Unit 5 Review, pages 620-627
Knowledge
1. (c)
2. (d)
3. (a)
4. (a)
5. (b)
6. (d)
7. (d)
8. (b)
9. (b)
10. (c)
11. (a)
12. (c)
13. (b)
14. (d)
15. False. When a liquid freezes, its molecules have only vibrational motion.
16. True
17. True
18. False. The density of a mixture of argon and nitrogen gases is greater than the density of a
sample of pure nitrogen gas at the same pressure.
19. False. Just prior to the 2008 Olympic Games in Beijing, China, the Chinese government
spent a great deal of money planting trees, blocking traffic, and relocating factories to improve
the air quality of the city.
20. True
21. False. The dangerous gas that might seep into a home’s basement, through cracked exterior
walls, is radon.
22. False. Children have a greater risk of respiratory irritation and health effects from air
pollution because they breathe more rapidly and inhale more pollutants per kilogram of body
mass.
23. True
24. False. When two variables are directly proportional to each other, dividing them yields a
constant answer.
25. False. According to Avogadro’s law, equal volumes of gas at the same temperature and
pressure must consist of equal amounts of entities.
26. True
27. False. Hydrogen can be cooled sufficiently to liquefy it by suddenly decreasing its pressure
greatly.
28. False. When a gas is collected over water, the partial pressure of the gas is less than the
pressure of the atmosphere.
29. False. Gas hydrates form under low temperatures and high pressures.
30. (a) (iii)
(b) (v)
(c) (ii)
(d) (i)
(e) (iv)
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Unit 5: Gases and Atmospheric Chemistry
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31. (a) A gas takes the shape of the entire container.
(b) A solid cannot be compressed into a smaller volume even with great pressure.
(c) The entities of a liquid and a gas can undergo vibrational, rotational, and translational motion.
(d) A solid retains its shape and volume independent of any container into which it is placed.
(e) The entities of a solid have a high degree of order.
(f) A gas can easily be compressed into a smaller volume.
(g) A liquid and a gas can be readily poured.
(h) The entities of a solid can only undergo vibrational motion.
(i) A liquid takes the shape of the container but has a definite volume.
32. More than 99 % of the atmosphere’s water vapour is found in the troposphere.
33. Sulfur dioxide, SO2(g), is the gaseous air pollutant that contributes to the formation of acid
rain and is produced by the burning of fossil fuels containing sulfur impurities.
34. (a) Photochemical smog is formed from ozone, other gases, and particulate matter.
(b) Sunlight is also required for the formation of photochemical smog.
35. Gases and particulate matter are chemical pollutants. Biological pollutants include
microscopic organisms and contaminants produced by living organisms.
36. (a) In high-altitude training, athletes exercise at a high elevation where less oxygen is inhaled
per breath. The athletes want their bodies to compensate for the lower oxygen levels by
producing more red blood cells. Then, when the athletes return to lower elevations to compete,
their blood will be able to carry more oxygen, giving the athletes better endurance.
(b) High-altitude training is not always successful. Some research shows benefits, other research
does not. The lack of oxygen experienced during training at high altitude may actually
compromise strength and endurance to such an extent that, overall, athletic performance at low
elevations shows no improvement despite the greater availability of oxygen.
37. Temperature is the result of the motion of entities at the molecular level. The coldest possible
temperature, which is chosen to be zero kelvins, must therefore be reached when entities stop
moving. It is impossible for entities to slow down any further if they are stopped, so their
temperature cannot go any lower once they become motionless. Thus there can be no
temperature below 0 K, meaning there can be no negative absolute temperature.
38. The volume of a tire is more or less constant. Assuming no leaks, the amount of air inside
also remains constant. If a tire is cooled, the gaseous entities inside will slow down. As a result,
they will hit the interior walls of the tire less frequently and with less kinetic energy, causing the
pressure to fall slightly inside the tire.
39. Answers may vary. Sample answer: The law of combining volumes says that the volumes in
which gases react will always be in whole-number ratios and the volumes of gaseous products
will be in whole-number ratios to the volumes of the reactants. For example, 0.1 L of hydrogen
gas, H2(g), will react with 0.1 L of chlorine gas, Cl2(g), to produce 0.2 L of hydrogen chloride
gas, HCl(g): H 2 (g) + Cl2 (g) ! 2 HCl(g) . The reactant volumes are in a 1 : 1 ratio and the ratio
of the volume of either of the reactants with the product is 1 : 2.
40. Ideal gases cannot be liquefied because there are, in theory, no attractive forces between the
entities in an ideal gas. Since butane can be liquefied, it cannot be an ideal gas.
41. When atoms move at a particular speed, they have a certain frequency. The frequency of
lasers can be set to match the frequency of the moving atoms precisely. When tuned this way, the
lasers slow the atoms down and cause their temperature to fall.
42. Kleppner and Greytak chose hydrogen for their experiments because hydrogen atoms have
small mass and weak attractive forces.
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Unit 5: Gases and Atmospheric Chemistry
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43. For mixed gases to obey the law of partial pressures they must not react and their pressures
must be expressed in the same units.
44. (a) Water and methane do not usually mix because water is polar and methane is non-polar.
The hydrogen bonds within water usually hold the molecules closely together with a much
stronger attraction than the weak London dispersion forces that act between methane and water
molecules.
(b) Gas hydrates only form under low temperatures and high pressures, such as those on the sea
floor. The breakdown of organic matter under these conditions results in the formation of cages
of water molecules with methane molecules trapped inside.
45. First, the proposal to extract gas hydrates from the sea floor is controversial because
extracting gas hydrates from the sea floor might cause significant environmental damage.
Second, if people believe that gas hydrates, which are a large potential supply of energy, can be
extracted, they may not feel as much urgency to reduce our dependence on fossil fuels. Thus, the
production of greenhouse gases would continue to increase.
Understanding
46. Compared to the entities of liquids and solids, the entities of a gas have much greater
freedom of motion. They are widely spread out and can travel larger distances before colliding
with other entities. This allows the entities of one gas to move easily into and among the entities
of another gas. In addition, the intermolecular forces in gases are weak, so attraction between
entities does not interfere with the mixing process. For these reasons, gases mix quickly.
47. Greenhouse gases trap heat in the atmosphere by absorbing radiation coming from Earth’s
surface and re-emitting the radiation back toward the surface. This results in a warmer Earth
because, without the greenhouse gases, much of the radiation from Earth’s surface would escape
into space, resulting in planetary cooling. Without naturally occurring greenhouse gases, our
planet would be too cold to support life as we know it. In modern times, however, human
activities are adding greenhouse gases to the atmosphere, raising greenhouse gas concentrations
above what would naturally be present. The extra greenhouse gases are causing a steady increase
in global temperatures, which will have a significant negative impact on world climates. Thus
greenhouse gases are good because our planet would be a “deep freeze” without them, but they
are also bad because too-high concentrations of them may lead to very adverse climate changes.
48. The percentage composition by volume of oxygen, O2(g), in the atmosphere is 20.95 %.
The volume of oxygen in 200 cm3 of air can now be found.
VO = 200 cm 3 ! 20.95 %
2
= 200 cm 3 ! 0.2095
VO = 40 cm 3
2
The volume of oxygen O2(g), in the 200 cm3 of air is 40 cm3.
49. If a chamber is filled with argon, it can contain no oxygen or moisture that might cause the
documents to deteriorate. Argon is a noble gas and is therefore unreactive, so argon will not
harm documents.
50. Ultraviolet radiation contributes to the formation of ozone in smog by taking part in a twostep reaction process. First, nitrogen dioxide, which is a common component of polluted air, is
decomposed by UV rays:
NO2(g) → NO(g) + O(g)
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The oxygen atoms produced by this reaction are very reactive and react with oxygen molecules
in the air to produce ozone:
O(g) + O2(g) → O3(g)
At midday on a sunny day, sunlight is more intense, so there is more UV radiation to drive the
reactions that form ozone in smog. Thus the concentration of ozone in smog tends to peak near
midday.
51. Propane camp stoves, like all devices that use open flames, have the potential to produce
carbon monoxide. In a closed tent, carbon monoxide from a camp stove could accumulate to
potentially lethal levels. The sleeping occupants of the tent could be rendered unconscious before
they even realized they were in danger, with fatal consequences. The lit stove would also be a
fire hazard if left unattended while the campers slept.
