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Graham's Law
Introduction
Gases diffuse through each other. The rate of diffusion is dependent on
total pressure, temperature, and molecular masses of the molecules.
Graham discovered that gases diffuse and effuse (through a small orifice)
at a rate that is inversely proportional to the square root of the mass. The
more massive molecules diffuse slower.
This can be derived from Kinetic Energy
KE = ½ mv2
At a constant temperature KEa = KEb
½ mava2= ½ mbvb2
multiply both sides by 2
mava2= mbvb2
Get all masses on right side, and all velocities on the left
va2 ma
----- = -------
vb2 mb
If you take the square root of both sides, you get graham’s law.
 Remember,
the mass in the kinetic energy equation is the mass of the individual
molecule. This is the molecular mass.
Each individual molecule diffuses by moving in a random, straight line
motion until it collides with another molecule. This can be a molecule of
the same gas or a molecule of the medium it is diffusing through. When a
collisions occur, the molecules bounce off in different directions and a
transfer of energy can also occur.
The result of all of these collisions and random motion, is the gas
diffuses through the medium. The closer together the molecules of the
medium, the slower the diffusion because there are more collisions.
Diffusion will take place fastest in a vacuum because there is no medium to keep
the molecules from moving in their original directions.
The gas is dispersed through the complete volume and not just at the outer
edge.
The distance, rate, and time refer to the outer edge of the diffusing
gas. The velocity is the average velocity of the gases. The faster
the average velocity of the molecules, the faster they diffuse.
d
v = ----- When this is substituted into Graham's Law, you get the
following equations.
t
This equation is used when the gases diffuse for the same amount of
time. |
This equation is used when the gases diffuse over the same distance. |
Graham's Law Hints
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1.
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You must use
the molecular mass instead of the actual mass of the chemical. If
the mass is given, ignore the mass.
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2.
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When solving
for molecular mass, square both sides of the equation, then solve for the
unknown Molecular Mass.
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| 3. |
Use the symbols
of the gases instead of A and B from the beginning.
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| 4. |
When working
with rate of diffusion, velocities of molecules, or distance; gas on top on
left side is on the bottom on the right side of the equation.
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| 5. |
When working
with time; gas on top on left side is on the top on the rights side of the
equation.
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| 6. |
If measuring
distance, time must be constant. If measuring time, distance must be
constant.
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| 7. |
Lightest gas
diffuses the fastest, has the highest velocities of molecules, travels the
farthest at a constant time, and takes the smallest time to travel a set
distance |
Table of Contents
Graham's Law Problems
| 1. |
How
much time does it take 3.67 grams of sulfur trioxide to diffuse the same
distance 1.23 grams of ammonia diffuses in 33.4 seconds ?
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| 2. |
What
is the molecular mass of a gas that diffuses 11.9 cm during the same time
helium diffuses 96.3 cm ?
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| 3. |
Chlorine
diffuses 5.67 cm/min. How fast does argon diffuses under the same
conditions ?
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| 4. |
How
far does 7.34 grams of oxygen diffuse in the same time 3.44 grams of
iodine diffuses
61.9 cm ?
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| 5. |
What
is the molecular mass of a gas that diffuses 4.89 cm/min under the
same conditions that ammonia diffuses 1.39 cm/min ?
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| 6. |
What
is the molecular mass of a gas that diffuses in 33.5 seconds the same
distance hexane diffuses in 28.4 sec ?
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Table of Contents
Graham's
Law Answers
| 1. |
How
much time does it take 3.67 grams of sulfur trioxide to diffuse the same
distance 1.23 grams of ammonia diffuses in 33.4 seconds ?
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|
tSO3
=
?
tNH3 = 33.4 s
MMSO3 = 80.1
g/mol MMNH3 = 17.0
g/mol
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Identify
the variables including the unknown.
If a mass is given, ignore the mass.
Find the Molecular Mass of any known chemical.
