1The properties of gases

Although gases are simple, both to describe and in terms of their internal structure, they are of immense importance. We spend our whole lives surrounded by gas in the form of air and the local variation in its properties is what we call the ‘weather’. To under-stand the atmospheres of this and other planets we need to understand gases. As we breathe, we pump gas in and out of our lungs, where it changes com-position and temperature. Many industrial processes involve gases, and both the outcome of the reaction and the design of the reaction vessels depend on knowledge of their properties.

Equations of state

We can specify the state of any sample of substance by giving the values of the following properties (all of which are defi ned in the Foundations section):

V, the volume of the samplep, the pressure of the sampleT, the temperature of the samplen, the amount of substance in the sample

However, an astonishing experimental fact is that these four quantities are not independent of one another. For instance, we cannot arbitrarily choose to have a sample of 0.555 mol H2O in a volume of 10 dm3 at 100 kPa and 500 K: it is found experimen-tally that that state simply does not exist. If we select the amount, the volume, and the temperature, then we fi nd that we have to accept a particular pressure (in this case, close to 230 kPa). This experimental generalization is summarized by saying the substance obeys an equation of state, an equation of the form

p = f(n,V,T) Equation of state (1.1)

This expression tells us that the pressure depends on (‘is a function of’) the amount, volume, and

Equations of state 18

1.1 The perfect gas equation of state 19

1.2 Using the perfect gas law 22

1.3 Mixtures of gases: partial pressures 23

The molecular model of gases 26

1.4 The pressure of a gas according to the kinetic model 27

1.5 The average speed of gas molecules 27

1.6 The Maxwell distribution of speeds 28

1.7 Diffusion and effusion 30

1.8 Molecular collisions 32

Real gases 33

1.9 Molecular interactions 33

1.10 The critical temperature 34

1.11 The compression factor 36

1.12 The virial equation of state 36

1.13 The van der Waals equation of state 37

1.14 The liquefaction of gases 40

FURTHER INFORMATION 1.1 41

CHECKLIST OF KEY CONCEPTS 42

ROAD MAP OF KEY EQUATIONS 43

QUESTIONS AND EXERCISES 43

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EQUATIONS OF STATE 19

temperature and that 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 an explicit expression for the pressure in terms of the other variables. However, certain equations of state are known. In particular, the equation of state of a low-pressure gas is known, and proves to be very simple 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 starting point for problems in chemical engineering, and even 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 was among the fi rst results to be established in physical chemistry. The original experiments were carried out by Robert Boyle in the seventeenth century and there was a revival of interest later in the century when people began to fl y in balloons. This techno-logical progress demanded more knowledge about the response of gases to changes of pressure and temperature and, like technological advances in other fi elds today, that interest stimulated a lot of experiments.

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

pV = nRT Perfect gas equation of state (1.2a)

This equation has the form of eqn 1.1 when re -arranged into

pnRTV

= (1.2b)

The gas constant R is currently an experimentally determined quantity obtained by evaluating R = pV/nRT as the pressure is allowed to approach zero. For calculations, we shall normally use the approximate value 8.3145 J K−1 mol−1. Values of R in a variety of units are given in Table 1.1.

The perfect gas equation of state—more briefl y, the ‘perfect gas law’—is so called because it is an idealization of the equations of state that gases actu-ally obey. Specifi cally, it is found that all gases obey the equation ever more closely as the pressure is reduced towards zero. That is, eqn 1.2 is an example of a limiting law, a law that becomes increasingly valid in a certain limit, in this case as the pressure is reduced to zero. The law is obeyed exactly in the limit of zero pressure.

A hypothetical substance that obeys eqn 1.2 at all pressures, not just at very low pressures, is called a perfect gas. From what has just been said, an actual gas, which is termed a real gas, behaves more and more like a perfect gas as its pressure is reduced towards zero. In practice, normal atmospheric pres-sure at sea level (p ≈ 100 kPa) is already low enough for most real gases to behave almost perfectly, and unless stated otherwise we shall always assume in this text that the gases we encounter behave like a perfect gas. The reason why a real gas behaves differently from a perfect gas can be traced to the attractions and repulsions that exist between actual molecules and which are absent in a perfect gas (Chapter 15).

A note on good practice A perfect gas is widely called an ‘ideal gas’ and the perfect gas equation of state is commonly called ‘the ideal gas equation’. We use ‘perfect gas’ to imply the absence of molecular interactions; we use ‘ideal’; in Chapter 6 to denote mixtures in which all the molecular interactions are the same but not necessarily zero.

The perfect gas law is based on and summarizes three sets of experimental observations. One is Boyle’s law:

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

Mathematically:

pV

∝1

At constant temperature Boyle’s law (1.3)

We can easily verify that eqn 1.2 is consistent with Boyle’s law: by treating n and T as constants, the perfect gas law becomes pV = constant, and hence p ∝ 1/V. Boyle’s law implies that if we compress (reduce the volume of) a fi xed amount of gas at con-stant temperature into half its original volume, then its pressure will double. Figure 1.1 shows the graph obtained by plotting experimental values of p against

Table 1.1

The gas constant in various units

R = 8.314 47 J 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

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CHAPTER 1 THE PROPERTIES OF GASES20

V for a fi xed amount of gas at different temperatures and the curves predicted by Boyle’s law. Each curve is a hyperbola (see The chemist’s toolkit 1.1 for a discussion of graphs) and called an isotherm because it depicts the variation of a property (in this case, the

pressure) at a single constant temperature. It is hard, from this graph, to judge how well Boyle’s law is obeyed. However, when we plot p against 1/V, we get straight lines, just as we would expect from Boyle’s law (Fig. 1.2).

Increasingtemperature

Volume, V

Pre

ssu

re, p

Fig. 1.1 The volume of a gas decreases as the pressure on it is increased. For a sample that obeys Boyle’s law and that is kept at constant temperature, the graph show-ing the dependence is a hyperbola, as shown here. Each curve corresponds to a single temperature, and hence is an isotherm. The isotherms are hyperbolas (The chemist’s toolkit 1.1).

1/Volume, 1/V

Pre

ssu

re, p

Observed

Perfect gas

Fig. 1.2 A good test of Boyle’s law is to plot the pressure against 1/V (at constant temperature), when a straight line should be obtained. This diagram shows that the observed pressures approach a straight line (the blue line) as the vol-ume is increased and the pressure reduced. A perfect gas would follow the straight line at all pressures; real gases obey Boyle’s law in the limit of low pressures.

The graphs of xy = a, with a a constant, or y = a/x are hyper-bolas. Examples are the isotherms in Fig. 1.1. The graph of y = ax2 is a parabola; it plays a role in the discussion of molecular vibrations. These two ‘conic sections’ are illustrated in Sketch 1.1.

The graph of y = mx + b is a straight line of slope m and passing through y = b at x = 0, the ‘intercept with the y-axis’ (Sketch 1.2). In the special case b = 0, y = mx and the straight line passes through the origin y = 0 when x = 0. In that case we say that ‘y is proportional to x’ (y ∝ x) the constant of proportionality being m. The intercept with the x-axis occurs when x = −b/m (because then y = m(−b/m) + b = 0).

The horizontal and vertical axes of graphs are labelled with pure numbers. Thus, if the units of x and y are denoted ‘units of x’ and ‘units of y’, the plot should be of y/(units of y) against x/(units of x). It follows that the slope of a graph is dimensionless. For a graph of the form y/(units of y) = b + mx/(units of x), both m and b are pure numbers. To interpret them in terms of physical quantities we multiply both sides of this expression by ‘units of y’, and obtain

y b ybb )yi f

Interpretationof the interceptee

I

units ofunits of

� ������� �� �� � ����+ ×

yx

nterpretationnnof the slope� ������� �� �� � ����

× x

The chemist’s toolkit 1.1 Graphs

x

y

0

y = mx + b

Intercept, bSlope, m

x = –b/m

Sketch 1.2 The characteristics of a straight-line graph of y against x with y = mx + b.

Sketch 1.1 The characteristics of (a) hyperbolas and (b) parabolas.

y

y

x xa1/2

a1/2

01

a

a

1–1

(a) (b)

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EQUATIONS OF STATE 21

A note on good practice In general, it is best to plot data as straight lines to test the validity of a theory because it is diffi cult to distinguish between subtly different shapes of a curve by eye. Instead, the mathematical function that repre-sents the theory is rearranged so that the resulting graph has the form of a straight line, y = mx + b.

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

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

The equation for this linear dependence is

V = A + Bθ Constant pressure Charles’s law (1.4a)

where θ (theta) is the temperature on the Celsius scale and A and B are constants that depend on the amount of gas and the pressure. Figure 1.3 shows typical plots of volume against temperature for a series of samples of gases at different pressures and confi rms that (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 the same very low temperature (−273.15 °C, in fact), regardless of the identity of the gas. Because a vol-ume cannot be negative, this common temperature must represent the absolute zero of temperature, a temperature below which it is impossible to cool an object. Indeed, as explained in Foundations 0.10, the ‘thermodynamic’ scale ascribes the value T = 0 to this absolute zero of temperature. In terms of the thermo-dynamic temperature, therefore, Charles’s law takes the simpler form

V ∝ T Constant pressure Charles’s law (1.4b)

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

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

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

That is, 1.00 dm3 of oxygen at 100 kPa and 300 K contains 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 if we double the number of molecules, but keep the temperature and pressure constant, then the volume of the sample will double. We can therefore write

V ∝ n Constant temperature and pressure Avogadro’s

principle (1.5)

This result follows easily from eqn 1.2 if we treat p and T as constants. Avogadro’s suggestion is a prin-ciple rather than a law (a direct summary of experi-ence), because it is based on a model of a gas, in this case as a collection of molecules. Even though there is no longer any doubt that molecules exist, this relation remains a principle rather than a law.

The molar volume, Vm, the volume a substance occupies per mole of molecules, was introduced in Foundations 0.4:

VVnm = Defi nition Molar volume (1.6a)

For a perfect gas, n = pV/RT, so in this case

VV

pV RT p RTm / /= = =

=n pV RT V a x/ Cancel 1/( / )� � 1==x a/� RT

p

Perfect gas Molar volume (1.6b)

To interpret this expression, note that

• Because eqn 1.6b makes no reference to the identity of the gas, provided the gas is behaving perfectly, its molar volume is the same whatever its chemical identity at the same temperature and pressure.

The data in Table 1.2 show that this conclusion is approximately true for most gases under normal conditions (normal atmospheric pressure of about 100 kPa and room temperature).

Temperature, /°C

Volu

me,

V

Observed

Perfect gas

Increasingpressure

–273.15θ

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

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CHAPTER 1 THE PROPERTIES OF GASES22

Chemists have found it convenient to report much of their data on gases at a particular set of ‘standard’ conditions.1 By standard ambient temperature and pressure (SATP) they mean a temperature of 25 °C (more precisely, 298.15 K) and a pressure of exactly 1 bar (100 kPa). The standard pressure is denoted pa, so pa = 1 bar exactly. The molar volume of a perfect gas at SATP is 24.79 dm3 mol−1, as can be verifi ed by substituting the values of the temperature and pressure into eqn 1.6b. This value implies that at SATP, 1 mol of perfect gas molecules occupies about 25 dm3 (a cube of about 30 cm on a side). An earlier set of standard conditions, which is still encountered, 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.

1.2 Using the perfect gas law

Here we review two elementary applications of the perfect gas equation of state:

• The prediction of the pressure of a gas given its temperature, its chemical amount, and the vol-ume it occupies.