52. An idling car or bus engine operates at lower temperatures than normal, resulting in more
incomplete combustion and thus results in more pollutants.
53. The water balloon does not break when dropped onto the bed of nails because the pressure
exerted by the tip of each nail on the balloon’s rubber skin is not sufficient to puncture it. This
results from the fact that pressure is force per unit area. The force is the weight of the balloon.
When the balloon is dropped on a single nail, the weight falls entirely on the tiny area of the
single nail tip, which results in a large pressure that easily forces the nail tip through the
balloon’s rubber skin. When the balloon is dropped onto the bed of nails, however, the combined
of area of many nail tips touching the balloon is much larger. A larger area results in a lower
pressure at each nail tip.
54. Given: initial Celsius temperature, t1 = 22 °C
final Celsius temperature, t2 = 37 °C
volume, V1 = 450 cm3
The amount of the gas and the gas pressure remain constant.
Required: final volume, V2
Analysis: Apply Charles’ law to the situation.
V1 V2
=
T1 T2
Solution:
Step 1. Convert temperature values to kelvins.
T = t + 273
T1 = t1 + 273
= 22 + 273
T1 = 295 K
T2 = t2 + 273
= 37 + 273
T2 = 310 K
Step 2. Rearrange the Charles’ law equation to isolate the unknown variable.
VT
V2 = 1 2
T1
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Unit 5: Gases and Atmospheric Chemistry
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Step 3. Substitute the given values (including units) into the equations and solve.
450 cm 3 ! 310 K
V2 =
295 K
V2 = 470 cm 3
Statement: The volume of the exhaled air will be 470 cm3.
55. Given: initial Celsius temperature, t1 = 30.0 °C
initial volume, V1 = 45 cm3
final volume, V2 = 35 cm3
The amount of the gas and the gas pressure remain constant.
Required: final Celsius temperature, t2
Analysis: Apply Charles’ law to the situation.
V1 V2
=
T1 T2
Solution:
Step 1. Convert temperature value to kelvins.
T = t + 273
T1 = t1 + 273
= 30.0 + 273
T1 = 303.0
Step 2. Rearrange the Charles’ law equation to isolate the unknown variable.
TV
T2 = 1 2
V1
Step 3. Substitute the given values (including units) into the equations and solve.
T2 =
303.0 K ! 35 cm 3
45 cm 3
T2 = 235.67 K [two extra digits carried]
T2 = t2 + 273
t2 = T2 " 273
= 235.67 " 273
t2 = "37 ºC
Statement: The temperature at which the sample occupies 35 cm3 is –37 ºC.
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Unit 5: Gases and Atmospheric Chemistry
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56. Given: initial Celsius temperature, t1 = 95 °C
final Celsius temperature, t2 = 20 °C
volume, V1 = 3.00 × 103 m3
The amount of the gas and the gas pressure remain constant.
Required: final volume, V2
Analysis: Apply Charles’ law to the situation.
V1 V2
=
T1 T2
Solution:
Step 1. Convert temperature values to kelvins.
T = t + 273
T1 = t1 + 273
= 95 + 273
T1 = 368 K
T2 = t2 + 273
= 20 + 273
T2 = 293 K
Step 2. Rearrange the Charles’ law equation to isolate the unknown variable.
VT
V2 = 1 2
T1
Step 3. Substitute the given values (including units) into the equations and solve.
3.00 ! 103 m 3 ! 293 K
V2 =
368 K
V2 = 2.39 ! 103 m 3
Statement: The volume of the air would occupy 2.39 × 103 m3.
57. Given: initial pressure, P1 = 90.7 kPa
initial volume, V1 = 30.0 cm3
initial temperature, t1 = 18 °C
final pressure, P2 = 96.0 kPa
final temperature, t2 = 28 °C
The amount of gas remains constant.
Required: final volume, V2
Analysis: Use the combined gas law.
P1V1 P2V2
=
T1
T2
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Unit 5: Gases and Atmospheric Chemistry
U5-9
Solution:
Step 1. Convert temperature values to kelvins.
T1 = t1 + 273
= 18 + 273
T1 = 291 K
T2 = t2 + 273
= 28 + 273
T2 = 301 K
Step 2. Rearrange the equation to isolate the unknown variable.
PV T
V2 = 1 1 2
P2T1
Step 3. Substitute the given values (including units) into the equation and solve.
V2 =
90.7 kPa ! 30.0 cm 3 ! 301 K
96.0 kPa ! 291 K
V2 = 29.3 cm 3 L
Statement: The final volume of the gas is 29.3 cm3.
58. The first student is correct. At a given temperature, all molecules will have the same average
kinetic energy. Thus larger (heavier) molecules will have a slower average speed and smaller
(lighter) molecules will have a higher average speed. Given the same volume, heavier gas
molecules can exert the same overall force on the walls of their container by virtue of their
greater mass motion.
59. (a) 2 NH 3 (aq) ! N 2 (g) + 3 H 2 (g)
(b) Given the same pressure and temperature, volumes of gases are in the same ratio as the
amounts. The coefficients in the balanced chemical equation for the decomposition indicate that
for each mole of nitrogen gas produced, 3 moles of hydrogen gas will be produced. So the ratio
of the volume of nitrogen gas to hydrogen gas produced is 1 : 3.
60. Given: mNe = 10.0 g ; STP conditions
Required: volume of neon, V
22.4 L
Analysis: V = n !
1 mol
20.18 g
M Ne =
1 mol
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Unit 5: Gases and Atmospheric Chemistry
U5-10
Solution:
Step 1. Determine the amount of neon gas in the sample by multiplying the mass of neon.
1 mol
n = mNe !
16.04 g
= 10.0 g !
1 mol
20.18 g
n = 0.496 mol
Step 2. Determine the volume of neon gas in the sample.
22.4 L
V =n!
1 mol
22.4 L
= 0.496 mol !
1 mol
V = 11.1 L
Statement: The volume of 10.0 g of neon is 11.1 L at STP.
61. Given: t = 27 °C
P = 90.0 kPa
R = 8.314 kPa i L i mol–1 i K–1
mSO (g) = 8.40 g
3
M SO (g) = 80.07 g/mol
3
Required: volume of the gas, V
Analysis: Use the ideal gas law.
PV = nRT
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 27 + 273
T = 300 K
Step 2. Convert mass into amount by using the appropriate conversion factor involving molar
mass.
1 mol
n = mSO !
3
80.07 g
1 mol
= 8.40 g !
80.07 g
n = 0.10491 mol [two extra digits carried]
Step 3. Rearrange the ideal gas law equation to isolate the required variable.
nRT
V=
P
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Unit 5: Gases and Atmospheric Chemistry
U5-11
Step 4. Solve the equation (including units).
0.10491 mol ! 8.314 kPa i L i mol–1 i K –1 ! 300 K
V=
90.0 kPa
V = 2.91 L
Statement: The volume of the gas would be 2.91 L at the given conditions of temperature and
pressure.
62. Given: pressure, P = 90.2 kPa
temperature, t = 30 °C
R = 8.314 kPa i L i mol–1 i K–1
M NH (g) = 17.04 g/mol
3
Required: density of pure ammonia, d NH
3
To find the density, you first have to determine the mass, m, and the volume, V.
Analysis: Assume there is exactly 1 mol of pure ammonia in a given sample.
First use the ideal gas law to determine the volume of the sample.
PV = nRT
Next, convert 1 mol of pure ammonia gas to mass.
m
Finally, use the expression d =
to calculate the density of the propane gas.
V
Solution:
Step 1. Convert temperature value to kelvins.
T = t + 273
= 30 + 273
T = 303 K
Step 2. Rearrange the ideal gas law equation to isolate the unknown variable, V.
nRT
V=
P
Step 3. Substitute known values into the equation and solve.
–1
V=
1 mol ! 8.314 kPa i L i mol i K –1 ! 303 K
90.2 kPa
V = 27.9 L [two extra digits carried]
Step 4. Convert amount of NH3 into mass of NH3 using the appropriate conversion factor
involving molar mass.
17.04 g
mNH = nNH !
3
3
1 mol
17.04 g
= 1 mol !
