Black plate (7,1)
CH AP TER
2
Properties of Gases
The study of the behavior of gases has given rise to numerous chemical and physical
theories. In many ways, the gaseous state is the easiest to investigate. In this chapter,
we examine several gas laws based on experimental observations, introduce the concept of temperature, and discuss the kinetic theory of gases.
2.1 Some Basic Definitions
Before we discuss the gas laws, it is useful to define a few basic terms that will be used
throughout the book. We often speak of the system in reference to a particular part of
the universe in which we are interested. Thus, a system could be a collection of oxygen molecules in a container, a NaCl solution, a tennis ball, or a Siamese cat. Having
defined a system, we call the rest of the universe the surroundings. There are three
types of systems (Figure 2.1). An open system is one that can exchange both mass and
energy with its surroundings. A closed system is one that does not exchange mass
with its surroundings but can exchange energy. An isolated system is one that can
exchange neither mass nor energy with its surroundings. To completely define a system, we need to understand certain experimental variables, such as pressure, volume,
temperature, and composition, that collectively describe the state of the system.
A system is separated from
the surroundings by definite
boundaries such as walls or
surfaces.
Water vapor
Heat
Heat
(a)
(b)
(c)
Figure 2.1
(a) An open system allows the exchange of both mass and energy; (b) a closed system allows
the exchange of energy but not mass; and (c) an isolated system allows exchange of neither
mass nor energy.
7
Black plate (8,1)
8
Chapter 2: Properties of Gases
Most of the properties of matter may be divided into two classes: extensive
properties and intensive properties. Consider, for example, two beakers containing the
same amounts of water at the same temperature. If we combine these two systems by
pouring the water from one beaker to the other, we find that the volume of the water
is doubled and so is its mass. On the other hand, the temperature and the density of
the water do not change. Properties whose values are directly proportional to the
amount of the material present in the system are called extensive properties; those
that do not depend on the amount are called intensive properties. Extensive properties include mass, area, volume, energy, and electrical charge. As mentioned, temperature and density are both intensive properties; so are pressure and electrical
potential.
2.2 An Operational Definition of Temperature
PA',VA'
PA'',VA''
P
PA''',VA'''
PA'''',VA''''
V
Figure 2.2
Plot of pressure versus volume at
constant temperature for a given
amount of a gas. Such a graph is
called an isotherm.
Temperature is a very important quantity in many branches of science, and not surprisingly, it can be defined in di¤erent ways. Daily experience tells us that temperature is a measure of coldness and hotness, but for our purposes we need a more
precise operational definition. Consider the following system of a container of gas A.
The walls of the container are flexible so that its volume can expand and contract.
This is a closed system that allows heat but not mass to flow into and out of the
container. The initial pressure ðPÞ and volume ðV Þ are PA and VA . Now we bring the
container in contact with a similar container of gas B at PB and VB . Heat exchange
will take place until thermal equilibrium is reached. At equilibrium, the pressure and
volume of A and B will be altered to PA0 ; VA0 and PB0 ; VB0 . It is possible to remove
container A temporarily, readjust its pressure and volume to PA00 and VA00 , and still
have A in thermal equilibrium with B at PB0 and VB0 . In fact, an infinite set of such
values ðPA0 ; VA0 Þ; ðPA00 ; VA00 Þ; ðPA000 ; VA000 Þ; . . . can be obtained that will satisfy the equilibrium conditions. Figure 2.2 shows a plot of these points.
For all these states of A to be in thermal equilibrium with B, they must have the
same value of the variable we call temperature. It follows that if two systems are in
thermal equilibrium with a third system, they must also be in thermal equilibrium
with each other. This statement is generally known as the zeroth law of thermodynamics. The curve in Figure 2.2 is the locus of all the points that represent the states
that can be in thermal equilibrium with system B. Such a curve is called an isotherm,
or ‘‘same temperature.’’ At another temperature, a di¤erent isotherm is obtained.
2.3 Ideal Gases
In this section we shall briefly examine the various gas laws formulated for the
behavior of an ideal gas.
Boyle’s Law
In a 1662 study of the physical behavior of gases, the English chemist Robert
Boyle (1627–1691) found that the volume ðV Þ of a given amount of gas at constant
temperature is inversely proportional to its pressure ðPÞ:
Vm
1
P
or
PV ¼ constant
ð2:1Þ
Black plate (9,1)
2.3 Ideal Gases
Equation 2.1 is known as Boyle’s law. A plot of P versus V at a given temperature
gives a hyperbola, which is the isotherm illustrated in Figure 2.2.
Boyle’s law is used to predict the pressure of a gas when its volume changes
and vice versa. Letting the initial values of pressure and volume be P1 and V1 and the
final values of pressure and volume be P2 and V2 , we have
P1 V1 ¼ P2 V2
ðconstant n and temperatureÞ
ð2:2Þ
where n is the number of moles of the gas present.
Charles’ and Gay-Lussac’s Law
Boyle’s law depends on the amount of gas and the temperature of the system
remaining constant. But suppose the temperature changes. How does a change in
temperature a¤ect the volume and pressure of a gas? Let us first look at the e¤ect of
temperature on the volume of a gas. The earliest investigators of this relationship
were the French physicists Jacques Alexandre Charles (1746–1823) and Joseph Louis
Gay-Lussac (1778–1850). Their studies showed that, at constant pressure, the volume
of a gas sample expands when heated and contracts when cooled. The quantitative
relations involved in changes in gas temperature and volume turn out to be remarkably consistent. For example, we observe an interesting phenomenon when we study
the temperature–volume relationship at various pressures. At any given pressure,
the plot of volume versus temperature yields a straight line. By extending the line to
zero volume, we find the intercept on the temperature axis to be 273:15 C. At any
other pressure, we obtain a di¤erent straight line for the volume–temperature plot,
but we get the same zero-volume temperature intercept at 273:15 C (Figure 2.3).
In
V
cr
ea
sin
P1
gp
e
P4
sur
P3
res
P2
273.15 °C
Figure 2.3
Plots of the volume of a given amount of gas versus
temperature ðtÞ at di¤erent pressures. All gases ultimately
condense if they are cooled to low enough temperatures.
When these lines are extrapolated, they all converge at
the point representing zero volume and a temperature
of 273.15 C.
t
(In practice, we can measure the volume of a gas over only a limited temperature
range, because all gases condense at low temperatures to form liquids.)
In 1848, the Scottish mathematician and physicist William Thomson (Lord
Kelvin, 1824–1907) realized the significance of this phenomenon. He identified
273:15 C as absolute zero, which is theoretically the lowest attainable temperature.
Then he set up an absolute temperature scale, now called the kelvin temperature
scale, with absolute zero as the starting point. On the kelvin scale, one kelvin (K) is
equal in magnitude to one degree Celsius. The only di¤erence between the absolute
temperature scale and the Celsius scale is that the zero position is shifted. The relation between the two scales is
T=K ¼ t= C þ 273:15
Dividing the symbol by the
unit gives us a pure number.
Thus, if T ¼ 298 K, then
T/K ¼ 298.
9
Black plate (10,1)
10
Chapter 2: Properties of Gases
Important points on the two scales match up as follows:
The normal freezing point and
normal boiling point are measured at 1 atm pressure.
Absolute zero
Freezing point of water
Boiling point of water
Kelvin Scale
Celsius Scale
0K
273.15 K
373.15 K
273.15 C
0 C
100 C
In most cases, we shall use 273 instead of 273.15 as the term relating the two scales.
By convention, we use T to denote absolute (kelvin) temperature and t to indicate
temperature on the Celsius scale. As we shall soon see, the absolute zero of temperature has major theoretical significance; absolute temperatures must be used in gas
law problems and thermodynamic calculations.
At constant pressure, the volume of a given amount of gas is directly proportional to the absolute temperature:
V mT
or
V
¼ constant
T
ð2:3Þ
Equation 2.3 is known as Charles’ law, or the law of Charles and Gay-Lussac. An
alternative form of Charles’ law relates the pressure of a given amount of gas to its
temperature at constant volume:
PmT
or
P
¼ constant
T
ð2:4Þ
Equations 2.3 and 2.4 permit us to relate the volume–temperature and pressure–
temperature values of a gas in states 1 and 2 as follows:
V1 V2
¼
T1 T2
ðconstant n and PÞ
ð2:5Þ
P1 P2
¼
T1 T2
ðconstant n and V Þ
ð2:6Þ
A practical consequence of Equation 2.6 is that automobile tire pressure should be
checked only when the car has been idle for a while. After a long drive, tires become
quite hot and the air pressure in them rises.
Avogadro’s Law
Another important gas law was formulated by Amedeo Avogadro in 1811. He
proposed that equal volumes of gases at the same temperature and pressure contain
the same number of molecules. This concept means that
V mn
Black plate (11,1)
2.3 Ideal Gases
or
V
¼ constant
n
ðconstant T and PÞ
ð2:7Þ
Equation 2.7 is known as Avogadro’s law.
The Ideal-Gas Equation
According to Equations 2.1, 2.3, and 2.7, the volume of a gas depends on the
pressure, temperature, and number of moles as follows:
Vm
1
P
ðconstant T and nÞ
ðBoyle’s lawÞ
V mT
ðconstant P and nÞ
ðCharles’ lawÞ
V mn
ðconstant T and PÞ
ðAvogadro’s lawÞ
Therefore, V must be proportional to the product of these three terms, that is,
Vm
nT
P
V ¼R
nT
P
or
PV ¼ nRT
(2.8)
where R, a proportionality constant, is the gas constant. Equation 2.8 is called the
ideal-gas equation. The ideal-gas equation is an example of an equation of state,
which provides the mathematical relationships among the properties that define the
state of the system, such as P; T, and V .
The value of R can be obtained as follows. Experimentally, it is found that 1
mole of an ideal gas occupies 22.414 L at 1 atm and 273.15 K (a condition known as
standard temperature and pressure, or STP). Thus,
R¼
ð1 atmÞð22:414 LÞ
¼ 0:08206 L atm K1 mol1
ð1 molÞð273:15 KÞ
To express R in units of J K1 mol1 , we use the conversion factors
1 atm ¼ 1:01325 10 5 Pa
1 L ¼ 1 103 m 3
and obtain
R¼
ð1:01325 10 5 N m2 Þð22:414 103 m 3 Þ
ð1 molÞð273:15 KÞ
¼ 8:314 N m K1 mol1
¼ 8:314 J K1 mol1
ð1 J ¼ 1 N mÞ
11
Black plate (12,1)
12
Chapter 2: Properties of Gases
From the two values of R, we can write
0:08206 L atm K1 mol1 ¼ 8:314 J K1 mol1
or
1 L atm ¼ 101:3 J
and
1 J ¼ 9:87 103 L atm
Example 2.1
Air entering the lungs ends up in tiny sacs called alveoli, from which oxygen di¤uses
into the blood. The average radius of the alveoli is 0.0050 cm, and the air inside contains
14 mole percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and the
temperature is 37 C, calculate the number of oxygen molecules in one of the alveoli.
