Free Study Guide for
Cracolice • Peters
Introductory Chemistry: An Active Learning Approach
Second Edition
www.brookscole.com/chemistry
Chapter 14
Combined Gas Law Applications
Chapter 14–Assignment A: Gases Revisited
In Section 4.3, you learned that there are four measurable properties of a gas: pressure,
volume, temperature, and quantity. In Chapter 4, quantity was constant. In Chapter 14,
quantity will be allowed to vary.
The three big ideas given below are review items from Chapter 4. This assignment is
presented for the purpose of reviewing these items.
1)
The Volume–Temperature (Charles') Law states that at constant pressure, the
volume of a fixed quantity of a gas is directly proportional to the absolute
temperature, V µ T. (Review Section 4.4, if necessary.)
2)
The Volume–Pressure (Boyle's) Law states that at constant temperature, the
volume of a fixed quantity of a gas is inversely proportional to its pressure,
V µ 1/P. (Review Section 4.5, if necessary).
3)
The Volume–Temperature and Volume–Pressure Laws can be coupled as the
Combined Gas Laws (Review Section 4.6, if necessary):
P1V1 P2V2
=
T1
T2
Learning Procedures
Study
Section 14.1. †
Strategy
Review Chapter 4, if necessary.
Answer
No end-of-chapter questions are given in Chapter 14 for this assignment.
You may wish to practice with questions from Chapter 4, if necessary.
†
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Chapter 14
Combined Gas Law Applications
Chapter 14–Assignment B: Molar Volume, Density, and Molar Mass
Three ratios, molar volume, density, and molar mass, are important for understanding
relationships among the four measurable properties of a gas. We examine these three ratios
in this assignment. Look for these important ideas:
1)
The molar volume of a gas is the volume occupied by one mole of gas at a given
temperature and pressure.
2)
One mole of any gas at standard temperature and pressure, 0°C and 1 atm, occupies
22.4 liters.
3)
Density, by definition, is mass per volume. Gas densities are usually measured in
grams per liter.
4)
Molar mass, by definition, is the mass of one mole of a substance. Molar mass is
constant, no matter the temperature and pressure.
Learning Procedures
Study
Sections 14.2–14.4. Focus on Goals 1–6 as you study.
Strategy
The emphasis in this assignment is on solving numeric problems. Practice
by solving lots of them.
Answer
Questions, Exercises, and Problems 1–13. Check your answers with those
at the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 1–13.
Chapter 14–Assignment C:
Pressure (STP)
Gas Stoichiometry at Standard Temperature and
FLEXTEXT OPTION
Chapter 9–Assignment B has the same title as this Assignment and the same Goal. If your
instructor did not assign Assignment 9–B with Chapter 9, Assignment 14–C should be
studied now. If Assignment 9–B was included in your study of Chapter 9, you may omit
this Assignment, although you might find a brief review helpful. Your ability to satisfy Goal
7 should help you decide if a review is necessary.
The main new idea in this section is:
1)
The stoichiometry path can be expanded to include gases. The molar volume of a
gas at STP, 22.4 L/mol, provides a conversion between gas volume and moles of
that gas.
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Study Guide for Introductory Chemistry: An Active Learning Approach
Learning Procedures
Study
Section 14.5. Focus on Goal 7 as you study.
Strategy
22.4 L per mole is a dimensional analysis conversion factor that can be used
to convert between the volume of a gas at STP and the number of particles of
that gas, counted in moles. 22.4 L/mol can be used only for ideal gases at
STP. If your stoichiometry skills are rusty, review Section 9.1.
Answer
Questions, Exercises, and Problems 14–15. Check your answers with those
at the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 14–15.
Chapter 14–Assignment D: Gas Stoichiometry at non-STP Conditions
Molar Volume Method
FLEXTEXT OPTION
Sections 14.6 and 14.7 offer alternative ways to solve gas stoichiometry problems at given
temperatures and pressures. Assignment D is also presented in alternative ways, each keyed
to one of the sections. If Section 14.6 is assigned, use this option for Assignment D and
disregard the next option (Combined Gas Equation Method). If Section 14.7 is assigned,
disregard this option and use the next option for Assignment D.