1 mol
mNH = 17.04 g
3
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-12
Step 5. Use the relationship d NH =
3
m
to determine the density of NH3.
V
m
3
V
17.04 g
=
27.9 L
d NH = 0.6 g/L
d NH =
3
Statement: The density of pure ammonia at these conditions is 0.6 g/L.
63. James Dewar and Heike Kamerlingh Onnes were racing to liquefy helium in the early 1900s.
Helium had just been discovered and both Dewar and Kamerlingh Onnes had difficulty obtaining
it, which presented a roadblock to their progress. Dewar actually worked next door to the
discoverers of helium and should have had an “inside track” to obtaining some of the gas for his
research. Dewar, however, had been highly critical of his colleagues and they consequently
refused to give him any helium. Had Dewar been more civil to his colleagues, he might have
won the race to liquefy helium by getting plentiful samples early on. As it turned out, Dewar
eventually did obtain helium, but he still lost the race because of human error: one of Dewar’s
assistants accidentally released all of their stored helium into the air. This story demonstrates
how human failings can greatly affect the progress of science.
64. Given: Ptotal = 100 kPa
PN = 40 kPa
2
PCO = PCO
2
Required: PCO and PCO
2
Analysis: Use the law of partial pressures.
Solution:
Substitute the values for PN and Ptotal in the equation for the law of partial pressures.
2
PCO + PCO + PN = Ptotal
2
2
PCO + PCO + 40 kPa = 100 kPa
2
PCO + PCO = 100 kPa ! 40 kPa
2
PCO + PCO = 60 kPa
2
Let PCO = PCO in this equation and solve for PCO .
2
PCO + PCO = 60 kPa
2PCO = 60 kPa
PCO = 30 kPa
Since PCO = PCO and PCO = 30 kPa , PCO = 30 kPa .
2
2
Statement: The partial pressures of PCO and PCO are both 30 kPa.
2
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Unit 5: Gases and Atmospheric Chemistry
U5-13
65. (a) Given: partial pressure of methane, P1 = 80 kPa
partial pressure of ethane, P2 = 40 kPa
partial pressure of propane, P3 = 30 kPa
Required: total pressure, Ptotal
Analysis: Apply Dalton’s law of partial pressures.
Ptotal = P1 + P2 + P3
Solution: Substitute known values into the equation and solve.
Ptotal = P1 + P2 + P3
Ptotal
= 80 kPa + 40 kPa + 30 kPa
= 150 kPa
Statement: The total pressure of the mixture is 150 kPa.
(b) The increase in pressure upon the injection of the additional methane is
pressure increase = 200 kPa ! 150 kPa
pressure increase = 50 kPa
The partial pressures of the other gases remain the same: 40 kPa for ethane and 30 kPa for
propane. Therefore, the partial pressure of the methane must increase to
(200 kPa – 70 kPa) = 130 kPa.
66. Given: atmospheric pressure, Patm = 85 kPa
temperature, t = 26 °C
vapour pressure of water at 26 °C (Table 1), PH O = 3.36 kPa
2
Required: partial pressure of oxygen gas, PO
2
Analysis: Use the equation for Dalton’s law of partial pressures.
Ptotal = Patm = PO + PH O
2
2
Solution:
PO = Patm ! PH O
2
2
= 85 kPa ! 3.36 kPa
PO = 82 kPa
2
Statement: The partial pressure of the oxygen gas is 82 kPa.
67. An airbag must inflate quickly to form a cushion during the first instant of a collision.
However, as the passenger’s body begins to press into the deployed airbag, the gas inside must
be able to escape through the cover. This way, the airbag can deflate gradually and “give” with
the impact of the passenger’s body, making the impact less violent.
68. Given: mass of potassium chlorate, mKClO = 100 g
3
molar mass of potassium chlorate, M KClO = 122.55 g/mol
3
pressure, P = 89 kPa
temperature, t = 23 °C
Required: volume of oxygen gas, VO
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2
Unit 5: Gases and Atmospheric Chemistry
U5-14
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 23 + 273
T = 296 K
Step 2. Write the balanced chemical equation, listing the given and required quantities with the
appropriate units underneath.
2 KClO3 (s) ! 2 KCl(s) + 3 O 2 (g)
100 g
VO
2
Step 3. Convert the mass of the given substance, mKClO , into an amount, nKClO , using a
3
3
conversion factor derived from the molar mass of potassium chlorate, KClO3.
1 mol
nKClO = 100 g !
3
122.55 g
nKClO = 0.816 mol
3
[two extra digits carried]
Step 4. Determine the amount of oxygen gas produced from the amount of potassium chlorate
used using the appropriate mole ratio derived from the balanced chemical equation.
3 molO
2
nO = 0.816 mol !
2
2 molKClO
3
nNO = 1.22 mol
2
Step 5. Determine the volume of the required substance, VO , using the ideal gas law equation,
–1
–1
2
PV = nRT. Remember that R = 8.314 kPa i L i mol i K .
PVO = nO RT
2
2
VO =
2
=
nO RT
2
P
(1.22 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 296 K
$
)
89 kPa
VO = 30 L
2
Statement: When 100 g of potassium chlorate decomposes, 30 L of oxygen gas is produced
(under the given conditions of temperature and pressure).
69. Given: volume of hydrogen, VH = 60 L
2
pressure, P = 100 kPa
temperature, t = 22 °C
Required: volume of ammonia, VNH
3
Analysis: Since all of the gases are measured at the same temperature and pressure, the volume
of ammonia can be calculated using a mole ratio from the chemical equation.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-15
Solution:
VNH = VH !
3
2
2 molNH
3 molH
= 60 L !
3
2
2 molNH
3 molH
3
2
VNH = 40 L
3
Statement: The volume of ammonia produced when 60 L of hydrogen reacts completely with
excess nitrogen is 40 L.
Analysis and Application
70. (a) Answers may vary. Sample answer: I predict that the violet colour from the dissolving
potassium permanganate will spread more rapidly in hot water than in cold water.
(b) In hot water, the water molecules move more rapidly than the molecules in cold water. As a
result, the molecules in the hot water will collide with the surface of the potassium permanganate
crystal more frequently and with more energy. This will cause the crystal to dissociate more
quickly. In addition, the rapidly moving water molecules will continue to collide with and thus
disperse the potassium and permanganate ions quickly throughout the solution. Consequently,
the violet colour will spread more rapidly in the hot water.
71. (a) Answers may vary. Sample answer: As the temperature rises, the pollen particles will
display more exaggerated motion: they will jiggle with greater velocity and displacement.
(b) The kinetic molecular theory states that the temperature of a sample of matter is directly
related to the average kinetic energy of the entities that compose the sample. This means that as
the temperature of the water rises, its molecules will move with more kinetic energy and thus
have a greater speed. The Brownian motion of the pollen grains is caused by water molecules
randomly colliding with the grains. At higher temperatures, the water molecules will collide with
the pollen grains with more force, causing them to be jiggled about more rapidly and over greater
distances.
72. Over the past 400 000 years, carbon dioxide concentrations have fluctuated between 175 ppm
and 300 ppm. Accordingly, this appears to be the typical extent of natural fluctuations. In the last
200 years, however, atmospheric carbon dioxide levels have steadily risen above the previous
upper limit of 300 ppm, reaching about 380 ppm. This increase is strongly correlated to human
activities such as the burning of fossil fuels for industrial activity and transportation. As a result,
it is reasonable to believe that the recent increase in atmospheric carbon dioxide is not the result
of a natural cycle.
73. The noble gases occur in nature in very low concentrations compared to many other
elements. In the air, for example, the percentage by volume of argon is less than 1 %, and the
percentages of helium and neon are about 1700 times smaller and 100 times smaller,
respectively. It is hard to realize that something is present when it occurs in such low
concentrations. In addition, the noble gases are odourless, colourless, tasteless, and unreactive.
Consequently, even if the noble gases were present in greater concentrations, their presence
would not be obvious to our human senses. They would only be found through scientific
measurements. In conclusion, the noble gases were found so late because there were few reasons
to suspect that they existed.