• The prediction of the change in pressure arising from changes in the conditions.

Calculations like these underlie more advanced considerations, including the way that meteorologists

1 Be very careful not to confuse these ‘standard conditions’ for gases with the ‘standard states’ of substances, which are introduced in Chapter 3.

understand the changes in the atmosphere that we call the weather (see Impact 1.1 following Section 1.3).

Table 1.2

The molar volumes of gases at standard ambient temperature and pressure (SATP: 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.

Example 1.1

Predicting the pressure of a sample of gas

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

Strategy For this calculation, arrange eqn 1.2a (pV = nRT ) into a form that gives the unknown (the pressure, p) in terms of the information supplied, which in this case is eqn 1.2b (p = nRT/V ). 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 Kelvin scale (by adding 273.15 to the Celsius temperature.

Solution The amount of N2 molecules (of molar mass 28.02 g mol−1) present is

nm

M( )

( ).

...

NN

gg mol2

21

1 2528 02

1 2528 02

= = =− mmol

The temperature of the sample is

T/K = 20 + 273.15, so T = (20 + 273.15) K

Therefore, from p = nRT/V,

p

n

=×( . . )1 25 28 02/ mol (8.314

� �������� ��������55 J K mol− − × +1 1 20 2) (

R� ���������� ����������

773 152 50 10 4 3

. )( . )

Km

T

V

� ������� �������

� ��× −

������� ��������

=× × +

× −( . . ) ( . ) ( . )

.1 25 28 02 8 3145 20 273 15

2 50 10 4

/ JJm3

.= × =

− = =1 1 1 103 3

4 35 10 4355

J m Pa kPa Pa

Pa kPa� �

Note how the units cancel like ordinary numbers. Had we required the pressure in different units we could, instead, have used Table 1.1 to select the appropriate value of R to match the information required.

A note on good practice It is best to postpone a numer-ical calculation to the last possible stage, and carry it out in a single step. This procedure avoids rounding errors.

Self-test 1.1

Calculate the pressure exerted by 1.22 g of carbon dioxide confi ned to a fl ask of volume 500 dm3 (ie, 5.00 × 102 dm3) at 37 °C.

Answer: 143 Pa

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EQUATIONS OF STATE 23

In some cases, we are given the pressure under one set of conditions and are asked to predict the pres-sure of the same sample under a different set of con-ditions. In that case, we use the perfect gas law as follows. Suppose the initial pressure is p1, the initial temperature is T1, and the initial volume is V1. Then by dividing both sides of eqn 1.2a by the tempera-ture, which gives pV/T = nR in general, we can write in this instance

p VT

nR1 1

1

=

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 1.2a tells us that

p VT

nR2 2

2

=

The nR on the right of these two equations is the same in each case, because R is a constant and the amount of gas molecules has not changed. It follows that we can combine the two equations into a single equation:

p VT

p VT

1 1

1

2 2

2

= Constant amount of gas

Combined gas equation (1.7)

This equation can be rearranged to calculate any one unknown (such as p2 for instance) in terms of the other variables.

Brief Illustration 1.1 The combined gas equation

Consider a sample of gas with an initial volume of 15 cm3 that has been heated from 25 °C to 1000 °C and had its pressure increased from 10.0 kPa to 150.0 kPa. We may rearrange eqn 1.7 and use it to calculate the fi nal volume

VpVT

Tp

pp

TT

V21 1

1

2

1

1

2

2

11= × = × ×

p p

10 0

1 2

= .

/

kPa� ����� ���� � ����

150 01000 273 15

2 1

.( . )

/

kPaK

×+

T T������ ��������� � ���� �

( . )25 273 1515 3

1

+×

Kcm

V����

= 4 3 3. cm

1.3 Mixtures of gases: partial pressures

Scientists are often concerned with mixtures of gases, such as when they are considering the properties of the 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. They need to be able to assess the contri-bution that each component of a gaseous mixture makes to the total pressure.

In the early nineteenth century, John Dalton carried out a series of experiments that led him to formulate what 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 same temperature:

p = pA + pB + . . . Perfect gases Dalton’s law (1.8)

In this expression, pJ is the pressure that the gas J would exert if it were alone in the container at the same temperature. Dalton’s law is strictly valid only for mixtures of perfect gases (or for real gases at such low pressures that they are behaving perfectly), but it can be treated as valid under most conditions we encounter.

Brief illustration 1.2 Dalton’s law

Suppose we were interested in the composition of inhaled and exhaled air, and we knew that a certain mass of carbon dioxide exerts a pressure of 5 kPa when present alone in a container, and that a certain mass of oxygen exerts 20 kPa when present alone in the same container at the same temperature. Then, when both gases are present in the container, the carbon dioxide in the mixture contributes 5 kPa to the total pressure and oxygen contributes 20 kPa; accord-ing to Dalton’s law, the total pressure of the mixture is the sum of these two pressures, or 25 kPa (Fig. 1.4).

Self-test 1.2

Calculate the fi nal pressure of a gas that is com-pressed from a volume of 20.0 dm3 to 10.0 dm3 and cooled from 100 °C to 25 °C if the initial pressure is 1.00 bar.

Answer: 1.60 bar

Self-test 1.3

Suppose some nitrogen, enough to generate a pres-sure of 10 kPa when there alone, is introduced into the mixture. What is the new total pressure?

Answer: 35 kPa

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

pJ = xJp Defi nition Partial pressure (1.9)

where xJ is the mole fraction of the gas J in the mixture. The mole fraction of J is the amount of J

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CHAPTER 1 THE PROPERTIES OF GASES24

molecules expressed as a fraction of the total amount of molecules in the mixture. In a mixture that con-sists of nA A molecules, nB B molecules, and so on (where the nJ are amounts in moles), the mole frac-tion of J (where J = A, B, . . .) is

xn

nJJ= n = nA + nB + . . . Defi nition Mole

fraction (1.10a)

Mole fractions are unitless because the unit mole in numerator and denominator cancel. For a binary mixture, one that consists of two species, this general expression becomes

xn

n nx

nn n

x xAA

A BB

B

A BA B=

+=

++ = 1

(1.10b)

When only A is present, xA = 1 and xB = 0. When only B is present, xB = 1 and xA = 0. When both are present in the same amounts, xA = 1/2 and xB = 1/2 (Fig. 1.5).

When it is typographically more convenient, we write amounts as n(J), mole fractions as x(J), and partial pressures as p(J).

Solution From eqn 0.2, the amounts of N2, O2, and Ar are

nm

MNN

N

gg mol

mol2

2

2

75 528 0

2 701= = =−.

..

nm

MOO

O

gg mol

mol2

2

2

23 232 0

0 7251= = =−.

..

nmMAr

Ar

Ar

gg mol

mol= = =−1 3

39 90 0331

..

.

The mole fraction of N2 molecules is, from eqn 1.10b,

xn

n n n

n

NN

N O Ar

mol2

2

2 2

2 70=

+ +=

� ������� �������

.22 70 0 725 0 033. . .mol mol mol+ +

= 0.780

By repeating the calculation for the other constituents, we fi nd that the mole factions of O2 and Ar in dry air are 0.210 and 0.009, respectively.

kPa kPa kPa

520

25

A B A+B

pA pB pA + pB

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

A

B

xA = 0.708

xB = 0.292

xA = 0.375

xB = 0.625

xA = 0.229

xB = 0.771

Fig. 1.5 A representation of the meaning of mole fraction. In each case, a small square represents one molecule of A (yellow squares) or B (green squares). There are 48 squares in each sample.

Example 1.2

Calculating mole fractions

A mass of 100.0 g of dry air consists of 75.5 g of N2, 23.2 g of O2, and 1.3 g of Ar. Express the composition of dry air in terms of mole fractions.

Strategy Begin by converting each mass to an amount in moles. Determine the mole fractions using eqn 1.10b as a ratio of the amount of each constituent to the total amount of substance.

Self-test 1.4

The mole fraction of NH3 in a sample of 10.0 mol of gas is 0.285. What mass of NH3 is present in the sample?

Answer: 79.8 g

For a mixture of perfect gases, we can identify the partial pressure of J with the contribution that J makes to the total pressure. Thus, if we introduce p = nRT/V into eqn 1.9, we get

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EQUATIONS OF STATE 25

pn

n

nRT

V

n RT

VJJ J= × =

p�

The quantity nJRT/V is the pressure that an amount nJ of perfect gas J would exert in the otherwise empty container. That is, for a perfect gas, the partial pressure of a gas in a mixture is the pressure it would exert if it were alone in the container (at the same temperature). If the gases are real, then their partial pressures are still given by eqn 1.9, for that defi nition applies to all gases, and the sum of these partial pressures is the total pressure (because the sum of all the mole fractions is 1); however, we can no longer say that each partial pressure is the pres-sure that the gas would exert when it is alone in the container.

Brief illustration 1.3 Partial pressures

From Example 1.2, we have x(N2) = 0.780, x(O2) = 0.210, and x(Ar) = 0.009 for dry air at sea level. It then follows from eqn 1.9 that when the total atmospheric pressure is 100 kPa, the partial pressure of nitrogen is

p(N2) = x(N2)p = 0.780 × (100 kPa) = 78.0 kPa

Similarly, for the other two components we fi nd p(O2) = 21.0 kPa and p(Ar) = 0.9 kPa. Provided the gases are perfect, these partial pressures are the pressures that each gas would exert if it were separated from the mixture and put in the same container on its own.

One of the most variable constituents of air is water vapour, and the humidity it causes. The presence of water vapour results in a lower density of air at a given temperature and pressure, as we may conclude from Avogadro’s prin-ciple. The numbers of molecules in 1 m3 of moist air and dry air are the same (at the same temperature and pressure), but the mass of an H2O molecule is less than that of all the other major constituents of air (the molar mass of H2O is 18 g mol−1, the average molar mass of air molecules 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 sea level, falling to −57 °C at the bottom of the tropopause at 11 km. This variation is much less pronounced when expressed on the Kelvin scale, ranging from 288 K to 216 K, an average of 268 K. If we suppose that the temperature has its average value all the way up to the edge of the troposphere, then the pressure varies with altitude, h, according to the barometric formula:

p = p0e−h/H

where p0 is the pressure at sea level and H is a constant approximately equal to 8 km. More specifi cally, H = RT/Mg, where M is the average molar mass of air, T is the tempera-ture, and g is the acceleration of free fall. The barometric formula fi ts the observed pressure distribution quite well even for regions well above the troposphere (Fig. 1.6). It implies that the pressure of the air and its density fall to half their sea-level value at h = H ln 2, or 6 km.

Local variations of pressure, temperature, and composition in the troposphere are manifest as ‘weather’. A small region of air is termed a parcel. First, we note that a parcel of warm

Self-test 1.5

The partial pressure of oxygen in air plays an important role in the aeration of water, to enable aquatic life to thrive, and in the absorption of oxygen by blood in our lungs (see Section 6.4). Calculate the partial pressures of a sample of gas consisting of 2.50 g of oxygen and 6.43 g of carbon dioxide with a total pressure of 88 kPa.

Answer: 31 kPa, 57 kPa

Impact on environmental science 1.1

The gas laws and the weather

The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases with the composition sum-marized in Table 1.3. The composition is maintained moder-ately constant by diffusion and convection (winds, particularly the local turbulence called eddies) but the pressure and temperature vary with altitude and with the local conditions, particularly in the troposphere (the ‘sphere of change’), the layer extending up to about 11 km.

