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Properties of Gases 1. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 2.

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Presentation on theme: "Properties of Gases 1. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 2."— Presentation transcript:

1 Properties of Gases 1

2 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 2

3 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 3

4 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 4

5 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 5

6 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 4. Avogadros Hypothesis 6

7 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 4. Avogadros Hypothesis 5. Ideal gas law 7

8 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 4. Avogadros Hypothesis 5. Ideal gas law 6. Daltons law of partial pressures 8

9 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 4. Avogadros Hypothesis 5. Ideal gas law 6. Daltons law of partial pressures 7. Grahams law of effusion 9

10 Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyles Law 3. Charles Law 4. Avogadros Hypothesis 5. Ideal gas law 6. Daltons law of partial pressures 7. Grahams law of effusion 8. Kinetic theory and real gases 10

11 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. 11

12 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are: 12

13 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are: 1. pressure 13

14 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are: 1. pressure 2. volume 14

15 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are: 1. pressure 2. volume 3. temperature 15

16 Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are: 1. pressure 2. volume 3. temperature 4. mass (or moles) 16

17 Pressure: Gases exert pressure on any surface with which they come into contact. That is, all gases expand uniformly to occupy whatever space is available. 17

18 Pressure: Gases exert pressure on any surface with which they come into contact. That is, all gases expand uniformly to occupy whatever space is available. Pressure is define by the equation: force pressure = area 18

19 Pressure: Gases exert pressure on any surface with which they come into contact. That is, all gases expand uniformly to occupy whatever space is available. Pressure is define by the equation: force pressure = area or F P = A 19

20 Units of pressure 20

21 Units of pressure distance velocity = time 21

22 Units of pressure distance velocity = time velocity acceleration = time 22

23 Units of pressure distance velocity = time velocity acceleration = time i.e. distance acceleration = time 2 23

24 To find the SI units of force: 24

25 To find the SI units of force: force = mass x acceleration 25

26 To find the SI units of force: force = mass x acceleration distance force = mass x time 2 26

27 To find the SI units of force: force = mass x acceleration distance force = mass x time 2 Now plug in the SI units for each quantity on the right-hand side of the equation. The SI unit of force is the newton (named after Newton) and abbreviated N. 27

28 To find the SI units of force: force = mass x acceleration distance force = mass x time 2 Now plug in the SI units for each quantity on the right-hand side of the equation. The SI unit of force is the newton (named after Newton) and abbreviated N. 1 N = 1 kg m s -2 28

29 The SI units of pressure are: Nm -2 29

30 The SI units of pressure are: Nm -2 The SI unit of pressure is called the pascal (named after Pascal) and is abbreviated Pa. 30

31 The SI units of pressure are: Nm -2 The SI unit of pressure is called the pascal (named after Pascal) and is abbreviated Pa. 1 Pa = 1 Nm -2 = 1 kg m -1 s -2 31

32 The SI units of pressure are: Nm -2 The SI unit of pressure is called the pascal (named after Pascal) and is abbreviated Pa. 1 Pa = 1 Nm -2 = 1 kg m -1 s -2 The most common unit of pressure is the non- SI unit, the atmosphere, abbreviated atm. 32

33 1 atm = 76 cm Hg 33

34 1 atm = 76 cm Hg = 760 mm Hg 34

35 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) 35

36 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition) 36

37 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition) 1 bar = 10 5 Pa 37

38 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition) 1 bar = 10 5 Pa Hence 1atm = 1.01325 bar 38

39 Atmospheric pressure is measured by a barometer. 39

40 The difference between a gas and a vapor: 40

41 The difference between a gas and a vapor: A gas is a substance normally in the gaseous state at ordinary temperatures and pressures. 41

42 The difference between a gas and a vapor: A gas is a substance normally in the gaseous state at ordinary temperatures and pressures. A vapor is the gaseous form of any substance that is a liquid or a solid at normal temperatures and pressures. 42

43 The difference between a gas and a vapor: A gas is a substance normally in the gaseous state at ordinary temperatures and pressures. A vapor is the gaseous form of any substance that is a liquid or a solid at normal temperatures and pressures. Thus, at room temperature and 1 atm, we speak of water vapor and helium gas. 43

44 Temperature: For the gas equations, the temperature is measured on the Kelvin scale – the units are K (kelvin). 44

45 Temperature: For the gas equations, the temperature is measured on the Kelvin scale – the units are K (kelvin). The absolute lowest temperature is zero degrees kelvin. 45

46 Temperature: For the gas equations, the temperature is measured on the Kelvin scale – the units are K (kelvin). The absolute lowest temperature is zero degrees kelvin. The temperature in K is determined from the temperature measured in o C by: degrees K = degrees o C + 273.15 46