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Unit 5: Gases and Atmospheric Chemistry
U5-16
74. (a) Answers may vary. Sample answer: I would advise the family to make sure that their
home is well ventilated for many months. The varnish on the hardwood floors, the solvents in the
paint, and the new window coverings might continue to off-gas potentially harmful VOCs. These
could include methanal, a known carcinogen, respiratory irritant, and allergen. New products can
off-gas significant quantities of dangerous substances for many weeks or even months. In fact,
off-gassing may not entirely cease for years. I would tell the family that if they can detect odours
similar to a “new car” smell or a gasoline-like smell, their ventilation should be increased.
(b) Even though the basement does not have materials that are off-gassing VOCs, vapours from
the upstairs could spread downward because gases move so readily through one another. In
addition, radon gas can seep into basements and pose a serious health risk. For both reasons, the
basement should be well ventilated as well.
(c) Mould can produce airborne fragments and spores that have the potential to cause respiratory
difficulties, rashes, or other allergic reactions in exposed individuals. If mould has grown in the
home and has simply been hidden by a coat of paint, it is likely to continue to grow and thus
pose a continuing health hazard. If the family thinks that the painting has been done in part to
hide mould, I would advise them to seek the advice of a professional regarding how the mould
could be eliminated.
75. Answers may vary. Sample answer: I predict that the bottle will crumple. Some of the hot
water will initially evaporate, producing water vapour that contributes to the gas pressure inside
the bottle. When the bottle is sealed and placed into the refrigerator, the gas entities in the bottle
will slow down due to the colder temperature. In addition, some of the water vapour molecules
will condense and become part of the liquid water in the bottle. Once there are fewer and slower
gas molecules inside the bottle, the pressure that the trapped gas exerts on the interior walls of
the bottle will decrease. The outside pressure will then exceed the inside pressure, causing the
bottle to collapse.
76. The entities of gas occupy the entire space inside the tire. The entities are in motion and
collide constantly with the inside of the tire. The combined effect of these collisions is a force
that the air exerts on the inside of the tire, which, when divided by the area of the walls, gives the
pressure in the tire. If enough air is pumped into the tire, the pressure becomes great enough to
inflate the tire to the point that it can support the weight of the car.
77. (a) The measurements for neon were taken at the highest constant pressure and the
measurements for krypton were taken at the lowest constant pressure. For a given amount of gas,
a higher pressure at a given temperature will result in a lower volume. The neon data points have
the smallest volumes of the three data sets, so the neon sample must have been measured at the
highest value of constant pressure. Similar reasoning explains why the krypton sample, with the
highest volumes of the three data sets, must have been measured at the lowest value of constant
pressure.
(b) Absolute zero is the same for all substances, so we would expect extrapolation of all three
data sets to yield –273 °C in the absence of experimental error.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-17
(c) For each gas, the contant a would not have the same value. The value of the constant is
unique to the specific conditions of the gas. Since the pressure was different for each set of
measurements, the value of a must be different for each set of measurements. Consider, for
example, the value of a for krypton compared with the corresponding value of a for neon at
50 °C. A temperature of 50 °C is 323 K.
V
aKr =
T
4.0 L
=
323 K
aKr = 0.012 LiK !1
V
T
0.9 L
=
323 K
= 0.003 LiK !1
aNe =
aNe
V
for both data sets, the denominator is the same temperature, 323 K. However, the
T
numerator, volume, is larger for krypton, so its value for a is greater than the value of a for neon.
78. (a) The pressure from the gas in the can will not compress the liquid into a smaller volume
because liquids are only slightly compressible and the pressure in the air space is not
exceptionally great.
(b) Given: P1 = 101.325 kPa
V1 = 90 cm3
V2 = 15 cm3
Required: P2
Analysis: Use Boyle’s law.
P1V1 = P2V2
Solution:
Step 1. Rearrange the equation to isolate the unknown variable.
PV
P2 = 1 1
V2
Step 2. Solve the equation (including units).
PV
P2 = 1 1
V2
In
=
101.325 kPa ! 90 cm 3
15 cm 3
P2 = 600 kPa
Statement: The gas pressure in the air space, 600 kPa, is not between 405 and 425 kPa, so it is
not at the correct value.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-18
79. Given: P1 = 220 kPa
t1 = 22 °C
t2 = 90 °C
Required: P2
Analysis: Use Gay-Lussac’s law.
P1 P2
=
T1 T2
Solution:
Step 1. Convert temperature values to kelvins.
T1 = t1 + 273
= 22 + 273
T1 = 295 K
T2 = t2 + 273
= 90 + 273
T2 = 363 K
Step 2. Rearrange the equation to isolate the unknown variable.
PT
P2 = 1 2
T1
Step 3. Solve the equation (including units).
220 kPa ! 363 K
P2 =
295 K
P2 = 270 kPa
Statement: This model of tire is a good choice. At 90 °C, the pressure in the tires rises to
270 kPa, and this value is less than the pressure at which the tires begin to make the car handle
poorly.
80. Answers may vary. Sample answer: I could find the mass of a completely emptied fuel can.
Then I could subtract this mass from a completely filled can to find the mass of the butane
present in a full can. Alternatively, I could check the label on the can to see if the mass of the
fuel in a full can is given. I could then measure the mass of a partially emptied can and subtract
from this value the mass of an empty can. This would give me the mass of the remaining butane.
These data would allow me to set up a proportion that I could solve to obtain the remaining burn
time:
5.5 h
remaining burn time
=
mass of butane in full can mass of butane in partially empty can
If the remaining burn time is greater than 15 min, there is enough butane in each can for the
investigation.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-19
81. Given: V = 50.0 L; SATP conditions
Required: mass of nitrogen gas, mN
2
Analysis: n = V !
1 mol
24.8 L
28.02 g
2
1 mol
Solution:
Step 1. Determine the amount of nitrogen gas that should be delivered.
1 mol
n=V !
24.8 L
1 mol
= 50.0 L !
24.8 L
n = 2.02 mol
Step 2. Determine the mass of nitrogen gas that should be delivered.
28.02 g
mN = n !
2
1 mol
28.02 g
= 2.02 mol !
1 mol
mN = 56.6 g
MN =
2
Statement: The mass of nitrogen that should be delivered into the airbag is 56.6 g.
82. (a) Flasks A, C, G, I, and J contain equal numbers of gaseous entities. Flasks D and F also
contain an equal number of entities, but this number is not the same as the number of entities in
flasks A, C, G, I, and J.
(b) Answers may vary. Sample answer: Avogadro’s law shows that equal volumes of gas at the
same pressure and temperature contain equal amounts of entities. Choosing 100 kPa and 25 °C to
be the reference pressure and temperature and determining the volume of each gas at these
conditions allows the volume of each gas to represent the number of entities in the gas. Flasks A,
C, D, F, H, and J are all at 100 kPa and 25 °C. The gas in flask B, however, must be doubled in
volume to have a pressure of 100 kPa, so its volume becomes 1.00 L at 100 kPa and 25 °C.
Similarly, the gas in flask G would have a volume of 0.50 L at 100 kPa and 25 °C. In addition,
the volume of the gas in flask I must increase by a factor of 5 to reduce the pressure to 100 kPa,
so it has a volume of 0.50 L at 100 kPa and 25 °C. The gas in flask E has a pressure of 100 kPa,
but its temperature is 50 °C. The volume this gas would have at 25 °C is found as follows:
Given: initial temperature, t1 = 50 °C
final temperature, t2 = 25 °C
volume, V1 = 0.50 L
The amount of the gas and the gas pressure remain constant.
Required: final volume, V2
Analysis: Apply Charles’ law to the situation.
V1 V2
=
T1 T2
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-20
Solution:
Step 1. Convert temperature values to kelvins.
T1 = t1 + 273
= 50 + 273
T1 = 323 K
T2 = t2 + 273
= 25 + 273
T2 = 298 K
Step 2. Rearrange the Charles’ law equation to isolate the unknown variable.
VT
V2 = 1 2
T1
Step 3. Substitute given values (including units) into the equations and solve.
0.50 L ! 298 K
V2 =
323 K
V2 = 0.46 L
Statement: The volume of the gas in flask E at 25 °C is 0.46 L
Examination of the volumes (after all have been adjusted as needed to 100 kPa and 25 °C) shows
that gases A, C, G, I, and J have a volume of 0.50 L and thus contain the same number of
entities. Similarly, gases D and F have a volume of 0.40 L and contain the same number of
entities.
