Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
Section 12.1: Avogadro’s Law and Molar Volume, pages 195–197
1. b
2. Table 1 Reactions and Ratios
Reaction
Ratio
2 H2 + O2 → 2 H2O
2:1:2
H2 + Cl2 → 2 HCl
1:1:2
N2 + 3 H2 → 2 NH3
1:3:2
3. The coefficients in the chemical equations in Table 1 also correspond to the mole ratio and volume ratio of
products and reactants.
4. If the pressure and temperature of a gas are kept constant, the amount of gas can be increased by increasing
the volume that the gas occupies using Avogadro’s law, which states that the volume of a gas and the number
of moles of gas present are directly proportional when temperature and pressure remain constant. To increase
the amount of gas in a balloon while keeping temperature and pressure constant, the size of the balloon must
be increased according to Avogadro’s law.
5. Figure 2
6. Given: n1 = 2.00 g; V1 = 2.00 L; V2 = 2.70 L
Required: mass of He added
Analysis: Use Avogadro’s law,
V1
n1
=
V2
n2
Solution:
Step 1: Convert mass of He to amount of He, n1.
n1 = 2.00 g !
1 mol
4.00 g
n1 = 0.500 mol
Step2: Rearrange Avogadro’s law to isolate n2, final amount of He.
n2 =
n1V2
V1
Step 3: Substitute values into the equation and solve.
n2 =
0.500 mol ! 2.70 L
2.00 L
n2 = 0.675 mol
Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12-1
Step 4: Calculate the number of moles of He added.
0.625 mol – 0.500 mol = 0.175 mol
Step 5: Calculate the mass of He added, m2.
mHe = 0.175 mol !
4.00 g
1 mol
mHe = 0.700 g
Statement: A mass of 0.700 g helium was added to the cylinder if the volume changed from 2.00 L to 2.70 L.
7. False. If one mole of hydrogen gas and one mole of oxygen gas are under the same pressure and at the same
temperature, they will occupy the same volume.
8. a
Section 12.2: Ideal Gases and the Ideal Gas Law, pages 198–200
1. True
2. Figure 1
3. When ideal gases collide with the walls of a container, these collisions are said to be elastic. The statement
means that there is no loss of kinetic energy.
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4. Given: V = 5.0 L; T = 25 °C; n = 10.0 g; R = 8.314 kPa•L•mol–1•K–1
Required: P
Analysis: PV = nRT
Solution:
Step 1: Convert temperature values to kelvins.
T = t + 273
= 25 + 273
T = 298
Step 2: Convert mass of gas to amount of gas.
n = 10.0 g !
1 mol
16.05 g
n = 0.62305 mol (two extra digits carried)
Step 3: Rearrange the ideal gas law equation to find the pressure of gas.
P=
nRT
V
Step 4: Substitute values into the equation and solve.
–1
P=
0.623 05 mol ! 8.314 kPa i L i mol i K –1 ! 298 K
5.0 L
P = 310 kPa
Statement: The pressure of the gas is 310 kPa.
5. The molar volume is a measurement of an ideal gas. No gas behaves in an ideal manner. As such, when the
molar volumes of actual gases are measured, they will be somewhat different than the molar volume of an
ideal gas.
6. c
7. False. Real gases do condense into solids and liquids.
8. Answers will vary. Sample answer: When real gases are compressed, their molecules move closer together
and the attraction between the molecules increases. When gases are cooled, the molecules move more slowly
and will be more attracted to each other. As the attractive forces between gas molecules increases, the
molecules will stick together prompting a state change, and the gas will become a liquid.
9. Answers may vary. Sample answer:
Figure 2
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Section 12.3: Chemistry Journal: The Science of Cold, page 201
1. Van der Waals discovered that in order to liquefy a gas, the pressure of the gas had to be lowered.
2. (a) Answers may vary. Sample answer: When the temperature and pressure of a gas are lowered, the
molecules in the gas are closer together and move much slower. Attractions between molecules become more
significant as the forces between molecules increases.
(b) As the gas molecules are pushed closer together, and as the kinetic energy decreases and the forces of
attraction become stronger, the molecules will be compressed. This compression will result in a change of state
or the liquefying of the gas.
3. (a) In this state of matter, the behaviour of atoms changed. Rather than acting like individual atoms, the
atoms exist in a state described as a “super atom.”
