The Gas Laws
Learning about the special behavior of
gases
Objective #4
Section 21.5, Note pack pg. 11
Graham’s Law of
Diffusion
• A key questions we need to be able to
answer is, “How fast do gases travel
in comparison to one another?”
• Diffusion - The tendency for gases
to travel from an area of higher
concentration to lower concentration
until an equilibrium is reached.
• Effusion – diffusion of gas molecules
through a pin hole.
Effusion – diffusion of gas
molecules through a pin hole.
Graham’s Law
• The ratio of velocity of a lighter gas
to a heavier gas is equal to the
square root of the inverse of their
molar masses.
Graham’s Law
• The ratio of velocity of a lighter gas
to a heavier gas is equal to the
square root of the inverse of their
molar masses.
Graham’s Law
• Show the mathematical proof that derives Graham’s
Law:
Kinetic energy = energy of movement
K.E. = ½ m
2
v
(m) = Mass (v) = Velocity
If 2 gas particles have the same K.E. (temp.), the
smaller particle would be moving faster.
Restate Graham’s Law
• Graham’s law: Lighter gases travel
faster than heavier gases at the same
temperature and pressure.
• Example: Compare the velocities of
Hydrogen and Oxygen.
We know that Hydrogen has a mass of 2
g/mol; Oxygen has a mass of 32g/mol.
Restate Graham’s Law
• Graham’s law: Lighter gases travel
faster than heavier gases at the same
temperature and pressure.
Hydrogen travels 4 x’s faster than oxygen.
Always list the lighter gas over the heavier one, then use the
square root of the inverse to find the rate.
Graham’s law: The ratio of velocity of a lighter
gas to a heavier gas is equal to the square root
of the inverse of their molar masses.
Lighter gas
Heavier gas’
mass
Heavier gas
Lighter gas’
mass
Reduce
Hydrogen travels 4 x’s faster
than oxygen.
Please explain how to find this to your
neighbor better than I did.
Graham’s Law: Example 1
pg. 12
• Place the following gases in order of
increasing average molecular speed at
25 Celsius.
– Ne, HBr, SO2, NF3, CO
Graham’s Law: Example 1
• Place the following gases in order of
increasing average molecular speed at
25 Celsius.
– Ne, HBr, SO2, NF3, CO
We need to know the masses of each gas
Graham’s Law: Example 1
• Place the following gases in order of
increasing average molecular speed at
25 Celsius.
– Ne, HBr, SO2, NF3, CO
20 81 64 71 28 (masses are g/mol)
Graham’s Law: Example 1
• Place the following gases in order of
increasing average molecular speed at
25 Celsius.
– Ne, HBr, SO2, NF3, CO
20 81
64 71 28 (masses are g/mol)
So, the correct order, from smallest mass to largest, would be…
Graham’s Law: Example 1
• Place the following gases in order of
increasing average molecular speed at
25 Celsius.
– Ne, HBr, SO2, NF3, CO
20 81
64 71 28 (masses are g/mol)
So, the correct order, from smallest mass to largest, would be…
Ne  CO  SO2  NF3  HBr
…but in order of increasing speed would be…
HBr  NF3  SO2  CO  Ne
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Lighter
Heavier
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Lighter  Rate He
Heavier  Rate N2
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Lighter  Rate He
Heavier  Rate N2
√28
√4
Graham’s law: The ratio of velocity of a lighter gas to a heavier gas
is equal to the square root of the inverse of their molar masses.