0.50
(c) The total mass of gas C divided by the mass of gas A is
, or 2 : 1. Since we assume that
0.25
gases C and A have the same number of entities (because they occupy equal volume at equal
pressure), each entity of C must have twice the mass of each entity of A. The mass ratio of the
entities is therefore 2 : 1.
(d) The volume of gas B is twice the volume of gas A at the same pressure and temperature, so
gas B has twice as many entities. An equal number of gas B entities, therefore, would have half
as much mass: 0.125 g. This means that the mass ratio of samples of A and B containing equal
0.25
numbers of entities is
0.25/0.125, or 2 : 1. This must be the same as the mass ratio of an
0.125
entity of A to an entity of B.
83. Given: mF = 0.76 g
2
M F = 38 g/mol
2
munkown = 0.34 g
Required: Munkown
Analysis: Use Avogadro’s law.
V1 V2
=
n1 n2
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-21
Solution:
First find the amount of fluorine, nF , by multiplying the mass of fluorine by its molar mass.
2
1 mol
38.00 g
1 mol
= 0.76 g !
38.00 g
nF = mF !
2
2
nF = 0.020 mol
2
Since the two gas samples have the same volume at the same pressure and temperature, they
must contain equal amounts of gas. Thus 0.020 mol of the unknown gas must be present. The
molar mass of the unknown gas can be determined from its mass and amount:
m
M unkown = unkown
nunkown
0.34 g
0.020 mol
= 17 g/mol
=
M unkown
Statement: The molar mass of the unknown gas is 17 g/mol.
84. (a) Given: R = 8.314 kPa i L i mol–1 i K–1
d = 3.46 g/L
P = 80.0 kPa
t = 27.0 °C
Required: molar mass of the gas, M
m
Analysis: Use the relationship d =
to find the mass of the gas.
v
m
d=
v
m = dv
Then use the ideal gas law to find the amount of the gas.
PV = nRT
Finally, use the mass and amount of gas to find its molar mass.
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 27.0 + 273
T = 300.0 K
Step 2. Assume there is 1 L of gas and find the mass of this gas.
m = dv
g
= 3.46
!1 L
L
m = 3.46 g
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-22
Step 3. Rearrange the ideal gas law equation to isolate the unknown variable, n.
PV
n=
RT
Step 4. Substitute the known values into the equation and solve.
n=
80.0 kPa ! 1 L
8.314 kPa i L imol"1 i K "1 ! 300.0 K
n = 0.0321 mol
Step 5. Use the relationship M = n/m to find the molar mass of the gas.
m
M=
n
3.46 g
=
0.0321 mol
M = 108 g/mol
Statement: The molar mass of the gas is 108 g/mol.
(b) The molar mass of N2O5(g) can be calculated from atomic masses:
M = (2 × 14.01 g/mol) + (5 × 16.00 g/mol)
M = 108.02 g/mol
This value for the molar mass of N2O5(g) agrees perfectly with the experimental value. The molar
mass obtained in part (a) therefore supports the chemist’s conclusion that the unknown gas is
N2O5(g).
85. (a) Given: R = 8.314 kPa i L i mol–1 i K–1
d = 0.714 g/L
P = 93.2 kPa
t = 22 °C
Required: molar mass of the gas, M
Analysis: Use the relationship d = m/v to find the mass of the gas.
m
d=
v
m = dv
Then use the ideal gas law to find the amount of the gas.
PV = nRT
Finally, use the mass and amount of gas to find its molar mass.
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 22 + 273
T = 295 K
Step 2. Assume there is 1 L of gas and find the mass of this gas.
m = dv
g
= 0.714
!1 L
L
m = 0.714 g
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-23
Step 3. Rearrange the ideal gas law equation to isolate the unknown variable, n.
PV
n=
RT
Step 4. Substitute the known values into the equation and solve.
n=
93.2 kPa ! 1 L
8.314 kPa i L imol"1 i K "1 ! 295 K
n = 0.0380 mol
n
Step 5. Use the relationship M =
to find the molar mass of the gas.
m
m
M=
n
0.714 g
=
0.0380 mol
M = 18.8 g/mol
Statement: The average molar mass of the gas is 18.8 g/mol.
(b) Mmethane = (12.01 g/mol) + (4 × 1.01 g/mol) = 16.05 g/mol
Methane = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) = 30.08 g/mol
(c) From parts (a) and (b), Mavg = 18.8 g/mol, Mmethane = 16.05 g/mol, and Methane = 30.08 g/mol.
The decimal fraction (by volume) of the mixture that is ethane, x, can now be found.
M avg = xM ethane + (1 ! x) M methane
18.8 = 30x + (1 ! x)16
18.8 = 30x + 16 ! 16x
2.8 = 14x
2.8
x=
14
x = 0.200
The decimal fraction of the gas (by volume) that is ethane is x = 0.200.
(d) Since 0.200 is the fraction of the gas (by volume) that is ethane, the ethane concentration
must be 20.0 %. The methane concentration is therefore 80.0 %.
86. Given: V = 283 000 L
t = 18 ºC
R = 8.314 kPa i L i mol–1 i K–1
m = 60.0 kg
M He = 4.00 g/mol
Required: pressure of the helium, P
Analysis: Use the ideal gas law equation to calculate the pressure that 60.0 kg of helium can
provide.
PV = nRT
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-24
Solution:
Step 1. Convert the temperature value to kelvins and mass to grams.
T = t + 273
= 18 + 273
T = 291 K
m = 60.0 kg
= 60.0 kg !
103 g
1 kg
m = 6.00 ! 104 g
Step 2. Convert mass into amount using the appropriate conversion factor involving molar mass.
1 mol
nHe = mHe !
4.00 g
1 mol
= 6.00 ! 104 g !
4.00 g
nHe = 1.50 ! 104 mol
Step 3. Rearrange the ideal gas law equation to find the pressure of the gas.
nRT
P=
V
Step 4. Substitute known values into the equation and solve.
–1
1.50 ! 104 mol ! 8.314 kPa i L i mol i K –1 ! 291 K
P=
283 000 L
P = 128 kPa
Statement: Since 60.0 kg can provide 128 kPa of pressure at these conditions, it is enough to
inflate the balloon to a desired pressure of 100 kPa.
87. In order for the partial pressures of the gases in a mixture to add up to be perfectly equal to
the total pressure, the entities of the gas must act independently of one another. In other words,
the gases must act ideally: attractive forces among the entities must be negligible, as should the
volume of the entities themselves. We expect stronger intermolecular attraction to exist among
polar molecules than among non-polar molecules, so the gases composed of polar molecules
would likely behave less ideally. Thus we could expect the sum of the partial pressures in the
mixture of gases composed of polar molecules to deviate more from the actual total pressure.
88. (a) Given: partial pressure of helium, P1 = 45.0 kPa
partial pressure of neon, P2 = 20.0 kPa
partial pressure of argon, P3 = 30.0 kPa
Required: total pressure, Ptotal
Analysis: Apply Dalton’s law of partial pressures.
Ptotal = P1 + P2 + P3
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-25
Solution: Substitute known values into the equation and solve.
Ptotal = P1 + P2 + P3
Ptotal
= 45.0 kPa + 20.0 kPa + 30.0 kPa
= 95.0 kPa
Statement: The total pressure in the chamber is 95.0 kPa.
(b) Given: initial volume, V1 = 2.0 L
initial helium pressure, P1He = 45.0 kPa
initial neon pressure, P1Ne = 20.0 kPa
initial argon pressure, P1Ar = 30.0 kPa
final volume, V2 = 6.0 L
The amount of gas and the temperature remain constant.
Required: final pressure, P2
Analysis: Use Boyle’s law for each gas.
P1V1 = P2V2
Since the gases occupy both chambers, the final volume is the sum of the volumes of the
chambers, which is 6.0 L.
Solution:
Step 1. Rearrange the equation to isolate the unknown variable.
PV
P2 = 1 1
V2
Step 2. Substitute given values (including units) into the equations and solve.
P V
P2He = 1He 1
V2
45.0 kPa ! 2.0 L
6.0 L
= 15 kPa
P2He =
P2He
P2Ne =
P1NeV1
V2
20.0 kPa ! 2.0 L
6.0 L
= 6.7 kPa
P2Ne =
P2Ne
P2Ar =
P1ArV1
V2
30 kPa ! 2.0 L
6.0 L
= 10 kPa
P2Ar =
P2Ar
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-26
Statement: The partial pressures of helium, neon, and argon when the valve is opened are,
respectively, 15 kPa, 6.7 kPa, and 10 kPa.
