Chapter 11 Review, pages 568–573
Knowledge
1. (d)
2. (b)
3. (d)
4. (d)
5. (a)
6. (a)
7. (b)
8. (c)
9. (d)
10. (a)
11. (c)
12. False. When a substance changes from a solid to a liquid and then from a liquid to a gas, the
degree of order in its molecules decreases.
13. True
14. False. The layers of the atmosphere in which temperature rises with increasing altitude are
the thermosphere and the stratosphere.
15. False. Nitrogen gas in the atmosphere is converted by bacteria into a form that plants can
readily absorb through their roots.
16. False. When a fire occurs in a residential building, the main cause of death will typically be
smoke inhalation.
17. True
18. False. The greatest proportion of carbon monoxide air pollution in Ontario comes from the
transportation sector.
19. True
20. (a) (ii)
(b) (i)
(c) (iv)
(d) (iii)
21. (a) A solid is virtually incompressible.
(b) A liquid takes the shape of a container, but has a definite volume.
(c) A solid has definite shape and volume.
(d) A gas is easily compressed.
(e) A gas takes the shape and volume of the entire container.
(f) A liquid and a gas flow easily.
22. As thermal energy is transferred to a solid, its entities vibrate more and more rapidly.
Eventually, the attractive forces between the entities are broken allowing the entities to flow past
each other. As a result, a change in state occurs in which the solid turns into a liquid.
23. With the discovery of argon, a new column was added to the periodic table to accommodate
argon and the other noble gases.
24. Ozone is a bad thing when it is present in the troposphere, the lowest layer of the atmosphere.
Ozone in the troposphere poses significant health and environmental concerns due to its toxicity.
25. Dust mites live in linens, mattresses, pillows, and carpets and feed on dander.
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Chapter 11: The Gas State and Gas Laws
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26. Boyle’s law states that PV = constant. Since V = 6.00 L when P = 1.00 kPa for the sample of
gas, each product PV = 6.00.
Table 1 Observations of a Sample of Gas
Pressure (atm)
Volume (L)
1.00
6.00
1.50
4.00
2.00
3.00
3.00
2.00
4.00
1.50
6.00
1.00
Understanding
27. In order of decreasing strength of intermolecular attractions, the elements are iodine, I2,
bromine, Br2, and chlorine, Cl2. Since all three substances are at the same temperature, their
states of matter are an indication of the strength of their intermolecular forces. Solids generally
have the strongest attractive forces between their entities and gases have the least attractive
forces.
28. According to the state symbol (s), sucrose is a solid. The sucrose molecules are only capable
of vibrating back and forth because they are “locked” into an ordered, solid arrangement. The
methane and water molecules, though, have greater freedom of motion because these substances
are liquids and gases. Accordingly, the methane and water molecules can move through space
and rotate as well as vibrate. Methane is a gas, so its molecules have the most freedom to move
around.
29. Gasoline powered lawn mowers release carbon dioxide gas into the air. Electric lawn mowers
require electric power, which usually means that fossil fuels will be burned and carbon dioxide
gas will be released into the air to produce electricity. Carbon dioxide is a greenhouse gas that
contributes to global warming. Using a push mower would result in lower emissions of carbon
dioxide compared to gasoline and electric powered mowers and would help to reduce global
warming.
30. The ozone high in the atmosphere is concentrated in the stratosphere, but the ozone
concentration is quite small: only about 10 ppm. Yet this small concentration of ozone is able to
act as an effective radiation shield that protects Earth’s surface from solar UV rays. Thus ozone
must be very efficient at absorbing UV radiation.
31. The concentration of argon in air is 0.934 %. The volume of argon in 2.00 L of air can be
calculated:
0.934
V = 2.00 L !
100
V = 0.0187 L
So 0.0187 L, or 18.7 mL of argon could be obtained from 2.00 L of air.
32. A yellow colour in a flame is an indicator of incomplete combustion. The black deposit
produced by the flame is soot. Its presence demonstrates that incomplete combustion produces
particulate matter. Incomplete combustion in the environment contributes to air pollution by
releasing particulate matter.
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Chapter 11: The Gas State and Gas Laws
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33. Setting the air conditioning temperature a little higher means that the air conditioner will not
have to run as often to cool the house to the desired temperature. If the air conditioner runs less
often, it will use less electricity. If less electricity is used, the electrical generating station
providing the electricity will not have to burn as much fossil fuel and thus will emit a smaller
quantity of combustion products into the air.
34. (a) Reaction III requires UV radiation from the Sun to break the nitrogen dioxide apart into
nitrogen monoxide and oxygen. No solar UV radiation is available at night, so reaction III would
cease, which would also cause reaction IV to stop producing ozone.
