Module 6 Gases & Their Properties Assignment Outline (Chapter 11)

The test outline for Module 6 of Exam #4 covers all of Chapter 11 for the Corwin 4th edition text. Below is a Part by Part test outline with links to sample tests and answers plus text reference sections to study for that objective:

Module Six: The Gaseous State  (Corwin Chapter 11)

A1._____(12) Properties of Solids (13.4), Liquids(13.1), and Gases 13.1 Answers

A._____(10) Kinetic Molecular Theory-Section 11.10  Answers a

B._____(00) Discussion Real vs Ideal Gas Equation-Sect 11.10 Answer bc

C._____(05) Standard Conditions/Molar Volume-Sect 9.5 Answer bc

D._____(10) Gas Laws/Vocabulary-Sections 11.4, 11.5, 11.6, 11.7, 11.8, 11.9 Answers

E._____(20) Gas Law Problems- Sections 11.4, 11.5, 11.6, 11.7, 11.8, 11.9 Answers

F. _____(10) Volume-Volume Stoichiometry Problem-Section 10.6 Answers fg

G._____(10) Mass-Volume Stoichiometry Problem-Section 10.5 Answers fg

M._____(25) Multiple Choice Application Chapter 11 (answers at bottom)
______(102) Total = ______%

Don't forget to view the short movies clips as you study chapter 11 which are provided by a 2045C text (Kotz Chapter 12) at:

Hardcopy of the lecture notes from Chapter 12 of a CHM 2045C textbook (Kotz 6e Chapter 12) (Similar to Chapters 11) produced Dr. Andrea Wallace of Coastal Georgia Community College and edited by your instructor are posted on line at:
http://www.fccj.us/chm1025/AssignmentOutline/45Kotz6eChapter4.htm

Part A: Kinetic Molecular Theory-Section 11.10

The gas properties and laws discussed in Chapter 11 are based on the Kinetic Molecular theory.  The CHM 1025C texts list five or six basic assumptions. You will write these assumptions

1. Gases are composed of molecules*.  The distance between the molecules is very-very great compared to the size of the molecules themselves, and the total volume of the molecules is only a very-very small fraction of the entire space occupied by the gas.  Therefore, considering volume, we are primary considering empty space.  (This assumption explains why gases are highly compressed and have very low densities.)
(Gases are made up of very tiny molecules. The volume of a gas is mainly empty space).

2. No attractive forces exist between molecules in a gas.  (This is what keeps a gas from spontaneously becoming a liquid.)
(Gas molecules have no attraction for one another.)

3. The molecules of a gas are in a state of constant, rapid motion, colliding with each other and with the walls of the container in a perfectly random manner.  (This assumption explains why different gases normally mix completely.  The collisions between molecules and the walls of the container account for the pressure exerted by the gas.)
(Gas molecules demonstrate rapid motion, move in straight lines, and travel in random directions.)

4. All of these molecular collisions are perfectly elastic. As a result, the system as a whole experiences no loss of kinetic energy, the energy derived from the motion of a particle.
(Gas molecules undergo perfect elastic collisions.)

5. The average kinetic energy per molecule of a gas is proportional to the absolute temperature, and the average kinetic energy per molecule is the same at a given temperature and pressure for all gases.
(The average kinetic energy of gas molecules is proportional to the Kelvin temperature, that is KE is approximately T.)

When we think of molecules of elemental gases, we usually think of the diatomic gases such as nitrogen, oxygen, hydrogen, etc. The Nobel gases exist as monoatomic gases such as Helium, Neon, etc.

If you take CHM 2045C, these assumptions are condensed as follows:

(a) Gases consist of particles (molecules or atoms), whose separation is much greater than the size of the particles themselves.

(b) The particles of a gas are in continual, random, and rapid motion. As they move, they collide with one another and with the walls of their container, but they do so without energy loss.

(c) The average kinetic energy of gas particles is proportional to the gas temperature. All gases, regardless of molar mass, have the same average kinetic energy at the same temperature.

Part B: Discussion Real vs Ideal Gas Equation-Sect 12.9

If you have an understanding of the Kinetic Molecular Theory above then when you read section 11.1 you apply the KMT to gases in non ideal behavior. At STP gases behave ideally. But under extreme conditions which cause overcrowding (what are these conditions?), the KMT breaks down such that the Ideal gas Equation: PV=nRT has to be re-written to the Real Gas Equation. This leads to the following discussion questions:

(a) In the Real Gas Equation:   (P + an2/V2) (V - nb) = n RT a pressure correction factor was added. Why? (What assumptions of the kinetic theory breakdown under extreme conditions of temperature and pressure?)