47 The Gas Laws 47

48 The Gas Laws The first systematic and quantitative study of gas behavior was carried out by Boyle. 48

49 The Gas Laws The first systematic and quantitative study of gas behavior was carried out by Boyle. Boyle noticed that when the temperature and the amount of gas are held constant, the volume of a gas is inversely proportional to the applied pressure acting on the gas. 49

50 50

51 (const. T and n) This is a statement of Boyles Law. (V is the volume, P is the pressure, T is the temperature, and n is the moles of gas). 51

52 (const. T and n) This is a statement of Boyles Law. (V is the volume, P is the pressure, T is the temperature, and n is the moles of gas). C V = (const. T and n) P 52

53 (const. T and n) This is a statement of Boyles Law. (V is the volume, P is the pressure, T is the temperature, and n is the moles of gas). C V = (const. T and n) P (C is the proportionality constant). 53

54 (const. T and n) This is a statement of Boyles Law. (V is the volume, P is the pressure, T is the temperature, and n is the moles of gas). C V = (const. T and n) P (C is the proportionality constant). This is also a statement of Boyles Law. 54

55 We can write Boyles law as: PV = C (const. T and n) 55

56 We can write Boyles law as: PV = C (const. T and n) If the initial values of pressure and volume are P i and V i and if the conditions are changed to a final pressure P f and final volume V f then we can write: 56

57 We can write Boyles law as: PV = C (const. T and n) If the initial values of pressure and volume are P i and V i and if the conditions are changed to a final pressure P f and final volume V f then we can write: initial conditions P i V i = C 57

58 We can write Boyles law as: PV = C (const. T and n) If the initial values of pressure and volume are P i and V i and if the conditions are changed to a final pressure P f and final volume V f then we can write: initial conditions P i V i = C final conditions P f V f = C 58

59 Hence: P i V i = P f V f (const. T and n) This is also a statement of Boyles Law. 59

60 Hence: P i V i = P f V f (const. T and n) This is also a statement of Boyles Law. Problem Example: If 0.100 liters of a gas, originally at 760.0 torr, is compressed to a pressure of 800.0 torr, at a constant temperature, what is the final volume of the gas? 60

61 No mention is made about the moles of gas – so make the assumption that n remains constant. Since temperature is constant, we can use Boyles Law. P i V i = P f V f 61

62 No mention is made about the moles of gas – so make the assumption that n remains constant. Since temperature is constant, we can use Boyles Law. P i V i = P f V f This can be rearranged to read: P i V f = V i P f 62

63 No mention is made about the moles of gas – so make the assumption that n remains constant. Since temperature is constant, we can use Boyles Law. P i V i = P f V f This can be rearranged to read: P i V f = V i P f 760.0 torr V f = x 0.100 l 800.0 torr V f = 0.0950 l 63

64 Charles and Gay-Lussacs Law 64

65 Charles and Gay-Lussacs Law Charles (1787) and Gay-Lussac (1802) were the first investigators to study the effect of temperature on gas volume. Their studies showed that at constant pressure, and a fixed number of moles of gas, that the volume of a gas is directly proportional to the temperature of the gas. 65

66 Charles and Gay-Lussacs Law Charles (1787) and Gay-Lussac (1802) were the first investigators to study the effect of temperature on gas volume. Their studies showed that at constant pressure, and a fixed number of moles of gas, that the volume of a gas is directly proportional to the temperature of the gas. (const. P and n) This is a statement of Charles Law. 66

67 67

68 The proportionality can be turned into an equality: V = C T (const. P and n) 68

69 The proportionality can be turned into an equality: V = C T (const. P and n) This is a statement of Charles Law. (The prime is used to signify that this constant is different to the constant appearing in Boyles law). 69

70 If the initial values of temperature and volume are T i and V i and if the conditions are changed to a final temperature T f and final volume V f then we can write: 70

71 If the initial values of temperature and volume are T i and V i and if the conditions are changed to a final temperature T f and final volume V f then we can write: V i initial conditions = C T i 71

72 If the initial values of temperature and volume are T i and V i and if the conditions are changed to a final temperature T f and final volume V f then we can write: V i initial conditions = C T i V f final conditions = C T f 72

73 If the initial values of temperature and volume are T i and V i and if the conditions are changed to a final temperature T f and final volume V f then we can write: V i initial conditions = C T i V f final conditions = C T f Hence: V i V f = (const. P, n) T i T f This is a statement of Charles Law. 73

74 Problem Example: A gas occupying 4.50 x 10 2 ml is heated from 22.0 o C to 187 o C at constant pressure. What is the final volume of the gas? 74