Table 1.3

The composition of the Earth’s atmosphere

Substance Percentage

By 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: Winter:

7.0 × 10−6

2.0 × 10−6 1.2 × 10−5

3.3 × 10−6

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CHAPTER 1 THE PROPERTIES OF GASES26

air is less dense than the same parcel of cool air. As a parcel rises, it expands without transfer of heat from its surroundings: it uses energy to push back the surrounding atmosphere and as a result it cools. Cool air can absorb lower concentrations of water vapour than warm air, so the moisture forms clouds. Cloudy skies can therefore be asso-ciated with rising air and clear skies are often associated with descending air.

The motion of air in the upper altitudes may lead to an accu-mulation in some regions and a loss of molecules from other regions. The former result in the formation of regions of high pressure (‘highs’ or anticyclones) and the latter result regions of low pressure (‘lows’, depressions, or cyclones). These regions are shown as H and L on the weather map in Fig. 1.7. The lines of constant pressure—differing by 4 mbar (400 Pa, about 3 Torr)—marked on it are called isobars. The elon-gated regions of high and low pressure are known, respec-tively, as ridges and troughs.

The molecular model of gases

We remarked in Foundations (in the section on mat-ter) that a gas may be pictured as a collection of par-ticles in ceaseless, random motion (Fig. 1.8). 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 important functions of physical chemistry is to convert qualitative notions into quantitative state-ments that can be tested experimentally by making measurements and comparing the results with pre-dictions. Indeed, an important component of science as a whole is its technique of proposing a qualitative model and then expressing that model mathemati-cally. The ‘kinetic model’ (or the ‘kinetic molecular theory’, KMT) of gases is an excellent example of this procedure: the model is very simple, and the quantitative prediction (the perfect gas law) is experi-mentally verifi able.

The kinetic model of gases is based on three assumptions:

1. A gas consists of molecules in ceaseless random motion.

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

3. The molecules do not interact, except during collisions.

The assumption that the molecules do not interact unless they are in contact implies that the potential energy of the molecules (their energy due to their position) is independent of their separation and may be set equal to zero. The total energy of a sample of gas is therefore the sum of the kinetic energies (the

0 0.5 1Pressure, p/p0

24

18

12

6

0

Alt

itu

de,

h/k

m

Fig. 1.6 The variation of atmospheric pressure with altitude as predicted by the barometric formula.

Fig. 1.7 A typical weather map; this one for the North Atlantic in Feburary 2012. Regions of high pressure are denoted H and those of low pressure L. Pressures are in millibars (1 mbar = 100 Pa).

Fig. 1.8 The model used for discussing the molecular basis of the physical properties of a perfect gas. The pointlike molecules move randomly with a wide range of speeds and in random directions, both of which change when they collide with the walls or with other molecules.

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THE MOLECULAR MODEL OF GASES 27

energy due to motion) of all the molecules present in it. It follows that the faster the molecules travel (and hence the greater their kinetic energy), the greater the total energy of the gas.

1.4 The pressure of a gas according to the kinetic model

The kinetic model accounts for the steady pressure exerted by a gas in terms of the collisions the mole-cules make with the walls of the container. Each impact gives rise to a brief force on the wall, but as billions of collisions take place every second, the walls experience a virtually constant force, and hence the gas exerts a steady pressure. On the basis of this model, the pressure exerted by a gas of molar mass M in a volume V is

pnMv

V= rms

2

3Pressure, according to KMT (1.11)

See Further information 1.1 for a derivation of this equation. Here vrms is the root-mean-square speed(r.m.s. speed) of the molecules. This quantity is defi ned as the square root of the mean value of the squares of the speeds, v, of the molecules. That is, for a sample consisting of N molecules with speeds v1, v2, . . . , vN, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by ⟨. . .⟩), and fi nally take the square root of the result:

vrms = ⟨v2⟩1/2 Defi nitionRoot-mean-square speed (1.12)

v v vN

N=+ + +⎛

⎝⎜⎜

⎞

⎠⎟⎟

12

22 2 1 2/...

The r.m.s. speed might at fi rst sight seem to be a rather peculiar measure of the mean speeds of the molecules, but its signifi cance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass m travelling at a speed v is Ek = 12 mv2, which implies that the mean kinetic energy of a collection of molecules, ⟨Ek⟩, is the average of this quantity, or 1

2 mv2rms. It follows from the relation

12 mv2

rms = ⟨Ek⟩, that

vEmrms

k=⎛⎝⎜

⎞⎠⎟

21 2⟨ ⟩ /

(1.13)

Therefore, wherever vrms appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s. speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, ;, of the molecules:

; =+ + +v v v

NN1 2

... Defi nition Mean speed (1.14)

For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is

; =⎛⎝⎜

⎞⎠⎟ =

83

0 9211 2

π

/

.v vrms rms

(1.15)

For elementary purposes, and for qualitative argu-ments, we don’t need to distinguish between the two measures of average speed, but for precise work the distinction is important.

Brief illustration 1.4 Root-mean-square values

Cars pass a point travelling at 45.0 (5), 47.0 (7), 50.0 (9), 53.0 (4), 57.0 (5) km h−1, where the number of cars is given in parentheses. The r.m.s. speed of the cars is given by eqn 1.12 in the form

#rms

km h km h

=

× + ×+ ×

− −5 45 0 7 47 05

1 2 1 2( . ) ( . )( 557 0

5 7 5

1 2

1 2

. )

/

km h−

+ +

⎧

⎨⎪⎪

⎩⎪⎪

⎫

⎬⎪⎪

⎭⎪⎪

= 50.2 km h−1

Self-test 1.6

Calculate the mean speed of the cars (from eqn 1.14).

Answer: 50.0 km h−1

Eqn 1.11 can be rearranged into

pV = 13 nMv2

rms (1.16)

which is starting to resemble the equation pV = nRT. This conclusion is a major success of the kinetic model, for the model implies an experimentally verifi ed result.

1.5 The average speed of gas molecules

We now suppose that the expression for pV derived from the kinetic model, eqn 1.16, is indeed the equa-tion of state of a perfect gas, pV = nRT. That being so, we can equate the expression on the right to nRT, which gives

13 nMv2

rms = nRT

The n on each side now cancel to give

13 Mv2

rms = RT

The great usefulness of this expression is that we can rearrange it into a formula for the r.m.s. speed of the gas molecules at any temperature, fi rst by writing

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CHAPTER 1 THE PROPERTIES OF GASES28

v2rms = 3RT/M and then by taking the square root of

both sides:

vRTMrms =

⎛⎝⎜

⎞⎠⎟

31 2/

r.m.s. speed of molecules (1.17)

Brief illustration 1.5 The r.m.s. speed of molecules

Substitution of the molar mass of O2 (32.0 g mol−1, corresponding to 3.20 × 10−2 kg mol−1) and a tem-perature corresponding to 25 °C (that is, 298 K) into eqn 1.17 gives an r.m.s. speed for these molecules of

#rmsJK mol

=× − −3 8 3145 1 1( . )

R� ���������� ����������� � ��� ���

� ������

×× − −

(.

2983 20 10 2 1

K)kg mol

T

M����� ����������

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

= −

1 2

482

/

m s 11

(We used 1 J = 1 kg m2 s−2 to cancel units.) The same calculation for nitrogen molecules gives 515 m s−1. Both these values are not far off the speed of sound in air (346 m s−1 at 25 °C). That similarity is reasonable, because sound is a wave of pressure variation transmitted by the movement of molecules, so the speed of propaga-tion of a wave should be approximately the same as the speed at which molecules can adjust their locations.

1.6 The Maxwell distribution of speeds

So far, we have dealt only with the average speed of molecules in a gas. Not all molecules, however, travel at the same speed: some move more slowly than the average (until they collide, and get accelerated to a high speed, like the impact of a bat on a ball), and others may briefl y move at much higher speeds than the average, but be brought to a sudden stop when they collide. There is a ceaseless redistribution of speeds among molecules as they undergo collisions. Each molecule collides about once every nanosecond (1 ns = 10−9 s) or so in a gas under normal conditions.

The mathematical expression that tells us the probability, P, that the molecules in a sample of gas have a speed that lies in a particular range at any instant is called the distribution of molecular speeds. Thus, the distribution might tell us that at 20 °C 19 out of 1000 O2 molecules, corresponding to P = 0.019, have a speed in the 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, corresponding to P = 0.021, and so on. The precise form of the distribution was worked out by James Clerk Maxwell towards the end of the nineteenth century, and his expression is known as the Maxwell distribution of speeds. According to Maxwell, the probability P(v,v + ∆v) that the molecules have a speed in a narrow range between v and v + ∆v (for example, between 300 m s−1 and 310 m s−1, corresponding to v = 300 m s−1 and ∆v = 10 m s−1) is

P(v,v + ∆v) = ρ(v)∆v with

ρ(v) = 42

1 22 22π

πMRT

v Mv RT⎛⎝⎜

⎞⎠⎟ −

/

e /

Maxwell distribution of speeds (1.18)

(ρ is the Greek letter rho; for a review of exponential functions, see The chemist’s toolkit 1.2.) This for-mula was used to calculate the numbers quoted above. Its origin can be traced to the Boltzmann distribution (Foundations 0.11), because the fraction of molecules that have a particular speed v have a kinetic energy Ek = 1/2mv2 (there is no contribution from potential energy) and according to Boltzmann, that fraction is proportional to e−Ek/kT, which becomes fi rst e−mv2/2kT and then e−Mv2/2RT when m and k are both multiplied by Avogadro’s constant, NA:

ρ( ) /v mv kT∝ =−

==

e e2 2

E kTmN MkN Rk /

A

A� ����� ����� �−−Mv RT2 2/

Self-test 1.7

Calculate the rms speed of H2 molecules at 25 °C.

Answer: 1920 m s−1

The important conclusion to draw from eqn 1.17 is that

The r.m.s. speed of molecules in a gas is propor-tional to the square root of the temperature: vrms ∝ T1/2.

Because the mean speed is proportional to the r.m.s. speed, the same is true of the mean speed too. Therefore, doubling the thermodynamic tempera-ture (that is, doubling the temperature on the Kelvin scale) increases the mean and the r.m.s. speed of molecules by a factor of 21/2 = 1.414. . . .

Brief illustration 1.6 Molecular speeds

Cooling a sample of air from 25 °C (298 K) to 0 °C (273 K) reduces the original r.m.s. speed of the mole-cules by a factor of

#

#

#

rms

rms

K)298 K)

Krms

((

/

273 273298

1 2

=

∝T�

K

⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

1 2

0 957/

.

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

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THE MOLECULAR MODEL OF GASES 29

Note that the function ρ(v) is a ‘probability density’, not the probability itself, in the sense that the actual probability of molecules having a speed in the range v to v + ∆v is given by the product of ρ(v) and the width of the range. (This relation is like the mass of a volume of material being given by the product of its mass density and the size of the region, the volume.) We shall encounter several other examples of prob-ability densities elsewhere in the text.

Although eqn 1.18 looks complicated, its features can be picked out quite readily. One of the skills to develop in physical chemistry is the ability to inter-pret the message carried by equations. Equations

convey information, and it is far more important to be able to read that information than simply to remember the equation. Let’s read the information in eqn 1.18 piece by piece.

• Because P(v,v + ∆v) is proportional to the range of speeds ∆v, we see that the probability of the speed lying in the range ∆v increases in proportion to the width of the range. If at a given speed we double the range of interest (but still keep it narrow), then the probability of molecules having speeds lying in that range doubles too.