(c) The total pressure in the two-chamber system is the sum of the partial pressures:
Ptotal = 15 kPa + 6.7 kPa + 10 kPa
Ptotal = 32 kPa
The total pressure is 32 kPa.
89. (a) Given: nN = 1.0 mol
2
nAr = 0.20 mol
nO = 1.2 mol
2
V = 10.0 L
t = 22 °C
Required: PN , PAr , and PO
2
2
Analysis: Use the ideal gas law to calculate the pressure of each gas.
PV = nRT
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 22 + 273
T = 295 K
Step 2. Rearrange the ideal gas law equation to find the pressure of the gas.
nRT
P=
V
Step 3. Substitute known values for each gas into the equation and solve.
–1
1.0 mol ! 8.314 kPai L i mol i K –1 ! 295 K
PN =
2
10.0 L
2
PN = 2.5 ! 10 kPa
2
–1
0.20 mol ! 8.314 kPai L i mol i K –1 ! 295 K
PAr =
10.0 L
PAr = 49 kPa
–1
1.2 mol ! 8.314 kPai L i mol i K
PO =
2
10.0 L
2
PO = 2.9 ! 10 kPa
–1
! 295 K
2
Statement: In the tank, the partial pressure of nitrogen is 2.5 × 102 kPa, the partial pressure of
argon is 49 kPa, and the partial pressure of oxygen is 2.9 × 102 kPa.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-27
(b) The total pressure is the sum of the partial pressures:
Ptotal = 245 kPa + 49 kPa + 294 kPa
= 588 kPa
The total pressure in the tank is 577 kPa.
90. (a) Given: P = 101.3 kPa
V = 0.100 L
t = 25 °C
Required: nair
Analysis: Use the ideal gas law to find the amount of air.
PV = nRT
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 25 + 273
T = 298 K
Step 2. Rearrange the ideal gas law equation to find the amount of gas.
PV
n=
RT
Step 3. Substitute known values into the equation and solve.
101.3 kPa ! 0.100 L
n=
8.314 kPa i L imol–1 i K –1 ! 298 K
n = 0.0041 mol
Statement: The amount of air in the chamber initially is 0.0041 mol.
(b) Given: mass of sodium azide, mNaN = 1.00 g
3
molar mass of sodium azide, M NaN = 65.02 g/mol
3
Required: amount of nitrogen gas, VN
2
Solution:
Step 1. Write the balanced chemical equation, listing the given and required quantities.
2 NaN 3 (s) ! 2 Na(s) + 3 N 2 (g) + energy
1.00 g
VN
2
Step 2. Convert the mass of the given substance, mNaN , into an amount, nNaN .
3
nNaN = 1.00 g !
3
1 mol
65.02 g
nNaN = 0.015 380 mol
3
3
[two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-28
Step 3. Determine the amount of nitrogen gas produced from the amount of sodium azide
decomposed.
3 molN
2
nN = 0.015 380 mol !
2
2 molNaN
3
nN = 0.0231 mol
2
Statement: When 1.00 g of sodium azide decomposes, 0.0231 mol of nitrogen gas is produced.
(c) The total amount of gas in the chamber is the sum of the amount of air and the amount of
nitrogen gas produced from the decomposition of sodium azide.
ntotal = nair + nN
2
ntotal
= 0.0041 mol + 0.0231 mol
= 0.0272 mol
The total amount of gas in the chamber is 0.0272 mol.
(d) Given: V = 0.100 L
temperature, T = T1 + Tincrease
initial temperature, t1 = 25 ºC
temperature increase from explosion, tincrease = 585 ºC
R = 8.314 kPa i L i mol–1 i K–1
Required: pressure inside the chamber, P
Analysis: Use the ideal gas law equation to calculate the pressure inside the chamber.
PV = nRT
Solution:
Step 1. Convert the temperature values to kelvins and find the temperature, t.
T1 = t1 + 273
= 25 + 273
T1 = 298 K
Tincrease = tincrease
Tincrease = 585 K
T = T1 + Tincrease
= 298 + 585
T = 883 K
Step 2. Rearrange the ideal gas law equation to find the pressure of the gas.
nRT
P=
V
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-29
Step 3. Substitute known values into the equation and solve.
–1
0.0272 mol ! 8.314 kPai L i mol i K –1 ! 883 K
P=
0.100 L
P = 2000 kPa
Statement: The gaseous pressure inside the chamber after the detonation of the sodium azide is
2000 kPa, or 2000 atm.
(e) The equipment is likely to be damaged. The inner chamber can only withstand pressures up
to 1720 kPa and the detonation of 1.00 g of sodium azide is predicted to generate a pressure of
2000 kPa, which is above the safe limit.
91. (a) Zn(s) + 2 HCl(aq) ! ZnCl2 (aq) + H 2 (g)
2 Al(s) + 6 HCl(aq) ! 2 AlCl3 (aq) + 3 H 2 (g)
(b) Given: mass of zinc, mZn = 0.25 g
mass of aluminum, mAl = 0.25 g
molar mass of zinc, M Zn = 65.41 g/mol
molar mass of aluminum, M Al = 26.98 g/mol
temperature, t = 21 ºC
pressure, P = 92 kPa
Required: volume of hydrogen gas produced, VH
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 21 + 273
T = 294 K
Step 2. Write the balanced chemical equations, listing the given and required quantities.
Zn(s) + 2 HCl(aq) ! ZnCl2 (aq) + H 2 (g)
0.25 g
VH
2
2 Al(s) + 6 HCl(aq) ! 2 AlCl3 (aq) + 3 H 2 (g)
0.25 g
VH
2
Step 3. Convert the masses of the given substances, mZn and mAl , into amounts, nZn and nAl .
nZn = 0.25 g !
1 mol
65.41 g
nZn = 0.0038 mol
nAl = 0.25 g !
1 mol
26.98 g
nAl = 0.0093 mol
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-30
Step 4. For each reaction, determine the amount of hydrogen gas produced from the amount of
metal used.
For zinc:
1 molH
2
nH = 0.0038 mol !
2
1 molZn
nH = 0.0038 mol
2
For aluminum:
nH = 0.0093 mol !
2
3 molH
2
2 molAl
nH = 0.014 mol
2
Step 5. For each reaction determine the volume of the required substance, VH , using the ideal
gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVH = nH RT
2
2
2
VH =
2
nH RT
2
P
For zinc:
nH RT 103 mL
VH = 2
!
2
P
1L
–1
0.0038 mol " 8.314 kPa iLi mol i K –1 $ 294 K
#
%
103 mL
=
!
1L
92 kPa
(
)
(
)
103 mL
1 L
2
= 1.0 ! 10 mL
= 0.010 L !
VH
2
For aluminum:
nH RT 103 mL
VH = 2
!
2
P
1L
–1
0.014 mol " 8.314 kPa iLi mol i K –1 $ 294 K
#
%
103 mL
=
!
1L
92 kPa
(
)
(
)
103 mL
1 L
2
= 3.7 ! 10 mL
= 0.37 L !
VH
2
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
U5-31
Statement: When 0.25 g of zinc reacts with hydrochloric acid, 1.0 × 102 mL of hydrogen gas is
produced (under the given conditions of temperature and pressure). When 0.25 g of aluminum
reacts with hydrochloric acid, 3.7 × 102 mL of hydrogen gas is produced (under the given
conditions of temperature and pressure).
(c) The unknown metal is most likely zinc because the predicted volume of hydrogen it would
produce, 102 mL, is closest to the experimental volume of 104 mL.
(d) The hydrogen gas was collected by the downward displacement of water. Some water vapour
will unavoidably mix with the collected gas, increasing the total number of gas molecules. A
greater number of gas molecules will cause the mixed gas sample to have a greater volume than
the hydrogen alone would have. The student would correct for this by subtracting the partial
pressure of water from the pressure of gas in the collection vessel, and then calculating the
volume of hydrogen produced.
