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Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc. No part of this work may be reproduced without the written permission of the publisher. 90 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 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 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): P 1 V 1 T 1 = P 2 V 2 T 2 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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Page 1: Chapter 14 · PDF file  Chapter 14 ... instructor did not assign Assignment 9–B with Chapter 9, Assignment 14–C ... Chapter 14 concludes with a section on

Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.No part of this work may be reproduced without the written permission of the publisher.

90

Free Study Guide forCracolice • Peters

Introductory Chemistry: An Active Learning ApproachSecond Edition

www.brookscole.com/chemistry

Chapter 14Combined 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 ispresented for the purpose of reviewing these items.

1) The Volume–Temperature (Charles') Law states that at constant pressure, thevolume of a fixed quantity of a gas is directly proportional to the absolutetemperature, V µ T. (Review Section 4.4, if necessary.)

2) The Volume–Pressure (Boyle's) Law states that at constant temperature, thevolume 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 theCombined Gas Laws (Review Section 4.6, if necessary):

P1V1

T1

=

P2V2

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

Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.No part of this work may be reproduced without the written permission of the publisher.

91

Chapter 14–Assignment B: Molar Volume, Density, and Molar Mass

Three ratios, molar volume, density, and molar mass, are important for understandingrelationships among the four measurable properties of a gas. We examine these three ratiosin 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 giventemperature and pressure.

2) One mole of any gas at standard temperature and pressure, 0°C and 1 atm, occupies22.4 liters.

3) Density, by definition, is mass per volume. Gas densities are usually measured ingrams per liter.

4) Molar mass, by definition, is the mass of one mole of a substance. Molar mass isconstant, 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, doQuestions, Exercises, and Problems 1–13.

Chapter 14–Assignment C: Gas Stoichiometry at Standard Temperature andPressure (STP)

FLEXTEXT OPTIONChapter 9–Assignment B has the same title as this Assignment and the same Goal. If yourinstructor did not assign Assignment 9–B with Chapter 9, Assignment 14–C should bestudied now. If Assignment 9–B was included in your study of Chapter 9, you may omitthis Assignment, although you might find a brief review helpful. Your ability to satisfy Goal7 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

Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.No part of this work may be reproduced without the written permission of the publisher.

92

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 usedto convert between the volume of a gas at STP and the number of particles ofthat gas, counted in moles. 22.4 L/mol can be used only for ideal gases atSTP. If your stoichiometry skills are rusty, review Section 9.1.

Answer Questions, Exercises, and Problems 14–15. Check your answers with thoseat the end of the chapter.

Workbook If your instructor recommends the Active Learning Workbook, doQuestions, Exercises, and Problems 14–15.

Chapter 14–Assignment D: Gas Stoichiometry at non-STP ConditionsMolar Volume Method

FLEXTEXT OPTIONSections 14.6 and 14.7 offer alternative ways to solve gas stoichiometry problems at giventemperatures and pressures. Assignment D is also presented in alternative ways, each keyedto one of the sections. If Section 14.6 is assigned, use this option for Assignment D anddisregard 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 themolar 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, doQuestions, Exercises, and Problems 16–19.

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Chapter 14 Combined Gas Law Applications

Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.No part of this work may be reproduced without the written permission of the publisher.

93

Chapter 14–Assignment D: Gas Stoichiometry at non-STP ConditionsCombined Gas Equation Method

FLEXTEXT OPTIONSections 14.6 and 14.7 offer alternative ways to solve gas stoichiometry problems at giventemperatures and pressures. Assignment D is also presented in alternative ways, each keyedto one of the sections. If Section 14.7 is assigned, use this option for Assignment D anddisregard 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 molarvolume 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 oldconcepts.

The ideas to review are:

1) The combined gas laws equation is

P1V1

T1

=

P2V2

T2

If you know the volume of a gas at a certain temperature and pressure, the volume ata new temperature and pressure is given by algebraically rearranging the combinedgas laws equation to

V2 =

P1V1T2

P2T1

2) The coefficients in a chemical equation express the mole relationships between thedifferent substances in the reaction. The coefficients may be used in a dimensionalanalysis conversion from moles of one substance to moles of another.