• Equation 1.18 includes a decaying exponential func -

tion, the term 4π MRT2

1 2

π( ) /v2e−Mv2/2RT. Its presence

implies that the probability that molecules will be found with very high speeds is very small because e−ax2

becomes very small when ax2 is large.

• The factor M/2RT multiplying v2 in the exponent,

4π MRT2

1 2

π( ) /v2e−Mv2/2RT, is large when the molar

mass, M, is large, so the exponential factor goes most rapidly towards zero when M is large. That tells us that heavy molecules are unlikely to be found with very high speeds.

• The opposite is true when the temperature, T, is high: then the factor M/2RT in the exponent is small, so the exponential factor falls towards zero relatively slowly as v increases. This tells us that at high temperatures, there is a greater probability of the molecules having high speeds than at low temperatures.

• A factor v2 in 4π MRT2

1 2

π( ) /v2e−Mv2/2RT multiplies

the exponential. This factor, which comes from taking into account that high velocities can be achieved in more ways than low velocities, goes to zero as v goes to zero, so the probability of fi nding molecules with very low speeds will also be very small.

• The remaining factors (the term 4π(M/2πRT)1/2v2

e−Mv2/2RT) simply ensure that when we add together the probabilities over the entire range of speeds from zero to infi nity, then we get 1.

Figure 1.9 is a graph of the Maxwell distribution, and shows these features pictorially for the same gas (the same value of M) but different temperatures. As we deduced from the equation, we see that there is only a small probability that molecules in the sample have very low or very high speeds. However, the probability of molecules having very high speeds increases sharply as the temperature is raised, as the tail of the distribution reaches up to higher speeds.

The chemist’s toolkit 1.2 Exponential and Gaussian functions

In preparation for their occurrence throughout the text (and throughout physical chemistry), it will be useful to know the shapes of the functions e−ax and e−ax2

(Sketch 1.3).An exponential function, a function of the form e−ax,

starts off at 1 when x = 0 and decays toward zero, which it reaches as x approaches infi nity. This function approaches zero more rapidly when a is large than when it is small.

A Gaussian function, a function of the form e−ax2, also

starts off at 1 when x = 0 and decays to zero as x increases, however, its decay is initially slower but then plunges down to zero more rapidly than an exponential function.

The illustration also shows the behaviour of the two functions for negative values of x. The exponential func-tion e−ax rises rapidly to infi nity as x → −∞, but the Gaussian function is symmetrical about x = 0 and traces out a bell-shaped curve that falls to zero as x → ±∞.

0

00

1

1

1

2

2

2

–1–2–3–4–5

0

1

x

x 3

3

3

4

4

5

5

7

8

9

10

6

e–x

e–x2

Sketch 1.3 The exponential function, e−x, and the bell-shaped Gaussian function, e−x2

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

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CHAPTER 1 THE PROPERTIES OF GASES30

This feature plays an important role in the rates of gas-phase chemical reactions, for (as we shall see in Section 10.10), the rate of a reaction in the gas phase depends on the energy with which two molecules crash together, which in turn depends on their speeds.

Figure 1.10 is a plot of the Maxwell distribution for molecules with different molar masses at the same temperature. As can be seen, not only do heavy molecules have lower average speeds than light mole-cules at a given temperature, but they also have a signifi cantly narrower spread of speeds. That narrow

spread means that most molecules will be found with speeds close to the average. In contrast, light molecules (such as H2) have high average speeds and a wide spread of speeds: many molecules will be found travelling either much more slowly or much more quickly than the average. This feature plays an important role in determining the composition of planetary atmospheres, because it means that a signifi cant fraction of light molecules travel at suffi -ciently high speeds to escape from the planet’s gravitational attraction. The ability of light mole-cules to escape is one reason why hydrogen (molar mass 2.02 g mol−1) and helium (4.00 g mol−1) are very rare in the Earth’s atmosphere.

The Maxwell distribution has been verifi ed experi-mentally by passing a beam of molecules from an oven at a given temperature through a series of coax-ial slotted disks. The speed of rotation of the disks brings the slots into line for molecules travelling at a particular speed, so only molecules with that speed pass through and are detected. By varying the rota-tion speed, the shape of the speed distribution can be explored and is found to match that predicted by eqn 1.18. Although the selector measures the distri-bution of speeds in one dimension, collisions within the beam ensure that that distribution matches the distribution in three dimensions and that the experi-ment monitors the Maxwell distribution.

1.7 Diffusion and effusion

Diffusion is the process by which the molecules of different substances mingle with each other. The atoms of two solids diffuse into each other when the two solids are in contact, but the process is very slow. The diffusion of a solute through a liquid solvent is much faster but mixing needs to be encouraged by stirring or shaking (the process is then no longer pure diffusion). The diffusion of one gas into another is much faster. It accounts for the largely uniform com-position of the atmosphere, for if a gas is produced by a localized source (such as carbon dioxide from the respiration of animals, oxygen from photosyn-thesis by green plants, and pollutants from vehicles and industrial sources), then the molecules of gas will diffuse from the source and in due course be dis-tributed throughout the atmosphere. In practice, the process of mixing is accelerated by the bulk motion we experience as winds. The process of effusion is the escape of a gas through a small hole, as in a puncture in an infl ated balloon or tyre (Fig. 1.11).

The rates of diffusion and effusion of gases increase with increasing temperature, for both processes

Lowtemperature

Hightemperature

Speed

Nu

mb

er o

f m

ole

cule

s

Fig. 1.9 The Maxwell distribution of speeds and its variation with the temperature. Note the broadening of the distribution and the shift of the r.m.s. speed to higher values as the tem-perature is increased.

Heavy molecules

Lightmolecules

Speed

Nu

mb

er o

f m

ole

cule

s

Fig. 1.10 The Maxwell distribution of speeds also depends on the molar mass of the molecules. Molecules of low molar mass have a broad spread of speeds, and a signifi cant frac-tion may be found travelling much faster than the r.m.s. speed. The distribution is much narrower for heavy mole-cules, and most of them travel with speeds close to the r.m.s. value.

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THE MOLECULAR MODEL OF GASES 31

depend on the motion of molecules, and molecular speeds increase with temperature. The rates also decrease with increasing molar mass, for molecular speeds decrease with increasing molar mass. The dependence on molar mass, however, is simple only in the case of effusion. In effusion, only a single substance is in motion, not the two or more inter-mingling gases involved in diffusion.

The experimental observations on the dependence of the rate of effusion of a gas on its molar mass are summarized by Graham’s law of effusion, proposed by Thomas Graham in 1833:

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

Rate of effusion ∝11 2M / Graham’s law (1.19)

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

Brief illustration 1.7 Graham’s law

The rates (in terms of amounts of molecules) at which hydrogen (molar mass 2.02 g mol−1) and carbon diox-ide (44.01 g mol−1) effuse under the same conditions of pressure and temperature are in the ratio

Rate of effusion of HRate of effusion of C

2

OOCOH2

Rate( )( )

/ / /

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

∝1 1 2

2

2

1 2MMM

�

=

444 012 016

1

1

1 2.

.

/g molg mol

−

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

44.012.016

⎛⎝⎜

⎞⎠⎟

1/2

= 4.672

The mass of carbon dioxide that escapes in a given interval is greater than the mass of hydrogen, because although nearly 5 times as many hydrogen molecules escape, each carbon dioxide molecule has over 20 times the mass of a molecule of hydrogen.

(a)

(b)

Fig. 1.11 (a) Diffusion is the spreading of the molecules of one substance into the region initially occupied by another species. Note that molecules of both substances move, and each substance diffuses into the other. (b) Effusion is the escape of molecules through a small hole in a confi ning wall.

Self-test 1.8

Suppose 5.0 g of argon escapes by effusion; what mass of nitrogen would escape under the same conditions?

Answer: 4.2 g

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

The high rate of effusion of hydrogen and helium is one reason why these two gases leak from contain-ers and through rubber diaphragms so readily. The different rates of effusion through a porous barrier are employed in the separation of uranium-235 from the more abundant and less useful uranium-238 in the processing of nuclear fuel. The process depends on the formation of uranium hexafl uoride, a volatile solid. However, because the ratio of the molar masses of 238UF6 and 235UF6 is only 1.008, the ratio of the rates of effusion is only (1.008)1/2 = 1.004. Thousands of successive effusion stages are therefore required to achieve a signifi cant separation. The rate of effusion of gases was once used to determine molar mass by comparison of the rate of effusion of a gas or vapour with that of a gas of known molar mass. However, there are now much more precise methods available, such as mass spectrometry.

Graham’s law is explained by noting that the r.m.s. speed of molecules in a gas is inversely propor-tional to the square root of their molar mass (eqn 1.17). Because the rate of effusion through a hole in a container is proportional to the rate at which mole-cules pass through the hole, it follows that the rate should be inversely proportional to M1/2, which is in accord with Graham’s law.

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CHAPTER 1 THE PROPERTIES OF GASES32

1.8 Molecular collisions

The average distance that a molecule travels between collisions is called its mean free path, λ (lambda). The mean free path in a liquid is less than the diam-eter of the molecules, because a molecule in a liquid meets a neighbour even if it moves only a fraction of a diameter. However, in gases, the mean free paths of molecules can be several hundred molecular diam-eters. If we think of a molecule as the size of a tennis ball, then the mean free path in a typical gas would be about the length of a tennis court.

The collision frequency, z, is the average rate of collisions made by one molecule. Specifi cally, z is the average number of collisions one molecule makes in a given time interval divided by the length of the interval. It follows that the inverse of the collision frequency, 1/z, is the time of fl ight, the average time that a molecule spends in fl ight between two colli-sions (for instance, if there are 10 collisions per sec-ond, so the collision frequency is 10 s−1, then the average time between collisions is 1/10 of a second and the time of fl ight is 1/10 s). As we shall see, the collision frequency in a typical gas is about 109 s−1 at 1 atm and room temperature, so the time of fl ight in a gas is typically 1 ns.

Because speed is distance travelled divided by the time taken for the journey, the r.m.s. speed vrms, which we can loosely think of as the average speed, is the average length of the fl ight of a molecule between collisions (that is, the mean free path, λ) divided by the time of fl ight (1/z). It follows that the mean free path and the collision frequency are related by

vrmsdistance between collisions

time betwee=

nn collisionsmean free path=

l� ������� ��������

� ������ ������time of flight

1/z

= =λ

λ1/z

z

(1.20)

Therefore, if we can calculate either λ or z, then we can fi nd the other from this equation and the value of vrms given in eqn 1.17.

To fi nd expressions for λ and z we need a slightly more elaborate version of the kinetic model. The basic kinetic model supposes that the molecules are effectively point-like; however, to obtain collisions, we need to assume that two ‘points’ score a hit when-ever they come within a certain range d of each other, where d can be thought of as the diameter of the molecules (Fig. 1.12). The collision cross-section, σ (sigma), the target area presented by one molecule to

another, is therefore the area of a circle of radius d, so σ = πd2. When this quantity is built into the kinetic model, it is possible to show2 that

λσ

=kT

p Mean free path (1.21)

where k is Boltzmann’s constant. Table 1.4 lists the collision cross-sections of some common atoms and molecules. Similarly, by combining this expression with that in eqn 1.20,

zv p

kT=

=z #rms /l� σ rms Collision frequency (1.22)

2 See, for instance, our Physical Chemistry (2010).

Diameter, d

Radius, d

Fig. 1.12 To calculate features of a perfect gas that are related to collisions, a point is regarded as being surrounded by a sphere of diameter d. A molecule will hit another mole-cule if the centre of the former lies within a circle of radius d. The collision cross-section is the target area, pd2.