Evaluation
92. Answers may vary. Sample answer: I do not agree with my friend. It is true that people in the
forest regions might have warmer weather, but this could disrupt their lives more than they
would benefit. For example, warmer weather will cause the species of trees to change. People
whose lives are built around the harvesting of trees like pines and firs will either have to move or
change occupations. In addition, insects that currently cannot survive the cold in the boreal forest
could invade the forest as temperatures rise, killing many trees. This could lead to a greater fire
danger and pose a threat to people and structures in the forest. From a broader perspective, even
if some people welcomed rising temperatures in the northern regions of Canada, global warming
is likely to bring terrible droughts to some regions of the world and devastating storms and
floods to others. Overall, global warming is unlikely to lead to greater good for Earth’s
inhabitants. (Note: Different answers are acceptable as long as they are supported with specific
examples.)
93. Answers may vary. Sample answer: I would first need to find the quantity of methane
emitted by a typical beaver pond over a given period of time such as a year. Then I would need
to obtain an estimate of the number of beaver ponds over the entire world. This would entail
determining how many beaver ponds exist on average per given area of mid-boreal forest and
then finding the total mid-boreal forest area worldwide. I could multiply the number of beaver
ponds per given area by the total area of mid-boreal forest to get the total number of beaver
ponds. Next, I would have to multiply the total number of beaver ponds by the quantity of
methane emitted per pond per year to get the total yearly emission of methane. I could then
calculate the increase in the concentration of methane in the atmosphere per year due to beaver
ponds by dividing the total quantity of methane produced in a year by the total volume of the
atmosphere (assuming the methane is not removed by some means). Finally, I could compare the
calculated increase in the atmospheric methane concentration to historical levels of atmospheric
methane to see if the increase due to beaver ponds is significant.
94. (a) Trees and other plants convert carbon dioxide gas and water into food and plant tissue
during photosynthesis. The carbon from the carbon dioxide thus becomes incorporated into the
biomass of the plant.
(b) Carbon dioxide is a greenhouse gas that contributes to global warming. By sequestering
carbon by removing carbon dioxide from the air, plants reduce the concentration of greenhouse
gases in the atmosphere, which reduces the heat trapped in Earth’s atmosphere.
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Unit 5: Gases and Atmospheric Chemistry
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(c) A tree that is harvested for lumber is processed into structural wood products that continue to
sequester carbon. This allows a tree harvested for lumber to serve as a carbon storage bank for
much longer than a tree that goes through a natural life cycle ending with death and immediate
decay.
(d) Mature trees do not grow much and therefore do not sequester nearly as much carbon per
year as do younger, actively growing trees. Once trees are harvested and the space is replanted
with tree seedlings, the new trees, through rapid growth, sequester carbon that the mature trees
would not.
(e) Growing and harvesting trees for lumber is not a perfect way to sequester carbon. When the
tree is processed for lumber, some of its biomass will become waste that will decay and
eventually produce carbon dioxide and methane. In addition, the burning of fossil fuels to power
the logging equipment and to transport harvested lumber produces carbon dioxide. The
greenhouse gases from the decay of waste biomass and fossil fuel consumption by logging
activities will partially offset the carbon sequestered in replacement trees.
95. Answers may vary. Sample answer: I believe that the second partner’s plan is better, mainly
because it is important to have detectors just outside each bedroom. Carbon monoxide gas is
odourless and tasteless, so its presence is not immediately obvious. Exposed individuals might
not recognize that they are being poisoned until their symptoms are already serious. This would
be especially true for people who are sleeping. If someone who is sleeping is exposed to carbon
monoxide, he or she might not wake up before being rendered unconscious, which could result in
death. If the furnace or water heater starts venting carbon monoxide into the basement at night
and the detectors go off, the family sleeping upstairs might not hear the alarms. If the detectors
are right next to the bedrooms, however, and carbon monoxide starts to accumulate, the family
will be sure to wake up when the alarm sounds. This will give them a chance to leave the house
safely. I would modify the plan by putting only one detector in the basement in a strategic
location and move the other to a first-floor location, such as near the stairwell to the basement.
This might allow the sleeping family to hear an alarm from the first-floor detector if carbon
monoxide starts seeping upward from the basement. The family would then be alerted to the
danger even sooner.
96. Answers may vary. Sample answer: I believe that student 2’s results are more trustworthy.
Student 1 used a rubber balloon to hold the air sample. The proper extrapolation of the data
depends on the volume of the gas remaining constant, but a rubber balloon will expand or
contract with temperature changes. The likely changes in the volume of student 1’s air sample
due to changing temperature would introduce significant error into the observations. Student 2,
on the other hand, used a test tube that would not have changed in volume at various
temperatures, so the volume remained constant. In addition, student 1 only placed the bottom of
the balloon in the trays of water at different temperatures. Only the lower portion of the air in the
balloon was close to the water. As a result, the air in the balloon may not have fully reached the
same temperature as the water in each bath. This would cause student 1’s plot to be in error
because the temperatures of the points might not represent the actual temperature of the air for
each reading. Student 2 immersed the full length of the test tube in the water baths, making it
more likely that the air in the test tube actually reached the same temperature as the water in each
bath.
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Unit 5: Gases and Atmospheric Chemistry
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97. Avogadro’s hypothesis indicates that the amount of the air displaced and the amount of either
methane or helium inside the envelope will be the same:
PV
n=
Rt
90 kPa ! 60 000 L
=
8.314 kPa i L imol"1 i K "1 ! 288 K
n = 2260 mol
The relevant masses are:
1 kg
mair = M air ! nair !
1000 g
28.90 g
=
! 2260 mol
1 mol
1 kg
= 65 300 g !
1000 g
mair = 65.3 kg
mmethane = M methane ! nmethane !
=
16.05 g
1 mol
1 kg
1000 g
! 2260 mol
= 36 300 g !
1 kg
1000 g
mmethane = 36.3 kg
1 kg
1000 g
4.00 g
1 kg
=
! 2260 mol !
1000 g
1 mol
1 kg
= 9040 g !
1000 g
mhelium = M helium ! nhelium !
mhelium = 9.04 kg
The lifting power of the methane is
mair – mCH4 = 65.3 kg – 36.3 kg
= 29 kg
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
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The load the methane must lift is
mass instruments + mass envelope + mass gas = 20.0 kg + 4.0 kg + 36.3 kg
= 60.3 kg
Since the load the methane must lift, 60.3 kg, is greater than its lifting power, 29 kg, methane
will not be suitable for the weather balloon.
The lifting power of the helium is
mair – mHe = 65.3 kg – 9.04 kg
= 56.3 kg
The load the helium must lift is
mass instruments + mass envelope + mass gas = 20.0 kg + 4.0 kg + 9.04 kg
= 33.0 kg
Since the load the helium must lift, 33.0 kg, is less than its lifting power, 56.3 kg, helium will be
suitable for the weather balloon. Thus, of the two gases, only helium will work.
98. Answers may vary. Sample answer: I do not believe that the student’s planned procedure will
yield an accurate value for the density of carbon dioxide gas at room conditions. As the carbon
dioxide comes into contact with water during its production and during the downward
displacement, some will unavoidably dissolve into the water and not end up in the gas “bubble”
in the cylinder. Thus the volume of the collected gas will be less than it should be. In addition,
the flask also initially contains some air, at least some of which will be forced out of the flask
and into the inverted cylinder along with the carbon dioxide. Water vapour will also unavoidably
mix with the collected gas. Finally, not all of the carbon dioxide produced will pass into the
graduated cylinder: some will remain in the flask. For all of these reasons, the gas collected in
the inverted cylinder will not be pure carbon dioxide and will not accurately reflect the mass of
carbon dioxide produced, so the calculated density will not be accurate.
99. Analysis of Fuel System I: Aluminum and Ammonium Perchlorate
Analyze the reaction assuming 10 mol Al(s) and 6 mol NH4ClO4(s) react:
M NH ClO = 117.50 g/mol
4
4
M Al = 26.98 g/mol
1 kg
4
103 g
117.50 g
mNH ClO = nNH ClO ! M NH ClO !
4
4
4
4
4
= 6 molNH ClO !
4
4
1 molNH ClO
4
= 705.0 g !
4
1 kg
103 g
mNH ClO = 0.705 kg
4
4
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Unit 5: Gases and Atmospheric Chemistry
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mAl = nAl ! M Al !
= 10 molAl !