3) The molar volume of all ideal gases at standard temperature and pressure (STP) is22.4 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, ifnecessary, before starting this assignment. You may find a brief review ofSection 4.6 particularly helpful if it has been some time since you studiedChapter 4.

Answer Questions, Exercises, and Problems 16–19. Check your answers with thoseat the end of the chapter.

Workbook If your instructor recommends the Active Learning Workbook, doQuestions, 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 andproduced 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 convertbetween 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

T1

=

P2V2

T2

and the stoichiometry path. If you have mastered each of

these skills, you simply combine them to solve volume-volume gas stoichiometry problems.

Answer Questions, Exercises, and Problems 20–23. Check your answers with those at the end of the chapter.

Workbook If your instructor recommends the Active Learning Workbook, doQuestions, 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 atSTP is 22.4 L/mol. Memorize this relationship. The molar volume of a gas at any othertemperature and pressure can be found by a modification of the combined gas equationfrom Chapter 4:

V2/mol = V1/mol ¥

P1

P2

¥

T2

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 theirdefinitions:

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 isidentical to that used in mass stoichiometry. The stoichiometry pattern is applied in bothcases. The only difference is the quantity unit being converted to moles, or vice versa. In onecase it is grams, and in the other, it is gas volume at specified temperature and pressure.

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Chapter 14 Combined Gas Law Applications

Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.No part of this work may be reproduced without the written permission of the publisher.

95

Learning Procedures

Review your lecture and textbook notes.

the Chapter in Review and the Key Terms and Concepts, and read the StudyHints and Pitfalls to Avoid.

Answer Concept-Linking Exercises 1–2. Check your answers with those at the endof the chapter.

Questions, Exercises, and Problems 24–27. Include Questions 28–29 ifassigned by your instructor. Check your answers with those at the end of thechapter.

Workbook If your instructor recommends the Active Learning Workbook, doQuestions, Exercises, and Problems 24–26. Include Questions 27–30 ifassigned by your instructor.

Take the chapter summary test that follows. Check your answers with those at theend 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 densityis 0.175 g/L?

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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 massof 0.369 gram.

7) Carbon dioxide can be removed from a closed-container breathing apparatus byreaction with potassium superoxide:

4 KO2(s) + 2 CO2(g) Æ 2 K2CO3(s) + 3 O2(g)

Calculate the mass of potassium superoxide needed to remove an STP volume of10.0 L of carbon dioxide.

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Chapter 14 Combined Gas Law Applications

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97

8) Calculate the mass (in grams) of zinc that must react to produce 148 mL ofhydrogen 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 burncompletely 4.18 L of butane measured at 1.34 atm and 38°C? The gas-phasereaction is

2 C4H1 0(g) + 13 O2(g) Æ 8 CO2(g) + 10 H2O(g)

Answers to Chapter 14 Sample Test

1) Molar Volume Temperature PressureInitial Value (1) 22.4 L/mol 273 K 1.00 atmFinal Value (2) V2 L/mol –17 + 273 = 256 K 1.03 atm

EQUATION: V2/mol = V1/mol ¥

P1

P2

¥

T2

T1

= 22.4 L/mol ¥

1.00 atm1.03 atm

¥

256 K273 K

=

20.4 L/mol

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Study Guide for Introductory Chemistry: An Active Learning Approach

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98

2) GIVEN: MV = 23.2 L/mol; 1.25 L WANTED: mol

PER/PATH: L

23.2 L/molæ Æ æ æ æ æ mol

1.25 L ¥

1 mol23.2 L

= 0.0539 mol

3) GIVEN: 10.0 g He; 0.175 g He/L He WANTED: volume (assume L)

PER/PATH: g He

0.175 g He/L Heæ Æ æ æ æ æ æ æ L He

10.0 g He ¥

1 L He0.175 g He

= 57.1 L He

4) GIVEN: 22.4 L NH3/mol; 17.03 g NH3/mol WANTED: Density (g/L)

EQUATION: D ≡

mV

=

17.03 g NH3/mol22.4 L NH3/mol

= 0.760 g/L

5) GIVEN: 16.04 g/mol; 0.645 g/L WANTED: Molar volume (L/mol)

