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The Gas LawsThe Gas LawsThe Gas LawsThe Gas LawsLearning about the special behavior of Learning about the special behavior of
gasesgases
Intro. to Objective # Intro. to Objective # 44
Section 21.5, Note pack pg. 8Section 21.5, Note pack pg. 8
Avogardro’s Hypothesis
• Although gas molecules of different gases are different sizes, equal volumes of gases at the same temperature and pressure contain equal number of particles.
In other words…• Even though Chlorine gas
molecules are 35 times bigger than hydrogen gas molecules, equal numbers of the two gases would occupy the same volume at the same temperature and pressure.
Example 1Determine the volume occupied
by .202 mole of a gas at STP
Example 1Determine the volume occupied
by .202 mole of a gas at STP
.202 mol x 22.4 L 1 1 mol
Example 1Determine the volume occupied
by .202 mole of a gas at STP
.202 mol x 22.4 L = 4.52 L 1 1 mol
Example 2
What is the volume occupied by .742 mol of Argon at STP?
Example 2
What is the volume occupied by .742 mol of Argon at STP?
.742 mol Ar x 22.4L Ar 1 1 mol Ar
Example 2
What is the volume occupied by .742 mol of Argon at STP?
.742 mol Ar x 22.4L Ar = 16.62 L Ar
1 1 mol Ar
Example 3
• Determine the volume occupied by 14 grams of nitrogen gas at STP?
Example 3
• Determine the volume occupied by 14 grams of nitrogen gas at STP?14g N2 x 1 mol N2
1 28g N2
Example 3
• Determine the volume occupied by 14 grams of nitrogen gas at STP?14g N2 x 1 mol N2 = 0.5 mol N2
1 28g N2
Example 3
• Determine the volume occupied by 14 grams of nitrogen gas at STP?14g N2 x 1 mol N2 = 0.5 mol N2
1 28g N2
0.5 mol N2 x 22.4 L N2
1 1 mol N2
Example 3
• Determine the volume occupied by 14 grams of nitrogen gas at STP?14g N2 x 1 mol N2 = 0.5 mol N2
1 28g N2
0.5 mol N2 x 22.4 L N2 = 11.2 L N2
1 1 mol N2
Dalton’s Law of Partial Pressures
• Many gases, including air, are mixtures. Remember, the particles in a gas at the same temperature have the same average kinetic energy.
• Gas pressure depends only on the number of gas particles in a given volume and their average kinetic energy. The kind of particle is not important.
Dalton’s Law of Partial Pressures
• Define Partial Pressure -– “The contribution each gas in a mixture
makes to the total pressure” is the partial pressure exerted by that gas…
• Dalton’s Law - At constant volume and temp, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures.
•
• the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures.
For the examples shown, we will follow this
progression:Mass ofThe gas
# of Moles
Ideal GasLaw
Daltons Law of Partial
Pressures
Example 1 Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
Example 1 Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
If Ptotal = PO2 + PN2 + PCO2 + Pother and
Example 1 Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
If Ptotal = PO2 + PN2 + PCO2 + Pother and
101.3 kPa = ? + 79.1 kPa + 0.04 kPa + 0.94 kPa
Example 1 Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
If Ptotal = PO2 + PN2 + PCO2 + Pother and
101.3 kPa = ? + 79.1 kPa + 0.04 kPa + 0.94 kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or…
Example 1 Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
If Ptotal = PO2 + PN2 + PCO2 + Pother and
101.3 kPa = ? + 79.1 kPa + 0.04 kPa + 0.94 kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or…
PO2 = 101.3 kPa – (79.1 + 0.04 + 0.94)
Example 1 Air contains oxygen, nitrogen, carbon dioxide, and
trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.
If Ptotal = PO2 + PN2 + PCO2 + Pother and
101.3 kPa = ? + 79.1 kPa + 0.04 kPa + 0.94 kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or… PO2 = 101.3 kPa – (79.1 + 0.04 + 0.94)
PO2 = 21.22 kPa
For the examples shown, we will follow this
progression:Mass ofThe gas
# of
Moles
Ideal Gas
Law
Daltons
Law of Partial Pressures
Example 2Determine the total pressure of a gas mixture that
contains oxygen, nitrogen, and helium if the partial pressure of the gases are as follows:
• Oxygen = 20 kPa• Nitrogen = 46.7 kPa• Helium = 26.7 kPa
Example 2Determine the total pressure of a gas mixture that
contains oxygen, nitrogen, and helium if the partial pressure of the gases are as follows:
• Oxygen = 20 kPa• Nitrogen = 46.7 kPa• Helium = 26.7 kPaIf Ptotal = PO2 + PN2 + PHe and
Example 2Determine the total pressure of a gas mixture that
contains oxygen, nitrogen, and helium if the partial pressure of the gases are as follows:
• Oxygen = 20 kPa• Nitrogen = 46.7 kPa• Helium = 26.7 kPaIf Ptotal = PO2 + PN2 + PHe and __?__ kPa = 20 kPa + 46.7 kPa + 26.7 kPa
Example 2Determine the total pressure of a gas mixture that
contains oxygen, nitrogen, and helium if the partial pressure of the gases are as follows:
• Oxygen = 20 kPa• Nitrogen = 46.7 kPa• Helium = 26.7 kPaIf Ptotal = PO2 + PN2 + PHe and
__?__ kPa = 20 kPa + 46.7 kPa + 26.7 kPa
Then the total pressure is 93.4 kPa
For the examples shown, we will follow this
progression:Mass ofThe gas
# of Moles
Ideal Gas
Law
Daltons
Law of Partial Pressures
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol Ne (g),
and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol Ne (g),
and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
We will need to re-arrange the Ideal Gas Law for each of these:
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol Ne (g),
and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
We will need to re-arrange the Ideal Gas Law for each of these:
PHe = nRT
V PNe = nRT
V PAr = nRT
V
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol Ne (g),
and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
We will need to re-arrange the Ideal Gas Law for each of these:
PHe = nRT = (0.538 mol)x(8.31 kPaL/ molK )x(298K) = 190.3 kPa
V (7 Liters)PNe = nRT
VPAr = nRT
V
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol Ne (g),
and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
We will need to re-arrange the Ideal Gas Law for each of these:
PHe = nRT = (0.538 mol)x(8.31 kPaL/ molK )x(298K) = 190.3 kPa
V (7 Liters)PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = 111.4
kPa V (7 Liters)PAr = nRT
V
Example 3A mixture containing 0.538 mole of He (g), 0.315 mol
Ne (g), and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
We will need to re-arrange the Ideal Gas Law for each of these:PHe = nRT = (0.538 mol)x(8.31 kPaL/ molK )x(298K) = 190.3 kPa V (7 Liters)PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = 111.4 kPa V (7 Liters)PAr = nRT = (0.103 mol)x(8.31 kPaL/ molK )x(298K) = 36.4 kPa V (7 Liters)
Now we know how much pressure each gas is contributing to the mixture.