Table 1.4

Collision cross-sections of atoms and molecules

Species s /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.

9780199608119_C01.indd 329780199608119_C01.indd 32 9/21/12 4:37 PM9/21/12 4:37 PM

REAL GASES 33

Brief illustration 1.8 Mean free path

From the information in Table 1.4 we can calculate that the mean free path of O2 molecules in a sample of oxygen at SATP (25 °C, 1 bar) is

l =× − −( . )1 381 10 23 1J K

k� ���������� ����������××

× −

(

( . )

298

0 40 10 18 2

K)

m

T� ��� ���

� �������� �s�������� � ����� �����× ×(1 105 Pa)

p

=× −( .1 381 10 233

18 5 2

2980 40 10 1 10

) ( )( . ) ( )

×× × ×−

JPa m

.= × == =− −

−

1 1 1 103 9

1 0 10 107

J Pa m nm m

m� �

00 nm

Under the same conditions, the collision frequency is 6.2 × 109 s−1, so each molecule makes 6.2 billion colli-sions each second.

time to travel to its neighbour in a denser, higher-pressure gas. For example, although the collision fre-quency for an O2 molecule in oxygen gas at SATP is 6.2 × 109 s−1, at 2.0 bar and the same temperature the collision frequency is doubled, to 1.2 × 1010 s−1.

• Because eqn 1.22 shows that z ∝ vrms, and we know that vrms ∝ 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 molecules less frequently.

Real gases

So far, everything we have said applies to perfect gases, in which the average separation of the mole-cules is so great that they move independently of one another. In terms of the quantities introduced in the previous section, a perfect gas is a gas for which the mean free path, λ, of the molecules in the sample is much greater than d, the separation at which they are regarded as being in contact:

Condition for perfect-gas behaviour: λ >> d

As a result of this large average separation, a perfect gas is a gas in which the only contribution to the energy comes from the kinetic energy of the motion of the molecules and there is no contribution to the total energy from the potential energy arising from the interaction of the molecules with one another. However, in fact all molecules do interact with one another provided they are close enough together, so the ‘kinetic energy only’ model is only an approxima-tion. Nevertheless, under most conditions the criter-ion λ >> d (the separation of the molecules is much greater than their diameters; think tennis court com-pared with tennis ball) is satisfi ed and the gas can be treated as though it is perfect.

1.9 Molecular interactions

There are two types of contribution to the potential energy of interaction between molecules. At rela-tively large separations (a few molecular diameters), molecules attract each other. This attraction is responsible for the condensation of gases into liquids at low temperatures. At low enough temperatures the molecules of a gas have insuffi cient kinetic energy to escape from each other’s attraction and they stick

Self-test 1.9

Use eqns 1.17 and 1.22, together with the information in Table 1.4, to determine the collision frequency for Cl2 molecules in a sample of chlorine gas under the same conditions.

Answer: 7.3 × 109 s−1

Once again, we should interpret the essence of eqns 1.21 and 1.22 rather than trying to remember them.

• Because λ ∝ 1/p, the mean free path decreases as the pressure increases.

This decrease is a result of the increase in the number of molecules present in a given volume as the pres-sure is increased, so each molecule travels a shorter distance before it collides with a neighbour. For example, the mean free path of an O2 molecule decreases from 73 nm to 36 nm when the pressure is increased from 1.0 bar to 2.0 bar at 25 °C.

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

For instance, the collision cross-section of a benzene molecule (0.88 nm2) is about four times greater than that of a helium atom (0.21 nm2), and at the same pressure and temperature its mean free path is four times shorter.

• 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, each molecule takes less

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CHAPTER 1 THE PROPERTIES OF GASES34

together. Second, although molecules attract each other when they are a few diameters apart, as soon as they come into contact they repel each other. This repulsion is responsible for the fact that liquids and solids have a defi nite bulk and do not collapse to an infi nitesimal point.

Molecular interactions—the attractions and re -pulsions between molecules—give rise to a potential energy that contributes to the total energy of a gas. Because attractions correspond to a lowering of total energy as molecules get closer together, they make a negative contribution to the potential energy. On the other hand, repulsions make a positive contribution to the total energy as the molecules squash together. Figure 1.13 illustrates the general form of the vari-ation of the intermolecular potential energy. At large separations, the energy-lowering interactions are dominant, but at short distances the energy-raising repulsions dominate.

Molecular interactions affect the bulk properties of a gas and, in particular, their equations of state. For example, the isotherms of real gases have shapes that differ from those implied by Boyle’s law, particu-larly at high pressures and low temperatures when the interactions are most important. Figure 1.14 shows a set of experimental isotherms for carbon dioxide. They should be compared with the perfect-gas isotherms shown in Fig. 1.1. Although the experi-mental isotherms resemble the perfect-gas isotherms at high temperatures (and at low pressures, off the scale on the right of the graph), there are very

striking differences between the two at temperatures below about 50 °C and at pressures above about 1 bar.

1.10 The critical temperature

To understand the signifi cance of the isotherms in Fig. 1.14, let’s begin with the isotherm at 20 °C and follow the changes from A to F:

• At point A the sample of carbon dioxide is a gas.

• As the sample is compressed to B by pressing in a piston, the pressure increases broadly in agree-ment with Boyle’s law.

• The increase continues until the sample reaches point C.

• Beyond this point, we fi nd that the piston can be pushed in without any further increase in pres-sure, through D to E.

• The reduction in volume from E to F requires a very large increase in pressure.

This variation of pressure with volume is exactly what we expect if the gas at C condenses to a com-pact liquid at E. Indeed, if we could observe the sample we would see that:

• At C the liquid begins to condense to a liquid.

• The condensation is complete when the piston is pushed in to E.

• At E, the piston is resting on the surface of the liquid.

• The subsequent reduction in volume, from E to F, corresponds to the very high pressure needed to compress a liquid into a smaller volume.

Repulsions dominant

Attractions dominant

Po

ten

tial

en

erg

y 0

Separation

Fig. 1.13 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 is negative, the attractive interactions dominate. At large separ-ations (on the right) the potential energy is zero and there is no interaction between the molecules.

140

120

100

80

60

40

20

0

Pre

ssu

re, p

/atm

0 0.2 0.4 0.6Molar volume, Vm/(dm3 mol–1)

A

BCDE

FCritical point

50 °C

40 °C

0 °C

20 °C

31.04 °C (Tc)*

Fig. 1.14 The experimental isotherms of carbon dioxide at several temperatures. The critical isotherm is at 31.04 °C.

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REAL GASES 35

In terms of intermolecular interactions:

• The step from C to E corresponds to the molecules being so close on average that they attract each other and cohere into a liquid.

• The step from E to F represents the effect of trying to force the molecules even closer together when they are already in contact, and hence trying to overcome the strong repulsive interactions between them.

If we could look inside the container at point D, we would see a liquid separated from the remaining gas by a sharp surface (Fig. 1.15). At a slightly higher temperature (at 30 °C, for instance), a liquid forms, but a higher pressure is needed to produce it. It might be diffi cult to make out the surface because the remaining gas is at such a high pressure that its density is similar to that of the liquid. At the special temperature of 31.04 °C (304.19 K) the gaseous state of carbon dioxide appears to transform continuously into the condensed state and at no stage is there a visible surface between the two states of matter. At this temperature (which is 304.19 K for carbon dioxide but varies from substance to substance), called the critical temperature, Tc, and at all higher temperatures, a single form of matter fi lls the con-tainer at all stages of the compression and there is no separation of a liquid from the gas. We have to conclude that a gas cannot be condensed to a liquid by the application of pressure unless the temperature is below the critical temperature.

Figure 1.14 also shows that in the critical isotherm, the isotherm at the critical temperature, the volumes

at each end of the horizontal part of the isotherm have merged to a single point, the critical point of the gas. The pressure and molar volume at the critical point are called the critical pressure, pc, and critical molar volume, Vc, of the substance. Collectively, pc, Vc, and Tc are the critical constants of a substance. Table 1.5 lists the critical temperatures of some com-mon gases. The data there imply, for example, that liquid nitrogen cannot be formed by the application of pressure unless the temperature is below 126 K (−147 °C). The critical temperature is sometimes used to distinguish the terms ‘vapour’ and ‘gas’:

a vapour is the gaseous phase of a substance below its critical temperature (and which can therefore be liquefied by compression alone)

a gas is the gaseous phase of a substance above its critical temperature (and which cannot therefore be liquefied by compression alone)

Oxygen at room temperature is therefore a true gas; the gaseous phase of water at room temperature is a vapour.

Increasing temperature

T < Tc T ≈ TcT > Tc

Fig. 1.15 When a liquid is heated in a sealed container, the density of the vapour phase increases and that of the liquid phase decreases, as depicted here by the changing density of shading. There comes a stage at which the two densities are equal and the interface between the two fl uids disap-pears. This disappearance occurs at the critical temperature. The container needs to be strong: the critical temperature of water is at 373 °C and the vapour pressure is then 218 atm.

Table 1.5

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

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CHAPTER 1 THE PROPERTIES OF GASES36

The dense fl uid obtained by compressing a gas when its temperature is higher than its critical tem-perature is not a true liquid, but it behaves like a liquid in many respects: it has a density similar to that of a liquid, for instance, and can act as a solvent. However, despite its density, the fl uid is not strictly a liquid because it never possesses a surface that separ-ates it from a vapour phase. Nor is it much like a gas, because it is so dense. It is an example of a super-critical fl uid. Supercritical fl uids are currently being used as solvents. For example, supercritical carbon dioxide is used to extract caffeine in the manufacture of decaffeinated coffee where, unlike organic solvents, it does not result in the formation of an unpleasant and possibly toxic residue. Supercritical fl uids are also currently of great interest in industrial processes, for they can be used instead of chlorofl uorocarbons (CFCs) and hence avoid the environmental damage that CFCs are known to cause. Because supercritical carbon dioxide is obtained either from the atmos-phere or from renewable organic sources (by fermen-tation) its use does not increase the net load of atmospheric carbon dioxide.

1.11 The compression factor

A useful quantity for discussing the properties of real gases is the compression factor, Z, which is the ratio of the actual experimentally determined molar vol-ume of a gas, Vm, to the molar volume of a perfect gas, Vm

perfect, under the same conditions:

ZV

V= m

mperfect Defi nition Compression factor (1.23a)

For a perfect gas, Vm = Vmperfect, so Z = 1 and devi-

ations of Z from 1 are a measure of how far a real gas departs from behaving perfectly. The molar volume of a perfect gas is RT/p (recall eqn 1.6b), so we can rewrite the defi nition of Z as

ZV

RT ppVRT

= ==V RT pm

perfect /�m m

/ (1.23b)

When Z is measured for real gases, it is found to vary with pressure, as shown in Fig. 1.16. At low pressures, some gases (methane, ethane, and ammo-nia, for instance) have Z < 1. That is, their molar volumes 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 these conditions, the attractive interactions are dominant. The compression factor rises above 1 at high pres-sures, whatever the identity of the gas, and for some

gases (hydrogen in the illustration) Z > 1 at all pres-sures. The type of behaviour exhibited depends on the temperature. The observation that Z > 1 tells us that the molar volume of the gas is now greater than that expected for a perfect gas of the same tempera-ture and pressure, so the molecules are pushed apart slightly. This behaviour indicates that the repulsive forces are dominant. For hydrogen, the attractive interactions are so weak that the repulsive interac-tions dominate even at low pressures.

