1. The Gas Laws

2. The Gas Laws The gas laws describe HOW gases behave.
They can be predicted by theory.
The amount of change can be calculated with mathematical equations.

3. Standard Atmospheric Pressure
One atmosphere is equal to mm Hg, torr, or kPa (kilopascals).

4. Standard Atmospheric Pressure
Perform the following pressure conversions.
a) 144 kPa = _____ atm
(1.42)
b) 795 mm Hg = _____ atm
(1.05)

5. Standard Atmospheric Pressure
Perform the following pressure conversions.
c) 669 torr = ______ kPa
(89.2)
d) 1.05 atm = ______ mm Hg
(798)

6. Standard Atmospheric Pressure
Air pressure at higher altitudes, such as on a mountaintop, is slightly lower than air pressure at sea level.

7. Standard Atmospheric Pressure
Air pressure is measured using a barometer.

9. Pressure and the Number of Molecules
More molecules mean more collisions between the gas molecules themselves and more collisions between the gas molecules and the walls of the container.
Number of molecules is DIRECTLY proportional to pressure.

10. Pressure and the Number of Molecules
Doubling the number of gas particles in a basketball doubles the pressure.

11. Pressure and the Number of Molecules
Gases naturally move from areas of high pressure to low pressure because there is empty space to move in.

12. If you double the number of molecules,
1 atm

13. If you double the number of molecules, you double the pressure.
2 atm

14. As you remove molecules from a container,
4 atm

15. As you remove molecules from a container, the pressure decreases.
2 atm

16. As you remove molecules from a container, the pressure decreases until the pressure inside equals the pressure outside.
1 atm

17. Changing the Size (Volume) of the Container
In a smaller container, molecules have less room to move.
The molecules hit the sides of the container more often, striking a smaller area with the same force.

18. Changing the Size (Volume) of the Container
As volume decreases, pressure increases.
Volume and pressure are INVERSELY proportional.

19. As the pressure on a gas increases,
1 atm
As the pressure on a gas increases,
4 Liters

20. As the pressure on a gas increases, the volume decreases.
2 atm
2 Liters

21. Temperature and Pressure
Raising the temperature of a gas increases the pressure if the volume is held constant.
At higher temperatures, the particles in a gas have greater kinetic energy.

22. Temperature and Pressure
They move faster and collide with the walls of the container more often and with greater force, so the pressure rises.

23. 300 K
If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to K, one of 2 things happens.

24. Either the volume will increase to 2 liters at 1 atm,
600 K
300 K
Either the volume will increase to 2 liters at 1 atm,

25. 300 K
600 K
or the pressure will increase to 2 atm while the volume remains constant.

26. Ideal Gases In this unit we will assume the gases behave ideally.
Ideal gases do not really exist, but this makes the math easier and is a close approximation.

27. Kinetic Molecular Theory of Gases
Gas particles are much smaller than the spaces between them. The particles have negligible volume.
There are no attractive or repulsive forces between gas molecules.

28. Kinetic Molecular Theory of Gases
Gas particles are in constant, random motion. Until they bump into something (another particle or the side of a container), particles move in a straight line.

29. Kinetic Molecular Theory of Gases
No kinetic energy is lost when gas particles collide with each other or with the walls of their container.
All gases have the same kinetic energy at a given temperature.

30. Temperature
Temperature is a measure of the average kinetic energy of the particles in a sample of matter.

31. Ideal Gases There are no gases for which this is true.
Real gases behave more ideally at high temperature and low pressure.

32. Ideal Gases
At low temperature, the gas molecules move more slowly, so attractive forces are no longer negligible.
As the pressure on a gas increases, the molecules are forced closer together and attractive forces are no longer negligible.
Therefore, real gases behave more ideally at high temperature and low pressure.

33. Avogadro’s Law
Avogadro’s law states that equal volumes of different gases (at the same temperature and pressure) contain equal numbers of atoms or molecules.

34. Avogadro’s Law 2 Liters of Helium 2 Liters of Oxygen
has the same number of particles as ..

35. Avogadro’s Law
The molar volume for a gas is the volume that one mole occupies at 0.00ºC and 1.00 atm.
1 mole = 22.4 L at STP (standard temperature and pressure).
As a result, the volume of gaseous reactants and products can be expressed as small whole numbers in reactions.