(b) Reaction I represents the formation of nitrogen monoxide in the engines of cars and trucks.
On a day with very light traffic, much less nitrogen monoxide is produced to feed the subsequent
reactions.
35. Indoor air is confined within a building. As a result, pollutants escaping into indoor air tend
to remain in the building, continually increasing in concentration. Outdoor air, on the other hand,
is free to mix with the entire atmosphere. Polluted air over a city, for example, can mix with
unpolluted air from the surrounding countryside. The mixing of polluted outdoor air with
surrounding clean air reduces the concentration of the pollutants.
36. Radon forms in relatively large quantities in uranium-rich soils. Since radon is a gas, it can
diffuse through soil and also through cracks in basement floors and walls. The presence of radon
in a home is a serious health hazard.
37. Idling cars produce higher concentrations of pollutants than cars that are moving. When cars
are idling outside of a school, their pollutants are released in one area, close to where children
wait, walk, and play. Children are particularly susceptible to adverse health consequences from
the air pollutants emitted by idling cars. Children breathe more rapidly than adults and inhale
more pollutants per kilogram of body mass. The greater load of pollutants in a child’s body
results in an increased risk of negative health effects.
38. There would be many more entities in the sample of air at sea level than in the sample at
3000 m. This is because the density of the atmospheric gases diminishes with increasing altitude.
39. When a sample of water freezes, the molecules do not become motionless. They continue to
move, but they are limited to vibrational motion only as opposed to the rotational, vibrational,
and translational motion they had in the liquid state.
40. Popcorn kernels pop due to moisture in the kernels that vaporizes and creates pressure during
cooking. If popcorn kernels were stored in an open container, moisture could steadily evaporate
out of them until there was insufficient water remaining for them to pop. Storing the kernels in a
sealed container ensures that they retain their moisture and thus their ability to pop.
41. Given: initial Celsius temperature, t1 = 25 °C
final Celsius temperature, t2 = –5 °C
volume, V1 = 4.000 × 103 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
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Chapter 11: The Gas State and Gas Laws
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Solution:
Step 1. Convert temperature values to kelvins.
T = t + 273
T1 = t1 + 273
= 25 + 273
T1 = 298
T2 = t2 + 273
= –5 + 273
T2 = 268
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.
4.000 ! 103 L ! 268 K
V2 =
298 K
= 3600 L
V2 = 3.60 ! 103 L
Statement: The volume of propane at the cooler temperature will be 3.60 × 103 L.
42. Given: initial Celsius temperature, t1 = 27 °C
initial volume, V1 = 12.0 L
final volume, V2 = 10.6 L
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 values to kelvins.
T = t + 273
T1 = t1 + 273
= 27 + 273
T1 = 300
Step 2. Rearrange the Charles’ law equation to isolate the unknown variable.
TV
T2 = 1 2
V1
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Chapter 11: The Gas State and Gas Laws
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Step 3. Substitute given values (including units) into the equations and solve.
300 K ! 10.6 L
T2 =
12.0 L
T2 = 265 K
Step 4. Convert temperatures values back to Celsius.
T2 = t2 + 273
t2 = T2 ! 273
= 265 ! 273
t2 = !8 º C
Statement: The nitrogen must be cooled to –8 ºC to have a volume of 10.6 L.
43. When the soccer ball is taken outdoors, the temperature of the air inside it falls, so the
entities of the air slow down. They therefore strike the interior surface of the ball less frequently
and with less kinetic energy. This causes the pressure inside the ball to fall, making the ball feel
soft.
44. Given: initial pressure, P1 = 90.659 kPa
initial volume, V1 = 100 mL
initial temperature, t1 = 22 °C
final pressure, P2 = 101.325 kPa
final temperature, T2 = 273 K
The amount of gas remains constant.
Required: final volume, V2
Analysis: Use the combined gas law.
P1V1 P2V2
=
T1
T2
Solution:
Step 1. Convert temperature values to kelvins.
T1 = t1 + 273
= 22 + 273
T1 = 295 K
Step 2. Rearrange the equation to isolate the unknown variable.
PV T
V2 = 1 1 2
P2T1
Step 3. Substitute given values (including units) into the equation and solve.
90.659 kPa ! 100 mL ! 273 K
V2 =
101.325 kPa ! 295 K
V2 = 80 mL
Statement: The gas sample would have a volume of 8 × 10 mL at standard temperature and
pressure.
