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**Unformatted text preview: **Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 5 Gases
INTRODUCTION
UNDER CERTAIN CONDITIONS OF PRESSURE AND TEMPERATURE, MOST 5.1 SUBSTANCES THAT EXIST AS GASES SUBSTANCES CAN EXIST IN ANY ONE OF THE THREE STATES OF MATTER: 5.2 PRESSURE OF A GAS SOLID, LIQUID, OR GAS. 5.3 THE GAS LAWS 5.4 THE IDEAL GAS EQUATION 5.5 GAS STOICHIOMETRY 5.6 DALTON’S LAW OF PARTIAL PRESSURES 5.7 THE KINETIC MOLECULAR THEORY OF
GASES 5.8 DEVIATION FROM IDEAL BEHAVIOR WATER, FOR EXAMPLE, CAN BE SOLID ICE, LIQ- UID WATER, OR STEAM OR WATER VAPOR. THE PHYSICAL PROPERTIES OF A SUBSTANCE OFTEN DEPEND ON ITS STATE. GASES, THE SUBJECT OF THIS CHAPTER, ARE MUCH SIMPLER THAN LIQUIDS OR SOLIDS IN MANY WAYS. MOLECULAR MOTION IN GASES IS TOTALLY RANDOM, AND THE FORCES OF ATTRACTION BETWEEN GAS MOLECULES ARE SO SMALL THAT EACH MOLECULE MOVES FREELY AND ESSENTIALLY INDEPENDENTLY OF OTHER MOLECULES. SUBJECTED TO CHANGES IN TEMPERATURE AND PRESSURE, GASES BEHAVE MUCH MORE
PREDICTABLY THAN DO SOLIDS AND LIQUIDS. THE LAWS THAT GOVERN THIS BEHAVIOR HAVE PLAYED AN IMPORTANT ROLE IN THE DEVELOPMENT
OF THE ATOMIC THEORY OF MATTER AND THE KINETIC MOLECULAR THEORY OF GASES. 155 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 156 GASES 5.1 SUBSTANCES THAT EXIST AS GASES We live at the bottom of an ocean of air whose composition by volume is roughly 78
percent N2, 21 percent O2, and 1 percent other gases, including CO2. In the 1990s, the
chemistry of this vital mixture of gases has become a source of great interest because
of the detrimental effects of environmental pollution. The chemistry of the atmosphere
and polluting gases is discussed in Chapter 17. Here we will focus generally on the
behavior of substances that exist as gases under normal atmospheric conditions, which
are defined as 25°C and 1 atmosphere (atm) pressure (see Section 5.2).
Figure 5.1 shows the elements that are gases under normal atmospheric conditions. Note that hydrogen, nitrogen, oxygen, fluorine, and chlorine exist as gaseous diatomic molecules: H2, N2, F2, and Cl2. An allotrope of oxygen, ozone (O3), is also a
gas at room temperature. All the elements in Group 8A, the noble gases, are monatomic
gases: He, Ne, Ar, Kr, Xe, and Rn.
Ionic compounds do not exist as gases at 25°C and 1 atm, because cations and
anions in an ionic solid are held together by very strong electrostatic forces. To overcome these attractions we must apply a large amount of energy, which in practice means
strongly heating the solid. Under normal conditions, all we can do is melt the solid;
for example, NaCl melts at the rather high temperature of 801°C. In order to boil it,
we would have to raise the temperature to well above 1000°C.
The behavior of molecular compounds is more varied. Some — for example, CO,
CO2, HCl, NH3, and CH4 (methane) — are gases, but the majority of molecular compounds are liquids or solids at room temperature. However, on heating they are converted to gases much more easily than ionic compounds. In other words, molecular
compounds usually boil at much lower temperatures than ionic compounds do. There
is no simple rule to help us determine whether a certain molecular compound is a gas
under normal atmospheric conditions. To make such a determination we need to understand the nature and magnitude of the attractive forces among the molecules, called
intermolecular forces (discussed in Chapter 11). In general, the stronger these attractions, the less likely a compound can exist as a gas at ordinary temperatures.
Of the gases listed in Table 5.1, only O2 is essential for our survival. Hydrogen
sulfide (H2S) and hydrogen cyanide (HCN) are deadly poisons. Several others, such
as CO, NO2, O3, and SO2, are somewhat less toxic. The gases He, Ne, and Ar are
chemically inert; that is, they do not react with any other substance. Most gases are
colorless. Exceptions are F2, Cl2, and NO2. The dark-brown color of NO2 is sometimes
visible in polluted air. All gases have the following physical characteristics:
•
•
•
• NO2 gas. 5.2 Gases
Gases
Gases
Gases assume the volume and shape of their containers.
are the most compressible of the states of matter.
will mix evenly and completely when confined to the same container.
have much lower densities than liquids and solids. PRESSURE OF A GAS Gases exert pressure on any surface with which they come in contact, because gas molecules are constantly in motion. We humans have adapted so well physiologically to
the pressure of the air around us that we are usually unaware of it, perhaps as fish are
not conscious of the water’s pressure on them. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.2 157 PRESSURE OF A GAS 1A 8A H He
2A 3A 4A 5A 6A 7A Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3B 4B 5B 6B 7B 8B 1B 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn FIGURE 5.1 Elements that exist
as gases at 25 C and 1 atm. The
noble gases (the Group 8A elements) are monatomic species;
the other elements exist as diatomic molecules. Ozone (O3) is
also a gas. It is easy to demonstrate atmospheric pressure. One everyday example is the ability to drink a liquid through a straw. Sucking air out of the straw reduces the pressure
inside the straw. The greater atmospheric pressure on the liquid pushes it up into the
straw to replace the air that has been sucked out.
SI UNITS OF PRESSURE Pressure is one of the most readily measurable properties of a gas. In order to understand how we measure the pressure of a gas, it is helpful to know how the units of
measurement are derived. We begin with velocity and acceleration.
Velocity is defined as the change in distance with elapsed time; that is
velocity
TABLE 5.1 distance moved
elapsed time Some Substances Found as Gases at 1 atm and 25 C ELEMENTS A gas is a substance that is normally in the gaseous state at ordinary temperatures and pressures; a
vapor is the gaseous form of any
substance that is a liquid or a
solid at normal temperatures and
pressures. Thus, at 25 C and 1
atm pressure, we speak of water
vapor and oxygen gas. COMPOUNDS H2 (molecular hydrogen)
N2 (molecular nitrogen)
O2 (molecular oxygen)
O3 (ozone)
F2 (molecular fluorine)
Cl2 (molecular chlorine)
He (helium)
Ne (neon)
Ar (argon)
Kr (krypton)
Xe (xenon)
Rn (radon) HF (hydrogen fluoride)
HCl (hydrogen chloride)
HBr (hydrogen bromide)
HI (hydrogen iodide)
CO (carbon monoxide)
CO2 (carbon dioxide)
NH3 (ammonia)
NO (nitric oxide)
NO2 (nitrogen dioxide)
N2O (nitrous oxide)
SO2 (sulfur dioxide)
H2S (hydrogen sulfide)
HCN (hydrogen cyanide)* *The boiling point of HCN is 26 C, but it is close enough to qualify as a gas at ordinary atmospheric conditions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 158 GASES The SI unit for velocity is m/s, although we also use cm/s.
Acceleration is the change in velocity with time, or
acceleration change in velocity
elapsed time Acceleration is measured in m/s2 (or cm/s2).
The second law of motion, formulated by Sir Isaac Newton† in the late seventeenth century, defines another term, from which the units of pressure are derived,
namely, force. According to this law,
force
1 N is roughly equivalent to the
force exerted by Earth’s gravity
on an apple. mass acceleration In this context, the SI unit of force is the newton (N), where
1N 1 kg m/s2 Finally, we define pressure as force applied per unit area:
pressure force
area The SI unit of pressure is the pascal (Pa),‡ defined as one newton per square meter:
1 Pa 1 N/m2 ATMOSPHERIC PRESSURE
Upper
atmosphere Sea level
FIGURE 5.2 A column of air
extending from sea level to the
upper atmosphere. The atoms and molecules of the gases in the atmosphere, like those of all other matter, are subject to Earth’s gravitational pull. As a consequence, the atmosphere is much
denser near the surface of Earth than at high altitudes. (The air outside the pressurized
cabin of an airplane at 9 km is too thin to breathe.) In fact, the density of air decreases
very rapidly with increasing distance from Earth (Figure 5.2). Measurements show that
about 50 percent of the atmosphere lies within 6.4 km of Earth’s surface, 90 percent
within 16 km, and 99 percent within 32 km. Not surprisingly the denser the air is, the
greater the pressure it exerts. The force experienced by any area exposed to Earth’s atmosphere is equal to the weight of the column of air above it. Atmospheric pressure
is the pressure exerted by Earth’s atmosphere. The actual value of atmospheric pressure depends on location, temperature, and weather conditions.
How is atmospheric pressure measured? The barometer is probably the most familiar instrument for measuring atmospheric pressure. A simple barometer consists of
a long glass tube, closed at one end and filled with mercury. If the tube is carefully inverted in a dish of mercury so that no air enters the tube, some mercury will flow out
of the tube into the dish, creating a vacuum at the top (Figure 5.3). The weight of the
mercury remaining in the tube is supported by atmospheric pressure acting on the surface of the mercury in the dish. Standard atmospheric pressure (1 atm) is equal to the
pressure that supports a column of mercury exactly 760 mm (or 76 cm) high at 0°C at
sea level. In other words, the standard atmosphere equals a pressure of 760 mmHg,
where mmHg represents the pressure exerted by a column of mercury 1 mm high. The
† Sir Isaac Newton (1642–1726). English mathematician, physicist, and astronomer. Newton is regarded by many as one
of the two greatest physicists the world has known (the other is Albert Einstein). There was hardly a branch of physics to
which Newton did not make a significant contribution. His book Principia, published in 1687, marks a milestone in the history of science.
‡
Blaise Pascal (1623–1662). French mathematician and physicist. Pascal’s work ranged widely in mathematics and physics,
but his specialty was in the area of hydrodynamics (the study of the motion of fluids). He also invented a calculating machine. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.2 PRESSURE OF A GAS 159 mmHg unit is also called the torr, after the Italian scientist Evangelista Torricelli,† who
invented the barometer. Thus
1 torr 1 mmHg 1 atm 760 mmHg and
760 torr The relation between atmospheres and pascals (see Appendix 2) is 76 cm 1 atm Atmospheric
pressure 101,325 Pa
105 Pa 1.01325 and since 1000 Pa 1 kPa (kilopascal)
1 atm 102 kPa 1.01325 The following examples show the conversion from mmHg to atm and kPa.
FIGURE 5.3 A barometer for
measuring atmospheric pressure.
Above the mercury in the tube is
a vacuum. The column of mercury
is supported by atmospheric pressure. EXAMPLE 5.1 The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized
to protect the passengers. What is the pressure in atmospheres in the cabin if the
barometer reading is 688 mmHg?
Answer The pressure in atmospheres is calculated as follows:
pressure 1 atm
760 mmHg 688 mmHg
0.905 atm Similar problem: 5.14.
PRACTICE EXERCISE Convert 749 mmHg to atmospheres.
EXAMPLE 5.2 The atmospheric pressure in San Francisco on a certain day was 732 mmHg. What
was the pressure in kPa?
Answer The conversion equations show that
1 atm 105 Pa 1.01325 760 mmHg which allow us to calculate the atmospheric pressure in kPa:
pressure 1.01325 105 Pa
760 mmHg 732 mmHg
9.76 104 Pa 97.6 kPa Similar problem: 5.13.
PRACTICE EXERCISE Convert 295 mmHg to kilopascals. † Evangelista Torricelli (1608–1674). Italian mathematician. Torricelli was supposedly the first person to recognize the existence of atmospheric pressure. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 160 GASES FIGURE 5.4 Two types of
manometers used to measure gas
pressures. (a) Gas pressure is less
than atmospheric pressure. (b)
Gas pressure is greater than atmospheric pressure. Vacuum Gas h Gas Pgas = Ph h Pgas = Ph + Patm (a) (b) A manometer is a device used to measure the pressure of gases other than the atmosphere. The principle of operation of a manometer is similar to that of a barometer.
There are two types of manometers, shown in Figure 5.4. The closed-tube manometer
is normally used to measure pressures below atmospheric pressure [Figure 5.4(a)] while
the open-tube manometer is better suited for measuring pressures equal to or greater
than atmospheric pressure [Figure 5.4(b)].
Nearly all barometers and most manometers use mercury as the working fluid,
despite the fact that it is a toxic substance with a harmful vapor. The reason is that
mercury has a very high density (13.6 g/mL) compared with most other liquids. Since
the height of the liquid in a column is inversely proportional to the liquid’s density,
this property allows the construction of manageably small barometers and manometers. 5.3 THE GAS LAWS The gas laws we will study in this chapter are the product of countless experiments on
the physical properties of gases that were carried out over several centuries. Each of
these generalizations regarding the macroscopic behavior of gaseous substances represents a milestone in the history of science. Together they have played a major role
in the development of many ideas in chemistry.
THE PRESSURE-VOLUME RELATIONSHIP: BOYLE’S LAW In the seventeenth century, Robert Boyle† studied the behavior of gases systematically
and quantitatively. In one series of studies, Boyle investigated the pressure-volume re†
Robert Boyle (1627–1691). British chemist and natural philosopher. Although Boyle is commonly associated with the gas
law that bears his name, he made many other significant contributions in chemistry and physics. Although he was often at
odds with contemporary scientists, his book The Skeptical Chymist (1661) influenced generations of chemists. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.3 161 THE GAS LAWS FIGURE 5.5 Apparatus for
studying the relationship between
pressure and volume of a gas. In
(a) the pressure of the gas is
equal to the atmospheric pressure. The pressure exerted on the
gas increases from (a) to (d) as
mercury is added, and the volume of the gas decreases, as
predicted by Boyle’s law. The extra pressure exerted on the gas is
shown by the difference in the
mercury levels (h mmHg). The
temperature of the gas is kept
constant. The pressure applied to a gas is
equal to the gas pressure. lationship of a gas sample using an apparatus like that shown in Figure 5.5. In Figure
5.5(a) the pressure exerted on the gas by the mercury added to the tube is equal to atmospheric pressure. In Figure 5.5(b) an increase in pressure due to the addition of more
mercury results in a decrease in the volume of the gas and in unequal levels of mercury in the tube. Boyle noticed that when temperature is held constant, the volume (V)
of a given amount of a gas decreases as the total applied pressure (P) — atmospheric
pressure plus the pressure due to the added mercury — is increased. This relationship
between pressure and volume is evident in Figure 5.5(b), (c), and (d). Conversely, if
the applied pressure is decreased, the gas volume becomes larger. The results of several pressure-volume measurements are given in Table 5.2.