1 kg
103 g
26.98 g
1 molAl
1 kg
103 g
= 269.8 g !
mAl = 0.2698 kg
mtotal = mNH ClO + mAl
4
mtotal
4
= 0.705 kg + 0.2698 kg
= 0.975 kg
The reaction produces 12 mol of gaseous H2O and 3 mol N2 gas for every 10 mol Al and 6 mol
NH4ClO4 reacted.
Now we can calculate the amount of gas per kilogram of fuel.
12 mol + 3 mol
amount of gas per kg of fuel =
0.975 kg
= 15.4 mol/kg
The amount of gas per kg of fuel for Fuel System I is 15.4 mol/kg.
Analysis of Fuel System II: Hydrogen Combustion:
Analyze the reaction assuming 2 mol H2(l) and 1 mol O2(l) react:
M H = 2.02 g/mol
2
M O = 32.00 g/mol
2
1 kg
103 g
2.02 g
mH = nH ! M H !
2
2
2
= 2 molH !
2
= 4.04 g !
1 molH
2
1 kg
103 g
mH = 0.004 04 kg
2
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
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1 kg
103 g
32.00 g
mO = nO ! M O !
2
2
2
= 1 molO !
2
1 molAl
= 32.00 g !
1 kg
103 g
mO = 0.032 00 kg
2
mtotal = mH + mO
2
mtotal
2
= 0.004 04 kg + 0.032 00 kg
= 0.0360 kg
The reaction produces 2 mol of gaseous H2O for every 2 mol H2 gas and 1 mol O2 gas reacted.
Now we can calculate the amount of gas per kilogram of fuel.
2 mol
amount of gas per kg of fuel =
0.0360 kg
= 55.6 mol/kg
The amount of gas per kg of fuel for Fuel System II is 55.6 mol/kg.
Analysis of Fuel System III: Hydrazine Decomposition:
Analyze the reaction assuming 1 mol N2H4(l) decomposes:
M N H = 32.06 g/mol
2
mN
2 H4
4
= nN
2 H4
! MN
= 1 molN
2 H4
= 32.06 g !
mN
2 H4
2 H4
!
1 kg
103 g
32.06 g
!
1 molN
2 H4
1 kg
103 g
= 0.032 06 kg
The reaction produces 1 mol of N2 gas and 2 mol of H2 gas for every 1 mol N2H4 reacted.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
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Now we can calculate the amount of gas per kilogram of fuel.
1 mol + 2 mol
amount of gas per kg of fuel =
0.003206 kg
= 93.6 mol/kg
The amount of gas per kg of fuel for Fuel System III is 93.6 mol/kg.
These calculations show that Fuel System III (hydrazine decomposition) produces the greatest
amount of gas per kilogram of fuel consumed.
Reflect on Your Learning
100. Answers may vary. Sample answer: I ride a bicycle and I often have to pump up its tires.
Boyle’s law explains how the tire pump works. When the pump handle is pulled up, the cylinder
of the pump fills with air. When I push down on the pump and compress the air in the cylinder to
a much smaller volume, the pressure of the air rises, just as Boyle’s law predicts. The high
pressure of the air is what forces air into the tires to inflate them.
101. Answers may vary. Sample answer: I had always heard about greenhouse gases in a
negative context because of people’s worries about global warming. I never realized, however,
that without the greenhouse gases that are naturally present, our world would be in a “deep
freeze” and life as we know it might not exist. I was interested to find that human activities add
to the greenhouse gases that are naturally present rather than being the sole source of greenhouse
gases.
102. Answers may vary. Sample answer: I became very interested in how gases are used as
anesthesia in medical procedures, particularly how the anesthetics are mixtures of gases whose
partial pressures must be very accurately measured. I can find out more by doing Internet
research, but more importantly, I would like to interview a practicing anesthesiologist to get an
impression of what a career in anesthesiology would be like and to learn more of the science
first-hand from an expert.
Research
103. Answers may vary. Students’ reports should focus on these details and may include other
factors that influence the ability of atmospheric molecules to escape into space.
Sample answer: The gaseous entities in a planetary atmosphere can escape if they reach escape
velocity in a region of sufficient altitude that molecular collisions do not impede the entities from
flying into space. More massive planets have stronger gravity, which means escape velocities are
higher, thus making it less likely that atmospheric entities will drift off into space. As a result,
more massive planets have a greater chance of keeping an atmosphere. Relatively small
planetary bodies and even moons can hold atmospheres, however, they are quite cold, because
the entities in their atmospheres have lower average speeds. In addition, the composition of a
planetary atmosphere is affected by the mass of its entities. Entities with greater molar masses
have lower speeds at a given temperature and these entities are likely to be enriched in an
atmosphere whereas entities with lower molar masses tend to be relatively depleted.
Copyright © 2011 Nelson Education Ltd.
Unit 5: Gases and Atmospheric Chemistry
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104. Answers may vary. Students’ reports should include some of the surprising findings of
Stedman’s research with regard to which vehicles are the worst polluters and whether or not
indoor, treadmill-type testing is really an effective means of emissions testing. Sample answer:
Dr. Stedman’s system, called FEAT (fuel efficiency automobile test), shines an infrared beam
through the exhaust plumes of passing cars. In real time, spectral analyzers on the other side of
the road measure the particular wavelengths of infrared light absorbed by the exhaust to detect
and quantify CO2, CO, hydrocarbons, and NO. At first, Stedman’s system was used to collect
data on which cars generated the most pollution. Later, the FEAT system was adapted in certain
locations to allow motorists to drive a car past a roadside station for an “on-the-fly” emissions
test.
105. Answers may vary. Sample answer: Mercury barometers are based on the original design of
Evangelista Torricelli. They consist of an evacuated tube filled partway with mercury, arranged
so that the open end of the tube is immersed in a reservoir of mercury exposed to the pressure of
the air. As the atmospheric pressure changes, the mercury level in the tube rises and falls.
Aneroid barometers consist of a sealed, partially evacuated metal can with flexible sides. As the
atmospheric pressure changes, the sides of the can are either pushed in or out depending on
whether the pressure has risen, or fallen, respectively. The motion of the walls is communicated
by electronics or mechanical linkages to a display that indicates pressure. Water barometers
generally consist of a reservoir of water that is sealed in a container with a headspace of air. A
spout is attached to the reservoir below the fluid line, and prior to sealing, the water is allowed to
seek its own level in the reservoir and spout. After the reservoir is sealed, changes in atmospheric
pressure cause the water in the spout to rise or fall. When barometer readings fall, inclement
weather is usually on the way. A rising barometer, on the other hand, usually signals fair weather
ahead.
106. Answers may vary. Students’ reports should relate these historical facts and also provide
specific examples of Avogadro’s reasoning in relation to the law of combining volumes.
Sample answer: John Dalton erroneously believed the chemical formula for water to be HO.
Knowing from his investigations that 8 g of oxygen combines with 1 g of hydrogen, Dalton drew
the conclusion that the atomic mass of oxygen is 8 times greater than the atomic mass of
hydrogen, when oxygen’s atomic mass is in fact 16 times greater. Since the determination of
atomic masses at that time depended on combining masses in series of related compounds,
Dalton’s error threw many results off. Gay-Lussac’s law of combining volumes, particularly as
applied to the formation of water, was at odds with water having a formula of HO; Gay-Lussac
observed that two volumes of hydrogen combine with one volume of oxygen, which suggests
that water is H2O. Dalton countered that Gay-Lussac’s measurements were inaccurate. Avogadro
reached the proper conclusion about the meaning of Gay-Lussac’s data in 1811, realizing that
equal volumes of gases at the same pressure and temperature have equal numbers of molecules.
Avogadro’s reasoning showed that the molecular formula for water is H2O, not HO, and further,
that a number of elemental gases such as hydrogen, oxygen, and nitrogen must be composed of
diatomic molecules. A vigorous debate over who was correct continued for about 50 years.
Finally, at a scientific conference at Karlsruhe, Germany in 1860, Stanislao Cannizzaro, an
Italian like Avogadro, made such a clear presentation of Avogadro’s hypothesis and its
implications that the scientific community in Europe recognized Avogadro’s brilliance and
accepted the correct meaning of Gay-Lussac’s law of combining volumes with regard to molar
and atomic masses.
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