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Chapter 11: The Gas State and Gas Laws
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Analysis and Application
45. Even though the hydrogen sulfide molecules travel very rapidly, they do not have an
unobstructed path from the bottle to the nose of someone across the room. There are countless air
molecules that occupy the intervening space. This means that the hydrogen sulfide molecules
will constantly collide with air molecules and, consequently, follow a zig-zag path through the
air. The time spent by the hydrogen sulfide molecules zig-zagging greatly delays their arrival at a
point across the room.
46. (a) The balanced chemical equation for baking soda reacting with ethanoic acid is:
C2H4O2(l) + NaHCO3(s) → NaC2H3O2(aq) + CO2(g) + H2O(l)
(b) The reaction forms carbon dioxide gas, which does not support combustion. Carbon dioxide
has a greater molar mass than nitrogen and oxygen, which make up most of the air, so carbon
dioxide gas has a greater density than air. As the reaction proceeds, the carbon dioxide gradually
fills the glass and remains in it due to its greater density. When the glass is tipped, the carbon
dioxide, being fluid, will pour through the air almost like water and fill up the beaker containing
the burning candle. In the beaker, the carbon dioxide pushes out the oxygen so the flame cannot
continue to burn. Therefore, the candle goes out.
47. Diffusion would be most rapid for two gases and least rapid for two solids. In gases, the
entities are free to undergo translational motion, so they move through one another relatively
rapidly. In contrast, the entities of a solid are held in place by chemical bonds. They have only
vibrational motion and do not move through space. As a result, we predict that there will be
almost no movement of the entities from one solid into another. Liquid entities do undergo
translational motion, but the molecules are much more crowded compared to the entities in a gas.
The crowding leads to more collisions between entities, which reduces the rate at which the
entities of one liquid will move through another liquid. Consequently, two liquids will have an
intermediate rate of diffusion.
48. First, find the total amount of air by multiplying the amount concentration of air molecules
by the volume of the stratosphere:
mol air
1 ! 10"3
! 2 ! 1022 L = 2 ! 1019 mol air
L
Next, determine the amount of ozone gas by multiplying the amount of air by the ratio of ozone
molecules to air molecules:
3 mol ozone
2 ! 1019 mol air !
= 6 ! 1013 mol ozone
6
10 mol air
Next, find the total mass of ozone by multiplying the amount of ozone by the conversion factor
related to the molar mass of ozone:
48 g
6 ! 1013 mol ozone !
= 2.88 ! 1015 g
1 mol ozone
Finally, convert this mass to tonnes:
3 ! 1015 g !
1 kg
3
10 g
!
1t
3
10 kg
= 3 ! 109 t
The approximate mass of ozone in the stratosphere is 3 × 109 t.
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Chapter 11: The Gas State and Gas Laws
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49. Sir William Ramsay passed a gaseous mixture of nitrogen and argon obtained from air over
hot magnesium to produce magnesium nitride. This removed the nitrogen to leave only argon.
The chemical equation for this reaction is: 3 Mg(s) + N2(g) → Mg3N2(s)
When the students burned magnesium in air in their laboratory exercise, magnesium
nitride might also have been produced. Using the molar mass values of 23.31 g/mol for
magnesium and 14.01 g/mol for nitrogen, the mass percentage of magnesium in magnesium
nitride can be found:
mass of magnesium
% magnesium =
! 100 %
total mass
3 ! 24.31
=
! 100 %
(3 ! 24.31) + (2 ! 14.01)
72.93
! 100 %
100.95
% magnesium = 72.24 %
If magnesium nitride were present in the solid product of the students’ reactions, its higher mass
percentage of magnesium would increase the average mass percentage of magnesium in the solid
product. This could make the class results significantly greater than 60.3 %.
50. If the temperature is very high, such as in the middle of the day, gasoline will evaporate
faster. If more gasoline evaporates as a car’s gas tank is being filled, more VOCs will enter and
pollute the atmosphere. Fueling in the early morning or late evening, when temperatures are
lower, will result in less gasoline evaporating and a smaller quantity of VOCs entering the air.
51. Answers may vary. Sample answer: I would encourage my family to do outdoor activities
such as jogging, gardening, or sports practice in the morning because an AQHI of 3 means that
air quality is ideal for outdoor activities. An AQHI of 7, on the other hand, means that air quality
is quite poor and outdoor activities are discouraged. I would recommend to my family that they
stay indoors in the late afternoon and refrain from driving anywhere if possible because car
exhaust could make the air quality even worse for everyone.