The units for the variable determine the
equation that will be used.
s is the unit for the variable for t. t is
directly related to square root of Molecular Mass. The same
chemical must be on the top of both sides.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. You will need a ( ) inside the square root. |
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| 2. |
What
is the molecular mass of a gas that diffuses 11.9 cm during the same time
helium diffuses 96.3 cm ?
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MMX
=
?
MMHe = 4.00 g/mol
dX = 11.9
cm
dHe = 96.3 cm
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Identify
the variables including the unknown.
Find the Molecular Mass of any known chemical.
An unknown chemical is labeled as X.
The units for the variable determine the
equation that will be used.
cm is the unit for the variable for d.
d is inversely related to square root of Molecular Mass. The chemical
on the top on one side, is on the bottom on the other side of the
equation.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Since the Molecular MassX is inside
the square root, square both sides of the equation.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. The ( ) do not have to be punched into the
calculator. They are necessary to show that the unit is squared in
addition to the number. |
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| 3. |
Chlorine
diffuses 5.67 cm/min. How fast does argon diffuses under the same
conditions ?
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rateCl2
= 5.67
cm/min
rateAr = ?
MMCl2 = 70.9
g/mol
MMAr = 39.1 g/mol
|
Identify
the variables including the unknown.
If a mass is given, ignore the mass.
Find the Molecular Mass of any known chemical.
The units for the variable determine the
equation that will be used.
cm/min is the unit for the variable for rate.
rate is inversely related to square root of Molecular Mass. The
chemical on the top of one side of the equation is on the bottom on the
other side.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. You will need a ( ) inside the square root.
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| 4. |
How
far does 7.34 grams of oxygen diffuse in the same time 3.44 grams of
iodine diffuses
61.9 cm ?
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dO2
=
?
dI2 = 61.9 cm
MMO2 = 32.0
g/mol MMI2
= 254 g/mol
|
Identify
the variables including the unknown.
If a mass is given, ignore the mass.
Find the Molecular Mass of any known chemical.
The units for the variable determine the
equation that will be used.
cm is the unit for the variable for distance.
distance is inversely related to square root of Molecular Mass. The
chemical on the top of one side of the equation is on the bottom on the
other side.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. You will need a ( ) inside the square root.
|
|
|
| 5. |
What
is the molecular mass of a gas that diffuses 2.89 cm/min under the
same conditions that ammonia diffuses 1.39 cm/min ?
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MMX
=
?
MMNH3 = 17.0 g/mol
rateX = 2.89
cm/min rateNH3
= 1.39 cm/min
|
Identify
the variables including the unknown.
Find the Molecular Mass of any known chemical.
An unknown chemical is labeled as X.
The units for the variable determine the
equation that will be used.
cm/min is the unit for the variable for Rate.
Rate is inversely related to square root of Molecular Mass. The chemical
on the top on one side, is on the bottom on the other side of the
equation.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Since the Molecular MassX is inside
the square root, square both sides of the equation.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. The ( ) do not have to be punched into the
calculator. They are necessary to show that the unit is squared in
addition to the number. |
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| 6. |
What
is the molecular mass of a gas that diffuses in 33.5 seconds the same
distance hexane diffuses in 28.4 sec ?
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MMX
=
?
MMC6H14 = 84.0 g/mol
tX = 33.5
s
tC6H14 = 28.4 s
|
Identify
the variables including the unknown.
Find the Molecular Mass of any known chemical.
An unknown chemical is labeled as X.
The units for the variable determine the
equation that will be used.
s is the unit for the variable for t.
t is directly related to square root of Molecular Mass. The chemical
on the top on one side, is also on the top on the other side of the
equation.
Write the equation using the symbols for the
chemicals instead of A and B or 1 and 2.
Since the Molecular MassX is inside
the square root, square both sides of the equation.
Solve the equation in symbols for the unknown
term.
Substitute values and punch as shown into the
calculator. The ( ) do not have to be punched into the
calculator. They are necessary to show that the unit is squared in
addition to the number. |
Table of Contents
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