1.12 The virial equation of state

We can use the deviation of Z from its ‘perfect’ value of 1 to construct an empirical (observation based) equation of state. To do so we suppose that, for real gases, the relation Z = 1 is only the fi rst term of a lengthier expression and write instead

ZB

VC

V= + + +1

m m2 (1.24a)

The coeffi cients B, C, . . . are called virial coeffi cients; B is the second virial coeffi cient, C, the third, and so on; the unwritten A = 1 is the fi rst. The word ‘virial’ comes from the Latin word for force, and it refl ects the fact that intermolecular forces are now signifi cant. The virial coeffi cients, which are also denoted B2, B3, etc. in place of B, C, etc., vary from gas 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

Co

mp

ress

ion

fac

tor,

Z

00

1

2

200 400 600 800Pressure, p/atm

H2

CH4

C2H4

NH3

Perfect gas

Fig. 1.16 The variation of the compression factor, Z, with pressure for several gases at 0 °C. A perfect gas has Z = 1 at all pressures. Of the gases shown, hydrogen shows positive deviations at all pressures (at this temperature); all the other gases show negative deviations initially but positive devi-ations at high pressures. The negative deviations are a result of the attractive interactions between molecules and the positive deviations are a result of the repulsive interactions.

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REAL GASES 37

volumes) and supposing that it is the fi rst term of a more complicated expression, is quite common in physical chemistry. The limiting expression is the fi rst approximation to the true expression, whatever that may be, and the additional terms progressively take into account the secondary effects that the limit-ing expression ignores.

The most important additional term on the right in eqn 1.24a is the one proportional to B (because under most conditions C/V2

m << B/Vm and C/V2m can

be neglected). In that case

ZB

V≈ +1

m

(1.24b)

From the graphs in Fig. 1.16, 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, and ammonia (so that for them Z < 1). However, in all cases Z rises again as the gas is compressed further (corresponding to small values of Vm and high pressures in Fig. 1.16), indicating that C/V2

m is positive. The values of the virial coeffi cients for many gases are known from measurements of Zover a range of molar volumes and using mathemat-ical software to fi t the data to eqn 1.24a by varying the coeffi cients until a good match is obtained.

To convert eqn 1.24a into an equation of state, we combine it with eqn 1.23b (Z = pVm/RT), which gives

pVRT

BV

CV

m

m m2= + + +1 ...

Now multiply both sides by RT/Vm and obtain

pRTV

BV

CV

= + + +⎛

⎝⎜⎜

⎞

⎠⎟⎟

m m m21 ...

Next, replace Vm by V/n throughout to get p as a function of n, V, and T:

pnRTV

nBV

n CV

= + + +⎛

⎝⎜⎜

⎞

⎠⎟⎟1

2

2... Virial equation

of state (1.25)

Equation 1.25 is the virial equation of state. When the molar volume is very large, the terms B/Vm and C/V2

m are both very small, and only the 1 inside the parentheses survives. In this limit, the virial equation of state approaches that of a perfect gas.

Brief illustration 1.9 The virial equation of state

The molar volume of NH3 is 1.00 dm3 mol−1 at 36.2 bar and 473 K. Assuming that under these conditions the virial equation of state may be written as p = (RT/Vm) × (1 − B/Vm), so

BpVRT

V= −⎛⎝⎜

⎞⎠⎟m

m1

The value of the second virial coeffi cient at this tem-perature is therefore

B =× ×( .

.

36 2 105

36 2

Pa) (1.0

bar� ������� �������00 10 m mol

dm mol

× − −

−

3 3 1

1 00 3 1

)

.� ����������� ������������

( . ) (8 3145 47311 1J K mol K)− − ×

−

⎛

⎝⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

× 1.00 × 10−3 m3 mol−1

= −79.5 × 10−6 m3 mol−1 = −79.5 cm3 mol−1

Self-test 1.10

The second virial coeffi cient for NH3 is −45.6 cm3

mol−1 at a temperature of 573 K. Determine the pres-sure at which the molar volume is 1.00 dm3 mol−1 at this temperature.

Answer: 45.6 bar

1.13 The van der Waals equation of state

Although it is the most reliable equation of state, the virial 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 physicist Johannes van der Waals, is only an approximate equation of state but it has the advantage of showing how the intermolecular interactions contribute to the deviations of a gas from the perfect gas law. We can view the van der Waals equation as another example of taking a soundly based qualitative idea and building up a mathematical expression that can be tested quantitatively.

The repulsive interaction between two molecules implies that they cannot come closer than a certain distance. Therefore, instead of being free to travel anywhere in a volume V, the actual volume in which the molecules can travel is reduced to an extent pro-portional to the number of molecules present and the volume they each exclude (Fig. 1.17). We can

r 2r

Excluded volume

Fig. 1.17 When two molecules, each of radius r and volume Vmol = 4

3 pr3 approach each other, the centre of one of them cannot penetrate into a sphere of radius 2r and therefore volume 8Vmol surrounding the other molecule.

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CHAPTER 1 THE PROPERTIES OF GASES38

therefore model the effect of the repulsive, volume-excluding forces by changing V in the perfect gas equation to V − nb, where b is the proportionality constant between the reduction in volume and the amount of molecules present in the container (see the following Derivation).

Reduction in pressure = ×⎛⎝⎜

⎞⎠⎟a

nV

2

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

pnRT

V nba

nV

=−

−⎛⎝⎜

⎞⎠⎟

2

van der Waals equation of state (1.26a)

This expression is the van der Waals equation of state. To show its resemblance to the perfect gas equation pV = nRT, eqn 1.26a is sometimes rear-ranged by bringing the term in a to the left, to give p + an2/V2, and then multiplying both sides by V − nb:

panV

V nb nRT+⎛

⎝⎜⎜

⎞

⎠⎟⎟ − =

2

2 ( ) (1.26b)

We have built the van der Waals equation by using physical arguments about the volumes of molecules and the effects of forces between them. It can be derived in other ways, but the present method has the advantage of showing how to derive the form of an equation out of general ideas. The derivation also has the advantage of keeping imprecise the signifi -cance of the van der Waals parameters, the constants a and b: they are much better regarded as empirical parameters than as precisely defi ned molecular prop-erties. The van der Waals parameters depend on the gas, but are taken as independent of temperature (Table 1.6). It follows from the way we have con-structed the equation that a (the parameter repre-senting the role of attractions) can be expected to be large when the molecules attract each other strongly, whereas b (the parameter representing the role of

Derivation 1.1

The molar volume of a gas described by the van der Waals equation

The volume of a sphere of radius R is 43 pR3. Figure 1.17

shows that the closest distance of two hard-sphere mole-cules of radius r, and volume Vmolecule = 43 pr 3, is 2r. Therefore, the excluded volume is 4

3 p(2r )3 = 8(43 pr 3), or 8Vmolecule. The

volume excluded per molecule is one-half this volume, or 4Vmolecule, so b ≈ 4VmoleculeNA.

So far, the perfect gas equation of state changes from p = nRT/V to

pnRT

V nb=

−

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

The effect of the attractive interactions between molecules is to reduce the pressure that the gas exerts. We can model the effect by supposing that the attrac-tion experienced by a given molecule is proportional to the concentration, n/V, of molecules in the con-tainer. Because the attractions slow the molecules down, the molecules strike the walls less frequently and strike it with a weaker impact. (This slowing does not mean that the gas is cooler close to the walls: the simple relation between T and r.m.s. speed in eqn 1.17 is valid only in the absence of intermolecular forces.) We can therefore expect the reduction in pressure to be proportional to the square of the molar concentration, one factor of n/V refl ecting 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

Table 1.6

van der Waals parameters of gases

Substance a/(102 kPa dm6 mol−2)

b/(10−2 dm3 mol−1)

Air 1.4 0.039Ammonia, NH3 4.114 3.71Argon, Ar 1.320 3.20Carbon dioxide, CO2 3.119 4.29Ethane, C2H6 5.435 6.51Ethene, C2H4 4.545 5.82Helium, He 0.0337 2.38Hydrogen, H2 0.2388 2.65Nitrogen, N2 1.347 3.87Oxygen, O2 1.359 3.19Xenon, Xe 4.135 5.16

9780199608119_C01.indd 389780199608119_C01.indd 38 9/21/12 4:37 PM9/21/12 4:37 PM

REAL GASES 39

repulsions) can be expected to be large when the molecules are large.

The reliability of the van der Waals equation can be judged by comparing the isotherms it predicts, which are shown in Fig. 1.18, with the experimental isotherms already shown in Fig. 1.14. Apart from the waves below the critical temperature they do re -semble experimental isotherms quite well. The waves, which are called van der Waals loops, are unrealistic because they suggest that under some conditions compression results in a decrease of pressure. The loops are therefore trimmed away and replaced by horizontal lines (Fig. 1.19). The van der Waals param- eters in Table 1.6 were found by fi tting the calculated curves to experimental isotherms.

Two important features of the van der Waals equation should be noted. First, perfect-gas iso-therms are obtained from the van der Waals equa-tion at high temperatures and low pressures. To confi rm this remark, we need to note that when the temperature is high, RT may be so large that the fi rst term on the right in eqn 1.26a greatly exceeds the second, so the latter may be ignored. Furthermore, at low pressures, the molar volume is so large that V − nb can be replaced by V. Hence, under these conditions (of high temperature and low pressure), eqn 1.26a reduces to p = nRT/V, the perfect gas equation. Second, and as shown in the following Derivation, the critical constants are related to the van der Waals coeffi cients as follows:

Vc = 3b, TaRb

pabc c= =

827 27 2, (1.27)

The fi rst of these relations shows that the critical volume is about three times the volume occupied by the molecules themselves.

0.1 1 10Reduced volume, V/Vc

1.5

1.5

1 1

0.5

0

Red

uce

d p

ress

ure

, p/p

c

0.8

Fig. 1.18 Isotherms calculated by using the van der Waals equation of state. The axes are labelled with the ‘reduced pressure’, p/pc, and ‘reduced volume’, V/Vc, where pc = a/27b2 and Vc = 3b. The individual isotherms are labelled with the ‘reduced temperature’, T/Tc, where Tc = 8a/27Rb. The isotherm labelled 1 is the critical isotherm (the isotherm at the critical temperature).

0.1 1 10Reduced volume, V/Vc

1.5

1

0.5

0

Red

uce

d p

ress

ure

, p/p

c

Fig. 1.19 The unphysical van der Waals loops are eliminated by drawing straight lines that divide the loops into areas of equal size. With this procedure, the isotherms strongly resemble the observed isotherms.

Example 1.3

Estimating the critical constants of a gas

Estimate the critical constants of carbon dioxide.

Strategy Treat carbon dioxide as a van der Waals gas and use eqn 1.27 with values of parameters taken from Table 1.6. Convert the parameters to base units before using them.

Answer The van der Waals parameters for CO2 are a = 3.460 dm6 bar mol−2 and b = 0.004 267 dm3 mol−1. To convert them to base units we write

a = =

−

−3 658 0 366 2

10 101 6 5

. .

(

dm bar mol

m) Pa� �558 6 2m Pa mol−

b = = ×

−

− −0 0429 4 29 103 1 5 3

10 1 3

. .

(

dm mol m mo

m)�ll−1

Then, by using eqn 1.27, we predict the values of the critical constants for CO2 to be

Vc = 3b = 3 × (4.29 × 10−5 m3 mol−1) = 1.29 × 10−4 m3 mol−1, or 0.129 dm3 mol−1

TaRbc

m Pa mol= =

××

−827

8 0 365827 8 3145

6 2( . )( . JJ K mol m mol− − − −× ×1 1 5 3 14 29 10) ( . )

= 304 K, or 31 °C

pabc

m Pa molm

= =× ×

−

−270 3658

27 4 29 102

6 2

5 3

.( . mol

MPa− =1 2 7 36)

.