ANSWER
The volume of one alveolus is
4
4
V ¼ pr 3 ¼ pð0:0050 cmÞ 3
3
3
¼ 5:2 107 cm 3 ¼ 5:2 1010 L ð1 L ¼ 10 3 cm 3 Þ
The number of moles of air in one alveolus is given by
n¼
PV
ð1:0 atmÞð5:2 1010 LÞ
¼
¼ 2:0 1011 mol
RT ð0:08206 L atm K1 mol1 Þð310 KÞ
Because the air inside the alveolus is 14% oxygen, the number of oxygen molecules is
2:0 1011 mol air 14% O2
6:022 10 23 O2 molecules
1 mol O2
100% air
¼ 1:7 10 12 O2 molecules
Dalton’s Law of Partial Pressures
So far, we have discussed the pressure–volume–temperature behavior of a pure
gas. Frequently, however, we work with mixtures of gases. For example, a chemist
researching the depletion of ozone in the atmosphere must deal with several gaseous
components. For a system containing two or more di¤erent gases, the total pressure
ðPT Þ is the sum of the individual pressures that each gas would exert if it were alone
and occupied the same volume. Thus,
PT ¼ P1 þ P2 þ PT ¼
X
i
Pi
(2.9)
Black plate (13,1)
2.3 Ideal Gases
P
where P1 ; P2 ; . . . are the individual or partial pressures of components 1; 2; . . . and
is the summation sign. Equation 2.9 is known as Dalton’s law of partial pressures
(after the English chemist and mathematician John Dalton, 1766–1844).
Consider a system containing two gases (1 and 2) at temperature T and volume
V . The partial pressures of the gases are P1 and P2 , respectively. From Equation 2.8,
P1 V ¼ n1 RT
or
P1 ¼
n1 RT
V
P2 V ¼ n2 RT
or
P2 ¼
n2 RT
V
where n1 and n2 are the numbers of moles of the two gases. According to Dalton’s
law,
PT ¼ P1 þ P2
¼ n1
RT
RT
þ n2
V
V
¼ ðn1 þ n2 Þ
RT
V
Dividing the partial pressures by the total pressure and rearranging, we get
P1 ¼
n1
PT ¼ x1 PT
n1 þ n2
P2 ¼
n2
PT ¼ x2 PT
n1 þ n2
and
where x1 and x2 are the mole fractions of gases 1 and 2. A mole fraction, defined as
the ratio of the number of moles of one gas to the total number of moles of all gases
present, is a dimensionless quantity. Furthermore, by definition, the sum of all the
mole fractions in a mixture must be unity, that is
X
xi ¼ 1
ð2:10Þ
i
In general, in a mixture of gases the partial pressure of the ith component, Pi , is
given by
Pi ¼ xi PT
(2.11)
How are partial pressures determined? A manometer can measure only the total
pressure of a gaseous mixture. To obtain partial pressures, we need to know the mole
fractions of the components. The most direct method of measuring partial pressures
is using a mass spectrometer. The relative intensities of the peaks in a mass spectrum
are directly proportional to the amounts, and hence to the mole fractions, of the
gases present.
The gas laws played a key role in the development of atomic theory, and many
practical illustrations of them appear in everyday life. The two brief examples below
13
Black plate (14,1)
14
Chapter 2: Properties of Gases
The pressure exerted by seawater is called the hydrostatic
pressure.
are particularly important to scuba divers. Seawater has a slightly higher density than
fresh water—approximately 1.03 g mL1 compared with 1.00 g mL1 . The pressure
exerted by a 33-ft (10-m) column of seawater is equivalent to 1 atm pressure. What
would happen if a diver were to rise to the surface rather quickly, holding his breath?
If the ascent started at 40 ft under water, the decrease in pressure from this depth to
the surface would be (40 ft/33 ft) 1 atm, or 1.2 atm. Assuming constant temperature, when the diver reached the surface, the volume of air trapped in his lungs would
have increased by a factor of ð1 þ 1:2Þ atm/1 atm, or 2.2 times! This sudden expansion of air could damage or rupture the membranes of his lungs, seriously injuring or
killing the diver.
Dalton’s law has a direct application to scuba diving. The partial pressure of
oxygen in air is approximately 0.2 atm. Because oxygen is essential for our survival, we may have a hard time believing that it could be harmful to breathe in more
than normal. In fact, the toxicity of oxygen is well documented.* Physiologically, our
bodies function best when the partial pressure of oxygen is 0.2 atm. For this reason,
the composition of the air in a scuba tank is adjusted when the diver is submerged.
For example, at a depth where the total pressure (hydrostatic plus atmospheric) is
4 atm, the oxygen content in the air supply should be reduced to 5% by volume
to maintain the optimal partial pressure ð0:05 4 atm ¼ 0:2 atmÞ. At a greater
depth, the oxygen content must be even lower. Although nitrogen would seem to be
the obvious choice for mixing with oxygen in a scuba tank because it is the major
component of air, it is not the best choice. When the partial pressure of nitrogen
exceeds 1 atm, a su‰cient amount will dissolve in the blood to cause nitrogen narcosis. Symptoms of this condition, which resembles alcohol intoxication, include lightheadedness and impaired judgment. Divers su¤ering from nitrogen narcosis have
been known to do strange things, such as dancing on the sea floor and chasing
sharks. For this reason, helium is usually employed to dilute oxygen in diving tanks.
Helium, an inert gas, is much less soluble in blood than nitrogen, and it does not
produce narcotic e¤ects.
2.4 Real Gases
The ideal-gas equation holds only for gases that have the following properties: (1) the
gas molecules possess negligible volume, and (2) there is no interaction, attractive or
repulsive, among the molecules. Obviously, no such gases exist. Nevertheless, Equation 2.8 is quite useful for many gases at high temperatures or moderately low pressures (a10 atm).
When a gas is being compressed, the molecules are brought closer to one another, and the gas will deviate appreciably from ideal behavior. One way to measure
the deviation from ideality is to plot the compressibility factor ðZÞ of a gas versus
pressure. Starting with the ideal-gas equation, we write
PV ¼ nRT
or
* At partial pressures above 2 atm, oxygen becomes toxic enough to produce convulsions and
coma. Years ago, newborn infants placed in oxygen tents often developed retrolental fibroplasia,
damage of the retinal tissues by excess oxygen. This damage usually resulted in partial or total
blindness.
Black plate (15,1)
2.4 Real Gases
15
N2
2.0
CH 4
He
1.5
Z
Ideal gas
1.0
0.5
0
200
400
600
800
1000
P /atm
Figure 2.4
Plot of the compressibility factor versus pressure for real gases and an ideal gas at 273 K.
Note that for an ideal gas Z ¼ 1, no matter how great the pressure.
Z¼
PV
PV
¼
nRT RT
(2.12)
where V is the molar volume of the gas ðV =nÞ or the volume of 1 mole of the gas at
the specified temperature and pressure. For an ideal gas, Z ¼ 1 for any value of P at
a given T. However, as Figure 2.4 shows, the compressibility factors for real gases
exhibit fairly divergent dependence on pressure. At low pressures, the compressibility
factors of most gases are close to unity. In fact, in the limit of P approaching zero, we
have Z ¼ 1 for all gases. This finding is expected because all real gases behave ideally
at low pressures. As pressure increases, some gases have Z < 1, which means that
they are easier to compress than an ideal gas. Then, as pressure increases further, all
gases have Z > 1. Over this region, the gases are harder to compress than an ideal
gas. These behaviors are consistent with our understanding of intermolecular forces.
In general, attractive forces are long-range forces, whereas repulsive forces operate
only within a short range (more on this topic in Chapter 13). When molecules are far
apart (for example, at low pressures), the predominant intermolecular interaction is
attraction. As the distance of separation between molecules decreases, the repulsive
interaction among molecules becomes more significant.
Over the years, considerable e¤ort has gone into modifying the ideal-gas equation for real gases. Of the numerous such equations proposed, we shall consider two:
the van der Waals equation and the virial equation of state.
Figure 2.5
The van der Waals Equation
The van der Waals equation of state (after the Dutch physicist Johannes Diderik
van der Waals, 1837–1923) attempts to account for the finite volume of individual
molecules in a nonideal gas and the attractive forces between them.
an 2
P þ 2 ðV nbÞ ¼ nRT
V
(2.13)
The pressure exerted by the individual molecules on the walls of the container depends on both the frequency of molecular collisions with the walls and the momentum imparted by the molecules to the walls. Both contributions are diminished by the
attractive intermolecular forces (Figure 2.5). In each case, the reduction in pressure
E¤ect of intermolecular forces
on the pressure exerted by a gas.
The speed of a molecule that is
moving toward the container
wall (red sphere) is reduced by
the attractive forces exerted by
its neighbors (gray spheres).
Consequently, the impact this
molecule makes with the wall is
not as great as it would be if no
intermolecular forces were present. In general, the measured
gas pressure is lower than the
pressure the gas would exert if
it behaved ideally.
Black plate (16,1)
16
Chapter 2: Properties of Gases
Table 2.1
van der Waals Constants and Boiling Points of Some Substances
Substance
a/atm L 2 mol2
b/L mol1
Boiling Point/K
He
Ne
Ar
H2
N2
O2
CO
CO2
CH4
H2 O
NH3
0.0341
0.214
1.34
0.240
1.35
1.34
1.45
3.60
2.26
5.47
4.25
0.0237
0.0174
0.0322
0.0264
0.0386
0.0312
0.0395
0.0427
0.0430
0.0305
0.0379
4.2
27.2
87.3
20.3
77.4
90.2
83.2
195.2
109.2
373.15
239.8
depends on the number of molecules present or the density of the gas, n=V , so that
n
n
reduction in pressure due to attractive forces m
V
V
¼a
n2
V2
where a is a proportionality constant.
Note that in Equation 2.13, P is the measured pressure of the gas and
ðP þ an 2 =V 2 Þ would be the pressure of the gas if there were no intermolecular forces
present. Because an 2 =V 2 must have units of pressure, a is expressed as atm L 2 mol2 .
To allow for the finite volume of molecules, we replace V in the ideal-gas equation
with ðV nbÞ, where nb represents the total e¤ective volume of n moles of the gas.
Thus, nb must have the unit of volume, and b has the units L mol1 . Both a and b are
constants characteristic of the gas under study. Table 2.1 lists the values of a and b
for several gases. The value of a is related to the magnitude of attractive forces. Using
the boiling point as a measure of the strength of intermolecular forces (the higher the
boiling point, the stronger the intermolecular forces), we see that there is a rough
correlation between the values of a and the boiling points of these substances. The
quantity b is more di‰cult to interpret. Although b is proportional to the size of the
molecule, the correlation is not always straightforward. For example, the value of b
for helium is 0.0237 L mol1 and that for neon is 0.0174 L mol1 . Based on these
values, we might expect that helium is larger than neon, which we know is not true.
The van der Waals equation is valid over a wider range of pressure and temperature than is the ideal-gas equation. Furthermore, it provides a molecular interpretation for the equation of state. At very high pressures and low temperatures, the van
der Waals equation also becomes unreliable.
The Virial Equation of State
Another way of representing gas nonideality is the virial equation of state. In this
relationship, the compressibility factor is expressed as a series expansion in inverse
powers of molar volume V :
Z ¼1þ
B
C
D
þ
þ þ
V V2 V3
(2.14)
Black plate (17,1)
2.4 Real Gases
where B; C; D, are called the second, third, fourth, . . . virial coe‰cients. The first
virial coe‰cient is 1. The second and higher virial coe‰cients are all temperature
dependent. For a given gas, they are evaluated from the P–V –T data of the gas by
a curve-fitting procedure using a computer. For an ideal gas, the second and higher
virial coe‰cients are zero and Equation 2.14 becomes Equation 2.8.
