1. 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

2. 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.

3. 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.

4. Example 1
Determine the volume occupied by .202 mole of a gas at STP

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

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

7. Example 2
What is the volume occupied by .742 mol of Argon at STP?

8. Example 2 What is the volume occupied by .742 mol of Argon at STP?
.742 mol Ar x 22.4L Ar
mol Ar

9. Example 2 What is the volume occupied by .742 mol of Argon at STP?
.742 mol Ar x 22.4L Ar = L Ar
mol Ar

10. Example 3
Determine the volume occupied by 14 grams of nitrogen gas at STP?

11. Example 3
Determine the volume occupied by 14 grams of nitrogen gas at STP?
14g N2 x 1 mol N2
g N2

12. Example 3
Determine the volume occupied by 14 grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
g N2

13. Example 3
Determine the volume occupied by 14 grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
g N2
0.5 mol N2 x L N2
mol N2

14. Example 3
Determine the volume occupied by 14 grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
g N2
0.5 mol N2 x L N2 = L N2
mol N2

15. 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.

16. 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.

17. the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures.

18. For the examples shown, we will follow this progression:
Mass of
The gas
 # of
Moles
 Ideal Gas
Law
Daltons
Law of Partial
Pressures

19. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at kPa if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 kPa, 0.04 kPa, and 0.94 kPa, respectively.

20. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 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 = PO PN PCO Pother and

21. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 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 = PO PN PCO Pother and
101.3 kPa = ? kPa kPa kPa

22. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 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 = PO PN PCO Pother and
101.3 kPa = ? kPa kPa kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or…

23. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 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 = PO PN PCO Pother and
101.3 kPa = ? kPa kPa kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or…
PO2 = kPa – ( )

24. Example 1
Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 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 = PO PN PCO Pother and
101.3 kPa = ? kPa kPa kPa
Then PO2 = Ptotal – (PN2 + PCO2 + Pothers) …or…
PO2 = kPa – ( )
PO2 = kPa

25. For the examples shown, we will follow this progression:
Mass of
The gas
 # of
Moles
 Ideal Gas
Law
Daltons
Law of Partial
Pressures

26. 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

27. 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 = PO PN PHe and

28. 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 = PO PN PHe and
__?__ kPa = 20 kPa kPa kPa

29. 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 = PO PN PHe and
__?__ kPa = 20 kPa kPa kPa
Then the total pressure is kPa

30. For the examples shown, we will follow this progression:
Mass of
The gas
 # of
Moles
 Ideal Gas
Law
Daltons
Law of Partial
Pressures

31. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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

32. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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:

33. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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

34. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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) = kPa
V (7 Liters)
PNe = nRT V
PAr = nRT

35. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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) = kPa
V (7 Liters)
PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = kPa
PAr = nRT V

36. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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) = kPa
V (7 Liters)
PNe = nRT = (0.315 mol)x(8.31 kPaL/ molK )x(298K) = kPa
PAr = nRT = (0.103 mol)x(8.31 kPaL/ molK )x(298K) = kPa
Now we know how much pressure each gas
is contributing to the mixture.

37. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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 = kPa
PNe = kPa
PAr = kPa
B. Calculate the total pressure of the mixture

38. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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 = kPa
PNe = kPa
PAr = kPa
B. Calculate the total pressure of the mixture
P total = PHe + PNe + PAr

39. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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 = kPa
PNe = kPa
PAr = kPa
B. Calculate the total pressure of the mixture
P total = PHe + PNe + PAr
P total = kPa kPa kPa

40. Example 3
A mixture containing mole of He (g), mol Ne (g), and 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 = kPa
PNe = kPa
PAr = kPa
B. Calculate the total pressure of the mixture
P total = PHe + PNe + PAr
P total = kPa kPa kPa
= kPa

41. For the examples shown, we will follow this progression:
Mass of
The gas
 # of
Moles
 Ideal Gas
Law
Daltons
Law of Partial
Pressures

42. 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.

43. 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.

44. 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.

45. 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 =
g CH4
2.5g C2H4 x 1 mol C2H4 =
g C2H4
2.5g C4H10 x 1 mol C4H10 =
g C4H10
Calculate the total pressure of the mixture.

46. 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 = mol CH4
g CH4
2.5g C2H4 x 1 mol C2H4 = mol C2H4
g C2H4
2.5g C4H10 x 1 mol C4H10 = mol C4H10
g 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.

47. 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.

48. 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.

49. 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.

50. 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) = kPa
V (2 Liters)
PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = kPa
V (2 Liters)
PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = kPa
Calculate the total pressure of the mixture.

51. 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) = kPa
V (2 Liters)
PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = kPa
V (2 Liters)
PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = kPa
Calculate the total pressure of the mixture.
Ptotal = PCH4 + PC2H4 + PC4H10

52. 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) = kPa
V (2 Liters)
PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = kPa
V (2 Liters)
PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = kPa
Calculate the total pressure of the mixture.
Ptotal = PCH4 + PC2H4 + PC4H10
= (186.7 kPa) + (106.5 kPa) + (51.5 kPa)

53. 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) = kPa
V (2 Liters)
PC2H4 = nRT = (0.089 mol)x(8.31 kPaL/molK)x(288K) = kPa
V (2 Liters)
PC4H10 = nRT = (0.043 mol)x(8.31 kPaL/molK)x(288K) = kPa
Calculate the total pressure of the mixture.
Ptotal = PCH4 + PC2H4 + PC4H10
344.7 kPa = (186.7 kPa) + (106.5 kPa) + (51.5 kPa)

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

55. The Gas Laws