The P V data recorded in Table 5.2 are consistent with this mathematical expression showing an inverse relationship:
1
P V where the symbol
write means proportional to. To change V to an equals sign, we must 1
P k1 (5.1a) where k1 is a constant called the proportionality constant. Equation (5.1a) is an expression of Boyle’s law, which states that the volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure. We can
rearrange Equation (5.1a) and obtain
PV
TABLE 5.2 724 V (arbitrary units) Back 1.50
Forward (5.1b) Typical Pressure-Volume Relationship Obtained by Boyle P (mmHg) PV k1 1.09 Main Menu 869 951 1.33
3 10 1.16 TOC 998 1.22
3 10 1.16 1230 1.16
3 10 1.16 Study Guide TOC 1893 0.94
3 10 1.2 2250 0.61
3 10 1.2 0.51
3 10 Textbook Website 1.1 103 MHHE Website 162 GASES This form of Boyle’s law says that the product of the pressure and volume of a gas at
constant temperature is a constant.
The concept of one quantity being proportional to another and the use of a proportionality constant can be clarified through the following analogy. The daily income
of a movie theater depends on both the price of the tickets (in dollars per ticket) and
the number of tickets sold. Assuming that the theater charges one price for all tickets,
we write
income (dollar/ticket) number of tickets sold Since the number of tickets sold varies from day to day, the income on a given day is
said to be proportional to the number of tickets sold:
income number of tickets sold
C number of tickets sold where C, the proportionality constant, is the price per ticket.
Figure 5.6 shows two conventional ways of expressing Boyle’s findings graphically. Figure 5.6(a) is a graph of the equation PV k1; Figure 5.6(b) is a graph of the
equivalent equation P k1 1/V. Note that the latter is a linear equation of the form
y mx b, where b 0.
Although the individual values of pressure and volume can vary greatly for a given
sample of gas, as long as the temperature is held constant and the amount of the gas
does not change, P times V is always equal to the same constant. Therefore, for a given
sample of gas under two different sets of conditions at constant temperature, we have
P1V1 k1
P2V2 P1V1 P2V2 or
(5.2) where V1 and V2 are the volumes at pressures P1 and P2, respectively.
One common application of Boyle’s law is predicting how the volume of a gas
will be affected by a change in pressure or how the pressure exerted by a gas will be
affected by a change in volume based on Equation (5.2). Example 5.3 illustrates the
procedure for solving problems with Boyle’s law. P P 0.6 atm
0.3 atm 2L 4L V 1
––
V (a)
(b)
FIGURE 5.6 Graphs showing variation of the volume of a gas sample with the pressure exerted on the gas, at constant temperature. (a) P versus V. Note that the volume of the gas doubles
as the pressure is halved; (b) P versus 1/ V. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.3 THE GAS LAWS 163 EXAMPLE 5.3 An inflated balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to
rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the
temperature remains constant, what is the final volume of the balloon?
Answer We use Equation (5.2):
P1V1 P2V2 where
INITIAL CONDITIONS P1 V1 FINAL CONDITIONS 1.0 atm
0.55 L P2
V2 0.40 atm
? Therefore
V2 A scientific research balloon. V1 P1
P2 0.55 L 1.0 atm
0.40 atm 1.4 L When pressure is reduced (at constant temperature), volume increases.
The final volume is greater than the initial volume, so the answer is reasonable. Comment Similar problem: 5.17. PRACTICE EXERCISE A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg.
Calculate the pressure of the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL. THE TEMPERATURE-VOLUME RELATIONSHIP: CHARLES’ AND GAY-LUSSAC’S LAW Capillary
tubing Mercury
Gas Low
temperature High
temperature FIGURE 5.7 Variation of the
volume of a gas sample with temperature, at constant pressure.
The pressure exerted on the gas
is the sum of the atmospheric
pressure and the pressure due to
the weight of the mercury. Boyle’s law depends on the temperature of the system remaining constant. But suppose the temperature changes: How does a change in temperature affect the volume
and pressure of a gas? Let us first look at the effect of temperature on the volume of
a gas. The earliest investigators of this relationship were French scientists, Jacques
Charles† and Joseph Gay-Lussac.‡ Their studies showed that, at constant pressure, the
volume of a gas sample expands when heated and contracts when cooled (Figure 5.7).
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 different straight line for the volume-temperature plot, but
we get the same zero-volume temperature intercept at 273.15 C (Figure 5.8). (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.)
† Jacques Alexandre Cesar Charles (1746–1823). French physicist. He was a gifted lecturer, an inventor of scientific apparatus, and the first person to use hydrogen to inflate balloons. ‡
Joseph Louis Gay-Lussac (1778–1850). French chemist and physicist. Like Charles, Gay-Lussac was a balloon enthusiast. Once he ascended to an altitude of 20,000 feet to collect air samples for analysis. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website GASES FIGURE 5.8 Variation of the
volume of a gas sample with
changing temperature, at constant pressure. Each line represents the variation at a certain
pressure. The pressures increase
from P1 to P4. All gases ultimately
condense (become liquids) if they
are cooled to sufficiently low temperatures; the solid portions of
the lines represent the temperature region above the condensation point. When these lines are
extrapolated, or extended (the
dashed portions), they all intersect at the point representing
zero volume and a temperature
of 273.15 C. 50 P1 40
V (mL) 164 P2 30
P3
20 –273.15°C P4 10
0
–300 –200 –100 0
t (°C) 100 200 300 400 In 1848 Lord Kelvin† realized the significance of this phenomenon. He identified
273.15 C as absolute zero, 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 (see Section 1.6). On the Kelvin scale, one kelvin
(K) is equal in magnitude to one degree Celsius. The only difference between the absolute temperature scale and the Celsius scale is that the zero position is shifted.
Important points on the two scales match up as follows:
Absolute zero: Under special experimental conditions, scientists have approached
to within a small fraction of a
kelvin of absolute zero. 0K 273.15 C Freezing point of water: 273.15 K 0C Boiling point of water: 373.15 K 100 C The conversion between C and K is given on p. 18. In most calculations we will use
273 instead of 273.15 as the term relating K and °C. By convention, we use T to denote absolute (kelvin) temperature and t to indicate temperature on the Celsius scale.
The dependence of the volume of a gas on temperature is given by
V
V The constant k2 is not the same as
the constant k1 in Equations (5.1a)
and (5.1b). k2T V
T or T k2 (5.3) where k2 is the proportionality constant. Equation (5.3) is known as Charles’ and GayLussac’s law, or simply Charles’ law, which states that the volume of a fixed amount
of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.
Just as we did for pressure-volume relationships at constant temperature, we can
compare two sets of volume-temperature conditions for a given sample of gas at constant pressure. From Equation (5.3) we can write
V1
T1
V1
T1 or k2 V2
T2 V2
T2 (5.4) † William Thomson, Lord Kelvin (1824–1907). Scottish mathematician and physicist. Kelvin did important work in almost
every branch of physics. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.3 THE GAS LAWS 165 where V1 and V2 are the volumes of the gases at temperatures T1 and T2 (both in
kelvins), respectively. In all subsequent calculations we assume that the temperatures
given in C are exact, so that they do not affect the number of significant figures.
The following example illustrates the use of Charles’ law.
EXAMPLE 5.4 A 452-mL sample of fluorine gas is heated from 22 C to 187 C at constant pressure. What is its final volume?
Answer We use Equation (5.4),
V1
T1 V2
T2 where
INITIAL CONDITIONS Remember to convert C to K
when solving gas-law problems. V1 T1 452 mL
(22 273) K Hence V2 FINAL CONDITIONS V2
T2 295 K ? (187 273) K 460 K T2
T1 V1 460 K
295 K 452 mL
705 mL
Similar problem: 5.21. Comment As you can see, when the gas is heated at constant pressure, it expands. PRACTICE EXERCISE A sample of carbon monoxide gas occupies 3.20 L at 125 C. Calculate the temperature at which the gas will occupy 1.54 L if the pressure remains constant.
THE VOLUME-AMOUNT RELATIONSHIP: AVOGADRO’S LAW
Avogadro’s name first appeared in
Section 3.2. The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle,
Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same
temperature and pressure, equal volumes of different gases contain the same number
of molecules (or atoms if the gas is monatomic). It follows that the volume of any
given gas must be proportional to the number of moles of molecules present; that is,
V n V k3n (5.5) where n represents the number of moles and k3 is the proportionality constant.
Equation (5.5) is the mathematical expression of Avogadro’s law, which states that
at constant pressure and temperature, the volume of a gas is directly proportional to
the number of moles of the gas present. From Avogadro’s law we learn that when two
gases react with each other, their reacting volumes have a simple ratio to each other.
If the product is a gas, its volume is related to the volume of the reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac). For example, consider the synthesis of ammonia from molecular hydrogen and molecular nitrogen:
3H2(g) Back Forward Main Menu TOC N2(g) 88n 2NH3(g) 3 mol 1 mol Study Guide TOC 2 mol Textbook Website MHHE Website 166 GASES FIGURE 5.9 Volume relationship of gases in a chemical reaction. The ratio of the volumes of
molecular hydrogen to molecular
nitrogen is 3:1, and that of ammonia (the product) to molecular
hydrogen and molecular nitrogen
combined (the reactants) is 2:4,
or 1:2. + N2 (g)
1 molecule
1 mole
1 volume +
+
+
+ 3H2 (g)
3 molecules
3 moles
3 volumes 2NH3 (g)
2 molecules
2 moles
2 volumes Since, at the same temperature and pressure, the volumes of gases are directly proportional to the number of moles of the gases present, we can now write
3H2(g)
3 volumes N2(g) 88n 2NH3(g)
1 volume 2 volumes The volume ratio of molecular hydrogen to molecular nitrogen is 3 1, and that of ammonia (the product) to molecular hydrogen and molecular nitrogen (the reactants) is
2 4, or 1 2 (Figure 5.9).
5.4 THE IDEAL GAS EQUATION Let us summarize the gas laws we have discussed so far:
Boyle’s law: V 1
P (at constant n and T ) Charles’ law: V T (at constant n and P) Avogadro’s law: V n (at constant P and T ) We can combine all three expressions to form a single master equation for the behavior of gases:
V nT
P
R nT
P or
PV nRT (5.6) where R, the proportionality constant, is called the gas constant. Equation (5.6), which
is called the ideal gas equation, describes the relationship among the four variables
P, V, T, and n. An ideal gas is a hypothetical gas whose pressure-volume-temperature
behavior can be completely accounted for by the ideal gas equation. The molecules of
an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container. Although there is no such thing in nature as
an ideal gas, discrepancies in the behavior of real gases over reasonable temperature
and pressure ranges do not significantly affect calculations. Thus we can safely use the
ideal gas equation to solve many gas problems.
Before we can apply the ideal gas equation to a real system, we must evaluate the
gas constant R. At 0 C (273.15 K) and 1 atm pressure, many real gases behave like an Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.4 THE IDEAL GAS EQUATION 167 FIGURE 5.10 A comparison of
the molar volume at STP (which is
approximately 22.4 L) with a
basketball. ideal gas. Experiments show that under these conditions, 1 mole of an ideal gas occupies 22.414 L, which is somewhat greater than the volume of a basketball, as shown
in Figure 5.10. The conditions 0 C and 1 atm are called standard temperature and
pressure, often abbreviated STP. From Equation (5.6) we can write
R The gas constant can be expressed
in different units (see Appendix 2). PV
nT
(1 atm)(22.414 L)
(1 mol)(273.15 K)
0.082057 L atm
K mol 0.082057 L atm/K mol The dots between L and atm and between K and mol remind us that both L and atm
are in the numerator and both K and mol are in the denominator. For most calculations, we will round off the value of R to three significant figures (0.0821 L atm/K
mol) and use 22.41 L for the molar volume of a gas at STP.
The following example shows that if we know the quantity, volume, and temperature of a gas, we can calculate its pressure using the ideal gas equation.
EXAMPLE 5.5 Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the
pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43
L at 69.5 C.
Answer Rearranging the ideal gas equation we write
P nRT
V
(1.82 mol)(0.0821 L atm/K mol)(69.5
5.43 L
9.42 atm Similar problem: 5.30. Back Forward Main Menu 273) K TOC Study Guide TOC Textbook Website MHHE Website 168 GASES PRACTICE EXERCISE Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54
atm and 76 C.
By using the fact that the molar volume of a gas occupies 22.41 L at STP, we can
calculate the volume of a gas at STP without using the ideal gas equation.
EXAMPLE 5.6 Calculate the volume (in liters) occupied by 7.40 g of CO2 at STP.
Answer Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP, we write
V 1 mol CO2
44.01 g CO2 7.40 g CO2 22.41 L
1 mol CO2 3.77 L Similar problems: 5.41, 5.42.
PRACTICE EXERCISE What is the volume (in liters) occupied by 49.8 g of HCl at STP?
The ideal gas equation is useful for problems that do not involve changes in P, V,
T, and n for a gas sample. At times, however, we need to deal with changes in pressure, volume, and temperature, or even in the amount of a gas. When conditions change,
we must employ a modified form of the ideal gas equation that takes into account the
initial and final conditions. We derive the modified equation as follows. From Equation
(5.6),
R The subscripts 1 and 2 denote the
initial and final states of the gas. P1V1
n1T1 (before change) and
R P2V2
n2T2 (after change) so that
P1V1
n1T1 P2V2
n2T2 If n1 n2, as is usually the case because the amount of gas normally does not change,
the equation then becomes
P1V1
T1 P2V2
T2 (5.7) Applications of Equation (5.7) are the subject of the following two examples.