An alternate form of the virial equation is given by a series expansion of the
compressibility factor in terms of the pressure, P:
Z ¼ 1 þ B 0P þ C 0P2 þ D 0P3 þ (2.15)
Because P and V are related, it is not surprising that relationships exist between B
and B 0 , C and C 0 , and so on. In each equation, the values of the coe‰cients decrease
rapidly. For example, in Equation 2.15, the magnitude of the coe‰cients is such that
B 0 g C 0 g D 0 so that at pressures between zero and 10 atm, say, we need to include
only the second term provided the temperature is not very low:
Z ¼ 1 þ B 0P
ð2:16Þ
Equations 2.13 and 2.14 or 2.15 exemplify two rather di¤erent approaches. The
van der Waals equation accounts for the nonideality of gases by correcting for the
finite molecular volume and intermolecular forces. Although these corrections do
result in a definite improvement over the ideal gas equation, Equation 2.13 is still
an approximate equation. The reason is that our present knowledge of intermolecular forces is insu‰cient to quantitatively explain macroscopic behavior. Of course,
we could further improve this equation by adding more corrective terms; indeed,
numerous other equations of state have been proposed since van der Waals first presented his analysis. On the other hand, Equation 2.14 is accurate for real gases, but it
does not provide us with any direct molecular interpretation. The nonideality of the
gas is accounted for mathematically by a series expansion in which the coe‰cients
B; C; . . . can be determined experimentally. These coe‰cients do not have any physical meaning, although they can be related to intermolecular forces in an indirect
way. Thus, our choice in this case is between an approximate equation that gives us
some physical insight or an equation that describes the gas behavior accurately (if the
coe‰cients are known), but tells us nothing about molecular behavior.
Example 2.2
Calculate the molar volume of methane at 300 K and 100 atm, given that the second
virial coe‰cient ðBÞ of methane is 0.042 L mol1 . Compare your result with that
obtained using the ideal-gas equation.
ANSWER
From Equation 2.14, neglecting terms containing C; D; . . . ;
Z ¼1þ
B
V
¼1þ
BP
RT
¼1þ
ð0:042 L mol1 Þð100 atmÞ
ð0:08206 L atm K1 mol1 Þð300 KÞ
¼ 1 0:17 ¼ 0:83
17
Black plate (18,1)
18
Chapter 2: Properties of Gases
V¼
¼
ZRT
P
ð0:83Þð0:08206 L atm K1 mol1 Þð300 KÞ
100 atm
¼ 0:20 L mol1
For an ideal gas,
V¼
¼
RT
P
ð0:08206 L atm K1 mol1 Þð300 KÞ
100 atm
¼ 0:25 L mol1
COMMENT
At 100 atm and 300 K, methane is more compressible than an ideal gas (Z ¼ 0:83
compared with Z ¼ 1) due to the attractive intermolecular forces between the CH 4
molecules.
2.5 Condensation of Gases and the Critical State
The condensation of gas to liquid is a familiar phenomenon. The first quantitative
study of the pressure–volume relationship of this process was made in 1869 by the
Irish chemist Thomas Andrews (1813–1885). He measured the volume of a given
amount of carbon dioxide as a function of pressure at various temperatures and
obtained a series of isotherms like those shown in Figure 2.6. At high temperatures,
the curves are roughly hyperbolic, indicating that the gas obeys Boyle’s law. As the
temperature is lowered, deviations become evident, and a drastically di¤erent behavior is observed at T4 . Moving along the isotherm from right to left, we see that
Critical point
P
T7
Figure 2.6
Isotherms of carbon dioxide at various temperatures
(temperature increases from T1 to T7 ). The critical
temperature is T5 . Above this temperature carbon
dioxide cannot be liquefied no matter how great the
pressure.
T6
T5
T4
Liquid and vapor
at equilibrium
T3
T2
T1
V
Black plate (19,1)
2.5 Condensation of Gases and the Critical State
P
(b)
Gas
(a)
(c)
Gas
Liquid
Liquid
T1
V
(a)
(b)
(c)
(d)
Figure 2.7
The liquefaction of carbon dioxide at T1 (see Figure 2.6). At (a), the first drop of the liquid
appears. From (b) to (c) the gas is gradually and completely converted to liquid at constant
pressure. Beyond (c) the volume decreases only slightly with increasing pressure because
liquids are highly incompressible. As temperature increases, the horizontal line becomes
shorter until it becomes a point at T5 , the critical temperature.
although the volume of the gas decreases with pressure, the product PV is no longer
a constant (because the curve is no longer a hyperbola). Increasing the pressure further, we reach a point that is the intersection between the isotherm and the dashed
curve on the right. If we could observe this process, we would note the formation of
liquid carbon dioxide at this pressure. With the pressure held constant, the volume
continues to decrease (as more vapor is converted to liquid) until all the vapor has
been condensed. Beyond this point (the intersection between the horizontal line and
the dashed curve on the left), the system is entirely liquid, and any further increase in
pressure will result in only a very small decrease in volume, because liquids are much
less compressible than gases (Figure 2.7).
The pressure corresponding to the horizontal line (the region in which vapor and
liquid coexist) is called the equilibrium vapor pressure or simply the vapor pressure of
the liquid at the temperature of the experiment. The length of the horizontal line
decreases with increasing temperature. At a particular temperature (T5 in this case),
the isotherm is tangential to the dashed curve and only one phase (the gas phase) is
present. The horizontal line is now at a point known as the critical point. The corresponding temperature, pressure, and volume at this point are called the critical temperature ðTc Þ, critical pressure ðPc Þ, and critical volume ðVc Þ, respectively. The critical temperature is the temperature above which no condensation can occur no matter
how great the pressure. The critical constants of several gases are listed in Table 2.2.*
Note that the critical volume is usually expressed as a molar quantity, called the
molar critical volume ðVc Þ, given by Vc =n, where n is the number of moles of the
substance present.
The phenomenon of condensation and the existence of a critical temperature are
direct consequences of the nonideal behavior of gases. After all, if molecules did not
attract one another, no condensation would occur and if molecules had no volume,
then we would not be able to observe liquids. As mentioned earlier, the nature of
molecular interaction is such that the force among molecules is attractive when they
* The van der Waals constants of a gas (see Equation 2.13) can be obtained from its critical constants. For mathematical details see the physical chemistry texts listed in Chapter 1.
19
Black plate (20,1)
20
Chapter 2: Properties of Gases
Table 2.2
Critical Constants of Some Substances
Substance
Pc /atm
Vc =L mol1
Tc =K
He
Ne
Ar
H2
N2
O2
CO
CO2
CH4
H2 O
NH3
SF6
2.25
26.2
49.3
12.8
33.6
50.8
34.5
73.0
45.4
217.7
109.8
37.6
0.0578
0.0417
0.0753
0.0650
0.0901
0.0764
0.0931
0.0957
0.0990
0.0560
0.0724
0.2052
5.2
44.4
151.0
32.9
126.1
154.6
132.9
304.2
190.2
647.6
405.3
318.7
are relatively far apart, but as they get closer to one another (for example, a liquid
under pressure) this force becomes repulsive, because of electrostatic repulsions between nuclei and between electrons. In general, the attractive force reaches a maximum at a certain finite intermolecular distance. At temperatures below Tc , it is possible to compress the gas and bring the molecules within this attractive range, where
condensation will occur. Above Tc , the kinetic energy of the gas molecules is such
that they will always be able to break away from this attraction and no condensation can take place. Figure 2.8 shows the critical phenomenon of sulfur hexafluoride
(SF6 ).
(a)
(b)
(c)
(d)
Figure 2.8
The critical phenomenon of sulfur hexafluoride (Tc ¼ 45:5 C; Pc ¼ 37:6 atm). (a) Below
the critical temperature, a clear liquid phase is visible. (b) Above the critical temperature,
the liquid phase disappears. (c) The substance is cooled just below its critical temperature.
(d) Finally, the liquid phase reappears.
Black plate (21,1)
2.6 Kinetic Theory of Gases
In recent years, there has been much interest in the practical applications of
supercritical fluids (SCF), that is, of the state of matters above the critical temperature. One of the most studied SCFs is carbon dioxide. Under appropriate conditions
of temperature and pressure, SCF CO2 can be used as a solvent for removing ca¤eine
from raw co¤ee beans and cooking oil from potato chips to produce crisp, oil-free
chips. It is also being used in environmental cleanups because it dissolves chlorinated
hydrocarbons. SCFs of CO2 , NH3 , and certain hydrocarbons such as hexane and
heptane are used in chromatography. SCF CO2 has been shown to be an e¤ective
carrier medium for substances such as antibiotics and hormones, which are unstable
at the high temperatures required for normal chromatographic separations.
2.6 Kinetic Theory of Gases
The study of gas laws exemplifies the phenomenological, macroscopic approach to
physical chemistry. Equations describing the gas laws are relatively simple, and experimental data are readily accessible. Yet studying gas laws gives us no real physical
insight into processes that occur at the molecular level. Although the van der Waals
equation attempts to account for nonideal behavior in terms of intermolecular interactions, it does so in a rather vague manner. It does not answer such questions as:
How is the pressure of a gas related to the motion of individual molecules, and why
do gases expand when heated at constant pressure? The next logical step, then, is
to try to explain the behavior of gases in terms of the dynamics of molecular motion.
To interpret the properties of gas molecules in a more quantitative manner, we turn
to the kinetic theory of gases.
The Model
Any time we try to develop a theory to account for experimental observations,
we must first define our system. If we do not understand all the properties of a system, as is usually the case, we must make a number of assumptions. Our model for
the kinetic theory of gases is based on the following assumptions:
1. A gas is made up of a great number of atoms or molecules, separated by
distances that are large compared to their size.
2. The molecules have mass, but their volume is negligibly small.
3. The molecules are constantly in random motion.
4. Collisions among molecules and between molecules and the walls of the
container are elastic; that is, kinetic energy may be transferred from one
molecule to another, but it is not converted to other forms of energy.
5. There is no interaction, attractive or repulsive, between the molecules.
Assumptions 2 and 5 should be familiar from our discussion of ideal gases. The difference between the ideal-gas laws and the kinetic theory of gases is that for the latter,
we shall use the foregoing assumptions in an explicit manner to derive expressions for
macroscopic properties, such as pressure and temperature, in terms of the motion of
individual molecules.