EQUATION: MV =

MMD

=

16.04 gmol

¥

1 L0.645 g

= 24.9 L/mol

6) Molar Volume Temperature PressureInitial Value (1) 22.4 L/mol 273 K 760 torrFinal Value (2) V2 L/mol 22 + 273 = 295 K 819 torr

EQUATION: V2/mol = V1/mol ¥

P1

P2

¥

T2

T1

= 22.4 L/mol ¥

760 torr819 torr

¥

295 K273 K

=

22.5 L/mol

EQUATION: MM = D ¥ MV = 0.369!g0.460!L ¥

22.5!Lmol = 18.0 g/mol

7) GIVEN: 10.0 L CO2 at STP WANTED: mass KO2 (assume g)

PER/PATH: L CO2

22.4 L CO2/mol CO2æ Æ æ æ æ æ æ æ æ æ mol CO2

4 mol KO2/2 mol CO2æ Æ æ æ æ æ æ æ æ æ æ mol KO2

71.1 g KO2/mol KO2æ Æ æ æ æ æ æ æ æ æ g KO2

10.0 L CO2 ¥

1 mol CO2

22.4 L CO2

¥

4 mol KO2

2 mol CO2

¥

71.1 g KO2

mol KO2

= 63.5 g KO2

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Chapter 14 Combined Gas Law Applications

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99

8) Molar volume methodGIVEN: 767 torr; 24°C (297 K) WANTED: MV (L/mol)

Molar Volume Temperature PressureInitial Value (1) 22.4 L/mol 273 K 760 torrFinal Value (2) V2 L/mol 24 + 273 = 297 K 767 torr

EQUATION: V2/mol = V1/mol ¥

P1

P2

¥

T2

T1

= 22.4 L/mol ¥

760 torr767 torr

¥

297 K273 K

=

24.1 L/mol

PER/PATH: mL H2

1000 mL H2/L H2æ Æ æ æ æ æ æ æ æ L H2

24.1 L H2/mol H2æ Æ æ æ æ æ æ æ æ mol H2

1 mol Zn/1 mol H2æ Æ æ æ æ æ æ æ æ mol Zn

65.39 g Zn/mol Znæ Æ æ æ æ æ æ æ æ g Zn

148 mL H2 ¥

1 L H2

1000 mL H2

¥

1 mol H2

24.1 L H2

¥

1 mol Zn1 mol H2

¥

65.39 g Znmol Zn

= 0.402 g Zn

Combined gas equation methodVolume Temperature Pressure

Initial Value (1) 148 mL 24 + 273 = 297 K 767 torrFinal Value (2) V2 273 K 760 torr

EQUATION: V2 = V1 ¥

P1

P2

¥

T2

T1

= 148 mL ¥

767 torr760 torr

¥

273 K297 K

= 137 mL

GIVEN: 137 mL H2 WANTED: g Zn

PER/PATH: mL H2

1000 mL H2/L H2æ Æ æ æ æ æ æ æ æ L H2

24.1 L H2/mol H2æ Æ æ æ æ æ æ æ æ mol H2

1 mol Zn/1 mol H2æ Æ æ æ æ æ æ æ æ mol Zn

65.39 g Zn/mol Znæ Æ æ æ æ æ æ æ æ g Zn

137 mL H2 ¥

1 L H2

1000 mL H2

¥

1 mol H2

22.4 L H2

¥

1 mol Zn1 mol H2

¥

65.39 g Znmol Zn

= 0.400 g Zn

9) Volume Temperature PressureInitial Value (1) 4.18 L 38 + 273 = 311 K 1.34 atmFinal Value (2) V2 18 + 273 = 291 K 0.891 atm

V2 = V1 ¥

P1

P2

¥

T2

T1

= 4.18 L ¥

1.34 atm0.891 atm

¥

291 K311 K

= 5.88 L

GIVEN: 5.88 L C4H1 0 WANTED: volume O2 (assume L)

PER/PATH: L C4H1 0

13 L O2/2 L C4H10æ Æ æ æ æ æ æ æ æ L O2

5.88 L C4H1 0 ¥

13 L O2

2 L C4H10

= 38.2 L O2


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