EXAMPLE 5.7 A small bubble rises from the bottom of a lake, where the temperature and pressure
are 8 C and 6.4 atm, to the water ’s surface, where the temperature is 25 C and pressure is 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.4 Answer THE IDEAL GAS EQUATION We start by writing
INITIAL CONDITIONS FINAL CONDITIONS P1
V1
T1 We can use any appropriate units
for volume (or pressure) as long
as we use the same units on both
sides of the equation. 169 P2
V2
T2 6.4 atm
2.1 mL
(8 273) K 281 K 1.0 atm
?
(25 273) K 298 K The amount of the gas in the bubble remains constant, so that n1
the final volume, V2, we rearrange Equation (5.7) as follows:
V2 V1 P1
P2 T2
T1
6.4 atm
1.0 atm 2.1 mL n2. To calculate 298 K
281 K 14 mL
Similar problems: 5.33, 5.36. Thus, the bubble’s volume increases from 2.1 mL to 14 mL because of the decrease
in water pressure and the increase in temperature.
PRACTICE EXERCISE A gas initially at 4.0 L, 1.2 atm, and 66 C undergoes a change so that its final volume and temperature become 1.7 L and 42 C. What is its final pressure? Assume
the number of moles remains unchanged.
EXAMPLE 5.8 Argon is an inert gas used in light bulbs to retard the vaporization of the filament.
A certain light bulb containing argon at 1.20 atm and 18 C is heated to 85 C at constant volume. Calculate its final pressure (in atm).
The volume and amount of the gas are unchanged, so V1
n2. Equation (5.7) can therefore be written as Answer P1
T1 V2 and n1 P2
T2 Next we write
Electric light bulbs are usually
filled with argon. INITIAL CONDITIONS P1
T1
One practical consequence of this
relationship is that automobile tire
pressures should be checked only
when the tires are at normal temperatures. After a long drive (especially in the summer), tires
become quite hot, and the air
pressure inside them rises. 1.20 atm
(18 273) K FINAL CONDITIONS P2
T2 291 K ?
(85 273) K 358 K The final pressure is given by
P2 P1 T2
T1 1.20 atm 358 K
291 K 1.48 atm At constant volume, the pressure of a given amount of gas is directly
proportional to its absolute temperature. Therefore the increase in pressure is reasonable here. Comment Similar problem: 5.34. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 170 GASES PRACTICE EXERCISE A sample of oxygen gas initially at 0.97 atm is cooled from 21°C to
stant volume. What is its final pressure (in atm)? 68°C at con- DENSITY CALCULATIONS If we rearrange the ideal gas equation, we can calculate the density of a gas:
n
V P
RT The number of moles of the gas, n, is given by
m n M where m is the mass of the gas in grams and M is its molar mass. Therefore
m MV P
RT Since density, d, is mass per unit volume, we can write
m
V d PM
RT (5.8) Unlike molecules in condensed matter (that is, in liquids and solids), gaseous molecules are separated by distances that are large compared with their size. Consequently,
the density of gases is very low under atmospheric conditions. For this reason, gas densities are usually expressed in grams per liter (g/L) rather than grams per milliliter
(g/mL), as the following example shows.
EXAMPLE 5.9 Calculate the density of ammonia (NH3) in grams per liter (g/L) at 752 mmHg and
55 C.
Answer To convert the pressure to atmospheres, we write
P 752 mmHg 1 atm
760 mmHg 752
atm
760 Using Equation (5.8) and T
d 273 55 328 K, we have PM
RT
(752/760) atm(17.03 g/mol)
(0.0821 L atm/K mol)(328 K)
0.626 g/L In units of grams per milliliter, the gas density would be 6.26 x 10
g/mL, which is a very small number.
Comment Similar problem: 5.46. Back Forward Main Menu TOC Study Guide TOC Textbook Website 4 MHHE Website 5.4 THE IDEAL GAS EQUATION 171 PRACTICE EXERCISE What is the density (in g/L) of uranium hexafluoride (UF6) at 779 mmHg and 62 C?
THE MOLAR MASS OF A GASEOUS SUBSTANCE FIGURE 5.11 An apparatus
for measuring the density of a
gas. A bulb of known volume is
filled with the gas under study at
a certain temperature and pressure. First the bulb is weighed,
and then it is emptied (evacuated) and weighed again. The
difference in masses gives the
mass of the gas. Knowing the
volume of the bulb, we can calculate the density of the gas. From what we have said so far, you may have the impression that the molar mass of
a substance is found by examining its formula and summing the molar masses of its
component atoms. However, this procedure works only if the actual formula of the substance is known. In practice, chemists often deal with substances of unknown or only
partially defined composition. If the unknown substance is gaseous, its molar mass can
nevertheless be found thanks to the ideal gas equation. All that is needed is an experimentally determined density value (or mass and volume data) for the gas at a known
temperature and pressure. By rearranging Equation (5.8) we get
M dRT
P (5.9) In a typical experiment, a bulb of known volume is filled with the gaseous substance
under study. The temperature and pressure of the gas sample are recorded, and the total mass of the bulb plus gas sample is determined (Figure 5.11). The bulb is then evacuated (emptied) and weighed again. The difference in mass is the mass of the gas. The
density of the gas is equal to its mass divided by the volume of the bulb. Once we
know the density of a gas, we can calculate the molar mass of the substance using
Equation (5.9). Example 5.10 shows this calculation.
EXAMPLE 5.10 A chemist has synthesized a greenish-yellow gaseous compound of chlorine and
oxygen and finds that its density is 7.71 g/L at 36 C and 2.88 atm. Calculate the
molar mass of the compound and determine its molecular formula.
Answer We substitute in Equation (5.9):
M dRT
P
(7.71 g/L)(0.0821 L atm/K mol)(36
2.88 atm 273) K 67.9 g/mol Similar problems: 5.47, 5.48. We can determine the molecular formula of the compound by trial and error, using
only the knowledge of the molar masses of chlorine (35.45 g) and oxygen (16.00
g). We know that a compound containing one Cl atom and one O atom would have
a molar mass of 51.45 g, which is too low, while the molar mass of a compound
made up of two Cl atoms and one O atom is 86.90 g, which is too high. Thus the
compound must contain one Cl atom and two O atoms and have the formula ClO2,
which has a molar mass of 67.45 g.
PRACTICE EXERCISE The density of a gaseous organic compound is 3.38 g/L at 40 C and 1.97 atm. What
is its molar mass? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 172 GASES Since Equation (5.9) is derived from the ideal gas equation, we can also calculate
the molar mass of a gaseous substance using the ideal gas equation, as shown below.
EXAMPLE 5.11 Chemical analysis of a gaseous compound showed that it contained 33.0 percent
silicon and 67.0 percent fluorine by mass. At 35 C, 0.210 L of the compound exerted a pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g,
calculate the molecular formula of the compound.
Answer First we determine the empirical formula of the compound.
33.0 g Si nF 1 mol Si
28.09 g Si 1.17 mol Si 67.0 g F nSi 1 mol F
19.00 g F 3.53 mol F Therefore, the formula is Si1.17F3.53, or SiF3. Next, we calculate the number of moles
contained in 2.38 g of the compound. From the idea gas equation
n PV
RT
(1.70 atm)(0.210 L)
(0.0821 L atm/K mol)(308 K) 0.0141 mol Therefore, the molar mass of the compound is
M Similar problems: 5.47, 5.48. 2.38 g
0.0141 mol 169 g/mol The empirical molar mass of SiF3 (empirical formula) is 85.09 g. Therefore the molecular formula of the compound must be Si2F6, because 2 85.09, or 170.2 g, is
very close to 169 g.
PRACTICE EXERCISE A gaseous compound is 78.14 percent boron and 21.86 percent hydrogen. At 27 C,
74.3 mL of the gas exerted a pressure of 1.12 atm. If the mass of the gas was 0.0934
g, what is its molecular formula? 5.5 GAS STOICHIOMETRY In Chapter 3 we used relationships between amounts (in moles) and masses (in grams)
of reactants and products to solve stoichiometry problems. When the reactants and/or
products are gases, we can also use the relationships between amounts (moles, n) and
volume (V) to solve such problems (Figure 5.12). The following examples show how
the gas laws are used in these calculations. FIGURE 5.12 Stoichiometric
calculations involving gases. Back Forward Amount of
reactant (grams
or volume) Main Menu TOC Moles of
reactant Study Guide TOC Moles of
product Amount of
product (grams
or volume) Textbook Website MHHE Website 5.5 GAS STOICHIOMETRY 173 EXAMPLE 5.12 Calculate the volume of O2 (in liters) at STP required for the complete combustion
of 2.64 L of acetylene (C2H2) at STP:
2C2H2(g) 5O2(g) 88n 4CO2(g) 2H2O(l) Answer Since both C2H2 and O2 are gases measured at the same temperature and
pressure, according to Avogadro’s law their reacting volumes are related to their coefficients in the balanced equation; that is, 2 L of C2H2 react with 5 L of O2. Knowing
this ratio we can calculate the volume (in liters) of O2 that will react with 2.64 L
of C2H2. volume of O2 2.64 L C2H2 5 L O2
2 L C2H2 6.60 L O2 Similar problem: 5.95.
PRACTICE EXERCISE Assuming no change in temperature and pressure, calculate the volume of O2 (in
liters) required for the complete combustion of 14.9 L of butane (C4H10):
2C4H10(g) 13O2(g) 88n 8CO2(g) 10H2O(l) EXAMPLE 5.13 Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows:
2NaN3(s) 88n 2Na(s) 3N2(g) The nitrogen gas produced quickly inflates the bag between the driver and the windshield. Calculate the volume of N2 generated at 21 C and 823 mmHg by the decomposition of 60.0 g of NaN3.
From the balanced equation we see that 2 mol NaN3
number of moles of N2 produced by 60.0 g NaN3 is
Answer moles of N2 An air bag can protect the driver
in an automobile collision. 60.0 g NaN3 1 mol NaN3
65.02 g NaN3 3 mol N2. The 3 mol N2
2 mol NaN3 1.38 moles of N2 The volume of 1.38 moles of N2 can be obtained by using the ideal gas equation:
V nRT
P (1.38 mol)(0.0821 L atm/K mol)(294 K)
(823/760) atm
30.8 L Similar problem: 5.54.
PRACTICE EXERCISE The equation for the metabolic breakdown of glucose (C6H12O6) is the same as the
equation for the combustion of glucose in air:
C6H12O6(s) 6O2(g) 88n 6CO2(g) 6H2O(l) Calculate the volume of CO2 produced at 37 C and 1.00 atm when 5.60 g of glucose
are used up in the reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 174 GASES EXAMPLE 5.14 Aqueous lithium hydroxide solution is used to purify air in spacecrafts and submarines because it absorbs carbon dioxide according to the equation
2LiOH(aq) The air in submerged submarines
and space vehicles needs to be
purified continuously. CO2(g) 88n Li2CO3(aq) H2O(l) The pressure of carbon dioxide in a cabin having a volume of 2.4 105 L is 7.9
10 3 atm at 312 K. A solution of lithium hydroxide (LiOH) of negligible volume
is introduced into the cabin. Eventually the pressure of CO2 is reduced to 1.2
10 4 atm. How many grams of lithium carbonate are formed by this process?
First we calculate the number of moles of CO2 consumed in the reaction.
The drop in pressure, which is 7.9 10 3 atm 1.2 10 4 atm, or 7.8 10 3
atm, corresponds to the consumption of CO2. Using the ideal gas equation, we write
Answer n PV
RT
(7.8 10 3 atm)(2.4 105 L)
(0.0821 L atm/K mol)(312 K) 73 mol From the equation we see that 1 mol CO2 1 mol Li2CO3, so the amount of Li2CO3
formed is also 73 moles. Then, with the molar mass of Li2CO3 (73.89 g) we calculate its mass:
mass of Li2CO3 formed
Similar problem: 5.93. 73 mol Li2CO3
5.4 73.89 g Li2CO3
1 mol Li 2CO3 103 g Li2CO3 PRACTICE EXERCISE A 2.14-L sample of hydrogen chloride gas at 2.61 atm and 28 C is completely dissolved in 668 mL of water to form hydrochloric acid solution. Calculate the molarity of the acid solution. 5.6 As mentioned earlier, gas pressure
results from the impact of gas
molecules against the walls
of the container. DALTON’S LAW OF PARTIAL PRESSURES Thus far we have concentrated on the behavior of pure gaseous substances, but experimental studies very often involve mixtures of gases. For example, for a study of
air pollution, we may be interested in the pressure-volume-temperature relationship of
a sample of air, which contains several gases. In this case, and all cases involving mixtures of gases, the total gas pressure is related to partial pressures, that is, the pressures of individual gas components in the mixture. In 1801 Dalton formulated a law,
now known as Dalton’s law of partial pressures, which states that the total pressure
of a mixture of gases is just the sum of the pressures that each gas would exert if it
were present alone.
Consider a case in which two gases, A and B, are in a container of volume V. The
pressure exerted by gas A, according to the ideal gas equation, is
PA Back Forward Main Menu TOC Study Guide TOC nART
V Textbook Website MHHE Website 5.6 DALTON’S LAW OF PARTIAL PRESSURES 175 where nA is the number of moles of A present. Similarly, the pressure exerted by gas
B is
nBRT
V PB In a mixture of gases A and B, the total pressure PT is the result of the collisions of
both types of molecules, A and B, with the walls of the container. Thus, according to
Dalton’s law,
PT PA PB nART
V nBRT
V RT
(nA
V nB) nRT
V where n, the total number of moles of gases present, is given by n nA nB, and PA
and PB are the partial pressures of gases A and B, respectively. For a mixture of gases,
then, PT depends only on the total number of moles of gas present, not on the nature
of the gas molecules.