Pressure of a Gas
Using the model for the kinetic theory of gases, we can derive an expression
for the pressure of a gas in terms of its molecular properties. Consider an ideal gas
21
Black plate (22,1)
22
Chapter 2: Properties of Gases
z
Velocity vector v
vz
Figure 2.9
Velocity vector v and its components
along the x; y, and z directions.
vy
O
y
vx
x
vx 2
vy 2
A
made up of N molecules, each of mass m, confined in a cubic box of length l. At
any instant, the molecular motion inside the container is completely random. Let
us analyze the motion of a particular molecule with velocity v. Because velocity is a
vector quantity—it has both magnitude and direction—v can be resolved into three
mutually perpendicular components vx ; vy , and vz . These three components give the
rates at which the molecule is moving along the x; y, and z directions, respectively; v
is simply the resultant velocity (Figure 2.9). The projection of the velocity vector on
the xy plane is OA, which, according to Pythagoras’ theorem, is given by
OA 2 ¼ vx2 þ vy2
Similarly,
vx
After
vx
Before
Figure 2.10
Change in velocity upon collision of a molecule moving with
vx with the wall of the container.
v 2 ¼ OA 2 þ vz2
¼ vx2 þ vy2 þ vz2
ð2:17Þ
Let us for the moment consider the motion of a molecule only along the x direction. Figure 2.10 shows the changes that take place when the molecule collides
with the wall of the container (the yz plane) with velocity component vx . Because the
collision is elastic, the velocity after collision is the same as before but opposite in
direction. The momentum of the molecule is mvx , where m is its mass, so that the
change in momentum is given by
mvx mðvx Þ ¼ 2mvx
The sign of vx is positive when the molecule moves from left to right and negative
when it moves in the opposite direction. Immediately after the collision, the molecule
will take time l=vx to collide with the other wall, and in time 2l=vx the molecule will
strike the same wall again.* Thus, the frequency of collision between the molecule
and a given wall (that is, the number of collisions per unit time) is vx =2l, and the
change in momentum per unit time is ð2mvx Þðvx =2lÞ, or mvx2 =l. According to Newton’s second law of motion,
force ¼ mass acceleration
¼ mass distance time2
¼ momentum time1
* We assume that the molecule does not collide with other molecules along the way. A more rigorous treatment including molecular collision gives exactly the same result.
Black plate (23,1)
2.6 Kinetic Theory of Gases
Therefore, the force exerted by one molecule on one wall as a result of the collision is
mvx2 =l, and the total force due to N molecules is Nmvx2 =l. Because pressure is force/
area and area is l 2 , we can now express the total pressure exerted on one wall as
P¼
¼
F
A
Nmvx2 Nmvx2
¼
lðl 2 Þ
V
or
PV ¼ Nmvx2
ð2:18Þ
where V is the volume of the cube (equal to l 3 ). When we are dealing with a large
collection of molecules (for example, when N is on the order of 6 10 23 ), there is a
tremendous spread of molecular velocities. It is more appropriate, therefore, to replace vx2 in Equation 2.18 with the mean or average quantity, vx2 . Referring to Equation 2.17, we see that the relation between the average of the square of the velocity
components and the average of the square of the velocity, v 2 , is still
v 2 ¼ vx2 þ vy2 þ vz2
The quantity v 2 is called the mean-square velocity, defined as
v2 ¼
v12 þ v22 þ þ vN2
N
ð2:19Þ
When N is a large number, it is correct to assume that molecular motions along the
x; y, and z directions are equally probable. This means that
vx2 ¼ vy2 ¼ vz2 ¼
v2
3
and Equation 2.18 can now be written as
P¼
Nmv 2
3V
Multiplying the top and bottom by 2 and recalling that the kinetic energy of the
molecule Etrans is given by 12 mv 2 (where the subscript trans denotes translational motion; that is, motion through space of the whole molecule), we obtain
2N 1 2
2N
P¼
mv ¼
E trans
3V 2
3V
ð2:20Þ
This is the pressure exerted by N molecules on one wall. The same result can be
obtained regardless of the direction (x; y, or z) we describe for the molecular motion.
We see that the pressure is directly proportional to the average kinetic energy or,
more explicitly, to the mean-square velocity of the molecule. The physical meaning
of this dependence is that the larger the velocity, the more frequent the collisions and
the greater the change in momentum. Thus, these two independent terms give us the
quantity v 2 in the kinetic theory expression for the pressure.
23
Black plate (24,1)
24
Chapter 2: Properties of Gases
Kinetic Energy and Temperature
Let us compare Equation 2.20 with the ideal-gas equation (Equation 2.8):
PV ¼ nRT
¼
N
RT
NA
or
P¼
NRT
NA V
ð2:21Þ
where NA is the Avogadro constant. Combining the pressures in Equations 2.20 and
2.21, we get
2N
N RT
E trans ¼
3V
NA V
or
E trans ¼
The kinetic energy of 1 mole
of the gas is given by
ð3=2ÞNA kB T ¼ ð3=2ÞRT.
3 RT 3
¼ kB T
2 NA 2
(2.22)
where R ¼ kB NA and kB is the Boltzmann constant, equal to 1:380658 1023 J K1
[after the Austrian physicist Ludwig Eduard Boltzmann (1844–1906)]. (In most calculations, we shall round kB to 1:381 1023 J K1 .) We see that the mean kinetic
energy of one molecule is proportional to absolute temperature.
The significance of Equation 2.22 is that it provides an explanation for the temperature of a gas in terms of molecular motion. For this reason, random molecular
motion is sometimes referred to as thermal motion. It is important to keep in mind
that the kinetic theory is a statistical treatment of our model; hence, it is meaningless
to associate temperature with the kinetic energy of just a few molecules. Equation
2.22 also tells us that whenever two ideal gases are at the same temperature T, they
must have the same average kinetic energy. The reason is that E trans in Equation 2.22
is independent of molecular properties such as size or molar mass or amount of the
gas present, as long as N is a large number.
It is easy to see that v 2 would be a very di‰cult quantity to measure, if indeed
it could be measured at all. To do so, we would need to measure each individual
velocity, square it, and then take the average (see Equation 2.19). Fortunately, v 2 can
be obtained quite directly from other quantities. From Equation 2.22, we write
1 2 3 RT 3
mv ¼
¼ kB T
2
2 NA 2
so that
Note that kB refers to one
molecule and R refers to one
mole of such molecules;
kB ¼ R=NA .
v2 ¼
3RT
3kB T
¼
mNA
m
or
rffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffi
3RT
3kB T
2
v ¼ vrms ¼
¼
M
m
ðM ¼ mNA Þ
ð2:23Þ
Black plate (25,1)
2.7 The Maxwell Distribution Laws
where vrms is the root-mean-square velocity* and m is the mass (in kg) of one molecule; M is the molar mass (in kg mol1 ). Note that vrms is directly proportional to the
square root of temperature and inversely proportional to the square root of molar
mass of the molecule. Therefore, the heavier the molecule, the slower its motion.
2.7 The Maxwell Distribution Laws
The root-mean-square velocity gives us an average measure that is very useful in the
study of a large number of molecules. When we are studying, say, one mole of a gas,
it is impossible to know the velocity of each individual molecule for two reasons.
First, the number of molecules is so huge that there is no way we can follow all their
motions. Second, although molecular motion is a well-defined quantity, we cannot
measure its velocity exactly. Therefore, rather than concerning ourselves with individual molecular velocities, we ask this question: For a given system at some known
temperature, how many molecules are moving at velocities between v and v þ Dv at
any moment? Or, how many molecules in a macroscopic gas sample have velocities,
say, between 306.5 m s1 and 306.6 m s1 at any moment?
Because the total number of molecules is very large, there is a continuous spread,
or distribution, of velocities as a result of collisions. We can therefore make the velocity range Dv smaller and smaller, and in the limit it becomes dv. This fact has great
significance, because it enables us to replace the summation sign with the integral sign
in calculating the number of molecules whose velocities fall between v and v þ dv.
Mathematically speaking, it is easier to integrate than to sum a large series. This
distribution-of-velocities approach was first employed by the Scottish physicist James
Clerk Maxwell (1831–1879) in 1860 and later refined by Boltzmann. They showed
that for a system containing N ideal gas molecules at thermal equilibrium with its
surroundings, the fraction of molecules dN=N moving at velocities between vx and
vx þ dvx along the x direction is given by
1=2
dN
m
2
¼
emvx =2kB T dvx
N
2pkB T
ð2:24Þ
where m is the mass of the molecule, kB the Boltzmann constant, and T the absolute
temperature.
As mentioned earlier, velocity is a vector quantity. In many cases, we need to
deal only with the speed of molecules (c), which is a scalar quantity; that is, it has
magnitude but no directional properties. The fraction of molecules dN=N moving
between speeds c and c þ dc is given by
3=2
dN
m
2
¼ 4pc 2
emc =2kB T dc
N
2pkB T
¼ f ðcÞdc
ð2:25Þ
where f ðcÞ, the Maxwell speed distribution function, is given by
f ðcÞ ¼ 4pc 2
m
2pkB T
3=2
emc
2
=2kB T
(2.26)
* Because velocity is a vector quantity, the average molecular velocity, v, must be zero; there are
just as many molecules moving in the positive direction as there are in the negative direction. On the
other hand, vrms is a scalar quantity; that is, it has magnitude but no direction.
25
Black plate (26,1)
Chapter 2: Properties of Gases
Cl 2 (70.90 g mol
100 K
1
)
300 K
700 K
0
500
1000
Molecular speed (m s
(a)
1
1500
)
1
)
T
Number of molecules
N 2 (28.02 g mol
Number of molecules
26
N 2 (28.02 g mol
1
300 K
)
He (4.003 g mol
0
500
1000
1500
Molecular speed (m s
2000
1
1
)
2500
)
(b)
Figure 2.11
(a) The distribution of speeds for nitrogen gas at three di¤erent temperatures. At higher
temperatures, more molecules are moving at faster speeds. (b) The distribution of speeds for
three gases at 300 K. At a given temperature, the light molecules are moving faster, on the
average.
Figure 2.11 shows the dependence of the speed distribution curve on temperature and molar mass. At any given temperature, the general shape of a distribution
curve can be explained as follows. Initially, at small c values, the c 2 term in Equation
2.25 dominates so f ðcÞ increases with increasing c. At larger values of c, the term
2
emc =2kB T becomes more important. These two opposing terms cause the curve to
reach a maximum beyond which it decreases roughly exponentially with increasing c.
The speed corresponding to the maximum value of f ðcÞ is called the most probable
speed, c mp , because it is the speed of the largest number of molecules.
Figure 2.11a shows how the shape of the distribution curve is influenced by
temperature. At low temperatures the distribution has a rather narrow range. As the
temperature increases, the curve becomes flatter, meaning that there are now more
fast-moving molecules. This temperature dependence of the distribution curve has
important implications in chemical reaction rates. As we shall see in Chapter 9, in
order to react, a molecule must possess a minimum amount of energy, called activation energy. At low temperatures the number of fast-moving molecules is small; hence
most reactions proceed at a slow rate. Raising the temperature increases the number
of energetic molecules and causes the reaction rate to increase. In Figure 2.11b we see
that heavier gases have a narrower range of speed distribution than lighter gases at
the same temperature. This is to be expected considering that heavier gases move
slower, on the average, than lighter gases. The validity of the Maxwell speed distribution has been verified experimentally.
The usefulness of the Maxwell speed distribution function is that it enables us to
calculate average quantities. In fact, we can obtain three related expressions for speed
called the most probable speed (c mp ), as defined above, average speed (c), which is
defined as the sum of the speeds of all the molecules divided by the number of molecules, and root-mean-square speed (crms ):*
* For the derivation of c mp , see Problem 2.66; for the derivation of c, see the physical chemistry
texts listed in Chapter 1. Because the square of the average velocity is a scalar quantity, it follows
that v 2 ¼ c 2 ; hence vrms ¼ crms .