52. Answers may vary. Sample answer: The neighbour’s plan is very unwise. All open flames
have the potential to produce carbon monoxide, especially when the oxygen supply is limited. In
the closed environment inside a home, the flame of a propane burner will likely deplete the
available oxygen, leading to low oxygen levels and the probable production of carbon monoxide.
A buildup of carbon monoxide could threaten the lives of the occupants of the neighbour’s house
if he burns propane indoors using the grill.
53. Answers may vary. Sample answer: My checklist would read: (1) Have any manufactured
wood products (such as particleboard, laminate flooring, and plywood) been used in recent
improvements to the home? Such products are made with glues or adhesives that can be sources
of methanal, a toxic gas that is an irritant and possible allergen. In high levels, methanal may
even cause cancer. Recently installed pressed wood should be checked to see if it is the type that
releases methanal. (2) Are all fireplaces and wood stoves in good working order with
unobstructed chimneys? If fireplace and stove flues do not work properly or if chimneys are
constricted or blocked, combustion products might vent back into the living space. The
combustion products will likely contain carbon monoxide and possibly methanal, both of which
pose significant health risks. (3) Has the furnace been recently inspected and adjusted? An
improperly adjusted furnace will not burn fuel completely. Incomplete combustion can produce
carbon monoxide that might enter the living space of the home and cause very adverse health
=
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Chapter 11: The Gas State and Gas Laws
11-8
effects. (4) Are carbon monoxide detectors installed and, if so, are they in good working order?
A buildup of carbon monoxide in a living space is very dangerous because of its great toxicity
and because it is hard to detect without an electronic detector. Carbon monoxide detectors can
alert the occupants of a home of danger so they can exit. Without detectors, occupants could
easily die in their sleep if carbon monoxide builds up in a home. (5) Is the basement (if there is
one) well ventilated? If not, has the air in that space been tested for radon gas? Are there any
cracks in the basement’s exterior walls or the floor? Radon is a gas and can seep into the
basements of homes, especially if there are cracks in the foundation wall. Due to its radioactivity,
radon is a significant health risk. Proper ventilation can vent radon inside a living space to the
outdoors. If ventilation is not installed in a basement, at the least, a radon test should be done to
make sure radon levels are safe. (6) Are all baseboards and basement walls free of mould? Have
any parts of the house been recently flooded? Moulds can release spores. Upon inhaling these
spores, sensitive individuals can develop respiratory difficulties, skin rashes, or other allergic
reactions. Moulds are more likely to grow when moisture is present.
54. (a) Water, H2O, is the only product of the reaction that is not a pollutant. Carbon dioxide,
CO2, is a normal product of a properly tuned car engine, but it is a greenhouse gas and, as such,
is a type of pollutant. It contributes to global warming. Carbon monoxide, CO, is very toxic
because it binds strongly to hemoglobin in the blood and prevents normal oxygen transport in the
body, which can be fatal. Carbon, C, is a type of particulate matter that can readily enter the body
by inhalation and produce adverse health effects. Nitrogen monoxide, NO, is a reactive gas that
can produce nitrogen dioxide in the air, which in turn can form nitric acid and ozone. Nitric acid
is one of the causes of acid precipitation and ozone can cause health problems. The hydrocarbon
product, C2H4, is an example of a VOC. VOCs can react with nitrogen monoxide in polluted air
to form ozone and, in some cases, cause heath problems such as eye irritation, depression of the
central nervous system, and cancer.
(b) An idling car engine runs at a lower temperature than when the car is being driven at
highway speeds. The lower temperature results in incomplete combustion of fuel, which in turn
results in the production of more carbon monoxide and other pollutants.
55. A bad headache and very red cheeks and lips are symptoms of carbon monoxide poisoning.
To support such a diagnosis, the doctor would want to know if the patient has been in a situation
that could result in exposure to carbon monoxide. An idling car produces carbon monoxide that
can enter the passenger compartment, especially if the exhaust pipe leaks. Even bad traffic jams
can result in an accumulation of carbon monoxide that can cause adverse health effects.
Similarly, an old or poorly adjusted furnace in a home can produce carbon monoxide.
Accordingly, the doctor would ask the patient if she had been in any of these situations. If she
answers yes, he would strongly suspect that carbon monoxide is the problem.
56. Answers may vary. Sample answer: Our bodies contain blood and other liquids that resist
being compressed. Our blood pressure also creates an outward force in our tissues that can push
back against the external pressure of the atmosphere. Our chest cavities do not collapse because
our lungs are filled with atmospheric air that pushes outward and our skeletal structure is also
relatively strong. Similarly, our sinuses are filled with air that pushes outward. Overall, forces in
our body pushing from the inside out offset the pressure of the atmosphere pressing inward, so
we are not crushed.