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Lighter  Rate He
Heavier  Rate N2
√28
√4 =
= 5.29
2
Graham’s Law: Example 2
• Compare the rate of diffusion of
nitrogen gas to helium gas
Lighter  Rate He
Heavier  Rate N2
√28
√4
= 5.29
= 2
= 2.65
Therefore, Helium will diffuse 2.65
times faster than nitrogen
Graham’s Law: Example 3
• Compare the rate of diffusion of
argon to neon gas
Graham’s Law: Example 3
• Compare the rate of diffusion of
argon to neon gas
Lighter  Neon
Heavier  Argon
Graham’s Law: Example 3
• Compare the rate of diffusion of
argon to neon gas
Lighter  Neon √39.9
Heavier  Argon √20.2
Graham’s Law: Example 3
• Compare the rate of diffusion of
argon to neon gas
Lighter  Neon √39.9 = 6.32
Heavier  Argon √20.2 = 4.49
= 1.4
Graham’s Law: Example 3
• Compare the rate of diffusion of
argon to neon gas
Lighter  Neon √39.9 = 6.32
Heavier  Argon √20.2 = 4.49
= 1.4
Neon (lighter) will diffuse 1.4 times
faster than Argon (heavier).
Example 4
• A gas of unknown molecular mass was allowed to effuse through
a small opening under constant pressure conditions. It required
105 seconds for 1.0 Liter of the gas to effuse. Under identical
conditions it required 31 seconds for 1.0 liter of oxygen to
effuse. Calculate the molar mass of the unknown gas.
Example 4
• A gas of unknown molecular mass was allowed to effuse through a
small opening under constant pressure conditions. It required
105 seconds for 1.0 Liter of the gas to effuse. Under identical
conditions it required 31 seconds for 1.0 liter of oxygen to
effuse. Calculate the molar mass of the unknown gas.
Here’s what we know:
Unknown gas
Oxygen
Diffusion time = 105 seconds
Diffusion time = 31 seconds
Heavier
Lighter
Example 4
• A gas of unknown molecular mass was allowed to effuse through a
small opening under constant pressure conditions. It required
105 seconds for 1.0 Liter of the gas to effuse. Under identical
conditions it required 31 seconds for 1.0 liter of oxygen to
effuse. Calculate the molar mass of the unknown gas.
SO…
Rate (oxy)
Rate (unknown)
√Mass (X)
√Mass 32
Example 4
• A gas of unknown molecular mass was allowed to effuse through a
small opening under constant pressure conditions. It required
105 seconds for 1.0 Liter of the gas to effuse. Under identical
conditions it required 31 seconds for 1.0 liter of oxygen to
effuse. Calculate the molar mass of the unknown gas.
SO…
Rate (oxy)
√Mass (X) 
Rate (unknown)
√Mass 32 
Invert the times to get the
105
= √x_
difference in rate between them 31
√32
To get rid of the square root, we need to square both
sides.
= _x_
3.392= _x_
11.49
32
32
x = 367.7 g/mol molar mass
=3.39
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
“homonuclear diatomic molecule” = one of our 7 diatomic molecules
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
.355 x’s Oxy (which means it is slower than O2 = larger than O2 )
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
.355 x’s Oxy (which means it is slower than O2 = larger than O2 )
Rate = .355 = √32
1
√X
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
.355 x’s Oxy (which means it is slower than O2 = larger than O2 )
Rate = .355 = √32
1
√X
To get rid of the square root, we
need to square both sides
Example 5
• An unknown gas composed of homonuclear diatomic
molecules effuses at a rate that is only .355 time
that of Oxygen at the same temperature. Calculate
the molar mss of the unknown, and identify it.
.355 x’s Oxy (which means it is slower than O2 = larger than O2 )
Rate = .355 = √32
1
√X To get rid of the square root, we
need to square both sides
.3552 = √32
√X


0.126025 = 32
= X
Example 5
• An unknown gas composed of homonuclear diatomic molecules
effuses at a rate that is only .355 time that of Oxygen at the
same temperature. Calculate the molar mss of the unknown, and
identify it.
.355 x’s Oxy (which means it is slower than O2 = larger than O2 )
Rate = .355
1
=
.3552 = √32
√X
√32
√X


To get rid of the square root, we
need to square both sides
0.126025 = 32
= X
X = 253.92 = diatomic molecule, so each atom is
126.9 … the mass of
Iodine (I2)
The Gas Laws
Learning about the special behavior
of gases
Objective # 5
Note pack pg. 13
Further Applications of the
Gas Laws
The Gas Laws
Learning about the special behavior
of gases
Objective # 6
Gas Laws and Stoichiometry