Black plate (27,1)
2.7 The Maxwell Distribution Laws
rffiffiffiffiffiffiffiffiffiffi
2RT
¼
M
c mp
(2.27)
rffiffiffiffiffiffiffiffiffiffi
8RT
c¼
pM
crms
(2.28)
rffiffiffiffiffiffiffiffiffiffi
3RT
¼
M
(2.29)
Example 2.3
Calculate the values of c mp ; c, and crms for O2 at 300 K.
ANSWER
The constants are
R ¼ 8:314 J K1 mol1
M ¼ 0:03200 kg mol
T ¼ 300 K
1
The most probable speed is given by
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 8:314 J K1 mol1 300 K
c mp ¼
0:03200 kg mol1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 1:56 10 5 J kg1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 1:56 10 5 m 2 s2
¼ 395 m s1
Similarly, we can show that
c¼
rffiffiffiffiffiffiffiffiffiffi
8RT
¼ 446 m s1
pM
and
crms
rffiffiffiffiffiffiffiffiffiffi
3RT
¼ 484 m s1
¼
M
COMMENT
The calculation shows, and indeed it is generally true, that crms > c > c mp . That c mp is
the smallest of the three speeds is due to the asymmetry of the curve (see Figure 2.11).
The reason crms is greater than c is that the squaring process in Equation 2.19 is
weighted toward larger values of c.
27
Black plate (28,1)
28
Chapter 2: Properties of Gases
Finally, note that both the c and crms values of N2 and O2 are close to the speed
of sound in air. Sound waves are pressure waves. The propagation of these waves is
directly related to the movement of molecules and hence to their speeds.
2.8 Molecular Collisions and the Mean Free Path
Now that we have an explicit expression for the average speed, c, we can use it to
study some dynamic processes involving gases. We know that the speed of a molecule
is not constant but changes frequently as a result of collisions. Therefore, the question
we ask is: How often do molecules collide with one another? The collision frequency
depends on the density of the gas and the molecular speed, and therefore on the
temperature of the system. In the kinetic theory model, we assume each molecule to
be a hard sphere of diameter d. A molecular collision is one in which the separation
between the two spheres (measured from each center) is d.
Let us consider the motion of a particular molecule. A simple approach is to assume that at a given instant, all molecules except this one are standing still. In time t,
this molecule moves a distance ct (where c is the average speed) and sweeps out a
collision tube that has a cross-sectional area pd 2 (Figure 2.12). The volume of the
ct
Just
hit
Just
miss
Figure 2.12
The collision cross section and the collision tube.
Any molecule whose center lies within or touches the
tube will collide with the moving molecule (red sphere).
Hit
Collision tube
Miss
Molecule
in motion
Area
πd
2
cylinder is ðpd 2 ÞðctÞ. Any molecule whose center lies within this cylinder will collide
with the moving molecule. If there are altogether N molecules in volume V , then the
number density of the gas is N=V , the number of collisions in time t is pd 2 ctðN=V Þ,
and the number of collisions per unit time, or the collision frequency, Z1 , is
pd 2 cðN=V Þ. The expression for the collision frequency needs a correction. If we assume that the rest of the molecules are not frozen in position, we should replace c
with the average relative speed. Figure 2.13 shows three di¤erent collisions
for two
pffiffiffi
molecules. The relative speed for the case shown in Figure 2.13c is 2c, so that
90 °
(a)
(b)
(c)
Figure 2.13
Three di¤erent approaches for two colliding molecules. The situations shown in (a) and (b)
represent the two extreme cases, while that shown in (c) may be taken as the ‘‘average’’ case
for molecular encounter.
Black plate (29,1)
2.8 Molecular Collisions and the Mean Free Path
pffiffiffi 2 N
Z1 ¼ 2pd c
V
collisions s1
29
ð2:30Þ
This is the number of collisions a single molecule makes in one second. Because there
are N molecules in volume V and each makes Z1 collisions per second, the total
number of binary collisions, or collisions between two molecules, per unit volume per
unit time, Z11 , is given by
1
N
Z11 ¼ Z1
2
V
pffiffiffi
2 2 N 2
pd c
¼
V
2
ð2:31Þ
collisions m
3
s
1
The factor 1=2 is introduced in Equation 2.31 to ensure that we are counting each
collision between two molecules only once. The probability of three or more molecules colliding at once is very small except at high pressures. Because the rate of a
chemical reaction generally depends on how often reacting molecules come in contact
with one another, Equation 2.31 is quite important in gas-phase chemical kinetics.
We shall return to this equation in Chapter 9.
A quantity closely related to the collision number is the average distance traveled
by a molecule between successive collisions. This distance, called the mean free path,
l (Figure 2.14), is defined as
l ¼ ðaverage speedÞ ðaverage time between collisionsÞ
Because the average time between collisions is the reciprocal of the collision frequency, we have
l¼
c
c
1
¼ pffiffiffi
¼ pffiffiffi
2
2
Z1
2pd cðN=V Þ
2pd ðN=V Þ
ð2:32Þ
Notice that the mean free path is inversely proportional to the number density of the
gas ðN=V Þ. This behavior is reasonable because in a dense gas, a molecule makes
more collisions per unit time and hence travels a shorter distance between successive
collisions. The mean free path can also be expressed in terms of the gas pressure.
Assuming ideal behavior,
P¼
¼
nRT
V
ðN=NA ÞRT
V
N PNA
¼
V
RT
Equation 2.32 can now be written as
RT
l ¼ pffiffiffi
2pd 2 PNA
(2.33)
Figure 2.14
The distances traveled by a
molecule between successive
collisions. The average of these
distances is called the mean
free path.
Black plate (30,1)
30
Chapter 2: Properties of Gases
Example 2.4
The concentration of dry air at 1.00 atm and 298 K is about 2:5 10 19 molecules cm3 .
Assuming that air contains only nitrogen molecules, calculate the collision frequency, the
binary collision number, and the mean free path of nitrogen molecules under these
conditions. The collision diameter of nitrogen is 3.75 Å. (1 Å ¼ 108 cm.)
ANSWER
Our first step is to calculate the average speed of nitrogen. From Equation 2.28, we find
c ¼ 4:8 10 2 m s1 . The collision frequency is given by
pffiffiffi
Z1 ¼ 2pð3:75 108 cmÞ 2 ð4:8 10 4 cm s1 Þð2:5 10 19 molecules cm3 Þ
¼ 7:5 10 9 collisions s1
Note that we have replaced the unit ‘‘molecules’’ with ‘‘collisions’’ because, in the
derivation of Z1 , every molecule in the collision volume represents a collision. The
binary collision number is
Z11 ¼
¼
Z1 N
2 V
ð7:5 10 9 collisions s1 Þ
2:5 10 19 molecules cm3
2
¼ 9:4 10 28 collisions cm3 s1
Again, we converted molecules to collisions in calculating the total number of binary
collisions. Finally, the mean free path is given by
l¼
c
4:8 10 4 cm s1
¼
Z1 7:5 10 9 collisions s1
¼ 6:4 106 cm collision1
¼ 640 Å collision1
COMMENT
It is usually su‰cient to express mean free path in terms of distance alone rather than
distance per collision. Thus, in this example, the mean free path of nitrogen is 640 Å, or
6:4 106 cm.
2.9 Graham’s Laws of Diffusion and Effusion
Perhaps without thinking about it, we witness molecular motion on a daily basis. The
scent of perfume and the shrinking of an inflated helium rubber balloon are examples
of di¤usion and e¤usion, respectively. We can apply the kinetic theory of gases to
both processes.
The phenomenon of gas di¤usion o¤ers direct evidence of molecular motion.
Were it not for di¤usion, there would be no perfume industry, and skunks would be
just another cute, furry species. Removing a partition separating two di¤erent gases
in a container quickly leads to a complete mixing of molecules. These are spontaneous processes for which we shall discuss the thermodynamic basis in Chapter 4.
During e¤usion, a gas travels from a high-pressure region to a low-pressure one
Black plate (31,1)
2.9 Graham’s Laws of Diffusion and Effusion
Orifice
Vacuum
Figure 2.15
An e¤usion process in which molecules move through an opening (orifice) into an evacuated
region. The conditions for e¤usion are that the mean free path of the molecules is large
compared to the size of the opening and the wall containing the opening is thin so that no
molecular collisions occur during the exit. Also, the pressure in the right chamber must be
low enough so as not to obstruct the molecular movement through the hole.
through a pinhole or orifice (Figure 2.15). For e¤usion to occur, the mean free path
of the molecules must be large compared with the diameter of the orifice. This
ensures that a molecule is unlikely to collide with another molecule when it reaches
the opening but will pass right through it. It follows then that the number of molecules passing through the orifice is equal to the number that would normally strike an
area of wall equal to the area of the hole.
Although the basic molecular mechanisms for di¤usion and e¤usion are quite
di¤erent (the former involves bulk flow, whereas the latter involves molecular flow),
these two phenomena obey laws of the same form. Both laws were discovered by the
Scottish chemist Thomas Graham (1805–1869), the law of di¤usion in 1831 and the
law of e¤usion in 1864. These laws state that under the same conditions of temperature and pressure, the rates of di¤usion (or e¤usion) of gases are inversely proportional to the square roots of their molar masses. Thus, for two gases 1 and 2, we have
r1
¼
r2
rffiffiffiffiffiffiffiffi
M2
M1
(2.34)
where r1 and r2 are the rates of di¤usion (or e¤usion) of the two gases.
Suggestions for Further Reading
Books
Hirschfelder, J. O., C. F. Curtiss, and R. B. Bird, The
Molecular Theory of Gases and Liquids, John Wiley &
Sons, New York, 1954.
Hildebrand, J. H. An Introduction to Molecular Kinetic
Theory, Chapman & Hall, London, 1963 (Van
Nostrand Reinhold Company, New York).
Walton, A. J. The Three Phases of Matter, 2nd ed., Oxford
University Press, New York, 1983.
Tabor, D. Gases, Liquids, and Solids, 3rd ed., Cambridge
University Press, New York, 1996.
Articles
Gas Laws and Equations of State
‘‘The van der Waals Gas Equation,’’ F. S. Swinbourne, J.
Chem. Educ. 32, 366 (1955).
‘‘A Simple Model for van der Waals,’’ S. S. Winter, J.
Chem. Educ. 33, 459 (1959).
‘‘Comparisons of Equations of State in E¤ectively
Describing PVT Relations,’’ J. B. Ott, J. R. Goales,
and H. T. Hall, J. Chem. Educ. 48, 515 (1971).
‘‘Scuba Diving and the Gas Laws,’’ E. D. Cooke, J. Chem.
Educ. 50, 425 (1973).
31
Black plate (32,1)
32
Chapter 2: Properties of Gases
‘‘Derivation of the Ideal Gas Law,’’ S. Levine, J. Chem.
Educ. 62, 399 (1985).
‘‘The Ideal Gas Law at the Center of the Sun,’’ D. B.
Clark, J. Chem. Educ. 66, 826 (1989).
‘‘The Many Faces of van der Waals’s Equation of State,’’
J. G. Eberhart, J. Chem. Educ. 66, 906 (1989).
‘‘Does a One-Molecule Gas Obey Boyle’s Law?’’ G.