57. The pressure of the air outside the plane is much less than the pressure of the air inside the
plane. When the hole in the cabin wall is created, air inside the plane will rush outward, forming
a wind that blows the passengers out through the hole.
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Chapter 11: The Gas State and Gas Laws
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58. (a) Use the fact that 1 atm = 101.3 kPa to convert the pressure into units of kPa:
101.3 kPa
60.0 atm !
= 6080 kPa
1 atm
The pressure in the pipeline is 6080 kPa.
(b) Standard pressure is 101.325 kPa:
pressure in pipeline
6080 kPa
=
= 60.0
standard pressure
101.325 kPa
The pressure in the pipeline is 60 times greater than standard atmospheric pressure.
(c) Answers may vary. Sample answer: A tiny crack in the pipe would allow gas to leak out.
Methane, one of the major components of natural gas, is a VOC and also a greenhouse gas.
There is also the danger of explosion if a spark or flame were to come in contact with the gas.
59. The air exerts less pressure and is less dense at higher elevations. The air in the baseball
stadium in Denver, therefore, is relatively thin (less dense) due to Denver’s high elevation
compared to other baseball stadiums. Thinner air offers less resistance to a baseball flying
through it, meaning that batters’ hits will travel faster and farther, thus giving hitters a greater
chance of hitting home runs.
60. At normal temperatures, dry ice rapidly sublimates into carbon dioxide gas. If dry ice is
placed in a sealed plastic pop bottle, the gas it produces on sublimation will generate pressure. If
enough dry ice is placed in the bottle, the pressure generated can become great enough to
explode the bottle and send dangerous fragments of plastic flying in all directions. The sound of
the explosion might also cause hearing damage.
61. Answers may vary. Sample answer:
As the gas is cooled at relatively high temperatures, the entities will slow down, but will still
remain relatively far apart. Their separation and the speed of their movement will keep the weak
attractive forces and individual volumes from affecting their behaviour. Under these conditions,
the gas will steadily shrink in volume as the temperature is lowered. At relatively low
temperatures, however, the small volume the gas has reached will begin to crowd the entities
together. Their low speeds and small separation will allow the attractive forces to hold them
together and they will form a liquid. As the liquid is cooled further it will freeze into a solid.
Both the liquid and solid will shrink slightly in volume as the temperature falls. This is because
slower entities do not range as far in their motions and, therefore, take up less space. However,
the rates at which the liquid and solid contract will be much less than the rate at which the gas
contracts because the entities in the liquid and solid are so tightly packed together.
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Chapter 11: The Gas State and Gas Laws
11-10
62. (a) As the balloon moves upward, the air around it will become less dense and will exert less
pressure. If the pressure on the outside of the balloon diminishes, the pressure of the helium gas
inside the balloon will now be great enough to expand the balloon further. Consequently, the
balloon will expand as it rises. However, at high altitudes the air is cooler than at ground level,
which would tend to make the balloon shrink. Without more data we cannot predict whether the
balloon will expand or shrink, overall.
(b) If the balloon expands too much as it rises, it might pop. However, if the balloon is strong
enough and does not burst, it will not rise indefinitely. When the density of the surrounding air
becomes equal to the average density of the balloon, it will no longer rise. As gas escapes from it
over time, it might even descend back to Earth’s surface.
63. Given: initial pressure, P1 = 13 800 kPa
initial temperature, t1 = 26 °C
final temperature, t2 = 1400 °C
The volume and amount of gas remain constant.
Required: final pressure, P2
Analysis: Use Gay-Lussac’s law.
P1 P2
=
T1 T2
Solution:
Step 1. Convert temperature values to kelvins.
T1 = t1 + 273
= 26 + 273
T1 = 299 K
T2 = t2 + 273
= 1400 + 273
T2 = 1673 K
Step 2. Rearrange the equation to isolate the unknown variable.
PT
P2 = 1 2
T1
Step 3. Solve the equation (including units).
13 800 kPa ! 1673 K
P2 =
299 K
P2 = 77 200 kPa
Statement: The calculations show that when the tank is heated to 1400 °C by the fire, its internal
pressure rises to 77 200 kPa. This is greater than the pressure that the valve on the tank can
safely withstand (68 900 kPa), so the valve is in danger of failing.