36. Problem
How many moles are in 45.0 L of a gas at STP?
2.01 moles

37. Problem
How many liters are in moles of a gas at STP?
14.2 L

38. V = K x n (K is some constant) V / n = K
Avogadro’s Law
V = K x n (K is some constant)
V / n = K
Easier to use: V1 V2 = n1 n2

39. Example
Consider two samples of nitrogen gas. Sample 1 contains 1.5 mol and has a volume of 36.7 L. Sample 2 has a volume of 16.5 L at the same temperature and pressure. Calculate the number of moles of nitrogen in sample 2.

40. Example
Sample 1 contains 1.5 mol and has a volume of 36.7 L. Sample 2 has a volume of 16.5 L. Calculate the number of moles of nitrogen in sample 2. V1 V2
36.7 L
16.5 L = n1 n2
1.5 mol
n2 = 0.67 mol

41. Problem
If mol of argon gas occupies a volume of 652 mL at a particular temperature and pressure, what volume would mol of argon occupy under the same conditions?
V2 = 1140 mL

42. Problem
If 46.2 g of oxygen gas occupies a volume of 100. L at a particular temperature and pressure, what volume would 5.00 g of oxygen gas occupy under the same conditions?
V2 = 10.8 L

43. Boyle’s Law
At Boyle’s law states that the pressure and volume of a gas at constant temperature are inversely proportional.
Inversely proportional means as one goes up the other goes down.

44. Boyle’s Law

45. Boyle’s Law
P x V = K (K is some constant)
P1 V1 = P2 V2

46. Boyle’s Law
The P-V graph for Boyle’s law results in a hyperbola because pressure and volume are inversely proportional.

47. P V

48. Example
A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume?

49. Example
First, make sure the pressure and volume units in the question match.
A balloon is filled with 25 L of air at atm pressure. If the pressure is changed to 1.5 atm, what is the new volume?
THEY DO!

50. Example P1 V1 = P2 V2 V2 V2 = 17 L 1.0 atm (25 L) 1.5 atm
A balloon is filled with 25 L of air at atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? P1 V1 = P2 V2 V2
1.0 atm
(25 L)
1.5 atm
V2 = 17 L

51. Problem
A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to L?
P2 = 2.2 atm

52. Problem
A gas is collected in a 242 cm3 container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at standard pressure?
V2 = 209 cm3

53. Problem
A gas is collected in a 24.2 L container. The pressure of the gas in the container is determined to be 756 mm Hg. What is the pressure of this gas if the volume increases to 30.0 L?
P2 = 610. mm Hg

54. Charles’ Law
The volume of a gas is directly proportional to the Kelvin temperature if the pressure is held constant.
K = °C

55. Charles’ Law

56. V = K x T (K is some constant) V / T = K
Charles’ Law
V = K x T (K is some constant)
V / T = K
Easier to use: V1 V2 = T1 T2

57. Charles’ Law
The V-T graph for Charles’ law results in a straight line because pressure and volume are directly proportional.

58. V T

59. Example
What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure?

60. Example First, make sure the volume units in the question match.
What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure?
THEY DO!

61. Example
Second, make sure to convert degrees Celsius to Kelvin.
What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? °C 25
K =
K = 298 K

62. Example
What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? V1 V2
2.5 L
4.1 L = T1 T2
298 K
T2 = 489 K

63. Problem
What is the final volume of a gas that starts at 8.3 L and 17 ºC and is heated to 96 ºC?
V2 = 11 L

64. Problem
A 225 cm3 volume of gas is collected at 57 ºC. What volume would this sample of gas occupy at standard temperature?
V2 = 186 cm3

65. Problem
A 225 cm3 volume of gas is collected at 42 ºC. If the volume is decreased to 115 cm3, what is the new temperature?
T2 = 161 K

66. Gay-Lussac’s Law
The temperature and the pressure of a gas are directly related at constant volume.

67. P = K x T (K is some constant) P / T = K
Gay-Lussac’s Law
P = K x T (K is some constant)
P / T = K
Easier to use: P1 P2 = T1 T2

68. P T

69. Example
What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? Volume remains constant.

70. There is only one pressure unit!
Example
First, make sure the pressure units in the question match.
What is the pressure inside a L can of deodorant that starts at 25 ºC and atm if the temperature is raised to 100 ºC?
There is only one pressure unit!