Rhodes, J. Chem. Educ. 69, 16 (1992).
‘‘Equations of State,’’ M. Ross in Encyclopedia of Applied
Physics, G. L. Trigg, Ed., VCH Publishers, New
York, 1993, Vol. 6, p. 291.
‘‘Interpretation of the Second Virial Coe‰cient,’’ J.
Wisniak, J. Chem. Educ. 76, 671 (1999).
The Critical State
‘‘The Critical Temperature: A Necessary Consequence of
Gas Nonideality,’’ F. L. Pilar, J. Chem. Educ. 44, 284
(1967).
‘‘Supercritical Fluids: Liquid, Gas, Both, or Neither? A
Di¤erent Approach,’’ E. F. Meyer and T. P. Meyer,
J. Chem. Educ. 63, 463 (1986).
‘‘Past, Present, and Possible Future Applications of
Supercritical Fluid Extraction Technology,’’ C. L.
Phelps, N. G. Smart, and C. M. Wai, J. Chem. Educ.
73, 1163 (1996).
Kinetic Theory of Gases
‘‘Kinetic Energies of Gas Molecules,’’ J. C. Aherne, J.
Chem. Educ. 42, 655 (1965).
‘‘Kinetic Theory, Temperature, and Equilibrium,’’ D. K.
Carpenter, J. Chem. Educ. 43, 332 (1966).
‘‘Graham’s Laws of Di¤usion and E¤usion,’’ E. A. Mason
and B. Kronstadt, J. Chem. Educ. 44, 740 (1967).
‘‘The Cabin Atmosphere in Manned Space Vehicles,’’
W. H. Bowman and R. M. Lawrence, J. Chem. Educ.
48, 152 (1971).
‘‘The Assumption of Elastic Collisions in Elementary Gas
Kinetic Theory,’’ B. Rice and C. J. G. Raw, J. Chem.
Educ. 51, 139 (1974).
‘‘Velocity and Energy Distribution in Gases,’’ B. A.
Morrow and D. F. Tessier, J. Chem. Educ. 59, 193
(1982).
‘‘Applications of Maxwell-Boltzmann Distribution
Diagrams,’’ G. D. Peckham and I. J. McNaught, J.
Chem. Educ. 69, 554 (1992).
‘‘Misuse of Graham’s Laws,’’ S. J. Hawkes, J. Chem.
Educ. 70, 836 (1993).
‘‘Graham’s Law and Perpetuation of Error,’’ S. J. Hawkes,
J. Chem. Educ. 74, 1069 (1997).
General
‘‘The Lung,’’ J. H. Comroe, Sci. Am. February 1966.
‘‘The Invention of the Balloon and the Birth of Modern
Chemistry,’’ A. F. Scott, Sci. Am. January 1984.
‘‘Temperature, Cool but Quick,’’ S. M. Cohen, J. Chem.
Educ. 63, 1038 (1986).
‘‘Mountain Sickness,’’ C. S. Houston, Sci. Am. October
1992.
Problems
Ideal Gases
2.1 Classify each of the following properties as intensive or extensive: force, pressure (P), volume (V ), temperature (T), mass, density, molar mass, molar volume
(V ).
2.2 Some gases, such as NO2 and NF2 , do not obey
Boyle’s law at any pressure. Explain.
2.3 An ideal gas originally at 0.85 atm and 66 C was
allowed to expand until its final volume, pressure, and
temperature were 94 mL, 0.60 atm, and 45 C, respectively. What was its initial volume?
2.4 Some ballpoint pens have a small hole in the main
body of the pen. What is the purpose of this hole?
2.5 Starting with the ideal-gas equation, show how you
can calculate the molar mass of a gas from a knowledge
of its density.
2.6 At STP (standard temperature and pressure),
0.280 L of a gas weighs 0.400 g. Calculate the molar
mass of the gas.
2.7 Ozone molecules in the stratosphere absorb much
of the harmful radiation from the sun. Typically, the
temperature and partial pressure of ozone in the strato-
sphere are 250 K and 1:0 103 atm, respectively. How
many ozone molecules are present in 1.0 L of air under
these conditions? Assume ideal-gas behavior.
2.8 Calculate the density of HBr in g L1 at 733 mmHg
and 46 C. Assume ideal-gas behavior.
2.9 Dissolving 3.00 g of an impure sample of CaCO3 in
an excess of HCl acid produced 0.656 L of CO2 (measured at 20 C and 792 mmHg). Calculate the percent by
mass of CaCO3 in the sample.
2.10 The saturated vapor pressure of mercury is
0.0020 mmHg at 300 K and the density of air at 300 K
is 1.18 g L1 . (a) Calculate the concentration of mercury vapor in air in mol L1 . (b) What is the number of
parts per million (ppm) by mass of mercury in air?
2.11 A very flexible balloon with a volume of 1.2 L at
1.0 atm and 300 K is allowed to rise to the stratosphere,
where the temperature and pressure are 250 K and
3:0 103 atm, respectively. What is the final volume
of the balloon? Assume ideal-gas behavior.
2.12 Sodium bicarbonate (NaHCO3 ) is called baking
soda because when heated, it releases carbon dioxide
Black plate (33,1)
Problems
gas, which causes cookies, doughnuts, and bread to rise
during baking. (a) Calculate the volume (in liters) of
CO2 produced by heating 5.0 g of NaHCO3 at 180 C
and 1.3 atm. (b) Ammonium bicarbonate (NH 4 HCO3 )
has also been used as a leavening agent. Suggest one
advantage and one disadvantage of using NH 4 HCO3
instead of NaHCO3 for baking.
2.13 A common, non-SI unit for pressure is pounds per
square inch (psi). Show that 1 atm ¼ 14:7 psi. An automobile tire is inflated to 28.0 psi gauge pressure when
cold, at 18 C. (a) What will the pressure be if the tire is
heated to 32 C by driving the car? (b) What percentage
of the air in the tire would have to be let out to reduce
the pressure to the original 28.0 psi? Assume that the
volume of the tire remains constant with temperature.
(A tire gauge measures not the pressure of the air inside
but its excess over the external pressure, which is 14.7
psi.)
2.14 (a) What volume of air at 1.0 atm and 22 C is
needed to fill a 0.98-L bicycle tire to a pressure of 5.0
atm at the same temperature? (Note that 5.0 atm is the
gauge pressure, which is the di¤erence between the pressure in the tire and atmospheric pressure. Initially, the
gauge pressure in the tire was 0 atm.) (b) What is the
total pressure in the tire when the gauge reads 5.0 atm?
(c) The tire is pumped with a hand pump full of air at
1.0 atm; compressing the gas in the cylinder adds all the
air in the pump to the air in the tire. If the volume of
the pump is 33% of the tire’s volume, what is the gauge
pressure in the tire after 3 full strokes of the pump?
2.15 A student breaks a thermometer and spills most of
the mercury (Hg) onto the floor of a laboratory that
measures 15.2 m long, 6.6 m wide, and 2.4 m high.
(a) Calculate the mass of mercury vapor (in grams) in
the room at 20 C. (b) Does the concentration of mercury vapor exceed the air quality regulation of 0.050
mg Hg m3 of air? (c) One way to treat small quantities
of spilled mercury is to spray powdered sulfur over the
metal. Suggest a physical and a chemical reason for this
treatment. The vapor pressure of mercury at 20 C is
1:7 1:06 atm.
2.16 Nitrogen forms several gaseous oxides. One of
them has a density of 1.27 g L1 measured at 764
mmHg and 150 C. Write the formula of the compound.
2.17 Nitrogen dioxide (NO2 ) cannot be obtained in a
pure form in the gas phase because it exists as a mixture
of NO2 and N2 O4 . At 25 C and 0.98 atm, the density
of this gas mixture is 2.7 g L1 . What is the partial
pressure of each gas?
2.18 An ultra-high-vacuum pump can reduce the pressure of air from 1.0 atm to 1:0 1012 mmHg. Calculate the number of air molecules in a liter at this pressure and 298 K. Compare your results with the number
of molecules in 1.0 L at 1.0 atm and 298 K. Assume
ideal-gas behavior.
2.19 An air bubble with a radius of 1.5 cm at the bottom of a lake where the temperature is 8.4 C and the
pressure is 2.8 atm rises to the surface, where the temperature is 25.0 C and the pressure is 1.0 atm. Calculate
the radius of the bubble when it reaches the surface. Assume ideal-gas behavior. [Hint: The volume of a sphere
is given by ð4=3Þpr 3 , where r is the radius.]
2.20 The density of dry air at 1.00 atm and 34.4 C is
1.15 g L1 . Calculate the composition of air (percent by
mass) assuming that it contains only nitrogen and oxygen and behaves like an ideal gas. (Hint: First calculate
the ‘‘molar mass’’ of air, then the mole fractions, and
then the mass fractions of O2 and N2 .)
2.21 A gas that evolved during the fermentation of glucose has a volume of 0.78 L when measured at 20.1 C
and 1.0 atm. What was the volume of this gas at the
fermentation temperature of 36.5 C? Assume ideal-gas
behavior.
2.22 Two bulbs of volumes VA and VB are connected
by a stopcock. The number of moles of gases in the
bulbs are nA and nB , and initially the gases are at the
same pressure, P, and temperature, T. Show that the
final pressure of the system, after the stopcock has been
opened, equals P. Assume ideal-gas behavior.
2.23 The composition of dry air at sea level is 78.03%
N2 , 20.99% O2 , and 0.033% CO2 by volume. (a) Calculate the average molar mass of this air sample. (b) Calculate the partial pressures of N2 , O2 , and CO2 in atm.
(At constant temperature and pressure, the volume of a
gas is directly proportional to the number of moles of
the gas.)
2.24 A mixture containing nitrogen and hydrogen
weighs 3.50 g and occupies a volume of 7.46 L at 300 K
and 1.00 atm. Calculate the mass percent of these two
gases. Assume ideal-gas behavior.
2.25 The relative humidity in a closed room with a volume of 645.2 m 3 is 87.6% at 300 K, and the vapor
pressure of water at 300 K is 0.0313 atm. Calculate the
mass of water in the air. [Hint: The relative humidity is
defined as ðP=Ps Þ 100%, where P and Ps are the partial pressure and saturated partial pressure of water
vapor, respectively.]
2.26 Death by su¤ocation in a sealed container is normally caused not by oxygen deficiency but by CO2 poisoning, which occurs at about 7% CO2 by volume. For
what length of time would it be safe to be in a sealed
room 10 10 20 ft? [Source: ‘‘Eco-Chem,’’ J. A.
Campbell, J. Chem. Educ. 49, 538 (1972).]
2.27 A flask contains a mixture of two ideal gases, A
and B. Show graphically how the total pressure of the
system depends on the amount of A present; that is,
plot the total pressure versus the mole fraction of A. Do
the same for B on the same graph. The total number of
moles of A and B is constant.
2.28 A mixture of helium and neon gases is collected
over water at 28.0 C and 745 mmHg. If the partial
pressure of helium is 368 mmHg, what is the partial
pressure of neon? (Note: The vapor pressure of water at
28 C is 28.3 mmHg.)
33
Black plate (34,1)
34
Chapter 2: Properties of Gases
2.29 If the barometric pressure falls in one part of the
world, it must rise somewhere else. Explain why.
