Learning about the special behavior of gases The Gas Laws Learning about the special behavior of gases Intro. to Objective # 4 Section 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 1 Determine the volume occupied by .202 mole of a gas at STP

Example 1 Determine the volume occupied by .202 mole of a gas at STP .202 mol x 22.4 L 1 1 mol

Example 1 Determine 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 of The gas  # of Moles  Ideal Gas Law 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 of The gas  # of Moles  Ideal Gas Law Daltons Law of Partial Pressures

Example 2 Determine 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 2 Determine 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 If Ptotal = PO2 + PN2 + PHe and

Example 2 Determine 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 If Ptotal = PO2 + PN2 + PHe and __?__ kPa = 20 kPa + 46.7 kPa + 26.7 kPa

Example 2 Determine 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 If 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 of The gas  # of Moles  Ideal Gas Law Daltons Law of Partial Pressures

Example 3 A 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 3 A 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 3 A 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 PAr = nRT

Example 3 A 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 kPaL/ molK )x(298K) = 190.3 kPa V (7 Liters) PNe = nRT V PAr = nRT

Example 3 A 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 kPaL/ molK )x(298K) = 190.3 kPa V (7 Liters) PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = 111.4 kPa PAr = nRT V

Example 3 A 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 kPaL/ molK )x(298K) = 190.3 kPa V (7 Liters) PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = 111.4 kPa PAr = nRT = (0.103 mol)x(8.31 kPaL/ molK )x(298K) = 36.4 kPa Now we know how much pressure each gas is contributing to the mixture.

Example 3 A 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 3 A 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 P total = PHe + PNe + PAr

Example 3 A 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 P total = PHe + PNe + PAr P total = 190.3 kPa + 111.4 kPa + 36.4 kPa

Example 3 A 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 P 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 of The gas  # of Moles  Ideal Gas Law 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 2.5g 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 = 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 PC4H10 = nRT 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)_________________ PC4H10 = nRT = (0.043 mol)_________________ 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 kPaL/molK)x(288K) = V (2 Liters) PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = V (2 Liters) PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 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 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters) PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters) PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa 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 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters) PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters) PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa 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 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters) PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters) PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa 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 kPaL/molK)x(288K) = 186.7 kPa V (2 Liters) PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = 106.5 kPa V (2 Liters) PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = 51.5 kPa Calculate the total pressure of the mixture. Ptotal = PCH4 + PC2H4 + PC4H10 344.7 kPa = (186.7 kPa) + (106.5 kPa) + (51.5 kPa)

Page 10 It would be good to review the section Page 10 It would be good to review the section. Collecting Gas over Water

The Gas Laws