The big idea in this section is:
1)
A gas stoichiometry problem at non-STP conditions can be solved by finding the
molar volume of the gas and then following the stoichiometry path.
Learning Procedures
Study
Section 14.6. Focus on Goal 8 as you study.
Strategy
You combine two skills in this assignment: finding the molar volume of a
gas and stoichiometry. If you have learned each of these skills, you simply
combine them in this section. If you have trouble, review Section 14.2 on
molar volume and/or Section 9.1 on stoichiometry.
Answer
Questions, Exercises, and Problems 16–19. Check your answers with
those at the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 16–19.
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Chapter 14
Combined Gas Law Applications
Chapter 14–Assignment D: Gas Stoichiometry at non-STP Conditions
Combined Gas Equation Method
FLEXTEXT OPTION
Sections 14.6 and 14.7 offer alternative ways to solve gas stoichiometry problems at given
temperatures and pressures. Assignment D is also presented in alternative ways, each keyed
to one of the sections. If Section 14.7 is assigned, use this option for Assignment D and
disregard the previous option (Molar Volume Method). If Section 14.6 is assigned,
disregard this option and use the previous option for Assignment D.
In this section, we use the combined gas equation from Chapter 4 along with the molar
volume of a gas at STP from Assignment B and the stoichiometry pattern from Chapter 9.
There are no new concepts here; this assignment just introduces a new combination of old
concepts.
The ideas to review are:
1)
The combined gas laws equation is
P1V1 P2V2
=
T1
T2
If you know the volume of a gas at a certain temperature and pressure, the volume at
a new temperature and pressure is given by algebraically rearranging the combined
gas laws equation to
†
†
V2 =
P1V1T2
P2 T1
2)
The coefficients in a chemical equation express the mole relationships between the
different substances in the reaction. The coefficients may be used in a dimensional
analysis conversion from moles†of one substance to moles of another.
3)
The molar volume of all ideal gases at standard temperature and pressure (STP) is
22.4 L/mol.
Learning Procedures
Study
Section 14.7. Focus on Goal 8 as you study.
Strategy
This assignment combines the three ideas listed above. Review them, if
necessary, before starting this assignment. You may find a brief review of
Section 4.6 particularly helpful if it has been some time since you studied
Chapter 4.
Answer
Questions, Exercises, and Problems 16–19. Check your answers with those
at the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 16–19.
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Chapter 14–Assignment E: Volume–Volume Gas Stoichiometry
Chapter 14 concludes with a section on converting between volumes of gases reacting and
produced in a chemical reaction. The main idea in this section is:
1)
The ratio of volumes of gases in a reaction is the same as the ratio of moles,
provided that the gas volumes are measured at the same temperature and pressure.
Thus the coefficients in a balanced chemical equation can be used to convert
between volumes, as long as the volumes are at the same temperature and pressure.
Learning Procedures
Study
Section 14.8. Focus on Goal 9 as you study.
Strategy
Two skills are combined in this assignment: using the equation
P1V1 P2V2
=
and the stoichiometry path. If you have mastered each of
T1
T2
these skills, you simply combine them to solve volume-volume gas
stoichiometry problems.
Answer
†
Workbook
†Questions, Exercises, and Problems 20–23. Check your answers with
those at the end of the chapter.
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 20–23.
Chapter 14–Assignment F: Summary and Review
The first important idea presented in Chapter 14 is that the molar volume of an ideal gas at
STP is 22.4 L/mol. Memorize this relationship. The molar volume of a gas at any other
temperature and pressure can be found by a modification of the combined gas equation
from Chapter 4:
V2 /mol = V1 /mol ¥
P1
T
¥ 2
P2
T1
We next introduced the three important ratios that have physical significance for gases:
density, molar mass, and molar volume. To use these ratios, you must know their
definitions:
†
†
D ≡ m/V
MV ≡ V/mol
MM ≡ mass/mol
Problem solving with these ratios is much easier if you include units!
As you solve stoichiometry problems involving gases, be sure to recognize that the pattern is
identical to that used in mass stoichiometry. The stoichiometry pattern is applied in both
cases. The only difference is the quantity unit being converted to moles, or vice versa. In one
case it is grams, and in the other, it is gas volume at specified temperature and pressure.