(b) Using magnetic fields and laser cooling, scientists were able to trap gas molecules at a specific speed. At
this speed, the molecules had a certain frequency. By setting the laser at the frequency of the atoms, the gas
molecules could be slowed to a super atom.
4. (a) Answers may vary. Sample answer: Scientific research can require a number of different steps
undertaken over a number of years. Various scientists must often contribute to the larger understanding of a
subject.
(b) Answers may vary. Sample answer: Collaboration in research is needed to help science move forward.
Without the sharing of ideas, science and society as a whole will not be able to benefit from discovery.
Section 12.4: Gas Mixtures and the Law of Partial Pressures, pages 202–203
1. b
2. False. When water vapour is added to dry air, the pressure of the air will increase.
3. Partial pressure is the pressure that a gas in a mixture would exert if it were the only gas present in the same
volume at the same temperature.
4. 1. The gases in a mixture must not react.
2. The units of the gases must be consistent.
5. Kinetic molecular theory posits that pressure is caused by the collisions of the gas with the walls of the
container. If gases are mixed, the collisions will increase, totalling the pressure of each gas individually.
6. Given: Ptotal = 150 kPa; P1 = 100 kPa; P2 = 34 kPa
Required: partial pressure of helium, P3
Analysis: Apply Dalton’s law of partial pressures.
Ptotal = P1 + P2 + P3
Solution:
Step 1: Rearrange the equation to isolate the required variable.
P3 = Ptotal ! P1 ! P2
Step 2: Substitute values into the equation and solve.
P3 = 150 kPa ! 100 kPa ! 34 kPa
P3 = 16 kPa
Statement: The pressure of the helium is 16 kPa.
7. (a) When the temperature of the water is increased, the number of water vapour molecules in the air will
increase.
(b) The vapour pressure will increase as the temperature increases. As the number of water molecules in the air
increases, this will cause more collisions with the walls of the container. Using the kinetic molecular theory,
this will result in an increase in the pressure.
8. Given: Ptotal = 65 kPa; PH O = 2.20 kPa (vapour pressure at 19 °C)
2
Required: Pbutane
Analysis: Ptotal = Pbutane + PH O
2
Solution:
Step 1: Rearrange the equation to isolate the required variable.
Pbutane = Ptotal ! PH 2O
Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12-4
Step 2: Substitute values into the equation and solve.
Pbutane = 65 kPa ! 2.20 kPa
Pbutane = 62.8 kPa
Statement: The partial pressure of the butane in the container is 62.8 kPa.
Section 12.5: Reactions of Gases and Gas Stoichiometry, pages 204–205
1. a
2. False. When gases react, the volumes of the reactants and the products react in whole number ratios if the
temperature and pressure remain constant.
3. b
4. (a) 2 C8H18(g) + 25 O2(g) → 16 CO2 (g) + 18 H2O(g)
(b) Given: VC H = 120 mL
8
Required: VO
18
2
Analysis: Use the balanced equation to determine the volume ratio of oxygen to octane.
Solution:
Step 1: Write the balanced chemical equation, listing the given and required quantities.
2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
120 mL
Voxygen
Step 2: Convert volume of octane to volume of oxygen required.
VO2 = VC8 H18 !
25 mol O2
2 mol C8 H18
= 120 mL !
25 mol O2
2 mol C8 H18
VO2 = 1500 mL
Statement: The volume of oxygen required to burn 120 mL of octane would be 1500 mL.
5. False. If the pressure and temperature change during a reaction, the ideal gas law should be used to calculate
the volume.
6. Given: PH = 88 kPa ; VH = 200.0 L ; tH = 25 °C ; tO = 0 °C ; PO = 101.325 kPa
2
Required: VO
2
2
2
2
2
Analysis: PV = nRT
Solution
Step 1: Write the balanced equation.
2 H2(g) + O2(g) → 2 H2O(l)
Step 2: Convert temperature values to kelvins.
TH 2 = t + 273
= 25 + 273
TH 2 = 298
TO2 = tO2 + 273
TO2
= 0 + 273
= 273
Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12-5
Step 3: Convert volume of H2, VH , into an amount, nH , using the ideal gas law equation, PV = nRT.