Example 3A mixture containing 0.538 mole of He (g), 0.315
mol Ne (g), and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
PHe = 190.3 kPa
PNe = 111.4 kPa
PAr = 36.4 kPa
• B. Calculate the total pressure of the mixture
Example 3A mixture containing 0.538 mole of He (g), 0.315
mol Ne (g), and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
PHe = 190.3 kPa
PNe = 111.4 kPa
PAr = 36.4 kPa
• B. Calculate the total pressure of the mixtureP total = PHe + PNe + PAr
Example 3A mixture containing 0.538 mole of He (g), 0.315
mol Ne (g), and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
PHe = 190.3 kPa
PNe = 111.4 kPa
PAr = 36.4 kPa
• B. Calculate the total pressure of the mixtureP total = PHe + PNe + PAr
P total = 190.3 kPa + 111.4 kPa + 36.4 kPa
Example 3A mixture containing 0.538 mole of He (g), 0.315
mol Ne (g), and 0.103 mole Ar (g) is confided in a 7 Liter vessel at 25 Celsus.
• A. Calculate the partial pressure of each of the gases in the mixture
PHe = 190.3 kPa
PNe = 111.4 kPa
PAr = 36.4 kPa
• B. Calculate the total pressure of the mixtureP total = PHe + PNe + PAr
P total = 190.3 kPa + 111.4 kPa + 36.4 kPa
= 338.1 kPa
For the examples shown, we will follow this
progression:Mass ofThe gas
# of Moles
Ideal GasLaw
Daltons Law of Partial
Pressures
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture.
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture.Before we can find the pressure, we first need to convert the
masses into moles of each gas.
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture.Before we can find the pressure, we first need to convert the
masses into moles of each gas.2.5g CH4 1 2.5g C2H4 1 2.5g C4H10 1
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture.Before we can find the pressure, we first need to convert the
masses into moles of each gas.2.5g CH4 x 1 mol CH4 = 1 16g CH4
2.5g C2H4 x 1 mol C2H4 = 1 28g C2H4
2.5g C4H10 x 1 mol C4H10 = 1 58g C4H10
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture.Before we can find the pressure, we first need to convert the
masses into moles of each gas.2.5g CH4 x 1 mol CH4 = 0.156 mol CH4
1 16g CH4
2.5g C2H4 x 1 mol C2H4 = 0.089 mol C2H4
1 28g C2H4
2.5g C4H10 x 1 mol C4H10 = 0.043 mol C4H10
1 58g C4H10
Now we will use the number of moles to help us find the partial pressure of each gas.
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT V PC2H4 = nRT V PC4H10 = nRT V
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. We’ll use the # of moles we found as “n”.PCH4 = nRT = (0.156 mol)__________________ V PC2H4 = nRT = (0.089 mol)_________________ V PC4H10 = nRT = (0.043 mol)_________________ V
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT = (0.156 mol)x(8.31 kPaL/molK)x(288K) = V (2 Liters)PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = V (2 Liters)PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = V (2 Liters)
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT = (0.156 mol)x(8.31 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters)PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5
kPa V (2 Liters)PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa V (2 Liters)
• Calculate the total pressure of the mixture.
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT = (0.156 mol)x(8.31 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters)PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters)PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa V (2 Liters)
• Calculate the total pressure of the mixture.Ptotal = PCH4 + PC2H4 + PC4H10
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT = (0.156 mol)x(8.31 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters)PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters)PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa V (2 Liters)
• Calculate the total pressure of the mixture.Ptotal = PCH4 + PC2H4 + PC4H10
= (186.7 kPa) + (106.5 kPa) + (51.5 kPa)
Example 4:A mixture of 2.5 grams of each of CH4, C2H4, and C4H10 is
contained in a 2 liter flask as a temperature of 15o Celsius.• Calculate the partial pressure of each of the gases in the
mixture. Now we can find the partial pressure for each gas.PCH4 = nRT = (0.156 mol)x(8.31 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters)PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters)PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa V (2 Liters)
• Calculate the total pressure of the mixture.Ptotal = PCH4 + PC2H4 + PC4H10
344.7 kPa = (186.7 kPa) + (106.5 kPa) + (51.5 kPa)
Page 10It would be good to review the
section. Collecting Gas over Water
The Gas LawsThe Gas LawsThe Gas LawsThe Gas Laws