Chapter 14
Combined Gas Law Applications
Learning Procedures
Review
your lecture and textbook notes.
the Chapter in Review and the Key Terms and Concepts, and read the Study
Hints and Pitfalls to Avoid.
Answer
Concept-Linking Exercises 1–2. Check your answers with those at the end
of the chapter.
Questions, Exercises, and Problems 24–27. Include Questions 28–29 if
assigned by your instructor. Check your answers with those at the end of the
chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 24–26. Include Questions 27–30 if
assigned by your instructor.
Take
the chapter summary test that follows. Check your answers with those at the
end of this assignment.
Chapter 14 Sample Test
1)
What is the molar volume of fluorine gas at –17°C and 1.03 atm?
2)
The molar volume of hydrogen bromide gas at 14°C and 772 torr is 23.2 L/mol.
How many moles of gas are in a 1.25 L vessel at these conditions?
3)
What is the volume occupied by 10.0 g helium at a temperature at which its density
is 0.175 g/L?
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Study Guide for Introductory Chemistry: An Active Learning Approach
4)
What is the density (g/L) of ammonia at STP?
5)
What is the molar volume of methane, CH4 , when its density is 0.645 g/L?
6)
Find the molar mass of a gas if 0.460 L, measured at 819 torr and 22°C, has a mass
of 0.369 gram.
7)
Carbon dioxide can be removed from a closed-container breathing apparatus by
reaction with potassium superoxide:
4 KO2 (s) + 2 CO2 (g) Æ 2 K2 CO3 (s) + 3 O2 (g)
Calculate the mass of potassium superoxide needed to remove an STP volume of
10.0 L of carbon dioxide.
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Chapter 14
8)
Combined Gas Law Applications
Calculate the mass (in grams) of zinc that must react to produce 148 mL of
hydrogen gas at 767 torr and 24°C by the reaction
Zn(s) + 2 HCl(aq) Æ H2 (g) + ZnCl2 (aq)
9)
What volume of oxygen, measured at 0.891 atm and 18°C is needed to burn
completely 4.18 L of butane measured at 1.34 atm and 38°C? The gas-phase
reaction is
2 C4 H1 0(g) + 13 O2 (g) Æ 8 CO2 (g) + 10 H2 O(g)
Answers to Chapter 14 Sample Test
1)
Initial Value (1)
Final Value (2)
Molar Volume
22.4 L/mol
V2 L/mol
EQUATION: V2 /mol = V1 /mol ¥
Temperature
273 K
–17 + 273 = 256 K
Pressure
1.00 atm
1.03 atm
P1
T
1.00 atm 256 K
¥ 2 = 22.4 L/mol ¥
¥
=
P2
T1
1.03 atm 273 K
20.4 L/mol
97
†
†
†
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Study Guide for Introductory Chemistry: An Active Learning Approach
2)
GIVEN: MV = 23.2 L/mol; 1.25 L
WANTED: mol
L/mol
PER/PATH: L æ23.2
æ æ æ æÆ mol
1 mol
1.25 L ¥
= 0.0539 mol
23.2 L
†
GIVEN: 10.0 g He; 0.175 g He/L He
3)
WANTED: volume (assume L)
