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AR STANDARDS for ch. 10-11
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Chapter 11
Gases
Table of Contents
Section 4 Diffusion and Effusion
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Chapter 11
Section 4 Diffusion and Effusion
Objectives
• Describe the process of diffusion.
• State Graham’s law of effusion.
• State the relationship between the average molecular
velocities of two gases and their molar masses.
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Opener #20 - Wednesday, December 12, 2012
Pick up calculator & study guide questions. Have out notes
packet.
Be sure you have turned in computer simulation lab AND
yesterday’s pressure probe Boyle’s Law Lab.
1. A sample of a gas occupies 250. mL at 1.00 atm of pressure.
If the pressure increases to 2.00 atm while the
temperature stays the same,what is the new volume?
2. An engineer pumps 5.00 mol of carbon monoxide gas
into a cylinder that has a capacity of 20.0 L. What is the
pressure in kPa of CO inside the cylinder at 25°C?
3. Ammonia can react with oxygen to produce nitrogen and
water according to the following equation.
4NH3(g) + 3O2(g) --> 2N2(g) + 6H2O(l)
If 1.78 L of O2 reacts, what volume of nitrogen will be
produced? Assume that temperature and pressure remain
constant.
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Assignments
WED DEC 12
CW: Notes 11.4 - discussed together first.
CW: ch. 10.1-11 Study Guide Questions
HW: Final Exam Study Guide due Wed. Dec. 19th
Be sure you have turned in your 2 labs with ALL questions
answered from simulation lab and pressure probe lab.
Make sure all of your notes are completed including
problems.
TEST ch.10.1-11 is FRIDAY.
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Chapter 11
Section 4 Diffusion and Effusion
Diffusion and Effusion
• The constant motion of gas molecules causes them to
spread out to fill any container they are in.
• The gradual mixing of two or more gases due to their
spontaneous, random motion is known as diffusion.
• Effusion is the process whereby the molecules of a
gas confined in a container randomly pass through a
tiny opening in the container.
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Chapter 11
Visual Concepts
Comparing Diffusion and Effusion - skip
Click below to watch the Visual Concept.
http://my.hrw.com/sh/hc6_003036809x/studen
Visual Concept
t/ch11/sec04/vc00/hc611_04_v00fs.htm
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Section 4 Diffusion and Effusion
Chapter 11
Graham’s Law of Effusion
• Rates of effusion and diffusion depend on the relative
velocities of gas molecules. The velocity of a gas
varies inversely with the square root of its molar
mass.
• Recall that the average kinetic energy of the molecules in any
gas depends only the temperature and equals
.
1
2
mv
2
• For two different gases, A and B, at the same temperature,
the following relationship is true.
1
2
M Av A 
2
1
2
M Bv B
2
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Chapter 11
Section 4 Diffusion and Effusion
Graham’s Law of Effusion
• From the equation relating the kinetic energy of two
different gases at the same conditions, one can derive
an equation relating the rates of effuses of two gases
with their molecular mass:
rate of effusion of A
rate of effusion of B

MB
MA
• This equation is known as Graham’s law of effusion,
which states that the rates of effusion
of gases at the same temperature and pressure
are inversely proportional to the square roots of
their molar masses.
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Chapter 11
Visual Concepts
Graham’s Law of Effusion - show
Click below to watch the Visual Concept.
http://my.hrw.com/sh/hc6_003036809x/stud
Visual Concept
ent/ch11/sec04/vc01/hc611_04_v01fs.htm
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Chapter 11
Visual Concepts
Equation for Graham’s Law of Effusion - skip
Click below to watch the Visual Concept.
http://my.hrw.com/sh/hc6_003036809
Visual Concept
x/student/ch11/sec04/vc02/hc611_04_
v02fs.htm
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Chapter 11
Section 4 Diffusion and Effusion
Graham’s Law
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Chapter 11
Section 4 Diffusion and Effusion
Graham’s Law of Effusion, continued
Sample Problem J
Compare the rates of effusion of hydrogen and oxygen
at the same temperature and pressure.
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Chapter 11
Section 4 Diffusion and Effusion
Graham’s Law of Effusion, continued
Sample Problem J Solution
Given: identities of two gases, H2 and O2
Unknown: relative rates of effusion
Solution: The ratio of the rates of effusion of two gases
at the same temperature and pressure can be found
from Graham’s law.
rate of effusion of A
rate of effusion of B

MB
MA
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Section 4 Diffusion and Effusion
Chapter 11
Graham’s Law of Effusion, continued
Sample Problem J Solution, continued
Substitute the given values into the equation:
rate of effusion of A
rate of effusion of B

MB
MA

32.00 g/m ol
2.02 g/m ol

32.00 g/m ol
2.02 g/m ol
 3.98
Hydrogen effuses 3.98 times faster than oxygen.
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Practice Problems pg. 388 #1-3
Complete practice problems #1-3 pg. 388. Show work.
v
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ANSWERS
Practice Problems pg. 388 #1-3
Complete practice problems #1-3 pg. 388. Show work.
v
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Note to self - additional stoichiometry & ideal gas
problem with acetylene gas C2H2 etc.
See also worksheets, etc.
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SECTION PROBLEMS
Explore these interactive demonstrations to learn how to
use equations to solve science problems.
Section 4: Diffusion and Effusion
Stoichiometry of gases
Volume–Volume Calculations
Stoichiometry of Gases
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of 25
Rates of Effusion
© Copyright Pearson Prentice Hall
End Show
HIGHLIGHTS OF 11.4
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Online Self-Check Quiz
Complete the online Quiz and record answers.
Ask if you have any questions about your
answers.
click here for online Quiz 11.4
(7 questions)
You must be in the “Play mode” for the
slideshow for hyperlink to work.
Slide
of 25
© Copyright Pearson Prentice Hall
End Show
VIDEOS FOR ADDITIONAL INSTRUCTION
Additional Videos for
•Section 11.4: Diffusion and Effusion
•Graham's Law
Slide
of 28
© Copyright Pearson Prentice Hall
End Show
SCI LINKS FOR CHAPTER
Additional Student SCI LINKS for CHAPTER 11
The NSTA-sponsored SciLinks Web site contains links to accurate and upto-date science information on the Internet. Just click on the button below
to go to the SciLinks site at www.scilinks.org and log in. Then, type in the
SciLinks code for the topic you want to research. The following is a list of
the SciLinks codes for this chapter.
Chapter 11: Gases
Topic: Gases
SciLinks code: HC60638
Topic: Gas Laws
SciLinks code: HC60637
Topic: Torricelli
SciLinks code: HC61539
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of 28
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End Show
End of Chapter 11.4 Show
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