2.30 A piece of sodium metal reacts completely with
water as follows:
2NaðsÞ þ 2H2 OðlÞ ! 2NaOHðaqÞ þ H2 ðgÞ
The hydrogen gas generated is collected over water at
25.0 C. The volume of the gas is 246 mL measured at
1.00 atm. Calculate the number of grams of sodium
used in the reaction. (Note: The vapor pressure of water
at 25 C is 0.0313 atm.)
2.31 A sample of zinc metal reacts completely with an
excess of hydrochloric acid:
ZnðsÞ þ 2HClðaqÞ ! ZnCl2 ðaqÞ þ H2 ðgÞ
The hydrogen gas produced is collected over water at
25.0 C. The volume of the gas is 7.80 L, and the pressure is 0.980 atm. Calculate the amount of zinc metal in
grams consumed in the reaction. (Note: The vapor pressure of water at 25 C is 23.8 mmHg.)
2.32 Helium is mixed with oxygen gas for deep sea
divers. Calculate the percent by volume of oxygen gas
in the mixture if the diver has to submerge to a depth
where the total pressure is 4.2 atm. The partial pressure
of oxygen is maintained at 0.20 atm at this depth.
2.33 A sample of ammonia (NH3 ) gas is completely
decomposed to nitrogen and hydrogen gases over
heated iron wool. If the total pressure is 866 mmHg,
calculate the partial pressures of N2 and H2 .
2.34 The partial pressure of carbon dioxide in air varies
with the seasons. Would you expect the partial pressure
in the Northern Hemisphere to be higher in the summer
or winter? Explain.
2.35 A healthy adult exhales about 5:0 10 2 mL of a
gaseous mixture with each breath. Calculate the number of molecules present in this volume at 37 C and
1.1 atm. List the major components of this gaseous
mixture.
2.36 Describe how you would measure, by either chemical or physical means (other than mass spectrometry),
the partial pressures of a mixture of gases: (a) CO2 and
H2 , (b) He and N2 .
2.37 The gas laws are vitally important to scuba divers.
The pressure exerted by 33 ft of seawater is equivalent
to 1 atm pressure. (a) A diver ascends quickly to the
surface of the water from a depth of 36 ft without
exhaling gas from his lungs. By what factor would the
volume of his lungs increase by the time he reaches the
surface? Assume that the temperature is constant.
(b) The partial pressure of oxygen in air is about 0.20
atm. (Air is 20% oxygen by volume.) In deep-sea diving,
the composition of air the diver breathes must be
changed to maintain this partial pressure. What must
the oxygen content (in percent by volume) be when the
total pressure exerted on the diver is 4.0 atm?
2.38 A 1.00-L bulb and a 1.50-L bulb, connected by a
stopcock, are filled, respectively, with argon at 0.75 atm
and helium at 1.20 atm at the same temperature. Calculate the total pressure and the partial pressures of each
gas after the stopcock has been opened and the mole
fraction of each gas. Assume ideal-gas behavior.
2.39 A mixture of helium and neon weighing 5.50 g
occupies a volume of 6.80 L at 300 K and 1.00 atm.
Calculate the composition of the mixture in mass
percent.
Nonideal Gases
2.40 Suggest two demonstrations to show that gases do
not behave ideally.
2.41 Which of the following combinations of conditions
most influences a gas to behave ideally: (a) low pressure
and low temperature, (b) low pressure and high temperature, (c) high pressure and high temperature, and
(d) high pressure and low temperature.
2.42 The van der Waals constants of a gas can be
obtained from its critical constants, where a ¼
ð27R 2 Tc 2 =64Pc Þ and b ¼ ðRTc =8Pc Þ. Given that Tc ¼
562 K and Pc ¼ 48:0 atm for benzene, calculate its a
and b values.
2.43 Using the data shown in Table 2.1, calculate the
pressure exerted by 2.500 moles of carbon dioxide confined in a volume of 1.000 L at 450 K. Compare the
pressure with that calculated assuming ideal behavior.
2.44 Without referring to a table, select from the following list the gas that has the largest value of b in the
van der Waals equation: CH 4 , O2 , H2 O, CCl 4 , Ne.
2.45 Referring to Figure 2.4, we see that for He the
plot has a positive slope even at low pressures. Explain
this behavior.
2.46 At 300 K, the virial coe‰cients (B) of N2 and
CH 4 are 4.2 cm 3 mol1 and 15 cm 3 mol1 , respectively. Which gas behaves more ideally at this
temperature?
2.47 Calculate the molar volume of carbon dioxide at
400 K and 30 atm, given that the second virial coe‰cient (B) of CO2 is 0.0605 L mol1 . Compare your result with that obtained using the ideal-gas equation.
2.48 Consider the virial equation Z ¼ 1 þ B 0 P þ C 0 P 2 ,
which describes the behavior of a gas at a certain temperature. From the following plot of Z versus P, deduce
the signs of B 0 and C 0 (< 0; ¼ 0; > 0).
1.4
1.3
1.2
Z
1.1
1.0
0.9
0.8
200
400
600
800
P /atm
1000
Black plate (35,1)
Problems
Kinetic Theory of Gases
2.49 Apply the kinetic theory of gases to explain
Boyle’s law, Charles’ law, and Dalton’s law.
2.50 Is temperature a microscopic or macroscopic concept? Explain.
2.51 In applying the kinetic molecular theory to gases,
we have assumed that the walls of the container are
elastic for molecular collisions. Actually, whether these
collisions are elastic or inelastic makes no di¤erence as
long as the walls are at the same temperature as the gas.
Explain.
2.52 If 2:0 10 23 argon (Ar) atoms strike 4.0 cm 2 of
wall per second at a 90 angle to the wall when moving
with a speed of 45,000 cm s1 , what pressure (in atm)
do they exert on the wall?
2.53 A square box contains He at 25 C. If the atoms
are colliding with the walls perpendicularly (at 90 ) at
the rate of 4:0 10 22 times per second, calculate the
force and the pressure exerted on the wall given that the
area of the wall is 100 cm 2 and the speed of the atoms
is 600 m s1 .
2.54 Calculate the average translational kinetic energy
for a N2 molecule and for 1 mole of N2 at 20 C.
2.55 To what temperature must He atoms be cooled so
that they have the same vrms as O2 at 25 C?
2.56 The crms of CH 4 is 846 m s1 . What is the temperature of the gas?
2.57 Calculate the value of the crms of ozone molecules
in the stratosphere, where the temperature is 250 K.
2.58 At what temperature will He atoms have the same
crms value as N2 molecules at 25 C? Solve this problem
without calculating the value of crms for N2 .
Maxwell Speed Distribution
2.59 List the conditions used for deriving the Maxwell
speed distribution.
2.60 Plot the speed distribution function for (a) He, O2 ,
and UF6 at the same temperature, and (b) CO2 at
300 K and 1000 K.
2.61 Account for the maximum in the Maxwell speed
distribution curve (Figure 2.11) by plotting the following two curves on the same graph: (1) c 2 versus c and
2
(2) emc =2kB T versus c. Use neon (Ne) at 300 K for the
plot in (2).
2.62 A N2 molecule at 20 C is released at sea level to
travel upward. Assuming that the temperature is constant and that the molecule does not collide with other
molecules, how far would it travel (in meters) before
coming to rest? Do the same calculation for a He atom.
[Hint: To calculate the altitude, h, the molecule will
travel, equate its kinetic energy with the potential energy, mgh, where m is the mass and g the acceleration
due to gravity (9.81 m s2 ).]
2.63 The speeds of 12 particles (in cm s1 ) are 0.5, 1.5,
1.8, 1.8, 1.8, 1.8, 2.0, 2.5, 2.5, 3.0, 3.5, and 4.0. Find
(a) the average speed, (b) the root-mean-square speed,
and (c) the most probable speed of these particles.
Explain your results.
2.64 At a certain temperature, the speeds of six gaseous
molecules in a container are 2.0 m s1 , 2.2 m s1 , 2.6
m s1 , 2.7 m s1 , 3.3 m s1 , and 3.5 m s1 . Calculate
the root-mean-square speed and the average speed of
the molecules. These two average values are close to
each other, but the root-mean-square value is always
the larger of the two. Why?
2.65 The following diagram shows the Maxwell speed
distribution curves for a certain ideal gas at two di¤erent temperatures (T1 and T2 ). Calculate the value of T2 .
T1
300 K
f (c )
T2
0
500
?
1000 1500 2000
c /m s
1
2.66 Derive an expression for c mp . [Hint: Di¤erentiate
f ðcÞ with respect to c in Equation 2.26 and set the result
to zero.]
2.67 Calculate the values of crms ; c mp , and c for argon
at 298 K.
2.68 Calculate the value of c mp for C2 H6 at 25 C.
What is the ratio of the number of molecules with a
speed of 989 m s1 to the number of molecules with this
value of c mp ?
Molecular Collisions and the Mean Free Path
2.69 Considering the magnitude of molecular speeds,
explain why it takes so long (on the order of minutes) to
detect the odor of ammonia when someone opens a bottle of concentrated ammonia at the other end of a laboratory bench.
2.70 How does the mean free path of a gas depend on
(a) the temperature at constant volume, (b) the density,
(c) the pressure at constant temperature, (d) the volume
at constant temperature, and (e) the size of molecules?
2.71 A bag containing 20 marbles is being shaken vigorously. Calculate the mean free path of the marbles if
the volume of the bag is 850 cm 3 . The diameter of each
marble is 1.0 cm.
2.72 Calculate the mean free path and the binary number of collisions per liter per second between HI molecules at 300 K and 1.00 atm. The collision diameter of
35
Black plate (36,1)
36
Chapter 2: Properties of Gases
the HI molecules may be taken to be 5.10 Å. Assume
ideal-gas behavior.
2.73 Ultra-high-vacuum experiments are routinely performed at a total pressure of 1:0 1010 torr. Calculate
the mean free path of N2 molecules at 350 K under
these conditions. The collision diameter of N2 is 3.75 Å.
2.74 Suppose that helium atoms in a sealed container
all start with the same speed, 2:74 10 4 cm s1 . The
atoms are then allowed to collide with one another until
the Maxwell distribution is established. What is the
temperature of the gas at equilibrium? Assume that
there is no heat exchange between the gas and its
surroundings.
2.75 Compare the collision number and the mean free
path for air molecules at (a) sea level (T ¼ 300 K and
density ¼ 1:2 g L1 ) and (b) in the stratosphere (T ¼
250 K and density ¼ 5:0 103 g L1 ). The molar
mass of air may be taken as 29.0 g, and the collision
diameter is 3.72 Å.
2.76 Calculate the values of Z1 and Z11 for mercury
(Hg) vapor at 40 C, both at P ¼ 1:0 atm and at
P ¼ 0:10 atm. How do these two quantities depend on
pressure? The collision diameter of Hg is 4.26 Å.
Gas Diffusion and Effusion
2.77 Derive Equation 2.34 from Equation 2.23.
2.78 An inflammable gas is generated in marsh lands
and sewage by a certain anaerobic bacterium. A pure
sample of this gas was found to e¤use through an orifice in 12.6 min. Under identical conditions of temperature and pressure, oxygen takes 17.8 min to e¤use
through the same orifice. Calculate the molar mass of
the gas, and suggest what this gas might be.