In general, the total pressure of a mixture of gases is given by
PT P1 P2 P3 ... where P1, P2, P3, . . . are the partial pressures of components 1, 2, 3, . . . . To see how
each partial pressure is related to the total pressure, consider again the case of a mixture of two gases A and B. Dividing PA by PT, we obtain
PA
PT (nA nART/V
nB)RT/V nA
nA nB XA where XA is called the mole fraction of gas A. The mole fraction is a dimensionless
quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. It is always smaller than 1, except when A is
the only component present. In that case, nB 0 and XA nA/nA 1. We can express
the partial pressure of A as
PA XAPT PB XBPT Similarly, Note that the sum of the mole fractions for a mixture of gases must be unity. If only
two components are present, then
XA XB nA
nA nB
nB nA nB 1 If a system contains more than two gases, then the partial pressure of the ith component is related to the total pressure by Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 176 GASES Pi XiPT (5.10) where Xi is the mole fraction of substance i.
How are partial pressures determined? A manometer can measure only the total
pressure of a gaseous mixture. To obtain the partial pressures, we need to know the
mole fractions of the components, which would involve elaborate chemical analyses.
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.
From mole fractions and total pressure, we can calculate the partial pressures of
individual components, as Example 5.15 shows. A direct application of Dalton’s law
of partial pressures to scuba diving is discussed in the Chemistry in Action essay on
p. 179.
EXAMPLE 5.15 A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and
2.15 moles of xenon (Xe). Calculate the partial pressures of the gases if the total
pressure is 2.00 atm at a certain temperature.
Answer The mole fraction of Ne is
XNe nNe nNe
nAr nXe 4.46 mol
0.74 mol 4.46 mol 2.15 mol 0.607 From Equation (5.10) PNe XNePT
0.607 2.00 atm 1.21 atm Similarly, PAr 0.10 2.00 atm 0.20 atm and PXe 0.293 2.00 atm 0.586 atm Similar problem: 5.57.
PRACTICE EXERCISE A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole of ethane
(C2H6), and 0.116 mole of propane (C3H8). If the total pressure of the gases is 1.37
atm, what are the partial pressures of the gases?
Dalton’s law of partial pressures is useful for calculating volumes of gases collected over water. For example, when potassium chlorate (KClO3) is heated, it decomposes to KCl and O2:
2KClO3(s) 88n 2KCl(s) 3O2(g) The oxygen gas can be collected over water, as shown in Figure 5.13. Initially, the inverted bottle is completely filled with water. As oxygen gas is generated, the gas bubbles rise to the top and displace water from the bottle. This method of collecting a gas
is based on the assumptions that the gas does not react with water and that it is not appreciably soluble in it. These assumptions are valid for oxygen gas, but not for gases Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.6 FIGURE 5.13 An apparatus
for collecting gas over water. The
oxygen generated by heating
potassium chlorate (KClO3) in the
presence of a small amount of
manganese dioxide (MnO2),
which speeds up the reaction, is
bubbled through water and collected in a bottle as shown.
Water originally present in the
bottle is pushed into the trough
by the oxygen gas. TEMPERATURE
(°C) 0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100 Back 4.58
6.54
9.21
12.79
17.54
23.76
31.82
42.18
55.32
71.88
92.51
118.04
149.38
187.54
233.70
289.10
355.10
433.60
525.76
633.90
760.00 Forward Main Menu 177 Bottle being filled with oxygen gas KClO3 and MnO2 TABLE 5.3 Pressure of
Water Vapor at Various
Temperatures
WATER
VAPOR
PRESSURE
(mmHg) DALTON’S LAW OF PARTIAL PRESSURES Bottle filled with water
ready to be placed in
the plastic basin Bottle full of oxygen gas such as NH3, which dissolves readily in water. The oxygen gas collected in this way
is not pure, however, because water vapor is also present in the bottle. The total gas
pressure is equal to the sum of the pressures exerted by the oxygen gas and the water
vapor:
PT PO2 PH2O Consequently, we must allow for the pressure caused by the presence of water vapor
when we calculate the amount of O2 generated. Table 5.3 shows the pressure of water
vapor at various temperatures. These data are plotted in Figure 5.14.
Example 5.16 shows how to use Dalton’s law to calculate the amount of a gas
collected over water.
EXAMPLE 5.16 Oxygen gas generated by the decomposition of potassium chlorate is collected as
shown in Figure 5.13. The volume of oxygen collected at 24 C and atmospheric
pressure of 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at 24 C is 22.4 mmHg.
Answer Our first step is to calculate the partial pressure of O2. We know that
PT PO2 PH2O Therefore TOC Study Guide TOC Textbook Website MHHE Website 178 GASES FIGURE 5.14 The pressure of
water vapor as a function of temperature. Note that at the boiling
point of water (100 C) the pressure is 760 mmHg, which is exactly equal to 1 atm. 800
760 P (mmHg) 600 400 200 0 20 40 60
t (°C) PO2 80 PT 100 PH2O 762 mmHg 22.4 mmHg 740 mmHg
740 mmHg 1 atm
760 mmHg 0.974 atm From the ideal gas equation we write
PV nRT m
RT
M where m and M are the mass of O2 collected and the molar mass of O2, respectively. Rearranging the equation we get
PVM
RT m (0.974 atm)(0.128 L)(32.00 g/mol)
(0.0821 L atm/K mol)(273 24) K
0.164 g Similar problem: 5.62.
PRACTICE EXERCISE Hydrogen gas generated when calcium metal reacts with water is collected as shown
in Figure 5.13. The volume of gas collected at 30 C and pressure of 988 mmHg is
641 mL. What is the mass (in grams) of the hydrogen gas obtained? The pressure
of water vapor at 30 C is 31.82 mmHg. 5.7 THE KINETIC MOLECULAR THEORY OF GASES The gas laws help us to predict the behavior of gases, but they do not explain what
happens at the molecular level to cause the changes we observe in the macroscopic
world. For example, why does a gas expand upon heating? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.7 THE KINETIC MOLECULAR THEORY OF GASES 179 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Scuba Diving and the Gas Laws
Scuba diving is an exhilarating sport, and, thanks in
part to the gas laws, it is also a safe activity for trained
individuals who are in good health. (“Scuba” is an
acronym for self-contained underwater breathing apparatus.) Two applications of the gas laws to this popular pastime are the development of guidelines for returning safely to the surface after a dive and the
determination of the proper mix of gases to prevent a
potentially fatal condition during a dive.
A typical dive might be 40 to 65 ft, but dives to
90 ft are not uncommon. Because seawater has a
slightly higher density than fresh water — about 1.03
g/mL, compared with 1.00 g/mL — the pressure exerted by a column of 33 ft of seawater is equivalent
to 1 atm pressure. Pressure increases with increasing
depth, so at a depth of 66 ft the pressure of the water will be 2 atm, and so on.
What would happen if a diver rose to the surface
from a depth of, say, 20 ft rather quickly without
breathing? The total decrease in pressure for this
change in depth would be (20 ft/33 ft) 1 atm, or
0.6 atm. When the diver reached the surface, the volume of air trapped in the lungs would have increased
by a factor of (1 0.6) atm/1 atm, or 1.6 times. This
sudden expansion of air can fatally rupture the membranes of the lungs. Another serious possibility is that
an air embolism might develop. As air expands in the
lungs, it is forced into tiny blood vessels called capillaries. The presence of air bubbles in these vessels can
block normal blood flow to the brain. As a result, the
diver might lose consciousness before reaching
the surface. The only cure for an air embolism is recompression. For this painful process, the victim is
placed in a chamber filled with compressed air. Here
bubbles in the blood are slowly squeezed down to
harmless size over the course of several hours to a day.
To avoid these unpleasant complications, divers know
they must ascend slowly, pausing at certain points to
give their bodies time to adjust to the falling pressure.
Our second example is a direct application of
Dalton’s law. Oxygen gas is essential for our survival,
so it is hard to believe that an excess of oxygen could Forward Main Menu TOC be harmful. Nevertheless, the toxicity of too much oxygen is well established. For example, newborn infants
placed in oxygen tents often sustain damage to the
retinal tissue, which can cause partial or total blindness.
Our bodies function best when oxygen gas has
a partial pressure of about 0.20 atm, as it does in the
air we breathe. The oxygen partial pressure is given
by
nO2
PO2 XO2PT
P
nO2 nN2 T
where PT is the total pressure. However, since volume
is directly proportional to the number of moles of gas
present (at constant temperature and pressure), we can
now write
VO2
PO2
P
VO2 VN2 T
Thus the composition of air is 20 percent oxygen gas
and 80 percent nitrogen gas by volume. When a diver
is submerged, the pressure of the water on the diver
is greater than atmospheric pressure. The air pressure
inside the body cavities (for example, lungs, sinuses)
must be the same as the pressure of the surrounding
water; otherwise they would collapse. A special valve in
le
b
ila
a
Av
t
No
“ n”
io
rs
Ve
t
ex
T
e- A scuba diver. Study Guide TOC Textbook Website MHHE Website 180 GASES Chemistry in Action Chemistry in Action automatically adjusts the pressure of the air breathed
from a scuba tank to ensure that the air pressure equals
the water pressure at all times. For example, at a depth
where the total pressure is 2.0 atm, the oxygen content in air should be reduced to 10 percent by volume
to maintain the same partial pressure of 0.20 atm;
that is,
VO2
PO2 0.20 atm
2.0 atm
VO2 VN2
VO2
VO2 VN2 0.20 atm
2.0 atm 0.10 or 10% Although nitrogen gas may seem to be the obvious choice to mix with oxygen gas, there is a serious
problem with it. When the partial pressure of nitrogen
gas exceeds 1 atm, enough of the gas dissolves in the
blood to cause a condition known as nitrogen narcosis. The effects on the diver resemble those associated
with alcohol intoxication. Divers suffering 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 often used to dilute oxygen
gas. An inert gas, helium is much less soluble in blood
than nitrogen and produces no narcotic effects. In the nineteenth century, a number of physicists, notably Ludwig Boltzmann† and
James Clerk Maxwell,‡ found that the physical properties of gases can be explained in
terms of the motion of individual molecules. This molecular movement is a form of
energy, which we define as the capacity to do work or to produce change. In mechanics, work is defined as force times distance. Since energy can be measured as work,
we can write
energy work done
force distance The joule (J) is the SI unit of energy
1J 1 kg m2/s2
1Nm Alternatively, energy can be expressed in kilojoules (kJ):
1 kJ 1000 J As we will see in Chapter 6, there are many different kinds of energy. Kinetic energy
(KE) is the type of energy expended by a moving object, or energy of motion.
The findings of Maxwell, Boltzmann, and others resulted in a number of generalizations about gas behavior that have since been known as the kinetic molecular
theory of gases, or simply the kinetic theory of gases. Central to the kinetic theory are
the following assumptions:
1. 2. A gas is composed of molecules that are separated from each other by distances
far greater than their own dimensions. The molecules can be considered to be
“points”; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions, and they frequently
collide with one another. Collisions among molecules are perfectly elastic. In other † Ludwig Eduard Boltzmann (1844–1906). Austrian physicist. Although Boltzmann was one of the greatest theoretical physicists of all time, his work was not recognized by other scientists in his own lifetime. Suffering from poor health and great
depression, he committed suicide in 1906.
‡James Clerk Maxwell (1831–1879). Scottish physicist. Maxwell was one of the great theoretical physicists of the nineteenth century; his work covered many areas in physics, including kinetic theory of gases, thermodynamics, and electricity
and magnetism. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.7 3.
4. THE KINETIC MOLECULAR THEORY OF GASES 181 words, energy can be transferred from one molecule to another as a result of a collision. Nevertheless, the total energy of all the molecules in a system remains the
same.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of
the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy. The average kinetic energy of a molecule is given by
1
2 KE mu2 where m is the mass of the molecule and u is its speed. The horizontal bar denotes
an average value. The quantity u2 is called mean square speed; it is the average of
the square of the speeds of all the molecules:
u2 u2
1 u2
2 2
uN N where N is the number of molecules.
Assumption 4 allows us to write
KE
1
2
1
2 T mu2 T mu 2 CT (5.11) where C is the proportionality constant and T is the absolute temperature.
According to the kinetic molecular theory, gas pressure is the result of collisions
between molecules and the walls of their container. It depends on the frequency of collision per unit area and on how “hard” the molecules strike the wall. The theory also
provides a molecular interpretation of temperature. According to Equation (5.11), the
absolute temperature of a gas is a measure of the average kinetic energy of the molecules. In other words, the absolute temperature is an index of the random motion of
the molecules — the higher the temperature, the more energetic the molecules. Because
it is related to the temperature of the gas sample, random molecular motion is sometimes referred to as thermal motion.
APPLICATION TO THE GAS LAWS Although the kinetic theory of gases is based on a rather simple model, the mathematical details involved are very complex. However, on a qualitative basis, it is possible to use the theory to account for the general properties of substances in the gaseous
state. The following examples illustrate the range of its utility.
Compressibility of gases. Since molecules in the gas phase are separated by large
distances (assumption 1), gases can be compressed easily to occupy less volume.
• Boyle’s law. The pressure exerted by a gas results from the impact of its molecules
on the walls of the container. The collision rate, or the number of molecular collisions with the walls per second, is proportional to the number density (that is, number of molecules per unit volume) of the gas. Decreasing the volume of a given
amount of gas increases its number density and hence its collision rate. For this reason, the pressure of a gas is inversely proportional to the volume it occupies; as
volume decreases, pressure increases and vice versa.
• Charles’ law. Since the average kinetic energy of gas molecules is proportional to
the sample’s absolute temperature (assumption 4), raising the temperature increases
• Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 182 GASES Another way of stating
Avogadro’s law is that at the same
pressure and temperature, equal
volumes of gases, whether they
are the same or different gases,
contain equal numbers
of molecules. the average kinetic energy. Consequently, molecules will collide with the walls of
the container more frequently and with greater impact if the gas is heated, and thus
the pressure increases. The volume of gas will expand until the gas pressure is balanced by the constant external pressure (see Figure 5.7).