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Chapter 11: The Gas State and Gas Laws
11-11
64. When one of the masses is removed, there will be less weight pressing down on the gas,
which will make the pressure on the gas less. For a short time, the entities of the gas will exert a
greater upward pressure than the downward pressure caused by the piston. As a result, the piston
will begin moving upward. As it moves, the entities of the gas will have more space in which to
move. This will cause them to collide less frequently with the interior walls of the cylinder and,
therefore, they will exert less and less pressure as the piston continues to move up. However,
when the piston has moved upward to the point that the pressure of the gas sample has become
equal to the downward pressure from the piston, the piston will no longer move.
65. Given: initial pressure, P1 = 101 kPa
initial volume, V1 = 1.20 × 103 L
final volume, V2 = 40.0 L
The amount of gas and the temperature remain constant.
Required: final pressure, 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. Substitute given values (including units) into the equations and solve.
101 kPa ! 1.20 ! 103 L
P2 =
40.0 L
P2 = 3030 kPa
Statement: The pressure in the filled tank will be 3030 kPa, which is significantly less than the
upper limit of pressure the tank can withstand, which is 5050 kPa. Therefore the tank will be able
to hold the hydrogen safely.
Evaluation
66. Answers may vary. Sample answer: To get people to take the time and spend the money to
plant more trees, I would first have to convince them that there is a true need to reduce carbon
dioxide levels in the atmosphere. I would need data about the increase of atmospheric carbon
dioxide prior to and after the Industrial Revolution to illustrate that carbon dioxide levels are
clearly rising and that the probable source is human activities. I would also need historical
temperature data to show that global warming is real and strongly correlated to the increase of
atmospheric carbon dioxide. Next, I would need data to show that vegetation can sequester large
amounts of carbon by converting atmospheric carbon dioxide into plant tissue, so planting trees
can make a real difference in reducing global warming. People would most likely not want to
take the time to research which plants would be best for sequestering carbon, flourishing in our
local environment, and providing landscaping beauty without costing too much, so I would have
to do this research and decide on the best species to recommend.
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Chapter 11: The Gas State and Gas Laws
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67. Answers may vary. Sample answer: When solvents evaporate out of fresh paint on a surface,
they enter the air as vapours. If the solvents are organic, they can contribute to the VOC content
of the atmosphere. VOCs can affect health both directly and indirectly by the formation of
ozone. To select the most environmentally friendly paint, I would research the solvents in each
paint product to see which paint would produce the fewest VOCs upon drying. I would also
consider the solvents required for cleanup. If petroleum-based solvents must be used to clean
brushes and paint pans, this could add to the VOCs emitted from a painting project. The best
paint would clean up with water or some other solvent that does not produce VOCs. At the same
time, I would consider product effectiveness. The best paint would be durable and not require
frequent reapplication. A low-VOC paint that requires many repeated applications over time
might produce more VOCs overall than a higher VOC paint that lasts for a very long time after a
single application. In addition, frequent reapplication would be expensive. It would also be a
poor choice environmentally due to the pollution emitted by producing and transporting the
replacement paint. Finally, I would have to examine the suitability of a paint product for a
particular job. A house paint must be waterproof, for example, whereas a furniture stain must be
compatible with polishes.
68. (a) Answers may vary. Sample answer: The main advantage of an airtight house is that
energy savings will result from keeping cold air out. If cold air does not enter the house, the air
will stay warmer inside and the furnace will not have to run as often. Thus the furnace will
consume less fuel, which will reduce heating costs. It will also mean that fewer combustion
products will be emitted into the air, which will help to improve air quality. The problem with an
airtight house is that indoor air quality might deteriorate. Any volatile chemicals, such as
methanal off-gassed by new furniture or pressed wood products, will build up in the relatively
unventilated interior space. Similarly, combustion gases from fireplaces or furnaces, such as
carbon monoxide or methanal, may build up in the indoor air. The health hazards posed by these
noxious gases could be significant.
(b) Answers may vary. Sample answer: I would recommend a compromise strategy to my
relative. He should seal large cracks between window frames and walls and also use draft guards
where there is a wide gap between the bottom of a door and the floor. However, he should not
strive to seal all possible entry points for cold air so perfectly that no exchange of air with the
outdoors is possible. For example, I would not recommend that he use plastic covers on unused
electrical receptacles.
69. (a) The mass of air displaced by the fully inflated balloon will be the product of the density
of air and the volume of air displaced:
m = dV
kg
= 1.23 3 ! 2650 m 3
m
m = 3260 kg
The fully inflated balloon will displace 3260 kg of air.
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Chapter 11: The Gas State and Gas Laws
11-13
(b) Given: initial Celsius temperature, t1 = 90 °C
final Celsius temperature, t2 = 20 °C
volume, V1 = 2650 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
= 90 + 273
T1 = 363
T2 = t2 + 273
= 20 + 273
T2 = 293
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.
