4. ! To be able to understand
the basic principles of
Graham's Law.
To analyse To define To identify
the
difference diffusion the
of diffusion and applications
from effusion of Graham's
effusion Law
checklist
(Objectives)
5. Gases (Short)
Kinetic Molecular
Theory
Diffusion
Effusion
Thomas Graham
Graham’s Law
Let’s start the discussion
6. Before we start
We would like to point out that we strongly
disagree with this* and you should too.
*Only from this point and
until this report ends :D
Before we start
7. Topic: Just Gas
» Moles of gases are always in
gases rapid, random, straight line
endlessly bumping against each other and
motion,
hitting the walls of the container. This is
discussed in the Kinetic
Molecular
Theory .
The entire space in which molecules of a
gas move represents the volume of the
gas.
Topic: Just Gas
8. Topic: Kinetic molecular Theory
Kinetic Molecular Theory
Gases are very far from each
other.
There are no interaction
(attraction) between the
molecules.
They collide with each other
and with the walls of the
container in a perfectly
elasticKinetic molecular Theory
Topic: manner.
9. Topic: Diffusion
»The rate at which two gases mix;
Diffusion movement of gases from high
concentration to low
concentration; mixing of gases due
to molecular motion
Topic: Diffusion
10. Topic: Effusion
» movement of gases
Effusion from high pressure to
low pressure
Topic: Effusion
11. Topic: Diffusion and Effusion
Diffusion % Effusion
» Effusion is
» the process
Diffusion that occurs
is the when a gas is
process permitted
of to escape its
slowly container
mixing through a
two small
gases opening.
together. Topic: Diffusion and Effusion
12. Topic: Diffusion and Effusion
Diffusion % Effusion
» Diffusion » Effusion
movement movement
of gases of gases
from high from high
concentrati
pressure to
on to low
low
concentrati
on pressure
Topic: Diffusion and Effusion
13. Topic: Thomas Graham
Thomas graham
» 1805-1869
» Scottish
chemist
He formulated the law
of diffusion. Also
called Graham’s Law in
1846.
Topic: Thomas Graham
14. Topic: Graham’s Law
Graham’s law of diffusion
"The velocity of a gas at a certain
temperature is inversely
proportional to the square root of
itsTopic: Graham’smass."
molecular Law
15. Topic: Graham’s Law
Graham’s law of diffusion
Graham's law is a gas law which
relates the rate of diffusion or
effusion of a gas to its molar
mass.
This means light gasses effuse or
diffuse quickly and heavier gases
effuse or diffuse slowly.
Topic: Graham’s Law
16. Topic: Graham’s Law
Graham’s law of diffusion
This means light gasses effuse or diffuse
quickly and heavier gases effuse or
diffuse slowly.
HydrogenHelium
Helium is lighter
than Hydrogen
Topic: Graham’s Law
17. Topic: Graham’s Law
Graham’s law of diffusion
This gas law is derived directly from the kinetic
molecular theory. Consider two gas molecules:
a B
Each gas molecule has its corresponding mass, m, and
volume, v. If they are at the same absolute
temperature, then their kinetic energies are equal.
Topic: Graham’s Law
18. Topic: Graham’s Law
Graham’s law of diffusion
Each gas molecule has its corresponding mass, m, and
volume, v. If they are at the same absolute
temperature, then their kinetic energies are equal.
Topic: Graham’s Law
19. Topic: Graham’s Law
Graham’s law of diffusion
The mass can be related to the molecular weight of
the gas
"The velocity of a gas at a certain
temperature is inversely
proportional to the square root of
its molecular mass."
Topic: Graham’s Law
20. Topic: Graham’s Law
Graham’s law of diffusion
The mass can be related to the molecular weight of
the gas
Gases with high molecular weight tend to diffuse
more slowly than gases with low molecular weight
Higher molecular weight : Higher velocity:
Numerator Numerator
Lower molecular weight : Lower velocity:
Denominator Denominator
Topic: Graham’s Law
22. Sample problem
How much faster does O2 escape through a
porous container than SO2?
(Use the periodic table to get the molecular weight of gas)
Given: MWSO2 VO2
MWO2 = 32.0 g/mol MWO2 VSO2
MWSO2 = 64.0 g/mol
VO2 64 g/mol
Solution: =
VSO2 32 g/mol
Substituting these
values in the VO2 8
=
Equation for Graham’s VSO2 5.66
law of diffusion.
Sample Problem
23. Sample problem
How much faster does O2 escape through a
porous container than SO2?
(Use the periodic table to get the molecular weight of gas)
Solution:
VO2 8 VO2
= = 1.41 VO2 = 1.41 VSO2
VSO2 5.66 VSO2
Final answer: VO2 = 1.41 VSO2
This means that O2 diffuses 1.41 times as
fast as SO2.
Sample Problem
25. APPLICATIONS
1. When someone sautés meat with garlic and
onion, the volatile substances responsible for
the aroma of the spices vaporize and mix with
the gases in the air.
APPLICATIONS
26. APPLICATIONS
2. A balloon filled with air and a boy allows the gas
to escape, the gas molecules will diffuse among the
molecules of air and mixture of gases will occupy the
whole room. Thus, we say that a gas spreads
throughout the space available to it.
APPLICATIONS
28. Quiz
1. How much faster does O2 escape
through a porous medium than CO2
given the same conditions?
2. Determine how much faster He would
escape through a porous medium than N2
under the same conditions.
quiz
31. Answers
1. How much faster does O2 escape
through a porous medium than CO2 given
the same conditions?
Given: Final answer:
VO2 44 g/mol
MWO2 = 32.0 g/mol = VO2 = = V
1.17 CO2
MWCO2 = 44.0 g/mol VCO2 32 g/mol
VO2 6.63 (6.6332)
Solution: =
VCO2 5.66 (5.6568)
MWCO2 VO2 VO2
= 1.17
MWO2 VCO2 VCO2
Answers
32. Answers
2. Determine how much faster He would
escape through a porous medium than N2
under the same conditions.
Given: Final answer:
VHe 28 g/mol
MWN2 = 28.0 g/mol = VCO
VHe = 2.65 VN2 2
MWCO2 = 4.0 g/mol VN2 4 g/mol
VHe 5.29
Solution: =
VN2 2
MWN2 VHe VHe
= 2.65 VN2
MWHe VN 2 VN2
Answers