• Avogadro’s law. We have shown that the pressure of a gas is directly proportional
to both the density and the temperature of the gas. Since the mass of the gas is directly proportional to the number of moles (n) of the gas, we can represent density
by n /V. Therefore
P n
T
V For two gases, 1 and 2, we write
P1 n1T1
V1 C n1T1
V1 P2 n2T2
V2 C n2T2
V2 where C is the proportionality constant. Thus, for two gases under the same conditions of pressure, volume, and temperature (that is, when P1 P2, T1 T2, and
V1 V2), it follows that n1 n2, which is a mathematical expression of Avogadro’s
law.
• Dalton’s law of partial pressures. If molecules do not attract or repel one another
(assumption 3), then the pressure exerted by one type of molecule is unaffected by
the presence of another gas. Consequently, the total pressure is given by the sum
of individual gas pressures.
DISTRIBUTION OF MOLECULAR SPEEDS Back Forward Main Menu T1 Number of molecules FIGURE 5.15 Maxwell’s speed
distribution for a gas at (a) temperature T1 and (b) a higher temperature T2. Note that the curve
flattens out at the higher temperature. The shaded areas represent
the number of molecules traveling
at a speed equal to or greater
than a certain speed u1. The
higher the temperature, the
greater the number of molecules
moving at high speed. Number of molecules The kinetic theory of gases allows us to investigate molecular motion in more detail.
Suppose we have a large number of gas molecules, say, 1 mole, in a container. As long
as we hold the temperature constant, the average kinetic energy and the mean-square
speed will remain unchanged as time passes. As you might expect, the motion of the
molecules is totally random and unpredictable. At a given instant, how many molecules are moving at a particular speed? To answer this question Maxwell analyzed the
behavior of gas molecules at different temperatures.
Figure 5.15 shows typical Maxwell speed distribution curves for an ideal gas at
two different temperatures. At a given temperature, the distribution curve tells us the
number of molecules moving at a certain speed. The peak of each curve represents the T2 u1
Molecular speed u u1
Molecular speed u (a) (b) TOC Study Guide TOC Textbook Website MHHE Website 5.7 FIGURE 5.16 (a) Apparatus
for studying molecular speed distribution at a certain temperature.
The vacuum pump causes the
molecules to travel from left to
right as shown. (b) The spread of
the deposit on the detector gives
the range of molecular speeds,
and the density of the deposit is
proportional to the fraction of
molecules moving at different
speeds. THE KINETIC MOLECULAR THEORY OF GASES 183 To vacuum pump
Motor
Slow
molecules
Oven Fast
molecules Detector
Chopper with
rotating slit Average
molecules (a) Detector
(b) most probable speed, that is, the speed of the largest number of molecules. Note that
at the higher temperature [Figure 5.15(b)] the most probable speed is greater than that
at the lower temperature [Figure 5.15(a)]. Comparing parts (a) and (b), we see that as
temperature increases, not only does the peak shift toward the right, but the curve also
flattens out, indicating that larger numbers of molecules are moving at greater speeds.
The distribution of molecular speeds can be demonstrated with the apparatus
shown in Figure 5.16. A beam of atoms (or molecules) exits from an oven at a known
temperature and passes through a pinhole (to collimate the beam). Two circular plates
mounted on the same shaft are rotated by a motor. The first plate is called the “chopper” and the second is the detector. The purpose of the chopper is to allow small bursts
of atoms (or molecules) to pass through it whenever the slit is aligned with the beam.
Within each burst, the faster-moving molecules will reach the detector earlier than the
slower-moving ones. Eventually, a layer of deposit will accumulate on the detector.
Since the two plates are rotating at the same speed, molecules in the next burst will
hit the detector plate at approximately the same place as molecules from the previous
burst having the same speed. In time, the molecular deposition will become visible.
The density of the deposition indicates the distribution of molecular speeds at that particular temperature.
ROOT-MEAN-SQUARE SPEED How fast does a molecule move, on the average, at any temperature T? One way to
estimate molecular speed is to calculate the root-mean-square (rms) speed (urms),
which is an average molecular speed. One of the results of the kinetic theory of gases
is that the total kinetic energy of a mole of any gas equals 3 RT. Earlier we saw that
2 the average kinetic energy of one molecule is 1 mu2 and so we can write
2
NA ( 1 mu2)
2 where NA is Avogadro’s number. Since NAm
to give
u2 Back Forward Main Menu TOC Study Guide TOC 3
2 RT M, the above equation can be rearranged
3RT
M Textbook Website MHHE Website 184 GASES Taking the square root of both sides gives
u2 urms 3RT
M (5.12) Equation (5.12) shows that the root-mean-square speed of a gas increases with the
square root of its temperature (in kelvins). Because M appears in the denominator, it
follows that the heavier the gas, the more slowly its molecules move. If we substitute
8.314 J/K mol for R (see Appendix 2) and convert the molar mass to kg/mol, then urms
will be calculated in meters per second (m/s). This procedure is illustrated in Example
5.17.
EXAMPLE 5.17 Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in
m/s at 25 C.
We need Equation (5.12) for this calculation. For He the molar mass is
4.003 g/mol, or 4.003 10 3 kg/mol: Answer urms 3RT
M
3(8.314 J/K mol)(298 K)
106 J/kg 1.86 Using the conversion factor,
1J 1 kg m2/s2 we get
1.86 106 kg m2/kg s2 1.86 urms 106 m2/s2 1.36 103 m/s The procedure is the same for N2, which has a molar mass of 28.02 g/mol, or 2.802
10 2 kg/mol:
urms 3(8.314 J/K mol)(298 K)
2.65 105 m2/s2 515 m/s Because of its smaller mass, a helium atom, on the average, moves about
2.6 times faster than a nitrogen molecule at the same temperature (1360 515
2.64). Comment Similar problems: 5.71, 5.72. PRACTICE EXERCISE Calculate the root-mean-square speed of molecular chlorine in m/s at 20 C.
The calculation in Example 5.17 has an interesting relationship to the composition of Earth’s atmosphere. Earth, unlike, say, Jupiter, does not have appreciable
amounts of hydrogen or helium in its atmosphere. Why is this the case? A smaller Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.8 FIGURE 5.17 The path traveled by a single gas molecule.
Each change in direction represents a collision with another molecule. DEVIATION FROM IDEAL BEHAVIOR 185 planet than Jupiter, Earth has a weaker gravitational attraction for these lighter molecules. A fairly straightforward calculation shows that to escape Earth’s gravitational
field, a molecule must possess an escape velocity equal to or greater than 1.1 104
m/s. Because the average speed of helium is considerably greater than that of molecular nitrogen or molecular oxygen, more helium atoms escape from Earth’s atmosphere
into outer space. Consequently, only a trace amount of helium is present in our atmosphere. On the other hand, Jupiter, with a mass about 320 times greater than that of
Earth, retains both heavy and light gases in its atmosphere
The Chemistry in Action Essay on p. 186 describes a fascinating phenomenon involving gases at extremely low temperatures.
GAS DIFFUSION A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Diffusion always proceeds from a region of higher concentration to one of lower
concentration. Despite the fact that molecular speeds are very great, the diffusion
process takes a relatively long time to complete. For example, when a bottle of concentrated ammonia solution is opened at one end of a lab bench, it takes some time
before a person at the other end of the bench can smell it. The reason is that a molecule experiences numerous collisions while moving from one end of the bench to the
other, as shown in Figure 5.17. Thus, diffusion of gases always happens gradually, and
not instantly as molecular speeds seem to suggest. Furthermore, because the root-meansquare speed of a light gas is greater than that of a heavier gas (see Example 5.17), a
lighter gas will diffuse through a certain space more quickly than will a heavier gas.
Figure 5.18 illustrates gaseous diffusion. 5.8 DEVIATION FROM IDEAL BEHAVIOR The gas laws and the kinetic molecular theory assume that molecules in the gaseous
state do not exert any force, either attractive or repulsive, on one another. The other
assumption is that the volume of the molecules is negligibly small compared with that FIGURE 5.18 A demonstration
of gas diffusion. NH3 gas (from
a bottle containing aqueous ammonia) combines with HCl gas
(from a bottle containing hydrochloric acid) to form solid
NH4Cl. Because NH3 is lighter
and therefore diffuses faster, solid
NH4Cl first appears nearer the
HCl bottle (on the right). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 186 GASES Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Super Cold Atoms
What happens to a gas when cooled to nearly absolute zero? More than seventy years ago, Albert
Einstein, extending work by the Indian physicist
Satyendra Nath Bose, predicted that at extremely low
temperatures gaseous atoms of certain elements would
“merge” or “condense” to form a single entity and a
new form of matter. Unlike ordinary gases, liquids, and
solids, this super-cooled substance, which was named
the Bose-Einstein condensate (BEC), would contain no
individual atoms because the original atoms would
overlap one another, leaving no space in between.
Einstein’s hypothesis inspired an international effort to produce the BEC. But, as sometimes happens
in science, the necessary technology was not available until fairly recently, and so early investigations
were fruitless. Lasers, which use a process based on
another of Einstein’s ideas, were not designed specifically for BEC research, but they became a critical tool
for this work.
Finally, in 1995, physicists found the evidence
they had sought for so long. A team at the University
of Colorado was the first to report success. They created a BEC by cooling a sample of gaseous rubidium
(Rb) atoms to about 1.7 10 7 K using a technique
called “laser cooling,” a process in which a laser light
is directed at a beam of atoms, hitting them head on
and dramatically slowing them down. The Rb atoms
were further cooled in an “optical molasses” produced
by the intersection of six lasers. The slowest, coolest
atoms were trapped in a magnetic field while the
faster-moving, “hotter” atoms escaped, thereby removing more energy from the gas. Under these conditions, the kinetic energy of the trapped atoms was
virtually zero, which accounts for the extremely low
temperature of the gas. At this point the Rb atoms
formed the condensate, just as Einstein had predicted.
Although this BEC was invisible to the naked eye (it
measured only 5 10 3 cm across), the scientists
were able to capture its image on a computer screen
by focusing another laser beam on it. The laser caused
the BEC to break up after about 15 seconds, but that
was long enough to record its existence. Forward Main Menu TOC Maxwell velocity distribution of Rb atoms at about 1.7
10 7 K. The velocity increases from the center (zero) outward along the two axes. The red color represents the lowest number of Rb atoms and the white color the highest.
The average speed in the white region is about 0.5 mm/s. The figure shows the Maxwell velocity distribution* of the Rb atoms at this temperature. The colors
indicate the number of atoms having velocity specified by the two horizontal axes. The blue and white
portions represent atoms that have merged to form the
BEC.
*Velocity distribution differs from speed distribution in that velocity has
both magnitude and direction. Thus velocity can have both positive and
negative values but speed can have only zero or positive values. Study Guide TOC Textbook Website MHHE Website 5.8 Chemistry in Action Within weeks of the Colorado team’s discovery,
a group of scientists at Rice University, using similar
techniques, succeeded in producing a BEC with lithium
atoms. At present, we can only guess at the significance of these discoveries for the future. It is expected
that studies of the BEC will shed light on atomic properties that are still not fully understood (see Chapter DEVIATION FROM IDEAL BEHAVIOR 187 7) and on the mechanism of superconductivity (see the
Chemistry in Action essay on this topic in Chapter 11).
An additional benefit might be the development of better lasers. Other applications will depend on further
study of the BEC itself. Nevertheless, the discovery of
a new form of matter has to be one of the foremost
scientific achievements of the twentieth century. of the container. A gas that satisfies these two conditions is said to exhibit ideal behavior.
Although we can assume that real gases behave like an ideal gas, we cannot expect them to do so under all conditions. For example, without intermolecular forces,
gases could not condense to form liquids. The important question is: Under what conditions will gases most likely exhibit nonideal behavior?
Figure 5.19 shows PV/RT plotted against P for three real gases and an ideal gas
at a given temperature. This graph provides a test of ideal gas behavior. According to
the ideal gas equation (for 1 mole of gas), PV/RT equals 1, regardless of the actual gas
pressure. (When n 1, PV nRT becomes PV RT, or PV/RT 1.) For real gases,
this is true only at moderately low pressures ( 5 atm); significant deviations occur as
pressure increases. Attractive forces operate among molecules at relatively short distances. At atmospheric pressure, the molecules in a gas are far apart and the attractive
forces are negligible. At high pressures, the density of the gas increases; the molecules
are much closer to one another. Intermolecular forces can then be significant enough
to affect the motion of the molecules, and the gas will not behave ideally.
Another way to observe the nonideal behavior of gases is to lower the temperature. Cooling a gas decreases the molecules’ average kinetic energy, which in a sense
deprives molecules of the drive they need to break from their mutual attraction.
To study real gases accurately, then, we need to modify the ideal gas equation,
taking into account intermolecular forces and finite molecular volumes. Such an analy- FIGURE 5.19 Plot of PV/RT
versus P of 1 mole of a gas at
0 C. For 1 mole of an ideal gas,
PV/RT is equal to 1, no matter
what the pressure of the gas is.
For real gases, we observe various deviations from ideality at
high pressures. At very low pressures, all gases exhibit ideal behavior; that is, their PV/RT values
all converge to 1 as P approaches zero. CH4 NH3
PV
1.0
RT 0 Back Forward Main Menu H2 2.0 TOC Ideal gas 200 400 600 800 1000 1200
P (atm) Study Guide TOC Textbook Website MHHE Website 188 GASES sis was first made by the Dutch physicist J. D. van der Waals† in 1873. Besides being
mathematically simple, van der Waals’ treatment provides us with an interpretation of
real gas behavior at the molecular level.