2650 m 3 ! 293 K
V2 =
363 K
V2 = 2140 m 3
Statement: The new volume of air in the balloon at 20 °C will be 2140 m3.
(c) The mass of the air is the product of its density and its volume:
m = dV
= 1.23
kg
3
! 2140 m 3
m
m = 2630 kg
The mass of air in the balloon is 2630 kg.
(d) Answers may vary. Sample answer: The engineer’s design will not work properly.
The mass of the balloon plus the air it contains is:
760 kg + 2630 kg = 3390 kg
This is greater than the mass of air displaced (3250 kg), so the balloon will not lift off the
ground. For the balloon to float, the mass of air displaced must be more than the overall mass of
the balloon and its hot air. The engineer could use a larger envelope for her design. For example,
suppose the volume of the envelope is increased to 3500 m3. The mass that the balloon could
now lift can be determined as follows:
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Chapter 11: The Gas State and Gas Laws
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First find the mass of air displaced by the larger balloon:
m = dV
= 1.23
kg
3
! 3500 m 3
m
m = 4310 kg
The mass of air displaced by the larger balloon is 4310 kg.
Next, find the volume of that air, adjusted to 20 °C, rather than 90 °C:
Given: initial Celsius temperature, t1 = 90 °C
final Celsius temperature, t2 = 20 °C
volume, V1 = 3500 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
= 90 + 273
T1 = 363
T2 = t2 + 273
= 20 + 273
T2 = 293
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.
3500 m 3 ! 293 K
V2 =
363 K
V2 = 2830 m 3
Statement: The new volume of air in the balloon at 20 °C will be 2830 m3.
Next, determine the mass of this air in the balloon’s envelope.
m = dV
kg
= 1.23
! 2830 m 3
3
m
m = 3480 kg
The mass of air in the balloon’s envelope is 3480 kg.
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Chapter 11: The Gas State and Gas Laws
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The total mass of the balloon and the air in its envelope is now:
760 kg + 3480 kg = 4240 kg
The mass of air displaced by this larger balloon is 4310 kg. The larger balloon will therefore be
able to lift off the ground, so it is a better design.
70. Answers may vary. Sample answer: On the positive side, the student has properly followed
the GRASS problem-solving format, thus nicely organizing the work. The student has included
the original formula in symbols and has shown the proper algebraic rearrangement to find the
solution. Also, the student has shown the information from the problem substituted into the
rearranged formula. The clear statement at the end answers the question.
On the negative side, the student has arrived at an incorrect answer and also failed to
show proper unit labels when the information from the problem was substituted into the
rearranged formula. Specifically, the student failed to convert the given temperatures into
absolute temperatures. The temperatures should have been found in this manner:
T1 = t1 + 273
= 10 + 273
T1 = 283 K
T2 = t2 + 273
= 50 + 273
T2 = 323 K
So the correct solution is:
P2 =
P1T2
T1
0.800 kPa ! 323 K
283 K
P2 = 0.91 kPa
P2 =
The correct statement is: When the temperature of a fixed amount of gas at 0.800 kPa and
constant volume is raised from 10 ºC to 50 ºC, the new pressure is 0.91 kPa.
Therefore, the increase in pressure is much less than the student calculated.
Reflect on Your Learning
71. Answers may vary. Sample answer: I have always been curious about how a person can lie
on a bed of nails and not get puncture wounds in his skin. I now know that the weight of the
person’s body is distributed over a relatively large surface area that results from a very large
number of nail tips. Since pressure is force divided by area, the large total area makes the
pressure on each nail tip relatively small. As a result, each nail does not exert enough pressure to
break the skin.
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Chapter 11: The Gas State and Gas Laws
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72. Answers may vary. Sample answer: When I drive a car, I will not let it idle for more than 30
seconds after I start it, when I am waiting in traffic, or when I am parked and waiting to give
someone a ride. This will reduce the quantities of carbon monoxide, VOCs, nitrogen oxides, and
particulate matter my car emits into the air. At home, I will mow the lawn with a push mower
instead of a gasoline-powered mower. This, too, will reduce the amount of combustion pollutants
that my actions add to the air. Finally, I will check to see if my home has had a recent radon test
or has adequate ventilation to prevent a buildup of radon in the interior space. Radon is a
radioactive gas and poses a serious health danger, so it is very important to take steps to keep
radon from building up inside a house.