9780199608119_C01.indd 399780199608119_C01.indd 39 9/21/12 4:37 PM9/21/12 4:37 PM

CHAPTER 1 THE PROPERTIES OF GASES40

The experimental values are 0.094 dm3 mol−1, 304 K, and 7.375 MPa, respectively.

carbon dioxide). For gases with very low boiling points (such as oxygen and nitrogen, at −183 °C and −186 °C, respectively), such a simple technique is not prac -tic able unless an even colder bath is available.

One alternative and widely used commercial tech-nique makes use of the forces that act between mole-cules. We saw earlier that the r.m.s. speed of molecules in a gas is proportional to the square root of the

Self-test 1.11

The critical pressure and temperature of CH4 are 46.1 bar and 191 K, respectively. Determine the value of the van der Waals parameter b.

Answer: 0.0431 dm3 mol−1

Derivation 1.2

Relating the critical constants to the van der Waals parameters

For this derivation we need know some of the rules of calcu-lus related to differentiation as summarized in The chemist’s toolkit 1.3. We see from Fig. 1.18 that, for T < Tc, the calcu-lated isotherms oscillate, and each one passes through a minimum followed by a maximum. These extrema converge as T → Tc and coincide at T = Tc; at the critical point the curve has a fl at infl exion (1). From the properties of curves, we know that an infl exion of this type occurs when both the fi rst and second derivatives are zero. Hence, we can fi nd the critical temperature by calculating these derivatives and setting them equal to zero. First, we use Vm = V/n to write eqn 1.26a as

pRT

V ba

V=

−−

m m2

1

The fi rst and second derivatives of p with respect to Vm (with p playing the role of y in The chemist’s toolkit 1.3, and Vm playing the role of x ) are, respectively:

dd

ddm m m m m

pV

RTV b

aV

pV

RTV

= −−

+ =( ) (2 3

2

2

2 2−−

−b

aV)3 4

6

m

At the critical point T = Tc, Vm = Vc, and both derivatives are equal to zero:

−−

+ =−

−RT

V ba

VRT

V ba

Vc

c c

c

c c( ) ( )2 3 3 4

20

2 6== 0

Solving this pair of equations gives (as you should verify) the expressions for Vc and Tc in eqn 1.27. When they are inserted in the van der Waals equation itself, we fi nd the expression for pc given there too.

1.14 The liquefaction of gases

A gas may be liquefi ed by cooling it below its boiling point at the pressure of the experiment. For example, chlorine at 1 atm can be liquefi ed by cooling it to below −34 °C in a bath cooled with dry ice (solid

The chemist’s toolkit 1.3 Differentiation

An important result from calculus is that

ddxx

nxn

n= −1

Thus, if y = mx2 + b, then, because m and b are both constants, dy/dx = 2mx. In this instance, we see that the slope (which is equal to dy/dx) increases with x (Sketch 1.4). The expression for dxn/dx also applies when n is negative. So, for instance,

dd

dd dx x x x

xx

n1 1 1d 1

22

1

= = = = −= x 2−=−

( / ) ( )d( / ) ( )d11 dx x 1

Two very important relations are

dd

e ed

de

dd

ex x

f xx

fdax ax f xed

ef

=e eax fe e⎛⎝⎜⎛⎛⎝⎝

⎞⎠⎟⎞⎞⎠⎠

(efedxx f

=xx ⎜ ⎟)xxx xx )

Two other fundamental results are

dd

dx a bx

b ba

1 1 2db2 2d

= − = −+( ) ( ) (a bx a bx2 2d a bxdbx x a2 dx a bxbb )3

The ‘second derivative’ of a function d(dy/dx )/dx is denoted d2y/dx2, or equivalently (d2/dx2)y, and is the result of taking the derivative a second time by using the same rule given above. For instance, because the fi rst derivative of y = mx2 + b is 2mx, the second derivative of the same function is 2m. For Derivation 1.2 you also need to know that

dd

d2

2 2

21 2dx a bx xdd

b b= −

⎧⎨⎨⎧⎧⎨⎨⎩⎨⎨⎩⎩

⎫⎬⎫⎫⎬⎬⎭⎬⎬⎬⎬⎬⎬ =

( ) ( )2a bx a bx⎬⎭⎭⎭⎬⎬⎬⎬ 33

dy/dx < 0

dy/dx > 0

dy/dx = 0

dy/dx = 0

x

y

Sketch 1.4 A function and its slope; the slope at any point x is given by the derivative of the function at that point.

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FURTHER INFORMATION 1.1 41

temperature (eqn 1.17). It follows that reducing the r.m.s. speed of the molecules is equivalent to cooling the gas. If the speed of the molecules can be reduced to the point that neighbours can capture each other by their intermolecular attractions, then the cooled gas will condense to a liquid.

To slow the gas molecules, we make use of an effect similar to that seen when a ball is thrown into the air: as it rises it slows in response to the gravita-tional attraction of the Earth and its kinetic energy is converted into potential energy. Molecules attract each other, as we have seen (the attraction is not gravitational, but the effect is the same), and if we can cause them to move apart from each other, like a ball rising from a planet, then they should slow. It is very easy to move molecules apart from each other: we simply allow the gas to expand, which increases the average separation of the molecules. To cool a gas, therefore, we allow it to expand without allow-ing any heat to enter from outside. As it does so, the molecules move apart to fi ll the available volume, struggling as they do so against the attraction of their neighbours. Because some kinetic energy must be converted into potential energy to reach greater separ- ations, the molecules travel more slowly as their separation increases. Therefore, because the average speed of the molecules has been reduced, the gas is now cooler than before the expansion. This process of cooling a real gas by expansion through a nar -row opening called a ‘throttle’ is called the Joule–Thomson effect. The effect was fi rst observed and analysed by James Joule (whose name is commemor-ated in the unit of energy) and William Thomson

(who later became Lord Kelvin). The procedure works only for real gases in which the attractive interactions are dominant, because the molecules have to climb apart against the attractive force in order for them to travel more slowly. For molecules under conditions when repulsions are dominant (corresponding to Z > 1, where Z is the compression factor), the Joule–Thomson effect results in the gas becoming warmer.

In practice, the gas is allowed to expand several times by recirculating it through a device called a Linde refrigerator (Fig. 1.20). On each successive expansion the gas becomes cooler, and as it fl ows past the incoming gas, the latter is cooled further. After several successive expansions, the gas becomes so cold that it condenses to a liquid.

Heatexchanger

Compressor

Throttle

Liquid

Coldgas

Fig. 1.20 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 still further. Eventually, liquefi ed gas drips from the throttle.

Kinetic molecular theory

One of the essential skills of a physical chemist is the ability to turn simple, qualitative ideas into rigid, testable, quanti-tative theories. The kinetic model of gases is an excellent 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 the underlying phys-ical picture, in this case a swarm of mass points in ceaseless random motion. The key quantitative ingredients we need are the equations of classical mechanics, which are covered in Foundations.

Further information 1.1

We begin the derivation of eqn 1.11, p = nMv2rms/3V, by

considering the arrangement in Fig. 1.21. When a particle of mass m that is travelling with a component of velocity vx

parallel to the x-axis (vx > 0 corresponding to motion to the right and vx < 0 to motion to the left) collides with the wall on the right and is refl ected, its linear momentum changes from +m |vx| before the collision to −m|vx| after the collision (when it is travelling in the opposite direction at the same speed; |x | means the value of x with the sign discarded, so |−3| = 3, for instance). The x-component of the momentum therefore changes by 2m |vx| on each collision (the y- and z-components are unchanged). Many molecules collide with the wall in an interval ∆t, and the total change of

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momentum is the product of the change in momentum of each molecule multiplied by the number of molecules that reach the wall during the interval.

We need to calculate that number. Because a molecule with velocity component vx can travel a distance |vx|∆t along the x-axis in an interval ∆t, all the molecules within a distance |vx|∆t of the wall will strike it if they are travelling 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 number density, the number of particles divided by the total volume, is nNA/V (where n is the total amount in moles of molecules in the container of volume V and NA is Avogadro’s constant), so the num-ber 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 wall during the interval ∆t is nNAA |vx|∆t/2V.

The total momentum change in the interval ∆t is the product of the number of collisions we have just calculated and the change of momentum on each collision, 2m |vx|:

Momentum change A=nN A v tA

Vx| |vvx

2

Number of collissionsss Momentumchangeon� ������ �� �� � ����

× 2m x| |vx

one collision� ��� �� �� � ����

= =nmN Av

V

t

Vx x=

t nMAvA2 t 2MAt nMAvt nMAvt nMAv

M mN==

Next, to fi nd the force, we calculate the rate of change of momentum:

Force =nMAv t V/x

2

Momentumchange� ������ �� �� ��� ����

∆t

nMAv

Vx

Interval

=2

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

Pressure = =nMAv V/

A

nMvx2

Force

Area

� ��� � ��� ����

xx

V

2

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

pnM

Vx=

⟨ ⟩vx2

To write an expression of the pressure in terms of the root-mean-square speed, vrms, we begin by writing the speed of a single molecule, v, as v2 = v2

x + v2y + v2

z. Because the root-mean-square speed, vrms, is defi ned as vrms = ⟨v2⟩1/2 (eqn 1.12), it follows that

v2rms = ⟨v2⟩ = ⟨v2

x⟩ + ⟨v2y⟩ + ⟨v2

z⟩

However, because the molecules are moving randomly, all three averages are the same. It follows that v2

rms = 3⟨v2x⟩.

Equation 1.11 now follows by substituting ⟨v2x⟩ = 13 v2

rms into p = nM⟨v2

x⟩/V.

|vx|∆t

x

Area, A

WillWon’t

Fig. 1.21 The model used for calculating the pressure of a perfect 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 the molecule collides with the wall). All molecules within the shaded area will reach the wall in an interval Dt provided they are moving towards it.

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

2 The perfect-gas equation of state is based on Boyle’s law (p ∝ 1/V ), Charles’s law (V ∝ T ), and Avogadro’s principle (V ∝ n).

3 Dalton’s law states that the total pressure of a mixture of perfect gases is the sum of the

Checklist of key concepts

pressures that each gas would exert if it were alone in the container at the same temperature.

4 The partial pressure of any gas is defi ned as pJ = xJp, where xJ is its mole fraction in a mixture and p is the total pressure.

5 The kinetic model of gases expresses the proper-ties of a perfect gas in terms of a collection of mass points in ceaseless random motion.

CHAPTER 1 THE PROPERTIES OF GASES42

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QUESTIONS AND EXERCISES 43

6 The mean speed and root-mean-square speed of molecules are proportional to the square root of the (absolute) temperature and inversely propor-tional to the square root of the molar mass.

7 The properties of the Maxwell distribution of speeds are summarized in Figs 1.9 and 1.10.

8 Diffusion is the spreading of one substance through another; effusion is the escape of a gas through a small hole.

9 Graham’s law states that the rate of effusion is inversely proportional to the square root of the molar mass.

10 The Joule–Thomson effect is the cooling of gas that occurs when it expands through a throttle without the infl ux of heat.

Blue boxes den ote limitation to perfect gases.

Road map of key equations

pJ = xJp

p = pA + pB +...

pV = nRT/(V – nb) pV = nRT/(V – nb) – an2/V 2

P(v,v + ∆v) = 4π(M/2πRT )1/2 v2 exp(−Mv2/2RT )∆vMaxwell–Boltzmann distribution

pV = nRT (1 + nB/V + n 2C /V 2 +...)