Consider the approach of a particular molecule toward the wall of a container
(Figure 5.20). The intermolecular attractions exerted by its neighbors tend to soften the
impact made by this molecule against the wall. The overall effect is a lower gas pressure than we would expect for an ideal gas. Van der Waals suggested that the pressure
exerted by an ideal gas, Pideal, is related to the experimentally measured pressure, Preal,
by the equation
Pideal an2
V2 Preal observed correction
pressure
term FIGURE 5.20 Effect 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 (green
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 always
lower than the pressure the gas
would exert if it behaved ideally. where a is a constant and n and V are the number of moles and volume of the gas, respectively. The correction term for pressure (an2/V 2) can be understood as follows. The
intermolecular interaction that gives rise to nonideal behavior depends on how frequently any two molecules approach each other closely. The number of such “encounters” increases with the square of the number of molecules per unit volume, (n/V)2,
because the presence of each of the two molecules in a particular region is proportional
to n/V. The quantity Pideal is the pressure we would measure if there were no intermolecular attractions, and so a is just a proportionality constant.
Another correction concerns the volume occupied by the gas molecules. In the
ideal gas equation, V represents the volume of the container. However, each molecule
does occupy a finite, although small, intrinsic volume, so the effective volume of the
gas becomes (V nb), where n is the number of moles of the gas and b is a constant.
The term nb represents the volume occupied by n moles of the gas.
Having taken into account the corrections for pressure and volume, we can rewrite
the ideal gas equation as follows:
P an2
(V
V2 nb) nRT (5.13) corrected corrected
pressure volume Equation (5.13), relating P, V, T, and n for a nonideal gas, is known as the van der
Waals equation. The van der Waals constants a and b are selected to give the best possible agreement between Equation (5.13) and observed behavior of a particular gas.
Table 5.4 lists the values of a and b for a number of gases. The value of a indicates how strongly molecules of a given type of gas attract one another. We see that
helium atoms have the weakest attraction for one another, because helium has the smallest a value. There is also a rough correlation between molecular size and b. Generally, †Johannes Diderick van der Waals (1837–1923). Dutch physicist. Van der Waals received the Nobel Prize in Physics in
1910 for his work on the properties of gases and liquids. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5.8 TABLE 5.4 Van der Waals
Constants of Some
Common Gases
GAS He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
CCl4
NH3
H2O a
atm L2
mol2 00.034
00.211
01.34
02.32
04.19
00.244
01.39
01.36
06.49
03.59
02.25
20.4
04.17
05.46 b
L
mol 0.0237
0.0171
0.0322
0.0398
0.0266
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.138
0.0371
0.0305 DEVIATION FROM IDEAL BEHAVIOR 189 the larger the molecule (or atom), the greater b is, but the relationship between b and
molecular (or atomic) size is not a simple one.
The following example compares the pressure of a gas calculated using the ideal
gas equation and the van der Waals equation. EXAMPLE 5.18 Given that 3.50 moles of NH3 occupy 5.20 L at 47 C, calculate the pressure of the
gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation.
Answer (a) We have the following data:
V 5.20 L T (47 n 3.50 mol 273) K 320 K R 0.0821 L atm/K mol We substitute these values in the ideal gas equation:
nRT
V
(3.50 mol)(0.0821 L atm/K mol)(320 K)
5.20 L P 17.7 atm (b) From Table 5.4, we have
a 4.17 atm L2/mol2 b 0.0371 L/mol It is convenient to calculate the correction terms for Equation (5.13) first. They are
(4.17 atm L2/mol2)(3.50 mol)2
(5.20 L)2 an2
V2
nb (3.50 mol)(0.0371 L/mol) 1.89 atm 0.130 L Finally, substituting in the van der Waals equation, we write
(P 1.89 atm)(5.20 L 0.130 L)
P (3.50 mol)(0.0821 L atm/K mol)(320 K)
16.2 atm The actual pressure measured under these conditions is 16.0 atm. Thus,
the pressure calculated by the van der Waals equation (16.2 atm) is closer to the actual value than that calculated by the ideal gas equation (17.7 atm). Comment Similar problem: 5.79. PRACTICE EXERCISE Using the data shown in Table 5.4, calculate the pressure exerted by 4.37 moles of
molecular chlorine confined in a volume of 2.45 L at 38 C. Compare the pressure
with that calculated using the ideal gas equation. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 190 GASES SUMMARY OF
KEY EQUATIONS • PV (5.1) Boyle’s law. Constant T and n. (5.2) Boyle’s law. For calculating changes in pressure or
volume. (5.3) Charles’ law. Constant P and n. (5.4) Charles’ law. For calculating temperature or volume
changes. k3n (5.5) Avogadro’s law. Constant P and T. • PV nRT
PV
P2V2
• 11
T1
T2 (5.6) Ideal gas equation. (5.7) For calculating changes in pressure, temperature, or
volume when n is constant. (5.8) For calculating density or molar mass. (5.10) Dalton’s law of partial pressures. For calculating
partial pressures. (5.11) Relating the average kinetic energy of a gas to its
absolute temperature. (5.12) For calculating the root-mean-square speed of gas
molecules. • P1V1
V
T
V1
•
T1
• •V •d
• Pi
• KE
• urms
•P SUMMARY OF FACTS
AND CONCEPTS Back Forward k1
P2V2
k2
V2
T2 PM
RT
XiPT
1
2 mu2 CT 3RT
M
an2
(V
V2 nb) nRT (5.13) van der Waals equation. For calculating the pressure
of a nonideal gas. 1. At 25 C and 1 atm, a number of elements and molecular compounds exist as gases. Ionic
compounds are solids rather than gases under atmospheric conditions.
2. Gases exert pressure because their molecules move freely and collide with any surface
with which they make contact. Units of gas pressure include millimeters of mercury
(mmHg), torr, pascals, and atmospheres. One atmosphere equals 760 mmHg, or 760 torr.
3. The pressure-volume relationships of ideal gases are governed by Boyle’s law: Volume is
inversely proportional to pressure (at constant T and n).
4. The temperature-volume relationships of ideal gases are described by Charles’ and GayLussac’s law: Volume is directly proportional to temperature (at constant P and n).
5. Absolute zero ( 273.15 C) is the lowest theoretically attainable temperature. The Kelvin
temperature scale takes 0 K as absolute zero. In all gas law calculations, temperature must
be expressed in kelvins.
6. The amount-volume relationships of ideal gases are described by Avogadro’s law: Equal
volumes of gases contain equal numbers of molecules (at the same T and P).
7. The ideal gas equation, PV nRT, combines the laws of Boyle, Charles, and Avogadro.
This equation describes the behavior of an ideal gas.
8. Dalton’s law of partial pressures states that each gas in a mixture of gases exerts the same
pressure that it would if it were alone and occupied the same volume.
9. The kinetic molecular theory, a mathematical way of describing the behavior of gas molecules, is based on the following assumptions: Gas molecules are separated by distances
far greater than their own dimensions, they possess mass but have negligible volume, they
are in constant motion, and they frequently collide with one another. The molecules neither attract nor repel one another.
10. A Maxwell speed distribution curve shows how many gas molecules are moving at various speeds at a given temperature. As temperature increases, more molecules move at
greater speeds.
11. Gas diffusion demonstrates random molecular motion. Main Menu TOC Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 191 12. The van der Waals equation is a modification of the ideal gas equation that takes into account the nonideal behavior of real gases. It corrects for the fact that real gas molecules
do exert forces on each other and that they do have volume. The van der Waals constants
are determined experimentally for each gas. KEY WORDS
Absolute temperature scale,
p. 164
Absolute zero, p. 164
Atmospheric pressure, p. 158
Avogadro’s law, p. 165
Barometer, p. 158
Boyle’s law, p. 161
Charles’ and Gay-Lussac’s
law, p. 164 Charles’ law, p. 164
Dalton’s law of partial
pressures, p. 174
Diffusion, p. 185
Gas constant, p. 166
Ideal gas, p. 166
Ideal gas equation, p. 166
Joule (J), p. 180
Kelvin temperature scale,
p. 164 Kinetic energy (KE), p. 180
Kinetic molecular theory of
gases, p. 180
Manometer, p. 160
Mole fraction, p. 175
Newton (N), p. 158
Partial pressures, p. 174
Pascal (Pa), p. 158
Pressure, p. 158 Root-mean-square speed,
p. 183
Standard atmospheric pressure
(1 atm), p. 158
Standard temperature and
pressure (STP), p. 167
Van der Waals equation, p.
188 QUESTIONS AND PROBLEMS
SUBSTANCES THAT EXIST AS GASES
Review Questions 5.1 Name five elements and five compounds that exist as
gases at room temperature.
5.2 List the physical characteristics of gases.
PRESSURE OF A GAS
Review Questions 5.3 Define pressure and give the common units for pressure.
5.4 Describe how a barometer and a manometer are used
to measure pressure.
5.5 Why is mercury a more suitable substance to use in
a barometer than water?
5.6 Explain why the height of mercury in a barometer is
independent of the cross-sectional area of the tube.
Would the barometer still work if the tubing were
tilted at an angle, say 15 (see Figure 5.3)?
5.7 Would it be easier to drink water with a straw on top
of Mt. Everest or at the foot? Explain.
5.8 Is the atmospheric pressure in a mine that is 500 m
below sea level greater or less than 1 atm?
5.9 What is the difference between a gas and a vapor? At
25 C, which of the following substances in the gas
phase should be properly called a gas and which
should be called a vapor: molecular nitrogen (N2),
mercury?
5.10 If the maximum distance that water may be brought
up a well by a suction pump is 34 ft (10.3 m), how
is it possible to obtain water and oil from hundreds
of feet below the surface of Earth? Back Forward Main Menu TOC 5.11 Why is it that if the barometer reading falls in one
part of the world, it must rise somewhere else?
5.12 Why do astronauts have to wear protective suits when
they are on the surface of the moon?
Problems 5.13 Convert 562 mmHg to kPa and 2.0 kPa to mmHg.
5.14 The atmospheric pressure at the summit of Mt.
McKinley is 606 mmHg on a certain day. What is the
pressure in atm and in kPa?
THE GAS LAWS
Review Questions 5.15 State the following gas laws in words and also in the
form of an equation: Boyle’s law, Charles’ law,
Avogadro’s law. In each case, indicate the conditions
under which the law is applicable, and give the units
for each quantity in the equation.
5.16 Explain why a helium weather balloon expands as it
rises in the air. Assume that the temperature remains
constant.
Problems 5.17 A gas occupying a volume of 725 mL at a pressure
of 0.970 atm is allowed to expand at constant temperature until its pressure reaches 0.541 atm. What is
its final volume?
5.18 At 46 C a sample of ammonia gas exerts a pressure
of 5.3 atm. What is the pressure when the volume of
the gas is reduced to one-tenth (0.10) of the original
value at the same temperature? Study Guide TOC Textbook Website MHHE Website 192 GASES 5.19 The volume of a gas is 5.80 L, measured at 1.00 atm.
What is the pressure of the gas in mmHg if the volume is changed to 9.65 L? (The temperature remains
constant.)
5.20 A sample of air occupies 3.8 L when the pressure is
1.2 atm. (a) What volume does it occupy at 6.6 atm?
(b) What pressure is required in order to compress it
to 0.075 L? (The temperature is kept constant.)
5.21 A 36.4-L volume of methane gas is heated from 25 C
to 88 C at constant pressure. What is the final volume of the gas?
5.22 Under constant-pressure conditions a sample of hydrogen gas initially at 88 C and 9.6 L is cooled until
its final volume is 3.4 L. What is its final temperature?
5.23 Ammonia burns in oxygen gas to form nitric oxide
(NO) and water vapor. How many volumes of NO
are obtained from one volume of ammonia at the
same temperature and pressure?
5.24 Molecular chlorine and molecular fluorine combine
to form a gaseous product. Under the same conditions of temperature and pressure it is found that one
volume of Cl2 reacts with three volumes of F2 to yield
two volumes of the product. What is the formula of
the product?
THE IDEAL GAS EQUATION
Review Questions 5.25 List the characteristics of an ideal gas.
5.26 Write the ideal gas equation and also state it in words.
Give the units for each term in the equation.
5.27 What are standard temperature and pressure (STP)?
What is the significance of STP in relation to the volume of 1 mole of an ideal gas?
5.28 Why is the density of a gas much lower than that of
a liquid or solid under atmospheric conditions? What
units are normally used to express the density of
gases?
Problems 5.29 A sample of nitrogen gas kept in a container of volume 2.3 L and at a temperature of 32 C exerts a pressure of 4.7 atm. Calculate the number of moles of gas
present.
5.30 Given that 6.9 moles of carbon monoxide gas are present in a container of volume 30.4 L, what is the pressure of the gas (in atm) if the temperature is 62 C?
5.31 What volume will 5.6 moles of sulfur hexafluoride
(SF6) gas occupy if the temperature and pressure of
the gas are 128 C and 9.4 atm? Back Forward Main Menu TOC 5.32 A certain amount of gas at 25 C and at a pressure of
0.800 atm is contained in a glass vessel. Suppose that
the vessel can withstand a pressure of 2.00 atm. How
high can you raise the temperature of the gas without bursting the vessel?
5.33 A gas-filled balloon having a volume of 2.50 L at 1.2
atm and 25 C is allowed to rise to the stratosphere
(about 30 km above the surface of Earth), where the
temperature and pressure are 23 C and 3.00
10 3 atm, respectively. Calculate the final volume of
the balloon.
5.34 The temperature of 2.5 L of a gas initially at STP is
raised to 250 C at constant volume. Calculate the final pressure of the gas in atm.
5.35 The pressure of 6.0 L of an ideal gas in a flexible
container is decreased to one-third of its original pressure, and its absolute temperature is decreased by
one-half. What is the final volume of the gas?
5.36 A gas evolved during the fermentation of glucose
(wine making) has a volume of 0.78 L at 20.1 C and
1.00 atm. What was the volume of this gas at the fermentation temperature of 36.5 C and 1.00 atm pressure?
5.37 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?
5.38 The volume of a gas at STP is 488 mL. Calculate its
volume at 22.5 atm and 150 C.
5.39 A gas at 772 mmHg and 35.0 C occupies a volume
of 6.85 L. Calculate its volume at STP.
5.40 Dry ice is solid carbon dioxide. A 0.050-g sample of
dry ice is placed in an evacuated 4.6-L vessel at 30 C.
