equal size. With this procedure, the isotherms strongly resemble the observed isotherms.

TEOC01 11/4/04 5:08 PM Page 34

REAL GASES

1.14 The liquefaction of gases

A gas may be liquefied by cooling it below its boilingpoint at the pressure of the experiment. For ex-ample, chlorine at 1 atm can be liquefied by coolingit to below −34°C in a bath cooled with dry ice (solidcarbon dioxide). For gases with very low boilingpoints (such as oxygen and nitrogen, at −183°C and−186°C, respectively), such a simple technique is notpracticable unless an even colder bath is available.

One alternative and widely used commercial tech-nique makes use of the forces that act betweenmolecules. We saw earlier that the rms speed ofmolecules in a gas is proportional to the square rootof the temperature (eqn 1.15). It follows that redu-cing the rms speed of the molecules is equivalent tocooling the gas. If the speed of the molecules can bereduced to the point that neighbours can captureeach other by their intermolecular attractions, thenthe cooled gas will condense to a liquid.

To slow the gas molecules, we make use of an eCect similar to that seen when a ball is thrown intothe air: as it rises it slows in response to the gravita-tional attraction of the Earth and its kinetic energy isconverted into potential energy. Molecules attracteach other, as we have seen (the attraction is notgravitational, but the eCect is the same), and if wecan cause them to move apart from each other, likea ball rising from a planet, then they should slowdown. It is very easy to move molecules apart fromeach other: we simply allow the gas to expand, whichincreases the average separation of the molecules.To cool a gas, therefore, we allow it to expand with-out allowing any heat to enter from outside. As itdoes so, the molecules move apart to fill the avail-able volume, struggling as they do so against the attraction of their neighbours. Because some kineticenergy must be converted into potential energy toreach greater separations, the molecules travel moreslowly as their separation increases. Therefore, because the average speed of the molecules has beenreduced, the gas is now cooler than before the expan-

sion. This process of cooling a real gas by expansionthrough a narrow opening called a ‘throttle’ is knownas the Joule–Thomson eIect.7 The procedure worksonly for real gases in which the attractive inter-actions are dominant, because the molecules have toclimb apart against the attractive force in order forthem to travel more slowly. For molecules underconditions when repulsions are dominant (cor-responding to Z > 1), the Joule−Thomson eCect results in the gas becoming warmer.

In practice, the gas is allowed to expand severaltimes by recirculating it through a device called aLinde refrigerator (Fig. 1.19). On each successiveexpansion the gas becomes cooler, and as it flowspast the incoming gas, the latter is cooled further.After several successive expansions, the gas becomesso cold that it condenses to a liquid.

35

Heatexchanger

Compressor

Liquid

Coldgas

Throttle

Fig. 1.19 The principle of the Linde refrigerator. The gas is recirculated and cools the gas that is about to undergo expansion through the throttle. The expanding gas cools stillfurther. Eventually, liquefied gas drips from the throttle.

7 The eCect was first observed and analysed by James Joule(whose name is commemorated in the unit of energy) and WilliamThomson (who later became Lord Kelvin).

TEOC01 11/4/04 5:08 PM Page 35

CHAPTER 1: THE PROPERTIES OF GASES36

CHECKLIST OF KEY IDEAS

You should now be familiar with the following concepts:

® 1 An equation of state is an equation relating pressure, volume, temperature, and amount of a substance.

® 2 The perfect gas equation of state, pV = nRT, is alimiting law applicable as p → 0.

® 3 The perfect gas equation of state is based onBoyle’s law (p ∝ 1/V ), Charles’s law (V ∝ T ), andAvogadro’s principle (V ∝ n).

® 4 Dalton’s law states that the total pressure of amixture of perfect gases is the sum of the pres-sures that each gas would exert if it were alone inthe container at the same temperature.

® 5 The partial pressure of any gas is defined as pJ = xJ × p, where xJ is its mole fraction in a mixture and p is the total pressure.

® 6 The kinetic model of gases expresses the prop-erties of a perfect gas in terms of a collection ofmass points in ceaseless random motion.

® 7 The mean speed and root-mean-square (rms)speed of molecules is proportional to the squareroot of the (absolute) temperature and inverselyproportional to the square root of the molarmass.

® 8 The properties of the Maxwell distribution ofspeeds are summarized in Figs 1.8 and 1.9.

® 9 Diffusion is the spreading of one substancethrough another; effusion is the escape of a gasthrough a small hole.

® 10 Graham’s law states that the rate of effusion isinversely proportional to the square root of themolar mass.

® 11 The collision frequency, z, and mean free path, l,of molecules in a gas are related by c = lz.