0.175 g He/L He
†PER/PATH: g He
æ æ æ æ æ æ æÆ L He
1 L He
10.0 g He ¥
= 57.1 L He
0.175 g He
†
GIVEN: 22.4 L NH3 /mol; 17.03 g NH3 /mol
4)
WANTED: Density (g/L)
m 17.03 g NH 3 /mol
†
EQUATION: D ≡
=
= 0.760 g/L
V
22.4 L NH 3 /mol
5)
GIVEN: 16.04 g/mol; 0.645 g/L
WANTED: Molar volume (L/mol)
† †
MM 16.04 g
1L
EQUATION: MV =
=
¥
= 24.9 L/mol
D
mol
0.645 g
6)
Molar Volume
22.4 L/mol
†
V2 L/mol
Initial Value
† (1) †
Final Value (2)
EQUATION: V2 /mol = V1 /mol ¥
Temperature
273 K
22 + 273 = 295 K
Pressure
760 torr
819 torr
P1
T
760 torr
295 K
¥ 2 = 22.4 L/mol ¥
¥
=
P2
T1
819 torr
273 K
22.5 L/mol
0.369!g
22.5!L
EQUATION: MM = †
D ¥ MV
† g/mol
† = 0.460!L ¥ †mol = 18.0
7)
GIVEN: 10.0 L CO2 at STP
WANTED: mass KO2 (assume g)
L CO /mol CO
PER/PATH: L CO2 æ22.4
æ æ æ æ2 æ æ ææ2 Æ mol CO2
4 mol KO /2 mol CO 2
71.1 g KO /mol KO
æ æ æ æ æ2 æ æ æ æ
æÆ mol KO2 æ æ æ æ æ2 æ æ ææ2 Æ g KO2
10.0 L CO†
2 ¥
†
†
1 mol CO 2
4 mol KO 2
71.1 g KO 2
¥
¥
= 63.5 g KO2
22.4 L CO 2
2 mol CO 2
mol KO 2
†
†
†
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Chapter 14
8)
Molar volume method
GIVEN: 767 torr; 24°C (297 K)
Molar Volume
Initial Value (1)
22.4 L/mol
Final Value (2)
V2 L/mol
EQUATION: V2 /mol = V1 /mol ¥
Combined Gas Law Applications
WANTED: MV (L/mol)
Temperature
Pressure
273 K
760 torr
24 + 273 = 297 K
767 torr
P1
T
760 torr
297 K
¥ 2 = 22.4 L/mol ¥
¥
=
P2
T1
767 torr
273 K
24.1 L/mol
mL H 2 /L H 2
24.1 L H 2 /mol H 2
PER/PATH: mL H2 æ1000
ææ
æ æ æ ææÆ L†H2 æ æ æ
†æ æ æ ææÆ
†
†
mol Zn/1 mol H 2
65.39 g Zn/mol Zn
mol H2 æ1æ
mol
Zn
æ æ æ æ ææÆ
æ æ æ æ æ æ ææÆ g Zn
1 L H2
1 mol H 2
1 mol Zn 65.39 g Zn
†
148 mL H2†¥
¥
¥
¥
= 0.402 g Zn
1000 mL H 2
24.1 L H 2 1 mol H 2
mol Zn
†
†
Combined gas equation method
Volume
Temperature
Pressure
† Value (1)
† 148 mL † 24 + 273†= 297 K
Initial
767 torr
Final Value (2)
V2
273 K
760 torr
EQUATION: V2 = V1 ¥
GIVEN: 137 mL H2
P1
T
767 torr
273 K
¥ 2 = 148 mL ¥
¥
= 137 mL
P2
T1
760 torr
297 K
WANTED: g Zn
†
1000 mL H†/L H 2
24.1 L H 2 /mol H 2
PER/PATH: †
mL H†
2 æ æ æ æ æ 2æ æ
æÆ L H2 æ æ æ æ æ
æ ææÆ
mol Zn/1 mol H 2
65.39 g Zn/mol Zn
mol H2 æ1æ
mol
Zn
æ æ æ æ ææÆ
æ æ æ æ æ æ ææÆ g Zn
9)
1 L H2
1 mol H 2
1 mol Zn 65.39 g Zn
†
137 mL H2†¥
¥
¥
¥
= 0.400 g Zn
1000 mL H 2
22.4 L H 2 1 mol H 2
mol Zn
†
†
Volume
Temperature
Pressure
Initial Value (1)
4.18 L
38 + 273 = 311 K
1.34 atm
Final
0.891 atm
†Value (2)
† V2
† 18 + 273†= 291 K
V2 = V1 ¥
P1
T
1.34 atm
291 K
¥ 2 = 4.18 L ¥
¥
= 5.88 L
P2
T1
0.891 atm 311 K
GIVEN: 5.88 L C4 H1 0
WANTED: volume O2 (assume L)
†
L O 2 /2 †
L C H10
P†ER/P†
ATH: L C4 H 1 0 æ13
æææ
æ æ 4ææ
Æ L O2
13 L O 2
5.88 L C4 H1 0 ¥
= 38.2 L O2
2 L C 4 H10
†
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†