Calculate the pressure inside the vessel after all the
dry ice has been converted to CO2 gas.
5.41 At STP, 0.280 L of a gas weighs 0.400 g. Calculate
the molar mass of the gas.
5.42 At 741 torr and 44 C, 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass of the gas?
5.43 Ozone molecules in the stratosphere absorb much of
the harmful radiation from the sun. Typically, the temperature and pressure of ozone in the stratosphere are
250 K and 1.0 10 3 atm, respectively. How many
ozone molecules are present in 1.0 L of air under these
conditions?
5.44 Assuming that air contains 78 percent N2, 21 percent
O2, and 1 percent Ar, all by volume, how many molecules of each type of gas are present in 1.0 L of air
at STP?
5.45 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm
and 27.0 C. (a) Calculate the density of the gas in Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS grams per liter. (b) What is the molar mass of the gas?
Assume ideal behavior.
5.46 Calculate the density of hydrogen bromide (HBr) gas
in grams per liter at 733 mmHg and 46 C.
5.47 A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 120 C and
750 mmHg, 1.00 L of the gaseous compound weighs
2.30 g. What is the molecular formula of the compound?
5.48 A compound has the empirical formula SF4. At 20 C,
0.100 g of the gaseous compound occupies a volume
of 22.1 mL and exerts a pressure of 1.02 atm. What
is the molecular formula of the gas?
GAS STOICHIOMETRY
Problems 5.49 A compound of P and F was analyzed as follows:
Heating 0.2324 g of the compound in a 378-cm3 container turned all of it to gas, which had a pressure of
97.3 mmHg at 77 C. Then the gas was mixed with
calcium chloride solution, which turned all of the F
to 0.2631 g of CaF2. Determine the molecular formula of the compound.
5.50 A quantity of 0.225 g of a metal M (molar mass
27.0 g/mol) liberated 0.303 L of molecular hydrogen
(measured at 17 C and 741 mmHg) from an excess
of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.
5.51 What is the mass of the solid NH4Cl formed when
73.0 g of NH3 are mixed with an equal mass of HCl?
What is the volume of the gas remaining, measured
at 14.0 C and 752 mmHg? What gas is it?
5.52 Dissolving 3.00 g of an impure sample of calcium
carbonate in hydrochloric acid produced 0.656 L of
carbon dioxide (measured at 20.0 C and 792 mmHg).
Calculate the percent by mass of calcium carbonate
in the sample.
5.53 Calculate the mass in grams of hydrogen chloride
produced when 5.6 L of molecular hydrogen measured at STP react with an excess of molecular chlorine gas.
5.54 Ethanol (C2H5OH) burns in air:
C2H5OH(l) O2(g) CO2(g) H2O(l) Balance the equation and determine the volume of air
in liters at 35.0 C and 790 mmHg required to burn
227 g of ethanol. Assume that air is 21.0 percent O2
by volume. Back Forward Main Menu TOC 193 DALTON’S LAW OF PARTIAL PRESSURES
Review Questions 5.55 State Dalton’s law of partial pressures and explain
what mole fraction is. Does mole fraction have units?
5.56 A sample of air contains only nitrogen and oxygen
gases whose partial pressures are 0.80 atm and 0.20
atm, respectively. Calculate the total pressure and the
mole fractions of the gases.
Problems 5.57 A mixture of gases contains 0.31 mole CH4, 0.25
mole C2H6, and 0.29 mole C3H8. The total pressure
is 1.50 atm. Calculate the partial pressures of the
gases.
5.58 A 2.5-L flask at 15 C contains a mixture of N2, He,
and Ne at partial pressures of 0.32 atm for N2, 0.15
atm for He, and 0.42 atm for Ne. (a) Calculate the
total pressure of the mixture. (b) Calculate the volume in liters at STP occupied by He and Ne if the N2
is removed selectively.
5.59 Dry air near sea level has the following composition
by volume: N2, 78.08 percent; O2, 20.94 percent; Ar,
0.93 percent; CO2, 0.05 percent. The atmospheric
pressure is 1.00 atm. Calculate (a) the partial pressure of each gas in atm and (b) the concentration of
each gas in moles per liter at 0 C. (Hint: Since volume is proportional to the number of moles present,
mole fractions of gases can be expressed as ratios of
volumes at the same temperature and pressure.)
5.60 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? (Vapor pressure of water at 28 C
28.3 mmHg.)
5.61 A piece of sodium metal reacts completely with water as follows:
2Na(s) 2H2O(l) 2NaOH(aq) H2(g) 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. (Vapor pressure of water at 25 C
0.0313 atm.)
5.62 A sample of zinc metal reacts completely with an excess of hydrochloric acid:
Zn(s) 2HCl(aq) ZnCl2(aq) H2(g) The hydrogen gas produced is collected over water
at 25.0 C using an arrangement similar to that shown
in Figure 5.13. The volume of the gas is 7.80 L, and Study Guide TOC Textbook Website MHHE Website 194 GASES the pressure is 0.980 atm. Calculate the amount of
zinc metal in grams consumed in the reaction. (Vapor
pressure of water at 25 C 23.8 mmHg.)
5.63 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.
5.64 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.
KINETIC MOLECULAR THEORY OF GASES
Review Questions 5.65 What are the basic assumptions of the kinetic molecular theory of gases?
5.66 How does the kinetic molecular theory explain
Boyle’s law, Charles’ law, Avogadro’s law, and
Dalton’s law of partial pressures?
5.67 What does the Maxwell speed distribution curve tell
us? Does Maxwell’s theory work for a sample of 200
molecules? Explain.
5.68 Write the expression for the root-mean-square speed
for a gas at temperature T. Define each term in the
equation and show the units that are used in the calculation.
5.69 Which of the following statements is correct? (a) Heat
is produced by the collision of gas molecules against
one another. (b) When a gas is heated, the molecules
collide with one another more often.
5.70 Uranium hexafluoride (UF6) is a much heavier gas
than helium, yet at a given temperature, the average
kinetic energies of the samples of the two gases are
the same. Explain.
Problems 5.71 Compare the root-mean-square speeds of O2 and UF6
at 65 C.
5.72 The temperature in the stratosphere is
23 C.
Calculate the root-mean-square speeds of N2, O2, and
O3 molecules in this region.
5.73 The average distance traveled by a molecule between
successive collisions is called mean free path. For a
given amount of a gas, how does the mean free path
of a gas depend on (a) density, (b) temperature at constant volume, (c) pressure at constant temperature,
(d) volume at constant temperature, and (e) size of
the atoms?
5.74 At a certain temperature the speeds of six gaseous Back Forward Main Menu TOC molecules in a container are 2.0 m/s, 2.2 m/s, 2.6 m/s,
2.7 m/s, 3.3 m/s, and 3.5 m/s. Calculate the rootmean-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?
DEVIATION FROM IDEAL BEHAVIOR
Review Questions 5.75 Cite two pieces of evidence to show that gases do not
behave ideally under all conditions.
5.76 Under what set of conditions would a gas be expected
to behave most ideally? (a) High temperature and low
pressure, (b) high temperature and high pressure, (c)
low temperature and high pressure, (d) low temperature and low pressure.
5.77 Write the van der Waals equation for a real gas.
Explain the corrective terms for pressure and volume.
5.78 (a) A real gas is introduced into a flask of volume V.
Is the corrected volume of the gas greater or less than
V? (b) Ammonia has a larger a value than neon does
(see Table 5.4). What can you conclude about the relative strength of the attractive forces between molecules of ammonia and between atoms of neon?
Problems 5.79 Using the data shown in Table 5.4, calculate the pressure exerted by 2.50 moles of CO2 confined in a volume of 5.00 L at 450 K. Compare the pressure with
that predicted by the ideal gas equation.
5.80 At 27 C, 10.0 moles of a gas in a 1.50-L container
exert a pressure of 130 atm. Is this an ideal gas?
ADDITIONAL PROBLEMS 5.81 Discuss the following phenomena in terms of the gas
laws: (a) the pressure increase in an automobile tire
on a hot day, (b) the “popping” of a paper bag, (c) the
expansion of a weather balloon as it rises in the air,
(d) the loud noise heard when a light bulb shatters.
5.82 Under the same conditions of temperature and pressure, which of the following gases would behave most
ideally: Ne, N2, or CH4? Explain.
5.83 Nitroglycerin, an explosive compound, decomposes
according to the equation
4C3H5(NO3)3(s) 12CO2(g) 10H2O(g)
O2(g) 6N2(g) Calculate the total volume of gases when collected at
1.2 atm and 25 C from 2.6 102 g of nitroglycerin. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 5.84 5.85 5.86 5.87 5.88 What are the partial pressures of the gases under these
conditions?
The empirical formula of a compound is CH. At
200 C, 0.145 g of this compound occupies 97.2 mL
at a pressure of 0.74 atm. What is the molecular formula of the compound?
When ammonium nitrite (NH4NO2) is heated, it decomposes to give nitrogen gas. This property is used
to inflate some tennis balls. (a) Write a balanced equation for the reaction. (b) Calculate the quantity (in
grams) of NH4NO2 needed to inflate a tennis ball to
a volume of 86.2 mL at 1.20 atm and 22 C.
The percent by mass of bicarbonate (HCO 3 ) in a certain Alka-Seltzer product is 32.5 percent. Calculate
the volume of CO2 generated (in mL) at 37 C and
1.00 atm when a person ingests a 3.29-g tablet. (Hint:
The reaction is between HCO 3 and HCl acid in the
stomach.)
The boiling point of liquid nitrogen is 196 C. On
the basis of this information alone, do you think nitrogen is an ideal gas?
In the metallurgical process of refining nickel, the
metal is first combined with carbon monoxide to form
tetracarbonylnickel, which is a gas at 43 C:
Ni(s) 4CO(g) Ni(CO)4(g) This reaction separates nickel from other solid impurities. (a) Starting with 86.4 g of Ni, calculate the pressure of Ni(CO)4 in a container of volume 4.00 L.
(Assume the above reaction goes to completion.) (b)
At temperatures above 43 C, the pressure of the gas
is observed to increase much more rapidly than predicted by the ideal gas equation. Explain.
5.89 The partial pressure of carbon dioxide varies with seasons. Would you expect the partial pressure in the
Northern Hemisphere to be higher in the summer or
winter ? Explain.
5.90 A healthy adult exhales about 5.0 102 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.
5.91 Sodium bicarbonate (NaHCO3) is called baking soda
because when heated, it releases carbon dioxide gas,
which is responsible for the rising of cookies, doughnuts, and bread. (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 (NH4HCO3)
has also been used for the same purpose. Suggest one
advantage and one disadvantage of using NH4HCO3
instead of NaHCO3 for baking. Back Forward Main Menu TOC 195 5.92 A barometer having a cross-sectional area of 1.00 cm2
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 cm2
of Earth’s surface. Given that the density of mercury
is 13.6 g/mL 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 4 r2
where r is the radius of the sphere.)
5.93 Some commercial drain cleaners contain a mixture of
sodium hydroxide and aluminum powder. When the
mixture is poured down a clogged drain, the following reaction occurs:
2NaOH(aq) 2Al(s) 6H2O(l) 2NaAl(OH)4(aq)
3H2(g) The heat generated in this reaction helps melt away
obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate
the volume of H2 formed at STP if 3.12 g of Al are
treated with an excess of NaOH.
5.94 The volume of a sample of pure HCl gas was 189 mL
at 25 C and 108 mmHg. It was completely dissolved
in about 60 mL of water and titrated with a NaOH solution; 15.7 mL of the NaOH solution were required
to neutralize the HCl. Calculate the molarity of the
NaOH solution.
5.95 Propane (C3H8) burns in oxygen to produce carbon
dioxide gas and water vapor. (a) Write a balanced
equation for this reaction. (b) Calculate the number
of liters of carbon dioxide measured at STP that could
be produced from 7.45 g of propane.
5.96 Consider the following apparatus. Calculate the partial pressures of helium and neon after the stopcock
is open. The temperature remains constant at 16 C. He Ne 1.2 L
0.63 atm 3.4 L
2.8 atm 5.97 Nitric oxide (NO) reacts with molecular oxygen as
follows:
2NO(g) O2(g) 2NO2(g) Initially NO and O2 are separated as shown below.
When the valve is opened, the reaction quickly goes
to completion. Determine what gases remain at the Study Guide TOC Textbook Website MHHE Website 196 GASES end and calculate their partial pressures. Assume that
the temperature remains constant at 25 C. NO O2 2.00 L at
1.00 atm 4.00 L at
0.500 atm 5.98 Consider the apparatus shown below. When a small
amount of water is introduced into the flask by
squeezing the bulb of the medicinal dropper, water is
squirted upward out of the long glass tubing. Explain
this observation. (Hint: Hydrogen chloride gas is soluble in water.) tion 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/s) at which the target is
moving. (Hint: The circumference of a circle is given
by 2 r, 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 urms of Bi at 850 C.
Comment on the difference.
Rotating cylinder HCl gas Target
Bi atoms
H2O Slit Rubber
bulb
H2O 5.99 Describe how you would measure, by either chemical or physical means, the partial pressures of a mixture of gases of the following composition: (a) CO2
and H2, (b) He and N2.
5.100 A certain hydrate has the formula MgSO4 xH2O. A
quantity of 54.2 g of the compound is heated in an
oven to drive off the water. If the steam generated exerts a pressure of 3.55 atm in a 2.00-L container at
120 C, calculate x.
5.101 A mixture of Na2CO3 and MgCO3 of mass 7.63 g is
reacted with an excess of hydrochloric acid. The CO2
gas generated occupies a volume of 1.67 L at 1.24
atm and 26 C. From these data, calculate the percent
composition by mass of Na2CO3 in the mixture.
5.102 The following apparatus can be used to measure
atomic and molecular speed. Suppose that 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 different speeds will strike the
target at different positions. In time, a layer of the
metal will deposit on the target area, and the varia- Back Forward Main Menu TOC 5.103 If 10.00 g of water are introduced into an evacuated
flask of volume 2.500 L at 65 C, calculate the mass
of water vaporized. (Hint: Assume that the volume
of the remaining liquid water is negligible; the vapor
pressure of water at 65 C is 187.5 mmHg.)