71. Example
Second, make sure to convert degrees Celsius to Kelvin.
What is the pressure inside a L can of deodorant that starts at 25 ºC and atm if the temperature is raised to 100 ºC? °C 25
K =
K = 298 K

72. Example
What is the pressure inside a L can of deodorant that starts at 25 ºC and atm if the temperature is raised to 100 ºC? °C
100
K =
K = 373 K

73. Example P1 P2 = T1 T2 P2 = 1.5 atm 1.2 atm 298 K 373 K
What is the pressure inside a L can of deodorant that starts at 25 ºC and atm if the temperature is raised to 100 ºC? P1 P2
1.2 atm = T1 T2
298 K
373 K
P2 = 1.5 atm

74. Problem
A can of deodorant starts at 43 ºC and 1.2 atm. If the volume remains constant, at what temperature will the can have a pressure of 2.2 atm?
T2 = 579 K

75. Problem
A can of shaving cream starts at ºC and 1.30 atm. If the temperature increases to 37 ºC and the volume remains constant, what is the pressure of the can?
P2 = atm

76. Problem
A 12 ounce can of a soft drink starts at STP. If the volume stays constant, at what temperature will the can have a pressure of 2.20 atm?
T2 = 601 K

77. The Combined Gas Law P1 V1 P2 V2 = T1 T2
The gas laws may be combined into a single law, called the combined gas law, which relates two sets of conditions of pressure, volume, and temperature by the following equation. P1 V1 P2 V2 = T1 T2

78. Example
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume?

79. There is only one volume unit!
Example
First, make sure the volume units in the question match.
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume?
There is only one volume unit!

80. Example Second, make sure the pressure units in the question match.
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume?
They do!

81. Example
Third, make sure to convert degrees Celsius to Kelvin.
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? °C 25
K =
K = 298 K

82. Example
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? °C 75
K =
K = 348 K

83. Example P1 V1 P2 V2 = T1 T2 V2 = 4.9 L 4.8 atm (15 L) 17 atm 298 K
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? P1 V1 P2 V2
4.8 atm
(15 L)
17 atm = T1 T2
298 K
348 K
V2 = 4.9 L

84. Problem
If 6.2 L of gas at 723 mm Hg at 21 ºC is compressed to 2.2 L at mm Hg, what is the temperature of the gas?
T2 = 594 K

85. Problem
A sample of nitrogen monoxide has a volume of 72.6 mL at a temperature of 16 °C and a pressure of kPa. What volume will the sample occupy at 24 °C and 99.3 kPa?
V2 = 78.2 mL

86. Problem
A hot air balloon rises to an altitude of 7000 m. At that height the atmospheric pressure drops to 300. mm Hg and the temperature cools to - 33 °C. Suppose on the hot air balloon there was a small balloon filled to 1.00 L at sea level and a temperature of 27 °C. What would its volume ultimately be when it reached the height of 7000 m?
V2 = 2.03 L

87. Daltons’ Law of Partial Pressures
Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture, as shown below.
PTotal = P1 + P2 + P3 + …
The partial pressure is the contribution by that gas.

88. Example
On the next slide, determine the pressure in the fourth container if all of the gas molecules from the 1st three containers are placed in the 4th container.

89. 2 atm
1 atm
3 atm ?? 6
atm

90. Problem
What is the total pressure in a balloon filled with air if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg?
790 mm Hg

91. Example
In a second balloon the total pressure is 1.30 atm. What is the pressure of oxygen (in mm Hg) if the pressure of nitrogen is 720. mm Hg?

92. Example
The two gas units do not match. We must convert the 1.30 atm into mm Hg.
760 mm Hg
1.30 atm
988 mm Hg =
1 atm

93. Example PTotal = P1 + P2 + P3 + … 988 mm Hg = 720 mm Hg + Poxygen

94. Problem
A container has a total pressure of 846 torr and contains carbon dioxide gas and nitrogen gas. What is the pressure of carbon dioxide (in kPa) if the pressure of nitrogen is 50. kPa?
63 kPa

95. Problem
When a container is filled with moles of H2, 2 moles of O2 and moles of N2, the pressure in the container is 8.7 atm. The partial pressure of H2 is _____.
2.9 atm

96. Daltons’ Law of Partial Pressures
It is common to synthesize gases and collect them by displacing a volume of water.

97. Problem
Hydrogen was collected over water at 21°C on a day when the atmospheric pressure is 748 torr. The volume of the gas sample collected was 300. mL. The vapor pressure of water at 21°C is
18.65 torr. Determine the partial pressure of the dry gas.
torr

98. Problem
A sample of oxygen gas is saturated with water vapor at 27ºC. The total pressure of the mixture is 772 mm Hg and the vapor pressure of water is 26.7 mm Hg at 27ºC. What is the partial pressure of the oxygen gas?
745.3 mm Hg

99. Remember Ideal Gases Don’t Exist
Molecules do take up space.
There are attractive forces; otherwise, there would be no liquids.

100. The Ideal Gas Law P V = n R T
Pressure times volume equals the number of moles (n) times the ideal gas constant (R) times the temperature in Kelvin.

101. The Ideal Gas Law R = 0.0821 (L atm)/(mol K) R = 8.314 (L kPa)/(mol K)
R = 62.4 (L mm Hg)/(mol K)
The one you choose depends on the unit for pressure!

102. Example
How many moles of air are there in a 2.0 L bottle at 19 ºC and 747 mm Hg?
Choose the value of R based on the pressure unit.
Since mm Hg are use, R = 62.4.

103. Example
Second, make sure to convert degrees Celsius to Kelvin.
How many moles of air are there in a L bottle at 19 ºC and 747 mm Hg? °C 19
K =
K = 292 K

104. Example P V = n R T n = 0.082 mol 747 (2.0) 62.4 (292)
How many moles of air are there in a L bottle at 19 ºC and 747 mm Hg?
292 K P V = n R T
747
(2.0)
62.4
(292)
n = mol

105. Example
What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at 27 ºC?
Choose the value of R based on the pressure unit.
Since atm is requested, R =

106. Example
Second, make sure to convert degrees Celsius to Kelvin.
What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at ºC? °C 27
K =
K = 300. K

107. Example P V = n R T P = 5.2 atm (4.3) 0.90 (0.0821) (300.)
What is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at 27 ºC?
300. K P V = n R T
(4.3)
0.90
(0.0821)
(300.)
P = 5.2 atm

108. Example 1.8 g H2 1 mol H2 __ = 0.90 mol H2 __ 2.0 g H2
Next, convert grams to moles.
What is the pressure in atm exerted by g of H2 gas in a 4.3 L balloon at 300. K?
1.8 g H2 1
mol H2 __
= 0.90 mol H2 __
2.0
g H2

109. Example P V = n R T P = 5.2 atm (4.3) 0.90 (0.0821) (300.)
What is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at 27 ºC?
0.90 mol
300. K P V = n R T
(4.3)
0.90
(0.0821)
(300.)
P = 5.2 atm

110. Problem
Sulfur hexafluoride (SF6) is a colorless, odorless and very unreactive gas. Calculate the pressure (in atm) exerted by moles of the gas in a steel vessel of volume 5.43 L at 69.5 ºC.
P = 9.42 atm

111. Problem
Calculate the volume (in liters) occupied by 7.40 g of CO2 at STP.
V = 3.77 L

112. Example Next, you will have to change grams to moles. 1.8 g 1 mol
What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at ºC?
1.8 g
1 mol
0.90 mol =
2.0 g

113. Problem
A sample of nitrogen gas kept in a container of volume 2.30 L and at a temperature of 32 ºC exerts a pressure of 476 kPa. Calculate the number of moles of gas present.
n = mol

114. Problem
A 1.30 L sample of a gas has a mass of 1.82 g at STP. What is the molar mass of the gas?
31.4 g/mol

115. Problem
Calculate the mass of nitrogen gas that can occupy 1.00 L at STP.
28.0 g