(b) Also a volume correction factor was subtracted. Why? (What assumptions of the KMT breakdown under extreme conditions?)

The Answers are not posted but if you consult a CHM 2045C text, like Kotz 6th edition:

In Section 12.9 on page 576 the answer to the first question is found in the third paragraph!

Another assumption of the kinetic molecular theory is that collisions between the molecules are elastic-that is, that the atoms or molecules of the gas never stick to one another by some type of intermolecular force.  This is not true at extreme conditions of overcrowding. When a molecule is about to strike the wall of its container, other molecules in the vicinity exert a slight attraction for the molecule and pull it away from the wall. As a result of the intermolecular forces, molecules strike the wall with less force than they would in the absence of intermolecular attractive forces. Therefore, in a real gas, the observed pressure is less than the predicted pressure by the ideal gas law and a pressure correction factor is added to account for this pressure loss.

Also a volume correction factor was subtracted. Why? (What assumptions of the KMT breakdown under extreme conditions?)

In Section 12.9 on page 576 the answer is in the second paragraph!

The kinetic molecular theory and the ideal gas law are concerned with the volume available to the molecules to move about, not the volume of the molecules themselves. It is clear the volume occupied by the gas molecules is NOT negligible at high pressures (or extreme low temperatures. The available volume is less than the volume of the container. The volume the molecules occupy must be subtracted from the volume of the container to obtain the volume of free space the molecule can move.

A good multiple choice question is: under what conditions does ideal gas behavior break down?

Part C: Standard Conditions/Molar Volume-Sect 11.3

In section 11.1 the properties of gases are discussed on page 291. This includes the introduction to the concept of Gas pressure. Here is a summary:

1.     Gases have indefinite shape

2.     Gases can expand

3.     Gases can compress

4.     Gases have low density

5.     Gases mix completely with other gases in the same container.

Atmospheric Pressure is discussed in Section 11.2. The pressure that a gas exerts depends on how often and how hard these molecules strike the walls of the container:

1.     If the molecules collide more often, the gas pressure increases.

2.     If the molecules collide with more energy, the pressure increases. Table 11.1 lists the units of gas pressure under standard conditions: State standard conditions (STP) in three units of pressure (the last is your choice) and oC and K temperatures: _760__mmHg or _760__torre=  __1___atm = _29.9 in_  = _14.7 psi_= 101 kPa

__0_oC   =    _273__K

You should know the value of the gram molar volume constant to three significant figures. On page 234 an ideal gas occupies 22.414 L at STP.  See Table 9.1. Therefore you would put 22.4 in any of the following blanks:

What are the values for the Molar Gas Volume Constant for the following gases:

1 moleCO2 =__22.4__L CO2@STP         1 moleH2 =__22.4___L H2@STP

1 moleN2 =__22.4___L N2@STP           1 moleO2 =__22.4___L O2@STP

Calculate the value of R in the Ideal Gas Equation at STP:

If you substitute the values of the Molar Gas Constant into the ideal gas equation (PV=nRT) you can calculate the value of the constant R:

PV  = nRT   (you must enter Kelvin temperatures-not Celsius)

(1 atm) (22.4 L) = (1 mole) R (273 K)

R = 0.08206 L atm/mol K

R can include energy units such as Joules or calories:

 Values for the gas constant R Units Value L atm/mol K 0.08206 cal/mol K 1.987 J/mol K 8.314 m3 Pa/mol K 8.314 L torr/mol K 62.36

We usually use the first value: 0.08206 L atm/mol K in the calculation in Module 6.

However, many times your pressure is not given in atmospheres and if you do not have the above table, you may have to make conversions of the units of pressure from one unit to another. Here are a few example problems: Part D: Gas Laws/Vocabulary-Sections 11.3-11.10

In section 11.3 Variables affecting gas pressure are best described by the figure 11.4: For Part D you simply write a statement of the gas laws covered in chapter 11. In section 12.4 Boyle's Law is defined, Charles Law-Section 11.5, Gay-Lussac’s Law 11.6;  Combined Gas Law 11.7; Vapor pressure Concept 11.8; Dalton’s law 11.9; and the Ideal Gas Law 11.11.

Here are the statements:

Boyle’s Law (In words) Section 11.4 p 296

The volume of a gas is inversely proportional to the pressure when the temperature remains constant.