® 12 In real gases, molecular interactions affect theequation of state; the true equation of state is expressed in terms of virial coefficients B, C, . . . :p = (nRT/V )(1 + nB/V + n2C/V 2 + . . . ).

® 13 The van der Waals equation of state is an approximation to the true equation of state inwhich attractions are represented by a para-meter a and repulsions are represented by a parameter b: p = nRT/(V − nb) − a(n/V )2.

® 14 The Joule–Thomson effect is the cooling of gasthat occurs when it expands through a throttlewithout the influx of heat.

FURTHER INFORMATION 1.1

Kinetic molecular theory

One of the essential skills of a physical chemist is the ability to turn simple, qualitative ideas into rigid, testable,quantitative theories. The kinetic model of gases is an excel-lent example of this technique, as it takes the concepts set out in the text and turns them into precise expressions.As usual in model building, there are a number of steps,but each one is motivated by a clear appreciation of theunderlying physical picture, in this case a swarm of masspoints in ceaseless random motion. The key quantitativeingredients we need are the equations of classical mechanics.

So we begin with a brief review of velocity, momentum,and Newton’s second law of motion.

The velocity, v, is a vector, a quantity with both magni-tude and direction. The magnitude of the velocity vector is the speed, v, given by v = (vx

2 + vy2 + vz

2)1/2, where vx, vy,and vz are the components of the vector along the x-, y-,and z-axes, respectively (Fig. 1.20). The magnitude of eachcomponent, its value without a sign, is denoted | . . . |. Forexample, |vx | means the magnitude of vx. The linear momentum, p, of a particle of mass m is the vector p = mvwith magnitude p = mv. Newton’s second law of motionstates that the force acting on a particle is equal to the rate

TEOC01 11/4/04 5:08 PM Page 36

FURTHER INFORMATION 1.1

of change of the momentum, the change of momentum divided by the interval during which it occurs.

Now we begin the derivation of eqn 1.9 by consideringthe arrangement in Fig. 1.21. When a particle of mass mthat is travelling with a component of velocity vx parallelto the x-axis (vx > 0 corresponding to motion to the rightand vx < 0 to motion to the left) collides with the wall onthe right and is reflected, its linear momentum changesfrom +m |vx | before the collision to −m |vx | after the col-lision (when it is travelling in the opposite direction at the same speed). The x-component of the momentumtherefore changes by 2m |vx | on each collision (the y- andz-components are unchanged). Many molecules collide

with the wall in an interval ∆t, and the total change of momentum is the product of the change in momentum of each molecule multiplied by the number of moleculesthat reach the wall during the interval.

Next, we need to calculate that number. Because amolecule with velocity component vx can travel a distance|vx |∆t along the x-axis in an interval ∆t, all the moleculeswithin a distance |vx |∆t of the wall will strike it if they aretravelling towards it. It follows that if the wall has area A,then all the particles in a volume A × |vx |∆t will reach the wall (if they are travelling towards it). The numberdensity, the number of particles divided by the total volume,is nNA/V (where n is the total amount of molecules in thecontainer of volume V and NA is Avogadro’s constant), so the number of molecules in the volume A|vx |∆t is(nNA/V) × A|vx |∆t. At any instant, half the particles are moving to the right and half are moving to the left.Therefore, the average number of collisions with the wallduring the interval ∆t is nNAA|vx |∆t /V.

The total momentum change in the interval ∆t is theproduct of the number we have just calculated and thechange 2m |vx | :

where M = mNA. Next, to find the force, we calculate therate of change of momentum:

It follows that the pressure, the force divided by the area, is

Not all the molecules travel with the same velocity, so thedetected pressure, p, is the average (denoted ⟨ . . . ⟩) of thequantity just calculated:

To write an expression of the pressure in terms of the rms speed, c, we begin by writing the speed of a singlemolecule, v, as v2 = vx

2 + vy2 + vz

2. Because the rms speed, c, is defined as c = ⟨v2⟩1/ 2 (eqn 1.10), it follows that

c2 = ⟨v2⟩ = ⟨vx2⟩ + ⟨vy

2⟩ + ⟨vz2⟩

However, because the molecules are moving randomly, all three averages are the same. It follows that c2 = 3⟨vx

2⟩.Equation 1.9 follows immediately by substituting ⟨vx

2⟩ =c2 into p = nM⟨vx

2⟩ /V.13

pnM v

Vx =

⟨ ⟩2

Pressure =nMv

Vx2

ForceChange of momentum

Time interval = =

nMAv

Vx2

Momentum change A

A

= ×

= =

nN A v t

Vm v

nmAN v t

V

nMAv t

V

xx

x x

| || |

∆

∆ ∆

22

2 2

12

37

vy

vz

vx

v

Fig. 1.20 A vector v and its three components on a set ofperpendicular axes.

x

| |v tx D

Volume = | |v tAx D

Area A

Will

Won't

Fig. 1.21 The model used for calculating the pressure of aperfect gas according to the kinetic molecular theory. Here,for clarity, we show only the x-component of the velocity(the other two components are not changed when themolecule collides with the wall). All molecules within theshaded area will reach the wall in an interval ∆t provided theyare moving towards it.