Z = pVm /RT

pc = a/27b 2

Vc = 3bTc = 8a/27Rb

Molecularrepulsions

Molecularattractions

Empiricalextension

Virial equation

van der Waals equation

Collisions

Mixtures

Criticalconstants

Assessment of real/perfect characterCompression factor

Partialpressure

Mean free path,collision frequency

Experimentalobservations

Model,kinetic theory

Boltzmanndistribution

p ∝ 1/VBoyle’s law

pV = 1/3nMv rms2

vrms = z

p ∝ TCharles’s law

p ∝ nAvogadro’s

principle

λ

pV = nRT

vrms = (3RT/M)1/2

Discussion questions

1.1 Explain how the experiments of Boyle, Charles, and Avogadro led to the formulation of the perfect gas equation of state.

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

1.3 Use the kinetic model of gases to explain why light gases, such as H2 and He, are rare in the Earth’s atmosphere but heavier gases, such as O2, CO2, and N2, are abundant.

Questions and exercises

1.4 Provide a molecular interpretation for the variation of the rates of diffusion and effusion of gases with temperature.

1.5 Explain how the compression factor varies with pressure and temperature and describe how it reveals information about intermolecular interactions in real gases.

1.6 What is the signifi cance of the critical constants of a gas?

1.7 Justify the formulation of the van der Waals equation of state.

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CHAPTER 1 THE PROPERTIES OF GASES44

Exercises

Treat all gases as perfect unless instructed otherwise.

1.1 What pressure is exerted by a sample of nitrogen gas of mass 3.055 g in a container of volume 3.00 dm3 at 32 °C?

1.2 A sample of neon of mass 425 mg occupies 6.00 dm3 at 77 K. What pressure does it exert?

1.3 Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect, a sample was collected in a container of volume 300.0 cm3. At 14.5 °C its pressure is found to be 34.5 kPa. What amount (in moles) of NO has been collected?

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

1.5 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 compress it to 100 cm3 at this temperature?

1.6 You are warned not to dispose of pressurized cans by throwing them on to a fi re. The gas in an aerosol container exerts a pressure of 125 kPa at 18 °C. The container is thrown on a fi re, and its temperature rises to 700 °C. What is the pressure at this temperature?

1.7 Until we fi nd an economical way of extracting oxygen from sea water or lunar rocks, we have to carry it with us to inhospitable places, and do so in compressed form in tanks. A sample of oxygen at 101 kPa is compressed at constant temperature from 7.20 dm3 to 4.21 dm3. Calculate the fi nal pressure of the gas.

1.8 To what temperature must a sample of helium gas be cooled from 22.2 °C to reduce its volume from 1.00 dm3 to 100 cm3?

1.9 Hot-air balloons gain their lift from the lowering of den-sity of air that occurs when the air in the envelope is heated. To what temperature should you heat a sample of air, initially at 315 K, to increase its volume by 25 per cent?

1.10 At sea level, where the pressure was 104 kPa and the temperature 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?

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

1.12 Balloons were used to obtain much of the early infor-mation about the atmosphere and continue to be used today

to obtain weather information. In 1783, Jacques Charles used a hydrogen-fi lled balloon to fl y from Paris 36 km into the French countryside. What is the mass density of hydrogen relative to air at the same temperature and pressure? What mass of payload can be lifted by 10 kg of hydrogen, neglect-ing the mass of the balloon?

1.13 Atmospheric pollution is a problem that has received much attention. Not all pollution, however, is from industrial sources. Volcanic eruptions can be a signifi cant source of air pollution. The Kilauea volcano in Hawaii emits 200–300 t of SO2 per day. If this gas is emitted at 800 °C and 1.0 atm, what volume of gas is emitted?

1.14 A meteorological balloon had a radius of 1.5 m when released at sea level at 20 °C and expanded to a radius of 3.5 m when it had risen to its maximum altitude where the temperature was −25 °C. What is the pressure inside the balloon at that altitude?

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

1.16 The vapour pressure of water at blood temperature is 47 Torr. What is the partial pressure of dry air in our lungs when the total pressure is 760 Torr?

1.17 A determination of the density of a gas or vapour can provide a quick estimate of its molar mass even though for practical work mass spectrometry is far more precise. The density 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?

1.18 In an experiment to measure the molar mass of a gas, 250 cm3 of the gas was confi ned in a glass vessel. The pres-sure was 152 Torr at 298 K and the mass of the gas was 33.5 mg. What is the molar mass of the gas?

1.19 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K. Calculate (a) their partial pressures and (b) the total pressure.

1.20 Use the kinetic model to calculate the root-mean-square speed of (a) N2, (b) H2O molecules in the Earth’s atmosphere at 273 K.

1.21 Use the equipartition theorem, which is outlined in Foundations 0.12, to derive eqn 1.17.

1.22 Calculate the mean speed of (a) He atoms, (b) CH4 molecules at (i) 79 K, (ii) 315 K, (iii) 1500 K.

1.23 A 1.0 dm3 glass bulb contains 1.0 × 1023 H2 molecules. If the pressure exerted by the gas is 100 kPa, what is (a) the temperature of the gas, (b) the root-mean-square speeds of the molecules? (c) Would the temperature be different if they were O2 molecules?

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QUESTIONS AND EXERCISES 45

1.24 Synthesis gas (also known as ‘syngas’) consists of a mixture of hydrogen, H2, and carbon monoxide, CO. Calculate the relative rate, as molecules per second, at which the hydrogen and carbon monoxide escape from a leaking cylin-der of synthesis gas.

1.25 A cylinder containing gas for a carbon dioxide laser con-tains equal amounts of carbon dioxide, nitrogen, and helium. If 1.0 g of carbon dioxide leaks from the cylinder by effusion, what mass of nitrogen and helium escapes?

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

1.27 At what pressure does the mean free path of argon at 25 °C become comparable to 10 times the diameters of the atoms themselves? Take s = 0.36 nm2.

1.28 When studying the photochemical processes that can occur in the upper atmosphere, it is important to know how often atoms and molecules collide. At an altitude of 20 km the temperature is 217 K and the pressure 0.050 atm. What is the mean free path of N2 molecules? Take s = 0.43 nm2.

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

1.30 Calculate the total number of collisions per second in 1.0 dm3 of argon under the same conditions as in Exercise 1.29.

1.31 How many collisions per second does an N2 molecule make at an altitude of 20 km? (See Exercise 1.28 for data.)

1.32 How does the mean free path in a sample of a gas vary with temperature in a constant-volume container?

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

1.34 The critical point of ammonia, NH3, gas occurs at 111.3 atm, 72.5 cm3 mol−1, and 405.5 K. Calculate the com-pression factor at the critical point. What can you infer?

1.35 The virial equation of state may also be written as an expansion in terms of pressure: Z = 1 + B ′p + . . . . The critical constants for water, H2O, are 218.3 atm, 55.3 cm3 mol−1, and 647.4 K. Assuming that the expansion may be truncated after the second term, calculate the value of the second virial coeffi cient B′ at the critical temperature.

1.36 Calculate the pressure exerted by 1.0 mol C2H6 behaving as (a) a perfect gas, (b) a van der Waals gas when it is confi ned 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 1.6.

1.37 How reliable is the perfect gas law in comparison with the van der Waals equation? Calculate the difference in pres-sure of 10.00 g of carbon dioxide confi ned to a container of volume 100 cm3 at 25.0 °C between treating it as a perfect gas and a van der Waals gas.

1.38 A certain gas obeys the van der Waals equation with a = 0.50 m6 Pa mol−2. Its molar volume is found to be 5.00 × 10−4 m3 mol−1 at 273 K and 3.0 MPa. From this infor-mation calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure?

1.39 Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Hint: The expan-sion you will need is (1 − x)−1 = 1 + x + x2 + . . . .

1.40 Measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coeffi cients at 273 K. What are the values of a and b in the corresponding van der Waals equation of state? Hint: Use the expression for B derived in Exercise 1.39.

1.41 Show that there is a temperature at which the second virial coeffi cient, B, is zero for a van der Waals gas, and calcu-late its value for carbon dioxide. Hint: Use the expression for B derived in Exercise 1.39.

1.42 The critical constants of ethane are pc = 48.20 atm, Vc = 148 cm3 mol−1, and Tc = 305.4 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules.

Projects

The symbol ‡ indicates that calculus is required.

1.43‡ In the following you are invited to explore the Maxwell distribution of speeds in more detail. (a) Confi rm that the mean speed of molecules of molar mass M at a temperature T is equal to (8RT/pM )1/2. Hint : You will need an integral of the form ∫0

∞ x3e−ax2

dx = n!/2a2. (b) Confi rm that the root-mean-square speed of molecules of molar mass M at a tempera-ture T is equal to (3RT/M )1/2 and hence confi rm eqn 1.17.

Hint: You will need an integral of the form ∫0∞ x4e−ax2

dx = (3/8a2)(p/a)1/2. (c) Find an expression for the most probable speed of molecules of molar mass M at a temperature T. Hint: Look for a maximum in the Maxwell distribution (the maximum occurs as dF/ds = 0). (d) Estimate the fraction of N2 molecules at 500 K that has speeds in the range 290 to 300 m s−1.

1.44‡ Here we explore the van der Waals equation of state. Using the language of calculus, the critical point of a van der

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Waals gas occurs where the isotherm has a fl at infl exion, which is where dp/dVm = 0 (zero slope) and d2p/dV 2

m = 0 (zero curvature). (a) Evaluate these two expressions using eqn 1.26b, and fi nd expressions for 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 3/8.

1.45 The kinetic model of gases is valid when the size of the particles is negligible compared with their mean free path. It may seem absurd, therefore, to expect the kinetic theory and, as a consequence, the perfect gas law, to be applicable to the dense matter of stellar interiors. In the Sun, for instance, the density is 150 times that of liquid water at its centre and comparable to that of water about halfway to its surface. However, we have to realize that the state of matter is that of a plasma, in which the electrons have been stripped from the atoms of hydrogen and helium that make up the bulk of the matter of stars. As a result, the particles making up the plasma have diameters comparable to those of nuclei, or about 10 fm. Therefore, a mean free path of only 0.1 pm satisfi es the criterion for the validity of the kinetic model and the perfect gas law. We can therefore use pV = nRT as the equation of state for the stellar interior. (a) Calculate the pres-sure halfway to the centre of the Sun, assuming that the

interior consists of ionized hydrogen atoms, the temperature is 3.6 MK, and the mass density is 1.20 g cm−3 (slightly higher than the density of water). (b) Combine the result from part (a) with the expression for the pressure from kinetic model to show that the pressure of the plasma is related to its kinetic energy density rk = Ek/V, the kinetic energy of the molecules in a region divided by the volume of the region, by p = (2/3)rk. (c) What is the kinetic energy density halfway to the centre of the Sun? Compare your result with the (translational) kinetic energy density of the Earth’s atmosphere on a warm day (25 °C): 1.5 × 105 J m−3 (corresponding to 0.15 J cm−3). (d) A star eventually depletes some of the hydrogen 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 which result 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 half-way to the centre a red giant has a temperature of 3500 K, is composed primarily of fully ionized carbon atoms and elec-trons, and has a mass density of 1200 kg m−3. What is the pressure at this point? (e) If the red giant in part (d) consisted of neutral carbon atoms, what would be the pressure at the same point under the same conditions?

CHAPTER 1 THE PROPERTIES OF GASES46

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