V1P1=V2P2

Note the graphical relationship between Pressure and Volume: Look at example 11.3 page 298 for a typical Boyle’s Law Calculation for Part E. Below is another set of worked examples not in your book: Additional Boyle’s Law problems are at the end of the chapter 11, Page 318 Problems 17-20 (#15 and 16 are good multiple choice type questions.)

Charles Law (in words)     Section 11.5 p 300

The volume of a gas is directly proportional to the Kelvin temperature if the pressure remains constant.

V1  =  V2

T1      T2

Note the graphical relationship between Temperature and Volume: Why does the first graph not interest the origin, while the second one does?
(Send me an email with an answer.)

Look at Example 11.4 page 301 for a typical Part E Problem. Here is another set of worked examples not in your book: Gay-Lussac’s Law (in words)     Section 11.6 p 302

The pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.

P1  =  P2

T1      T2

Note the graphical relationship between Temperature and Pressure: It looks the same as Charles Law. Why?

Study Example 11.5 page 303 for a sample problem for Part E.  You should try Problems 29-32 at the end of the chapter on page 318 for additional Part E type problems.

Dalton’s Law of Partial Pressures (in words)  Section 11.9 p308-310

The total pressure of a gaseous mixture is equal to the sum of the individual pressure of each gas.

Ptotal =  P1  + P2  + P3 + …

In Section 11.8 the Vapor pressure concept leads to Dalton’s Law:

Define Vapor pressure: Vapor pressure is the pressure exerted by the gaseous vapor above a liquid (usually in a closed container) when the rates of evaporation and condensation are equal.  Look at Example 11.7 page 309 for a Dalton’s Law calculation. Also try problems 47-54 page 319 at the end of Chapter 11 for more sample of Part e type problem.

Here are more examples of work Dalton’s law application which are not in our book, which may be more of a 2045 problem or two:  Combined Gas Law Equation (write only the equation): Section 11.7 p 3043

V1P1   =   V2P2

T1           T2

Study example 11.6 pages 305-306 for a sample Part E Problem.

Ideal Gas Equation (write only the equation): Section 11.10 p 299-300

PV = nRT

Part E: Gas Law Problems- Sections 11.4-119 Answers

Part E asks you to do simple gas law calculations. Problem #1 is Boyle's law. In Section  11.4 Example 11.3 on page 2987 is a sample of the first problem. Work  Questions
15-20 p 318 for more Boyle’s law samples

Problem #2 is a Charles Law calculation. In Section 11.5 Example 11.4 p 301 is a Charles Law Application. Work Charles Law Questions 21-26

p 318

Problem #3 is a Gay-Lussac’s Law calculation. Example 11.5 on page 303 is a Gay-Lussac’s application. Work Gas Lussac Questions #27-#32 on page 318.

Problem #4 is a Dalton's law of Partial Pressure application. Section 11.9 covers Dalton's Law. Example 11.7 on page 309 is a sample of a Dalton's law application. Work Problems #47-#54 pages 319 for more samples of the forth gas law application problem in Part E. Be careful when the application includes a gas collected over water and apply the vapor pressure concept and Dalton’s Law.

Problem #5 is a combined gas law application calculation. Example 11.6 is a General Gas Law application in section 11.7 page 305.  Work exercises #33-#42 for end of the chapter exercises on page 319.

Problem #6 in Part E is an Ideal gas Law application explained in section 11.10 and 11.11. Example 11.9 on page 313 is an example. At the end of the chapter work exercises #63-66 on page 319.

On the next page is a more complicated Ideal Gas Calculation application. Part F: Volume-Volume Stoichiometry Problem -
Section 10.6

Volume-Volume stoichiometry applies directly Avogadro's Hypothesis. Go back to chapter 10 and read section 10.6. Example 10.8 in section 10.6 shows that in an all gas phase reaction (all reactants and products are gases) you can apply the coefficients that balance an equation as a direct volume-volume ratio (similar to the Mole Ratio). Work Problems #37-#48 on page 285 are additional exercises to work.  Part G: Mass-Volume Stoichiometry Problem-
Section 10.5

Mass-Volume Stoichiometric problems are covered in Section 10.5. Your road map is Figure 10.2 on page 281 to solve these problem types. Study Examples 10.6 and Example 10.7 on page 268-9 as examples of Part G problems.  At the end of the chapter work Problems #29-#36 on page 284. Road Map for a Volume Mass Problem, when the Temperature and pressure conditions are not STP: Revised Concept map for non-STP Conditions:  