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 4^{th} 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
TheorySection 11.10 Answers a
B._____(00) Discussion Real vs Ideal Gas EquationSect 11.10 Answer bc
C._____(05) Standard
Conditions/Molar VolumeSect 9.5 Answer bc
D._____(10) Gas Laws/VocabularySections 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) VolumeVolume Stoichiometry ProblemSection 10.6 Answers fg
G._____(10) MassVolume Stoichiometry ProblemSection 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:
http://www.fccj.us/chm2045/KotzMovies/chapter12/index.html
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
http://www.fccj.us/chm1025/AssignmentOutline/45Kotz6eChapter4.htm
To begin your study of Module 6, please
read Chapter 11.
Part A: Kinetic Molecular TheorySection 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*^{[1]}. The distance
between the molecules is veryvery great compared to the size of the
molecules themselves, and the total volume of the molecules is only a veryvery
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 EquationSect 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 rewritten to the Real Gas Equation. This leads to the following
discussion questions:
(a) In the Real Gas
Equation: (P + an^{2}/V^{2}) (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
6^{th} 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
elasticthat 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 VolumeSect 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 ^{o}C and K temperatures:
_760__mmHg
or _760__torre= __1___atm = _29.9 in_
= _14.7 psi_= 101 kPa
__0_^{o}C
= _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 mole_{CO2 }=__22.4__L _{CO2@STP
}1 mole_{H2 }=__22.4___L _{H2@STP}
_{ }1 mole_{N2 }=__22.4___L _{N2@STP
}1 mole_{O2 }=__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 temperaturesnot 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 
m^{3} 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/VocabularySections 11.311.10
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 LawSection 11.5, GayLussac’s
Law 11.6; Combined
Gas Law 11.7; Vapor pressure Concept 11.8;
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.
V_{1}P_{1}=V_{2}P_{2}
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 1720 (#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.
V_{1} = V_{2}
T_{1} T_{2}
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:
GayLussac’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.
P_{1} = P_{2}
T_{1} T_{2 }_{}
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 2932 at the end of the chapter on page 318 for additional Part E type
problems.
The total pressure of a gaseous mixture is
equal to the sum of the individual pressure of each gas.
P_{total} = P_{1} + P_{2} + P_{3} + …
In Section 11.8 the Vapor pressure concept
leads to
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
Here are more examples of work
Combined
Gas Law Equation (write only the equation): Section
11.7 p 3043
V_{1}P_{1 }= V_{2}P_{2}
T_{1} T_{2}
Study example 11.6 pages 305306 for a sample Part E
Problem.
Ideal
Gas Equation (write only the equation): Section
11.10 p 299300
PV = nRT
Part E: Gas Law Problems Sections 11.4119 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
1520 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 2126
p 318
Problem #3 is a GayLussac’s Law calculation. Example 11.5 on page 303 is a
GayLussac’s application. Work Gas Lussac Questions #27#32 on page 318.
Problem #4 is a
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 #6366 on page 319.
On the next page is a
more complicated Ideal Gas Calculation application.
Part F: VolumeVolume Stoichiometry Problem 
Section 10.6 Answers
VolumeVolume 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
volumevolume ratio (similar to the Mole Ratio). Work Problems #37#48 on page 285
are additional exercises to work.
Part G: MassVolume Stoichiometry Problem
Section 10.5 Answers
MassVolume 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 2689 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
nonSTP Conditions: