1. Chemeketa Community College
Gases
Chapter 12
Larry Emme
Chemeketa Community College

2. Properties of Gases

3. May be compressed
Expand to fill container
Low density
May be mixed
Constant, uniform pressure on container walls

4. The Kinetic- Molecular Theory

5. The Kinetic-Molecular Theory
KMT is based on the motions of gas particles.
A gas that behaves exactly as outlined by KMT is known as an ideal gas.
While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure (high temperature & low pressure).

6. Principle Assumptions of the KMT
Gases consist of tiny subatomic particles.
The distance between particles is large compared with the size of the particles themselves.
Gas particles have no attraction for one another.

7. Principle Assumptions of the KMT
Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container.

8. Principle Assumptions of the KMT
No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic.
The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.

9. Kinetic Energy

10. Kinetic Energy
All gases have the same kinetic energy at the same temperature.
As a result, lighter molecules move faster than heavier molecules. mH 2
= 2 mO 2
= 32

11. Parameters for Describing Gases
A measurable factor forming one of a set that defines a system (or sets the conditions of its operation).

12. Parameter Symbol Description Units Force per unit area mmHg (torr)
or atm
Pressure P
Volume of
container
Volume V
liter
No. of moles
of gas
No.of moles n
mole
Absolute temp
of gas
Temperature K T

13. Measurement of Pressure of Gases

14. The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.

15. Pressure equals force per unit area.

16. Evangelista Torricelli 1644

17. Measuring Pressure
Attempts to pump water out of flooded mines often failed because
H2O can’t be lifted more than 34 feet.

18. Measuring Pressure
Torricelli believed the reason was that the P of atmosphere could not hold anything heavier than a 34’ column of water.

19. Pet watering dish
Chicken watering dish

20. Denmark Visitor Centre
dismantled
in 2011
Feb. 4, 2013: A giant barometer, which is used to measure atmospheric pressure, is being built in the Engineering building at Portland State University.

22. The atmosphere would support a column of H2O
Measuring Pressure
The atmosphere would support a column of H2O
> 34 feet high.
1 Atm
34’ column
of water

23. Mercury used because it’s so dense.
Torricelli Barometer
Pressure of the atmosphere supports a
column of Hg 760 mm high.
1 atm =
760 mm Hg
760 torr
29.92 in Hg
14.7 lb/in2
101,325 Pa
vacuum
1 atm
Mercury used because it’s so dense.

24. Boyle’s Law 1662

25. Robert Boyle

26. At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

27. Graph of pressure versus volume
Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

28. The effect of pressure on the volume of a gas.

29. An 8. 00 L sample of N2 is at a pressure of 500. torr
An 8.00 L sample of N2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 1. Organize the given information:
V1 = 8.00 L
P1 = 500 torr
V2 = 3.00 L
P2 = ?

30. Step 2. Write and solve the equation for the unknown.
An 8.00 L sample of N2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 2. Write and solve the equation for the unknown.

31. Step 3. Put the given information into the equation and calculate.
An 8.00 L sample of N2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 3. Put the given information into the equation and calculate.

32. A tank of O2 contains 1500. ml of gas at a pressure of 350. torr
A tank of O2 contains ml of gas at a pressure of 350. torr. What volume would the gas occupy at torr?
Step 1. Organize the given information:
V1 = ml
P2 = torr
P1 = 350. torr
V2 = ?

33. Step 2. Write and solve the equation for the unknown.
A tank of O2 contains ml of gas at a pressure of 350. torr. What volume would the gas occupy at torr?
Step 2. Write and solve the equation for
the unknown.

34. = 525 ml Step 3. Put the given information into
A tank of O2 contains ml of gas at a pressure of 350. torr. What volume would the gas occupy at torr?
Step 3. Put the given information into
the equation and calculate.
= 525 ml

35. Charles’ Law 1787

36. Jacques Charles

37. December 1783

38. Volume-temperature relationship of methane (CH4).

39. Charles’ Law
At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

40. Effect of temperature on the volume of a gas
Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

41. Step 1. Organize the information (remember to make units the same):
A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?
Step 1. Organize the information (remember to make units the same):
V1 = 255 mL
T1 = 75oC = 348 K
V2 = ?
T2 = 250oC = 523 K

42. Step 2. Write and solve the equation for the unknown:
A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?
Step 2. Write and solve the equation for the unknown:

43. Step 3. Put the given information into the equation and calculate:
A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?
Step 3. Put the given information into the equation and calculate:
V1 = 255 mL
T1 = 75oC = 348 K
V2 = ?
T2 = 250oC = 523 K

44. A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).
Step 1. Organize the given information:
V1 = 16.0 L
V2 = 20.0 L
T1 = 500.K
T2 = ?

45. Step 2. Write and solve the equation for the unknown.
A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).
Step 2. Write and solve the equation for
the unknown.

46. Step 3. Put the given information into the equation and calculate:
A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).
Step 3. Put the given information into
the equation and calculate:
= 625 K

47. Gay-Lussac’s Law 1802

48. Joseph Gay-Lussac

49. Lower T Lower P Higher T Higher P
Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature.
The pressure of a gas in a fixed volume increases with increasing temperature.
Lower T
Lower P
Higher T
Higher P

50. The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

51. Step 1. Organize the information (remember to make units the same):
At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?
Step 1. Organize the information (remember to make units the same):
P1 = 21.5 atm
T1 = 40oC = 313 K
P2 = ?
T2 = 100oC = 373 K

52. Step 2. Write and solve the equation for the unknown:
At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?
Step 2. Write and solve the equation for the unknown:

53. Step 3. Put the given information into the equation and calculate:
At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?
Step 3. Put the given information into the equation and calculate:
P1 = 21.5 atm
T1 = 40oC = 313 K
P2 = ?
T2 = 100oC = 373 K

54. Combined Gas Laws

55. A combination of Boyle’s and Charles’ Laws or Charles’ and Gay-Lussac’s Laws.
Used when pressure and temperature change at the same time.
Solve the equation for any one of the 6 variables

56. A sample of hydrogen occupies 465 ml at at 0oC and 760 torr
A sample of hydrogen occupies 465 ml at at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?
Step 1. Organize the given information, putting temperature in Kelvins:
oC = K
0oC = 273 K
-15oC = 258 K

57. Step 1. Organize the given information:
A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?
Step 1. Organize the given information:
P1 = 760 torr P2 = 950 torr
V1 = 465 mL V2 = ?
T1 = 273 K T2 = 258 K

58. Step 2. Write and solve the equation for the unknown V2.
A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?
Step 2. Write and solve the equation for
the unknown V2.

59. Step 3 Put the given information into the equation and calculate.
A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?
Step 3 Put the given information into the equation and calculate.

60. Dalton’s Law of Partial Pressures

61. John Dalton

62. Ptotal = Pa + Pb + Pc + Pd + ….
Each gas in a mixture exerts a pressure that is independent of the other gases present.
The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.
Ptotal = Pa + Pb + Pc + Pd + ….

63. Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm
A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container?
Ptotal = PHe + PNe+ PAr
Ptotal = 0.5 atm atm atm = 2.40 atm

64. Collecting a Gas Sample Over Water
The pressure in the collection container is equal to the atmospheric pressure.
The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure.

65. Oxygen collected over water.

66. A sample of O2 was collected over water in a bottle at a temperature of 25oC when the atmospheric pressure was 760 torr. What is the pressure of the O2 alone? The vapor pressure of water at 25oC is 23.8 torr.

67. Temp ºC
Press (torr)
4.6 13
11.2 26
25.2 39
52.4 1
4.9 14
12.0 27
26.7 40
55.3 2
5.3 15
12.8 28
28.3 41
58.3 3
5.7 16
13.6 29
30.0 42
61.5 4
6.1 17
14.5 30
31.8 43
64.8 5
6.5 18
15.5 31
33.7 44
68.3 6
7.0 19
16.5 32
35.7 45
71.9 7
7.5 20
17.5 33
37.7 46
75.7 8
8.0 21
18.7 34
39.9 47
79.6 9
8.6 22
19.8 35
42.2 48
83.7 10
9.2 23
21.0 36
44.6 49
88.0 11
9.8 24
22.4 37
37.1 50
92.5 12
10.5 25
23.8 38
49.7 51
97.2

68. Temp ºC
Press (torr) 52
102.1 64
179.3 76
301.4 88
487.1 53
107.2 65
187.5 77
314.1 89
506.1 54
112.5 66
196.1 78
327.3 90
525.8 55
118.0 67
205.0 79
341.0 91
546.0 56
123.8 68
214.2 80
355.1 92
567.0 57
129.8 69
223.7 81
369.7 93
588.6 58
136.1 70
233.7 82
384.9 94
610.9 59
142.6 71
243.9 83
400.6 95
633.9 60
149.4 72
254.6 84
416.8 96
657.6 61
156.4 73
265.7 85
433.6 97
682.1 62
163.8 74
277.2 86
450.9 98
707.3 63
171.4 75
289.1 87
468.7 99
733.2
100
760.0

69. A sample of nitrogen was collected over water and occupies 300
A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of 23oC is 21.0 torr.
Step 1. Organize the given information, putting temperature in Kelvin and correcting for water vapor pressure:
0oC = 273 K
23oC = 296 K

70. Step 1. Organize the given information,
A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of 23oC is 21.0 torr.
Step 1. Organize the given information,
P1 = 729 torr P2 = 760 torr
V1 = 300. mL V2 = ?
T1 = 296 K T2 = 273 K

71. A sample of nitrogen was collected over water and occupies 300
A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of 23oC is 21.0 torr.
Step 2. Write and solve the equation for the unknown V2.

72. A sample of nitrogen was collected over water and occupies 300
A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of 23oC is 21.0 torr.
Step 3. Put the given information into the equation and calculate.

73. Avogadro’s Law 1811

74. Amadeo (Amedeo) Avogadro

75. Avogadro’s Law
Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

76. Standard Temperature and Pressure

77. Standard Temperature and Pressure
Selected common reference points of temperature and pressure.
Standard Conditions
Standard Temperature and Pressure
STP
K or 0.00oC
1 atm or 760 torr or 760 mm Hg

78. Volume of one mole of any gas at STP = 22.4 L.
22.4 L at STP is known as the molar volume of any gas.

80. Ultimate Combined Gas Law

81. A combination of Boyle’s , Charles’ and Avogadro’s Laws.
Used when pressure, temperature, and moles change at the same time.
Solve the equation for any one of the 8 variables

82. Step 1. Organize the given information:
A L sample of He at 300 K and 9.7 atm containing moles is heated to 310 K under a pressure of atm. An additional 1.27 moles of He are added to the container. What is the new volume?
Step 1. Organize the given information:
P1 = 9.7 atm
V1 = 8.74 L
T1 = 300 K
n1 = 3.45 moles
P2 = 15.0 atm
V2 = ?
T2 = 310 K
n2 = = 4.72

83. Step 2. Write and solve the equation for the unknown V2.
A L sample of He at 300 K and 9.7 atm containing moles is heated to 310 K under a pressure of atm. An additional 1.27 moles of He are added to the container. What is the new volume?
Step 2. Write and solve the equation for the unknown V2.

84. Step 3. Put the given information into the equation and calculate.
A L sample of He at 300 K and 9.7 atm containing moles is heated to 310 K under a pressure of atm. An additional 1.27 moles of He are added to the container. What is the new volume?
Step 3. Put the given information into the equation and calculate.
= 7.99 L

85. Law of Combining Volumes Provided a method for
AVOGADRO'S LAW
Explained Gay Lussac's
Law of Combining Volumes
Provided a method
for
the determination of
mole weights of gases
comparing densities of gases
of known mole weight
Served as a foundation
for the devolopment of the
Kinetic-Molecular Theory

86. Mole-Mass-Volume Relationships

87. Density of Gases

89. Density of Gases
grams
liters

90. Density of Gases
depends on T and P

91. The density of neon at STP is 0. 900 g/L
The density of neon at STP is g/L. What is the mole weight of neon?

92. 1 mole of any gas occupies 22.4 L at STP
The mole weight of SO2 is g/mole. Determine the density of SO2 at STP.
1 mole of any gas occupies 22.4 L at STP

93. Ideal Gas Equation

94. V a P nT

95. V a P nT

96. V a P nT

97. atmospheres nT
V a P

98. liters nT
V a P

99. moles nT
V a P

100. Kelvin nT
V a P

101. Ideal Gas Constant nT
V a P

103. nT
V a P

104. Step 1. Organize the given information.
A balloon filled with moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?
Step 1. Organize the given information.
Convert temperature to kelvins.
K = oC + 273
K = 25oC = 298K
Convert pressure to atmospheres.

105. Step 2. Write and solve the ideal gas equation for the unknown.
A balloon filled with moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?
Step 2. Write and solve the ideal gas equation for the unknown.
Step 3. Substitute the given data into the equation and calculate.
(25º + 273)
(750/760)

106. Mole-Weight Calculations

107. Determination of Molecular Weights Using the Ideal Gas Equation

108. Calculate the mole weight of an unknown gas, if 0
Calculate the mole weight of an unknown gas, if g occupies 250 mL at a temperature of 305 K and a pressure of atm.
V = 250 mL = L
g = g
P = atm
T = 305 K

109. Gas Stoichiometry

110. Definition
Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

111. All calculations are done at STP.
Gases are assumed to behave as ideal gases.
A gas not at STP is converted to STP.

112. Gas Stoichiometry
Primary conversions involved in stoichiometry.

113. Mole-Volume Calculations
Mass-Volume Calculations

114. What volume of oxygen (at STP) can be formed from 0
What volume of oxygen (at STP) can be formed from mole of potassium chlorate?
Step 1 Write the balanced equation
2 KClO3  2 KCl + 3 O2
Step 2 The starting amount is mol KClO3. The conversion is
moles KClO3  moles O2  liters O2

115. Step 3. Solve the problem in one continuous calculation.
What volume of oxygen (at STP) can be formed from mole of potassium chlorate?
Step 3. Solve the problem in one continuous calculation.
2 KClO3  2KCl + 3 O2

116. 2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
2 Al(s) HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 1 Calculate moles of H2.
grams Al  moles Al  moles H2

117. Step 2 Calculate liters of H2.
What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 2 Calculate liters of H2.
Convert oC to K: 30.oC = 303 K
Convert torr to atm:

118. Step 3. Solve the ideal gas equation for V
What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
Step 3. Solve the ideal gas equation for V
PV = nRT

119. Volume-Volume Calculations

120. Gay Lussac’s Law of Combining Volumes 1809
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers. N2
1 volume +
3 H2
3 volumes →
2 NH3
2 volumes

121. For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.
H2(g) Cl2(g)  HCl(g)
1 mole H2
1 mole Cl2
2 mole HCl
22.4 L
STP
22.4 L
STP
2 x L
STP
1 volume
1 volume
2 volumes
Y volume
Y volume
2Y volumes

122. What volume of nitrogen will react with 600
What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed?
N2(g) + 3H2(g)  2NH3(g)

123. Molecular Formula Calculations

124. Chapter 7 Review

125. A compound made of 30. 4% N and 69
A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.
N3O6

126. Step 1. Determine the Mole Weight:
An unknown liquid contains 14.3% H and 85.7% C by weight grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?
Step 1. Determine the Mole Weight:

127. mass = 29.4 grams V = 3.60 L T = 260 + 273 = 533 K P = 2.84 atm
An unknown liquid contains 14.3% H and 85.7% C by weight grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?
mass = 29.4 grams
V = 3.60 L
T = = 533 K
P = 2.84 atm

128. Step 2. Use the mole weight and percentages to find the formula:
An unknown liquid contains 14.3% H and 85.7% C by weight grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?
Step 2. Use the mole weight and
percentages to find the formula:
C9H18

129. Diffusion
The ability of two or more gases to mix spontaneously until they form a uniform mixture.
Stopcock closed No diffusion occurs
Stopcock open Diffusion occurs

130. Effusion
A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

131. Thomas Graham

132. Graham’s Law of Effusion/Diffusion
The rates of effusion or diffusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their mole weights.

133. What is the ratio of the rate of diffusion of CO to CO2?

134. It takes 30.0 sec for 10. ml of O2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

135. It takes 30.0 sec for 10. ml of O2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

136. Real Gases

137. Ideal Gas An ideal gas obeys the gas laws.
The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures.
The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.

138. Real Gases
Deviations from the gas laws occur at high pressures and low temperatures.
At high pressures the volumes of the real gas molecules are not negligible compared to the volume of the gas
At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.

139. van der Waals model
Real gases are often modeled by taking into account their molar weight and molar volume
Where P is the pressure, T is the temperature, R the ideal gas constant, and Vm the molar volume. a and b are parameters that are determined empirically for each gas, but are sometimes estimated from their critical temperature (Tc) and critical pressure (Pc) using these relations:

140. The End