TEOC01 11/4/04 5:08 PM Page 37

CHAPTER 1: THE PROPERTIES OF GASES38

DISCUSSION QUESTIONS

1.1 Explain how the experiments of Boyle, Charles, andAvogadro led to the formulation of the perfect gas equationof state.

1.2 Explain the term ‘partial pressure’ and why Dalton’s lawis a limiting law.

1.3 Provide a molecular interpretation for the variation of therates of diffusion and effusion of gases with temperature.

1.4 Describe the formulation of the van der Waals equationstate.

EXERCISES

Treat all gases as perfect unless instructed otherwise.The symbol ‡ indicates that calculus is required.

1.5 What pressure is exerted by a sample of nitrogen gas ofmass 2.045 g in a container of volume 2.00 dm3 at 21°C?

1.6 A sample of neon of mass 255 mg occupies 3.00 dm3 at122 K. What pressure does it exert?

1.7 Much to everyone’s surprise, nitrogen monoxide (NO) hasbeen found to act as a neurotransmitter. To prepare to studyits effect, a sample was collected in a container of volume250.0 cm3. At 19.5°C its pressure is found to be 24.5 kPa.What amount (in moles) of NO has been collected?

1.8 A domestic water-carbonating kit uses steel cylinders of carbon dioxide of volume 250 cm3. They weigh 1.04 kgwhen full and 0.74 kg when empty. What is the pressure ofgas in the cylinder at 20°C?

1.9 The effect of high pressure on organisms, including humans, is studied to gain information about deep-sea diving and anaesthesia. A sample of air occupies 1.00 dm3

at 25°C and 1.00 atm. What pressure is needed to compressit to 100 cm3 at this temperature?

1.10 You are warned not to dispose of pressurized cans bythrowing them on to a fire. The gas in an aerosol containerexerts a pressure of 125 kPa at 18°C. The container isthrown on a fire, and its temperature rises to 700°C. What is the pressure at this temperature?

1.11 Until we find an economical way of extracting oxygenfrom sea water or lunar rocks, we have to carry it with us toinhospitable places, and do so in compressed form in tanks.A sample of oxygen at 101 kPa is compressed at constanttemperature from 7.20 dm3 to 4.21 dm3. Calculate the finalpressure of the gas.

1.12 To what temperature must a sample of helium gas becooled from 22.2°C to reduce its volume from 1.00 dm3 to100 cm3?

1.13 Hot-air balloons gain their lift from the lowering of dens-ity of air that occurs when the air in the envelope is heated.To what temperature should you heat a sample of air, initiallyat 340 K, to increase its volume by 14 per cent?

1.14 At sea level, where the pressure was 104 kPa and thetemperature 21.1°C, a certain mass of air occupied 2.0 m3.To what volume will the region expand when it has risen to an altitude where the pressure and temperature are (a) 52 kPa, −5.0°C, (b) 880 Pa, −52.0°C?

1.15 A diving bell has an air space of 3.0 m3 when on thedeck of a boat. What is the volume of the air space when thebell has been lowered to a depth of 50 m? Take the meandensity of sea water to be 1.025 g cm−3 and assume that thetemperature is the same as on the surface.

1.16 A meteorological balloon had a radius of 1.0 m when released at sea level at 20°C and expanded to a radius of 3.0 m when it had risen to its maximum altitude where thetemperature was −20°C. What is the pressure inside the balloon at that altitude?

1.17 A gas mixture being used to simulate the atmosphereof another planet consists of 320 mg of methane, 175 mg of argon, and 225 mg of nitrogen. The partial pressure of nitrogen at 300 K is 15.2 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.

1.18 The vapour pressure of water at blood temperature is47 Torr. What is the partial pressure of dry air in our lungswhen the total pressure is 760 Torr?

1.19 A determination of the density of a gas or vapour canprovide a quick estimate of its molar mass even though forpractical work mass spectrometry is far more precise. Thedensity of a gaseous compound was found to be 1.23 g dm−3

at 330 K and 25.5 kPa. What is the molar mass of the compound?