73. Answers may vary. Sample answer: In the past, I never gave much thought to the indoor air
pollution in a home. I had heard about carbon monoxide posing health risks, but I now
understand better what appliances can produce carbon monoxide, such as improperly adjusted
furnace burners, outdoor heaters brought indoors, and obstructed fireplace vents. In the future, I
will be much more alert to potential carbon monoxide sources and routinely check for dangerous
conditions. We also need to check that our carbon monoxide detectors are working properly.
I also learned that off-gassing from new furnishings and wood floors can cause methanal vapours
to build up indoors. If I can detect an odour coming from a new product in my home, I will make
sure to increase the ventilation until the smell goes away.
74. Answers may vary. Sample answer: I always wondered what it meant when the meteorologist
on television talked about barometric pressure, saying something like, “The barometric pressure
is 101 kPa and falling.” I now understand that kilopascals express the force the air exerts per unit
surface area. I also know that the average pressure of the atmosphere at sea level is 101.325 kPa.
When the weather reporter gives a barometric pressure above 101 kPa, I will know that a highpressure system lies over our region. Similarly, if the reported pressure is less than 101 kPa, I
will know that a low-pressure system lies over our region.
Research
75. Argon is typically obtained as a desirable by-product of the fractional distillation of air. The
air is purified of contaminants, carbon dioxide, and water, then passed through progressive
cryogenic cooling cycles to obtain liquid air. The liquid air is fractionally distilled in towers to
obtain purified nitrogen, oxygen, and argon as cryogenic liquids. Among other things, argon is
used to fill incandescent light bulbs (to keep the filament from deteriorating), produce a blue
colour in neon-type lamps, and provide an inert atmosphere for various processes.
76. Lake Nyos is unusual, unfortunately, in that carbon dioxide from volcanic activity collects in
the lower portion of the lake water. Most lakes “turn over” periodically, thus mixing the contents
and bringing deep materials near the surface. Lake Nyos, however, does not tend to turn over, so
carbon dioxide accumulates in great quantities at the bottom of the lake. In August of 1986,
something caused the lake to turn over suddenly, resulting in the release of a very large quantity
of carbon dioxide that, due to a density greater than the density of air, flowed like a river of gas
over the surrounding area. The carbon dioxide asphyxiated virtually all living things within a 25
km radius of the lake. The human death toll was 1700. In the aftermath of the disaster, degassing
equipment was installed in an effort to prevent a future recurrence. Students’ reports should
relate the disaster to the properties of gases as explained by the kinetic molecular theory.
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Chapter 11: The Gas State and Gas Laws
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77. Luis and Walter Alvarez discovered unusually high levels of iridium in a sedimentary layer
found all over the world, the K-T boundary, which was laid down about 65 million years ago.
Iridium is rare in Earth’s crust, so the likely source of the iridium was meteoric material.
Dinosaur fossils and many other fossils are common in earlier geologic layers, but are absent
above the K-T layer. Putting two and two together, the father-and-son team theorized that a large
asteroid had struck the Earth, sending up a plume of dust and aerosols that blocked sunlight to
the point that photosynthetic organisms died out. This destroyed the base of the food chain,
leading to the extinction of the dinosaurs and many other species. Students’ reports should
expand on these ideas and also explain how other evidence gave even more credence to the
impact theory.
78. An overwhelming majority of sources say that modern cars, which have electronic ignition,
need no more than 30 seconds of engine warm-up before being driven. In fact, driving a car
slowly is the best way to warm it up because the engine’s operating temperature will rise more
rapidly and moving parts not in the engine will also warm up. Moreover, an idling engine
operates at a lower temperature, which results in less efficient combustion. The inefficient
combustion produces more pollutants and can cause deposits that actually harm the engine. In
extremely cold weather, longer idle times might be needed before a car can be driven, but the
best solution in such cases is to use an engine block heater. Students’ reports should expand on
all of these aspects of the issue.
79. The uncertainty principle of quantum mechanics states that it is impossible to determine the
exact position and momentum of any particle simultaneously. If all entities stopped moving
entirely at absolute zero, each entity’s momentum would be exactly zero. At the same time, each
entity’s position would be known exactly because the entities would not move out of their
existing positions. This would be a violation of the uncertainty principle. Consequently, all
entities retain a small, residual vibrational motion at absolute zero. The energy of these
vibrations cannot be extracted as thermal energy, so absolute zero is best defined as the
temperature at which no more thermal energy can be extracted from a system. The laws of
thermodynamics state that as absolute zero is approached by repeated cooling cycles, the work
needed to extract further thermal energy increases. It turns out that removing the very last bit of
thermal energy from a system would require an infinite amount of work, which, in practice, is
impossible to provide. Thus absolute zero is unattainable by experiment.
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Chapter 11: The Gas State and Gas Laws
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