2
2
PH 2VH 2 = nH 2 RTH 2
nH 2 =
=
PH 2VH 2
RTH 2
(8.314
(88
)(200.0 L )
kPa i L i mol i K )(298 K )
kPa
–1
–1
nH 2 = 7.104 mol (two extra digits carried)
Step 4: Determine the amount of O2 required for the amount of H2.
nO2 = nH 2 !
1 mol O2
2 mol H 2
= 7.104 mol !
1 mol O2
2 mol H 2
nO2 = 3.552 mol (two extra digits carried)
Step 5: Determine the volume of O2, VO , using the ideal gas law equation, PV = nRT.
2
PO2VO2 = nO2 RTO2
VO2 =
=
nO2 RTO2
PO2
(3.552
mol
)(8.314
)
kPa i L i mol –1 i K –1 (273 K )
101.325 kPa
VO2 = 80 mol
Statement: 80 L of oxygen will be required to react with 200.0 L of hydrogen at 25 °C and 88 kPa to make
water at STP.
Section 12.6: Explore Applications of Gases: Burning Snowballs, page 206
1. False. Gas hydrates form under low temperatures and high pressures.
2. Answers may vary. Sample answer:
Table 1 Benefits and Drawbacks of Gas Hydrates
Benefits
Drawbacks
• Gas hydrates may contain more organic carbon
• Use of gas hydrates is limited.
than all other known fossil fuels.
• They are hard to access as they are found on the ocean
• The amount of carbon in these hydrates would
floor.
provide a viable source of energy that would be
• Gas hydrate extraction may have a negative impact on
more potent than current fossil fuel resources.
aquatic ecosystems.
3. Answers may vary: Sample answer: In order to make burning gas hydrates more environmentally friendly,
scrubbers or carbon sequestration could be used to capture the greenhouse gases that result from the burning of
these resources. These tools would reduce the greenhouse gas emissions produced from burning gas hydrates.
4. Answers may vary: Sample answer: The report on a specific fossil fuel alternative should contain
information about both the benefits and drawbacks of the alternative. This information should be supported
with scientific data from credible sources. The report should also contain a comparison of the fossil fuel
alternative to existing resources currently being used.
Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12-6
Chapter 12 Questions, pages 208–209
1. b
2. d
3. (a) False. The volume of an ideal gas is insignificant (near zero) when compared to the volume of the
container.
(b) False. All liquids exist with a certain amount of vapour above their surface as a result of evaporation.
4. There are no attractive or repulsive forces between gas molecules. This statement refers to an ideal gas.
5. The standard molar volume of 22.4 L assumes that the gas is an ideal gas. Real gases do not behave like
ideal gases and therefore have a molar volume that is either slightly above or below the standard of 22.4 L.
6. When the temperature is decreased, the molecules in a gas begin to move more slowly. Kinetic energy
decreases and molecules experience forces of attraction when they collide.
7.
8. Given: PH = 2.9 kPa ; PHe = 7.2 kPa
2
Required: Ptotal
Analysis: Ptotal = PH + PHe
2
Solution: Substitute values into the equation and solve.
Ptotal = 2.9 kPa + 7.2 kPa
Ptotal =10.1 kPa
Statement: The total pressure is 1.01 kPa.
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Unit 5 Questions, pages 210–211
1. d
2. a
3. (a) False. The pH scale is used to identify acids and bases.
(b) true
4. If the temperature and pressure of the gas are held constant in a balloon, the only way to increase the volume
of the gas is to add more gas to the balloon.
5. As the temperature of the gas decreases, the molecules slow down and will not collide with the container,
reducing their kinetic energy. As the gas is cooled, the attractions between the molecules become stronger,
causing the molecules to interact with one another. If the temperature is reduced enough, the gas will undergo
a state change and become a liquid.
6. Table 1: Common Types of Carbon Sequestration
Standard
Ambient
temperature
0 °C
25 °C
pressure
101.325 kPa
100 kPa
7.
8. (a) When gas is bubbled into the water, the gas displaces the water.
(b) When all of the water is pushed out of the container, the collected gas and water vapour will remain in the
container.
(c) Using Dalton’s law of partial pressures. Subtract the partial pressure of the water vapour at 25 °C, 3.17 kPa,
from the total pressure of the container, 34 kPa. The pressure of the gas is: 34 kPa – 3.17 kPa = 30.83 kPa.
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