1.20 In an experiment to measure the molar mass of a gas,250 cm3 of the gas was confined in a glass vessel. The

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EXERCISES

pressure was 152 Torr at 298 K and the mass of the gas was 33.5 mg. What is the molar mass of the gas?

1.21 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and1.0 mol N2 at 273.15 K. Calculate (a) their partial pressuresand (b) the total pressure.

1.22 The composition of planetary atmospheres is determinedin part by the speeds of the molecules of the constituentgases, because the faster-moving molecules can reach escape velocity and leave the planet. Calculate the meanspeed of (a) He atoms, (b) CH4 molecules at (i) 77 K, (ii) 298 K,(iii) 1000 K.

1.23 ‡Use the Maxwell distribution of speeds to confirm that the mean speed of molecules of molar mass M at atemperature T is equal to (8RT/pM )1/ 2. (Hint. You will needan integral of the form ∫ 0

∞ x 3 e−ax 2dx = n!/2a2.)

1.24 ‡Use the Maxwell distribution of speeds to confirm that the rms speed of molecules of molar mass M at a temperature T is equal to (3RT/M )1/2 and hence confirm eqn 1.13. (Hint. You will need an integral of the form

∫ 0∞x 4 e−ax 2

dx = (3/8a2)(p/a)1/2.)

1.25 ‡Use the Maxwell distribution of speeds to find an expression for the most probable speed of molecules ofmolar mass M at a temperature T. (Hint. Look for a max-imum in the Maxwell distribution (the maximum occurs asdF/ds = 0).)

1.26 ‡Use the Maxwell distribution of speeds to estimatethe fraction of N2 molecules at 500 K that have speeds in therange 290 to 300 m s−1.

1.27 At what pressure does the mean free path of argon at 25°C become comparable to the diameter of a spher-ical vessel of volume 1.0 dm3 that contains it? Take s = 0.36 nm2.

1.28 At what pressure does the mean free path of argon at25°C become comparable to 10 times the diameter of theatoms themselves? Take s = 0.36 nm2.

1.29 When we are studying the photochemical processesthat can occur in the upper atmosphere, we need to know how often atoms and molecules collide. At an alti-tude of 20 km the temperature is 217 K and the pressure0.050 atm. What is the mean free path of N2 molecules?Take s = 0.43 nm2.

1.30 How many collisions does a single Ar atom make in 1.0 swhen the temperature is 25°C and the pressure is (a) 10 bar,(b) 100 kPa, (c) 1.0 Pa?

1.31 Calculate the total number of collisions per second in 1.0 dm3 of argon under the same conditions as in Exercise 1.30.

1.32 How many collisions per second does an N2 moleculemake at an altitude of 20 km? (See Exercise 1.29 for data.)

1.33 The spread of pollutants through the atmosphere isgoverned partly by the effects of winds but also by the natural tendency of molecules to diffuse. The latter dependson how far a molecule can travel before colliding with another molecule. Calculate the mean free path of diatomicmolecules in air using s = 0.43 nm2 at 25°C and (a) 10 bar, (b) 103 kPa, (c) 1.0 Pa.

1.34 How does the mean free path in a sample of a gas varywith temperature in a constant-volume container?

1.35 Calculate the pressure exerted by 1.0 mol C2H6 behav-ing as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 1000 K in 100 cm3. Use the data in Table 1.5.

1.36 How reliable is the perfect gas law in comparison withthe van der Waals equation? Calculate the difference inpressure of 10.00 g of carbon dioxide confined to a containerof volume 100 cm3 at 25.0°C between treating it as a perfectgas and a van der Waals gas.

1.37 Express the van der Waals equation of state as a virialexpansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. (Hint. The expansion you will need is (1 − x )−1 = 1 + x + x2 + . . . . Seriesexpansions are discussed in Appendix 2.)

1.38 Measurements on argon gave B = −21.7 cm3 mol−1 andC = 1200 cm6 mol−2 for the virial coefficients at 273 K. Whatare the values of a and b in the corresponding van der Waalsequation of state?

1.39 Show that there is a temperature at which the secondvirial coefficient, B, is zero for a van der Waals gas, and cal-culate its value for carbon dioxide. (Hint. Use the expressionfor B derived in Exercise 1.37.)

1.40 ‡The critical point of a van der Waals gas occurs wherethe isotherm has a flat inflexion, which is where dp/dVm = 0(zero slope) and d2p/dV 2

m = 0 (zero curvature). (a) Evaluatethese two expressions using eqn 1.23b, and find expressionsfor the critical constants in terms of the van der Waals parameters. (b) Show that the value of the compression factor at the critical point is .3

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