Learning about the special behavior of gases 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.

(m) = Mass (v) = Velocity Graham’s Law Show the mathematical proof that derives Graham’s Law: Kinetic energy = energy of movement K.E. = ½ m v2 (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.

Hydrogen travels 4 x’s faster than oxygen. 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. Heavier gas’ mass Reduce Lighter gas Heavier gas Lighter gas’ mass 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 √28 Heavier  Rate N2 √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 √28 = 5.29 Heavier  Rate N2 √4 = 2

Graham’s Law: Example 2 Compare the rate of diffusion of nitrogen gas to helium gas Lighter  Rate He √28 = 5.29 Heavier  Rate N2 √4 = 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 Here’s what we know: Unknown gas Oxygen Heavier Lighter 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 SO… Rate (oxy) √Mass (X) Rate (unknown) √Mass 32 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

Example 4 =3.39 3.392 11.49 SO… Rate (oxy) √Mass (X)  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_ = _x_ 32 32 x = 367.7 g/mol molar mass =3.39 3.392 11.49

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  0.126025 = 32 √X  = 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 .3552 = √32  0.126025 = 32 √X  = X X = 253.92 = diatomic molecule, so each atom is 126.9 … the mass of Iodine (I2)

The Gas Laws Objective # 5 Further Applications of the Gas Laws Learning about the special behavior of gases Objective # 5 Note pack pg. 13 Further Applications of the Gas Laws

Gas Laws and Stoichiometry The Gas Laws Learning about the special behavior of gases Objective # 6 Gas Laws and Stoichiometry