1. 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

2. 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?”

3. 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.

4. Effusion – diffusion of gas molecules through a pin hole.

5. 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.

6. 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.

7. (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.

8. 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.

9. 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.

10. 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.

11. 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

12. 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

13. Graham’s Law: Example 1
Place the following gases in order of increasing average molecular speed at 25 Celsius.
Ne, HBr, SO2, NF3, CO
(masses are g/mol)

14. Graham’s Law: Example 1
Place the following gases in order of increasing average molecular speed at 25 Celsius.
Ne, HBr, SO2, NF3, CO
(masses are g/mol)
So, the correct order, from smallest mass to largest, would be…

15. Graham’s Law: Example 1
Place the following gases in order of increasing average molecular speed at 25 Celsius.
Ne, HBr, SO2, NF3, CO
(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

16. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas

17. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas
Lighter
Heavier

18. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas
Lighter  Rate He
Heavier  Rate N2

19. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas
Lighter  Rate He √28
Heavier  Rate N √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.

20. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas
Lighter  Rate He √28 =
Heavier  Rate N √4 =

21. Graham’s Law: Example 2
Compare the rate of diffusion of nitrogen gas to helium gas
Lighter  Rate He √28 =
Heavier  Rate N √4 = 2
= 2.65
Therefore, Helium will diffuse 2.65 times faster than nitrogen

22. Graham’s Law: Example 3
Compare the rate of diffusion of argon to neon gas

23. Graham’s Law: Example 3
Compare the rate of diffusion of argon to neon gas
Lighter  Neon
Heavier  Argon

24. Graham’s Law: Example 3
Compare the rate of diffusion of argon to neon gas
Lighter  Neon √39.9
Heavier  Argon √20.2

25. Graham’s Law: Example 3
Compare the rate of diffusion of argon to neon gas
Lighter  Neon √ =
Heavier  Argon √ =
= 1.4

26. Graham’s Law: Example 3
Compare the rate of diffusion of argon to neon gas
Lighter  Neon √ =
Heavier  Argon √ =
= 1.4
Neon (lighter) will diffuse 1.4 times faster than Argon (heavier).

27. 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.

28. 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

29. 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

30. 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 = √x_
difference in rate between them √32
To get rid of the square root, we need to square both sides.
= _x_ = _x_
x = g/mol molar mass
=3.39
3.392
11.49

31. 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.

32. 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

33. 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 )

34. 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 = = √32 √X

35. 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 = = √32
√X To get rid of the square root, we
need to square both sides

36. 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 = = √32
√X To get rid of the square root, we
need to square both sides
.3552 = √  = 32
√X  = X

37. 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 = = √32
√X To get rid of the square root, we
need to square both sides
.3552 = √  = 32
√X  = X
X = = diatomic molecule, so each atom is
126.9 … the mass of Iodine (I2)

38. 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

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