2.79 Nickel forms a gaseous compound of the formula
Ni(CO)x . What is the value of x given the fact that
under the same conditions of temperature and pressure, methane (CH 4 ) e¤uses 3.3 times faster than the
compound?
2.80 In 2.00 min, 29.7 mL of He e¤use through a small
hole. Under the same conditions of temperature and
pressure, 10.0 mL of a mixture of CO and CO2 e¤use
through the hole in the same amount of time. Calculate
the percent composition by volume of the mixture.
2.81 Uranium-235 can be separated from uranium-238
by the e¤usion process involving UF6 . Assuming a
50 : 50 mixture at the start, what is the percentage of
enrichment after a single stage of separation?
2.82 An equimolar mixture of H2 and D2 e¤uses
through an orifice at a certain temperature. Calculate
the composition (in mole fractions) of the gas that
passes through the orifice. The molar mass of deuterium
is 2.014 g mol1 .
2.83 The rate ðre¤ Þ at which molecules confined to a
volume V e¤use through an orifice of area A is given by
ð1=4ÞnNA cA=V , where n is the number of moles of the
gas. An automobile tire of volume 30.0 L and pressure
1,500 torr is punctured as it runs over a sharp nail.
(a) Calculate the e¤usion rate if the diameter of the hole
is 1.0 mm. (b) How long would it take to lose half of
the air in the tire through e¤usion? Assume a constant
e¤usion rate and constant volume. The molar mass of
air is 29.0 g, and the temperature is 32.0 C.
Additional Problems
2.84 A barometer with a cross-sectional area of 1.00
cm 2 at sea level measures a pressure of 76.0 cm of mercury. The pressure exerted by this column of mercury is
equal to the pressure exerted by all the air on 1 cm 2 of
Earth’s surface. Given that the density of mercury is
13.6 g cm3 and the average radius of Earth is 6371
km, calculate the total mass of Earth’s atmosphere in
kilograms. (Hint: The surface area of a sphere is 4pr 2 ,
where r is the radius of the sphere.)
2.85 It has been said that every breath we take, on
average, contains molecules once exhaled by Wolfgang
Amadeus Mozart (1756–1791). The following calculations demonstrate the validity of this statement.
(a) Calculate the total number of molecules in the atmosphere. (Hint: Use the result from Problem 2.84 and
29.0 g mol1 as the molar mass of air.) (b) Assuming
the volume of every breath (inhale or exhale) is 500 mL,
calculate the number of molecules exhaled in each
breath at 37 C, which is the body temperature. (c) If
Mozart’s life span was exactly 35 years, how many molecules did he exhale in that period (given that an average person breathes 12 times per minute)? (d) Calculate
the fraction of molecules in the atmosphere that were
exhaled by Mozart. How many of Mozart’s molecules
do we inhale with each breath of air? Round your answer to one significant digit. (e) List three important
assumptions in these calculations.
2.86 A stockroom supervisor measured the contents of
a partially filled 25.0-gallon acetone drum on a day
when the temperature was 18.0 C and the atmospheric
pressure was 750 mmHg, and found that 15.4 gallons of
the solvent remained. After tightly sealing the drum, an
assistant dropped the drum while carrying it upstairs to
the organic laboratory. The drum was dented and its
internal volume was decreased to 20.4 gallons. What is
the total pressure inside the drum after the accident?
The vapor pressure of acetone at 18.0 C is 400 mmHg.
(Hint: At the time the drum was sealed, the pressure
inside the drum, which is equal to the sum of the pressures of air and acetone, was equal to the atmospheric
pressure.)
2.87 A relation known as the barometric formula is
useful for estimating the change in atmospheric pressure
with altitude. (a) Starting with the knowledge that
atmospheric pressure decreases with altitude, we have
dP ¼ rg dh, where r is the density of air, g is the acceleration due to gravity (9.81 m s2 ), and P and h are
Black plate (37,1)
Problems
the pressure and height, respectively. Assuming idealgas behavior and constant temperature, show that the
pressure P at height h is related to the pressure at sea
level P0 ðh ¼ 0Þ by P ¼ P0 egMh=RT . (Hint: For an ideal
gas, r ¼ PM=RT, where M is the molar mass.) (b) Calculate the atmospheric pressure at a height of 5.0 km,
assuming the temperature is constant at 5.0 C, given
that the average molar mass of air is 29.0 g mol1 .
2.88 In terms of the hard-sphere gas model, molecules
are assumed to possess finite volume, but there is no
interaction among the molecules. (a) Compare the P–V
isotherm for an ideal gas and that for a hard-sphere gas.
(b) Let b be the e¤ective volume of the gas. Write an
equation of state for this gas. (c) From this equation,
derive an expression for Z ¼ PV =RT for the hardsphere gas and make a plot of Z versus P for two values of T (T1 and T2 , T2 > T1 ). Be sure to indicate the
value of the intercepts on the Z axis. (d) Plot Z versus
T for fixed P for an ideal gas and for the hard-sphere
gas.
2.89 One way to gain a physical understanding of b in
the van der Waals equation is to calculate the
‘‘excluded volume.’’ Assume that the distance of closest
approach between two similar spherical molecules is the
sum of their radii ð2rÞ. (a) Calculate the volume around
each molecule into which the center of another molecule
cannot penetrate. (b) From your result in (a), calculate
the excluded volume for one mole of molecules, which
is the constant b. How does this compare with the sum
of the volumes of 1 mole of the same molecules?
2.90 You may have witnessed a demonstration in which
a burning candle standing in water is covered by an
upturned glass. The candle goes out and the water rises
in the glass. The explanation usually given for this phenomenon is that the oxygen in the glass is consumed by
combustion, leading to a decrease in volume and hence
the rise in the water level. However, the loss of oxygen
is only a minor consideration. (a) Using C12 H26 as the
formula for para‰n wax, write a balanced equation for
the combustion. Based on the nature of the products,
show that the predicted rise in water level due to the removal of oxygen is far less than the observed change.
(b) Devise a chemical process that would allow you to
measure the volume of oxygen in the trapped air. (Hint:
Use steel wool.) (c) What is the main reason for the
water rising in the glass after the flame is extinguished?
2.91 Express the van der Waals equation in the form of
Equation 2.14. Derive relationships between the van der
Waals constants (a and b) and the virial coe‰cients
(B; C, and D), given that
1
¼ 1 þ x þ x2 þ x3 þ 1x
jxj < 1
2.92 The Boyle temperature is the temperature at which
the coe‰cient B is zero. Therefore, a real gas behaves
like an ideal gas at this temperature. (a) Give a physical
interpretation of this behavior. (b) Using your result for
B for the van der Waals equation in Problem 2.91, calculate the Boyle temperature for argon, given that
a ¼ 1:345 atm L 2 mol2 and b ¼ 3:22 102 L mol1 .
2.93 Estimate the distance (in Å) between molecules of
water vapor at 100 C and 1.0 atm. Assume ideal-gas
behavior. Repeat the calculation for liquid water at
100 C, given that the density of water at 100 C is 0.96
g cm3 . Comment on your results. (The diameter of a
H2 O molecule is approximately 3 Å. 1 Å ¼ 108 cm.)
2.94 The following apparatus can be used to measure
atomic and molecular speed. A beam of metal atoms is
directed at a rotating cylinder in a vacuum. A small
opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at di¤erent speeds will strike the target at di¤erent
positions. In time, a layer of the metal will deposit on
the target area, and the variation in its thickness is
found to correspond to Maxwell’s speed distribution. In
one experiment, it is found that at 850 C, some bismuth
(Bi) atoms struck the target at a point 2.80 cm from the
spot directly opposite the slit. The diameter of the cylinder is 15.0 cm, and it is rotating at 130 revolutions per
second. (a) Calculate the speed (m s1 ) at which the target is moving. (Hint: The circumference of a circle is
given by 2pr, where r is the radius.) (b) Calculate the
time (in seconds) it takes for the target to travel 2.80
cm. (c) Determine the speed of the Bi atoms. Compare
your result in (c) with the crms value for Bi at 850 C.
Comment on the di¤erence.
Rotating cylinder
Target
Bi atoms
Slit
2.95 The escape velocity, v, from Earth’s gravitational
field is given by ð2GM=rÞ 1=2 , where G is the universal
gravitational constant (6:67 1011 m 3 kg1 s2 ), M is
the mass of Earth (6:0 10 24 kg), and r is the distance
from the center of Earth to the object, in meters. Compare the average speeds of He and N2 molecules in the
thermosphere (altitude about 100 km, T ¼ 250 K).
Which of the two molecules will have a greater tendency to escape? The radius of Earth is 6:4 10 6 m.
2.96 Calculate the ratio of the number of O3 molecules
with a speed of 1300 m s1 at 360 K to the number
with that speed at 293 K.
2.97 Calculate the collision frequency for 1.0 mole of
krypton (Kr) at equilibrium at 300 K and 1.0 atm pressure. Which of the following alterations increases the
collision frequency more: (a) doubling the temperature
at constant pressure or (b) doubling the pressure at con-
37
Black plate (38,1)
38
Chapter 2: Properties of Gases
stant temperature? (Hint: The collision diameter of Kr
is 4.16 Å.)
2.98 Apply your knowledge of the kinetic theory of
gases to the following situations. (a) Two flasks of volumes V1 and V2 (where V2 > V1 ) contain the same
number of helium atoms at the same temperature.
(i) Compare the root-mean-square (rms) speeds and
average kinetic energies of the helium (He) atoms in the
flasks. (ii) Compare the frequency and the force with
which the He atoms collide with the walls of their containers. (b) Equal numbers of He atoms are placed in
two flasks of the same volume at temperatures T1 and
T2 (where T2 > T1 ). (i) Compare the rms speeds of the
atoms in the two flasks. (ii) Compare the frequency and
the force with which the He atoms collide with the walls
of their containers. (c) Equal numbers of He and neon
(Ne) atoms are placed in two flasks of the same volume.
The temperature of both gases is 74 C. Comment on
the validity of the following statements: (i) The rms
speed of He is equal to that of Ne. (ii) The average kinetic energies of the two gases are equal. (iii) The rms
speed of each He atom is 1:47 10 3 m s1 .
2.99 Consider 1 mole each of gaseous He and N2 at the
same temperature and pressure. State which gas (if any)
has the greater value for: (a) c, (b) crms , (c) E trans ,
(d) Z1 , (e) Z11 , (f ) density, and (g) mean free path. The
diameter of N2 is 1.7 times that of He.
2.100 The root-mean-square velocity of a certain gaseous oxide is 493 m s1 at 20 C. What is the molecular
formula of the compound?
2.101 Calculate the mean kinetic energy ðE trans Þ in
joules of the following molecules at 350 K: (a) He,
(b) CO2 , and (c) UF6 . Explain your results.
2.102 A sample of neon gas is heated from 300 K to
390 K. Calculate the percent increase in its kinetic
energy.
2.103 A CO2 fire extinguisher is located on the outside
of a building in Massachusetts. During the winter
months, one can hear a sloshing sound when the extinguisher is gently shaken. In the summertime, there is
often no sound when it is shaken. Explain. Assume that
the extinguisher has no leaks and that it has not been
used.