5.104 Commercially, compressed oxygen is sold in metal
cylinders. If a 120-L cylinder is filled with oxygen to
a pressure of 132 atm at 22 C, what is the mass of
O2 present? How many liters of O2 gas at 1.00 atm
and 22 C could the cylinder produce? (Assume ideal
behavior.)
5.105 The shells of hard-boiled eggs sometimes crack due
to the rapid thermal expansion of the shells at high
temperatures. Suggest another reason why the shells
may crack.
5.106 Ethylene gas (C2H4) is emitted by fruits and is known
to be responsible for their ripening. Based on this information, explain why a bunch of bananas ripens
faster in a closed paper bag than in a bowl.
5.107 About 8.0 106 tons of urea [(NH2)2CO] are used
annually as a fertilizer. The urea is prepared at 200 C
and under high-pressure conditions from carbon dioxide and ammonia (the products are urea and steam).
Calculate the volume of ammonia (in liters) measured
at 150 atm needed to prepare 1.0 ton of urea.
5.108 Some ballpoint pens have a small hole in the main
body of the pen. What is the purpose of this hole? Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 5.109 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 will
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? (At constant temperature and pressure, the volume of a gas is directly proportional to
the number of moles of gases.) (Hint: See Chemistry
in Action essay on p. 179.)
5.110 Nitrous oxide (N2O) can be obtained by the thermal
decomposition of ammonium nitrate (NH4NO3). (a)
Write a balanced equation for the reaction. (b) In a
certain experiment, a student obtains 0.340 L of the
gas at 718 mmHg and 24 C. If the gas weighs 0.580
g, calculate the value of the gas constant.
5.111 Two vessels are labeled A and B. Vessel A contains
NH3 gas at 70 C, and vessel B contains Ne gas at the
same temperature. If the average kinetic energy of
NH3 is 7.1 10 21 J/molecule, calculate the meansquare speed of Ne atoms in m2/s2.
5.112 Which of the following molecules has the largest a
value: CH4, F2, C6H6, Ne?
5.113 The following procedure is a simple though somewhat crude way to measure the molar mass of a gas.
A liquid of mass 0.0184 g is introduced into a syringe
like the one shown below by injection through the
rubber tip using a hypodermic needle. The syringe is
then transferred to a temperature bath heated to 45 C,
and the liquid vaporizes. The final volume of the vapor (measured by the outward movement of the
plunger) is 5.58 mL and the atmospheric pressure is
760 mmHg. Given that the compound’s empirical formula is CH2, determine the molar mass of the compound. 5.115 5.116 5.117 5.118 5.119 Rubber tip 1 2 3 4 5 5.120 5.114 In 1995 a man suffocated as he walked by an aban- Back Forward Main Menu TOC 197 doned mine in England. At that moment there was a
sharp drop in atmospheric pressure due to a change
in the weather. Suggest what might have caused the
man’s death.
Acidic oxides such as carbon dioxide react with basic oxides like calcium oxide (CaO) and barium oxide (BaO) to form salts (metal carbonates). (a) Write
equations representing these two reactions. (b) A student placed a mixture of BaO and CaO of combined
mass 4.88 g in a 1.46-L flask containing carbon dioxide gas at 35 C and 746 mmHg. After the reactions
were complete, she found that the CO2 pressure had
dropped to 252 mmHg. Calculate the percent composition by mass of the mixture.
(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 the 5.0 atm is the
gauge pressure, which is the difference between the
pressure in the tire and atmospheric pressure. Before
filling, the pressure in the tire was 1.0 atm.) (b) What
is the total pressure in the tire when the gauge pressure reads 5.0 atm? (c) The tire is pumped by filling
the cylinder of a hand pump with air at 1.0 atm and
then, by compressing the gas in the cylinder, adding
all the air in the pump to the air in the tire. If the volume of the pump is 33 percent of the tire’s volume,
what is the gauge pressure in the tire after 3 full
strokes of the pump? Assume constant temperature.
The running engine of an automobile produces carbon monoxide (CO), a toxic gas, at the rate of about
188 g CO per hour. A car is left idling in a poorly
ventilated garage that is 6.0 m long, 4.0 m wide, and
2.2 m high at 20 C. (a) Calculate the rate of CO production in moles per minute. (b) How long would it
take to build up a lethal concentration of CO of 1000
ppmv (parts per million by volume)?
Interstellar space contains mostly hydrogen atoms at
a concentration of about 1 atom/cm3. (a) Calculate
the pressure of the H atoms. (b) Calculate the volume
(in liters) that contains 1.0 g of H atoms. The temperature is 3 K.
Atop Mt. Everest, the atmospheric pressure is 210
mmHg and the air density is 0.426 kg/m3. (a)
Calculate the air temperature, given that the molar
mass of air is 29.0 g/mol. (b) Assuming no change in
air composition, calculate the percent decrease in
oxygen gas from sea level to the top of Mt. Everest.
Relative humidity is defined as the ratio (expressed
as a percentage) of the partial pressure of water vapor in the air to the equilibrium vapor pressure (see
Table 5.3) at a given temperature. On a certain sum- Study Guide TOC Textbook Website MHHE Website 198 GASES 5.121 5.122 5.123 5.124 5.125 5.126 5.127 mer day in North Carolina the partial pressure of water vapor in the air is 3.9 103 Pa at 30 C. Calculate
the relative humidity.
Under the same conditions of temperature and pressure, why does one liter of moist air weigh less than
one liter of dry air? In weather forecasts, an oncoming low-pressure front usually means imminent rainfall. Explain.
Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that oxygen diffuses into the
blood. The average radius of the alveoli is 0.0050 cm
and the air inside contains 14% 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. (Hint: The volume of a
sphere of radius r is 4 r3.)
3
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 sulfur powder over the metal. Suggest a physical and a chemical reason for this action. The vapor pressure of mercury at 20 C is 1.7 10 6 atm.
Nitrogen forms several gaseous oxides. One of them
has a density of 1.33 g/L measured at 764 mmHg and
150 C. Write the formula of the compound.
Nitrogen dioxide (NO2) cannot be obtained in a pure
form in the gas phase because it exists as a mixture
of NO2 and N2O4. At 25 C and 0.98 atm, the density
of this gas mixture is 2.7 g/L. What is the partial pressure of each gas?
The Chemistry in Action essay on p 186 describes
the cooling of rubidium vapor to 1.7 10 7 K.
Calculate the root-mean-square speed and average kinetic energy of a Rb atom at this temperature.
Lithium hydride reacts with water as follows:
LiH(s) H2O(l) LiOH(aq) H2(g) During World War II, U.S. pilots carried LiH tablets.
In the event of a crash landing at sea, the LiH would
react with the seawater and fill their life belts and
lifeboats with hydrogen gas. How many grams of LiH
are needed to fill a 4.1-L life belt at 0.97 atm and
12 C?
5.128 The atmosphere on Mars is composed mainly of carbon dioxide. The surface temperature is 220 K and
the atmospheric pressure is about 6.0 mmHg. Taking Back Forward Main Menu TOC 5.129 5.130 5.131 5.132 these values as Martian “STP,” calculate the molar
volume in liters of an ideal gas on Mars.
Venus’s atmosphere is composed of 96.5 percent CO2,
3.5 percent N2, and 0.015 percent SO2 by volume.
Its standard atmospheric pressure is 9.0 106 Pa.
Calculate the partial pressures of the gases in pascals.
A student tries to determine the volume of a bulb like
the one shown on p. 171. These are her results: Mass
of the bulb filled with dry air at 23 C and 744 mmHg
91.6843 g; mass of evacuated bulb 91.4715 g.
Assume the composition of air is 78 percent N2, 21
percent O2, and 1 percent argon. What is the volume
(in milliliters) of the bulb? (Hint: First calculate the
average molar mass of air, as shown in Problem
3.130.)
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, and 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 1.47 103 m/s.
It has been said that every breath we take, on average, contains molecules that were 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 in Problem
5.92 and 29.0 g/mol 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 lifespan was exactly 35
years, what is the number of molecules he exhaled in
that period? (Given that an average person breathes
12 times per minute.) (d) Calculate the fraction of
molecules in the atmosphere that were breathed out Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS by Mozart. How many of Mozart’s molecules do we
breathe in with every inhale of air? Round off your
answer to one significant figure. (e) List three important assumptions in these calculations. Back Forward Main Menu TOC 199 Answers to Practice Exercises: 5.1 0.986 atm. 5.2 39.3 kPa. 5.3 4.46 103 mmHg. 5.4 192 K, or 81 C. 5.5 9.29 L.
5.6 30.6 L. 5.7 2.6 atm. 5.8 0.68 atm. 5.9 13.1 g/L. 5.10 44.1
g/mol. 5.11 B2H6. 5.12 96.9 L. 5.13 4.75 L. 5.14 0.338 M.
5.15 CH4: 1.29 atm; C2H6: 0.0657 atm; C3H8: 0.0181 atm.
5.16 0.0653 g. 5.17 321 m/s. 5.18 30.0 atm; 45.5 atm using the
ideal gas equation. Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY Out of Oxygen† I n September 1991 four men and four women entered the world’s largest glass
bubble, known as Biosphere II, to test the idea that humans could design and
build a totally self-contained ecosystem, a model for some future colony on another planet. Biosphere II (Earth is considered Biopshere I) was a 3-acre miniworld, complete with a tropical rain forest, savanna, marsh, desert, and working
farm that was intended to be fully self-sufficient. This unique experiment was to continue for 2 to 3 years, but almost immediately there were signs that the project could
be in jeopardy.
Soon after the bubble had been sealed, sensors inside the facility showed that the
concentration of oxygen in Biosphere II’s atmosphere had fallen from its initial level
of 21 percent (by volume), while the amount of carbon dioxide had risen from a level
of 0.035 percent (by volume), or 350 ppm (parts per million). Alarmingly, the oxygen
level continued to fall at a rate of about 0.5 percent a month and the level of carbon
dioxide kept rising, forcing the crew to turn on electrically powered chemical scrubbers, similar to those on submarines, to remove some of the excess CO2. Gradually the
CO2 level stabilized around 4000 ppm, which is high but not dangerous. The loss of
oxygen did not stop, though. By January 1993 — 16 months into the experiment — the
oxygen concentration had dropped to 14 percent, which is equivalent to the O2 concentration in air at an elevation of 4360 meters (14,300 ft). The crew began having
trouble performing normal tasks. For their safety it was necessary to pump pure oxygen into Biosphere II.
With all the plants present in Biosphere II, the production of oxygen should have
been greater as a consequence of photosynthesis. Why had the oxygen concentration
declined to such a low level? A small part of the loss was blamed on unusually cloudy
weather, which had slowed down plant growth. The possibility that iron in the soil was
reacting with oxygen to form iron(III) oxide or rust was ruled out along with several
other explanations for lack of evidence. The most plausible hypothesis was that microbes (microorganisms) were using oxygen to metabolize the excess organic matter
that had been added to the soils to promote plant growth. This turned out to be the
case.
Identifying the cause of oxygen depletion raised another question. Metabolism
produces carbon dioxide. Based on the amount of oxygen consumed by the microbes,
the CO2 level should be at 40,000 ppm, ten times what was measured. What happened
to the excess gas? After ruling out leakage to the outside world and reactions between
CO2 with compounds in the soils and in water, scientists found that concrete inside
Biosphere II was consuming large amounts of CO2!
Concrete is a mixture of sand and gravel held together by a binding agent which
is a mixture of calcium silicate hydrates and calcium hydroxide. The calcium hydrox- †
Adapted with permission from “Biosphere II: Out of Oxygen,” by Joe Alper, CHEM MATTERS, February, 1995, p. 8.
Copyright 1995 American Chemical Society. 200 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ide is the key ingredient in the CO2 mystery. Carbon dioxide diffuses into the porous
structure of concrete, then reacts with calcium hydroxide to form calcium carbonate
and water:
Ca(OH)2(s) CO2(g) CaCO3(s) H2O(l) Under normal conditions, this reaction goes on slowly. But CO2 concentrations in
Biosphere II were much higher than normal, so the reaction proceeded much faster. In
fact, in just over 2 years, CaCO3 had accumulated to a depth of more than 2 cm in
Biosphere II’s concrete. Some 10,000 m2 of exposed concrete was hiding 500,000 to
1,500,000 moles of CO2.
The water produced in the reaction between Ca(OH)2 and CO2 created another
problem: CO2 also reacts with water to form carbonic acid (H2CO3), and hydrogen
ions produced by the acid promote the corrosion of the reinforcing iron bars in the concrete, thereby weakening its structure. This situation was dealt with effectively by painting all concrete surfaces with an impermeable coating.
In the meantime the decline in oxygen (and hence also the rise in carbon dioxide) slowed, perhaps because there was now less organic matter in the soils and also
because new lights in the agricultural areas may have boosted photosynthesis. The project was terminated prematurely, and as of 1996, the facility was transformed into an
science education and research center.
The Biosphere II experiment is an interesting project from which we can learn a
lot about Earth and its inhabitants. If nothing else, it has shown us how complex Earth’s
ecosystems are and how difficult it is to mimic nature, even on a small scale.
CHEMICAL CLUES What solution would you use in a chemical scrubber to remove carbon dioxide?
Photosynthesis converts carbon dioxide and water to carbohydrates and oxygen
gas, while metabolism is the process by which carbohydrates react with oxygen to
form carbon dioxide and water. Using glucose (C6H12O6) to represent carbohydrates, write equations for these two processes.
3. Why was diffusion of O2 from Biosphere II to the outside world not considered a
possible cause for the depletion in oxygen?
4. Carbonic acid is a diprotic acid. Write equations for the stepwise ionization of the
acid in water.
5. What are the factors to consider in choosing a planet on which to build a structure
like Biosphere II?
1.
2. Vegetation in Biosphere II. 201 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...

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