75 Problem Example: A gas occupying 4.50 x 10 2 ml is heated from 22.0 o C to 187 o C at constant pressure. What is the final volume of the gas? There is no mention of the moles of gas changing, so assume that n is constant. Since pressure is constant, apply Charles law: 75

76 Problem Example: A gas occupying 4.50 x 10 2 ml is heated from 22.0 o C to 187 o C at constant pressure. What is the final volume of the gas? There is no mention of the moles of gas changing, so assume that n is constant. Since pressure is constant, apply Charles law: V i V f = T i T f 76

77 Problem Example: A gas occupying 4.50 x 10 2 ml is heated from 22.0 o C to 187 o C at constant pressure. What is the final volume of the gas? There is no mention of the moles of gas changing, so assume that n is constant. Since pressure is constant, apply Charles law: V i V f = T i T f Note that it is necessary to convert the temperatures given to units of K. 77

78 T i = 22.0 + 273.15 = 295.2 K T f = 187 + 273.15 = 4.60 x 10 2 K 78

79 T i = 22.0 + 273.15 = 295.2 K T f = 187 + 273.15 = 4.60 x 10 2 K T f V f = V i T i 4.60 x 10 2 K = 4.50 x 10 2 ml 2.952 x 10 2 K = 701 ml 79

80 Avogadros Law 80

81 Avogadros Law In 1811 Avogadro published a hypothesis which stated that at the same temperature and pressure, equal volumes of gases contain the same number of molecules (or atoms if the gas is monatomic). 81

82 Avogadros Law In 1811 Avogadro published a hypothesis which stated that at the same temperature and pressure, equal volumes of gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of molecules present. 82

83 It is more convenient to work in terms of moles. A mole contains 6.02214 x 10 23 items. 83

84 It is more convenient to work in terms of moles. A mole contains 6.02214 x 10 23 items. For example, one mole of argon contains 6.02214 x 10 23 atoms of argon, a mole of dinitrogen contains 6.02214 x 10 23 molecules of N 2. 84

85 It is more convenient to work in terms of moles. A mole contains 6.02214 x 10 23 items. For example, one mole of argon contains 6.02214 x 10 23 atoms of argon, a mole of dinitrogen contains 6.02214 x 10 23 molecules of N 2. A mole of apples is really, really,...really big! 85

86 It is more convenient to work in terms of moles. A mole contains 6.02214 x 10 23 items. For example, one mole of argon contains 6.02214 x 10 23 atoms of argon, a mole of dinitrogen contains 6.02214 x 10 23 molecules of N 2. A mole of apples is really, really,...really big! The mole is a convenient lab-sized unit for amount of substance. 86

87 Avogadros law can be written as: (const. P and T) 87

88 Avogadros law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = Cn (const. P and T) 88

89 Avogadros law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = Cn (const. P and T) This is a statement of Avogadros Law. 89

90 Avogadros law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = Cn (const. P and T) This is a statement of Avogadros Law. (The double prime is used to signify that this constant is different to the constants appearing in Boyles law and Charles law). 90

91 If the initial values of volume and moles are V i and n i and if the conditions are changed to a final volume V f and final moles n f then we can write: 91

92 If the initial values of volume and moles are V i and n i and if the conditions are changed to a final volume V f and final moles n f then we can write: V i initial conditions = C n i 92

93 If the initial values of volume and moles are V i and n i and if the conditions are changed to a final volume V f and final moles n f then we can write: V i initial conditions = C n i V f final conditions = C n f 93

94 If the initial values of volume and moles are V i and n i and if the conditions are changed to a final volume V f and final moles n f then we can write: V i initial conditions = C n i V f final conditions = C n f Hence: V i V f = (const. P, T) n i n f 94

95 If the initial values of volume and moles are V i and n i and if the conditions are changed to a final volume V f and final moles n f then we can write: V i initial conditions = C n i V f final conditions = C n f Hence: V i V f = (const. P, T) n i n f This is a statement of Avogadros Law. 95

96 Avogadros law means that, when two gases react with each other, their volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio. 96

97 Avogadros law means that, when two gases react with each other, their volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio. Example: 3 H 2(g) + N 2(g) 2 NH 3(g) 97

98 Avogadros law means that, when two gases react with each other, their volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio. Example: 3 H 2(g) + N 2(g) 2 NH 3(g) 3 volumes 1 volume 2 volumes 98

99 Ideal Gas Law 99

100 Ideal Gas Law Summary so far: 100

101 Ideal Gas Law Summary so far: 1 Boyles law: V (const. T and n) P 101

102 Ideal Gas Law Summary so far: 1 Boyles law: V (const. T and n) P Charles law V T (const. P and n) 102

103 Ideal Gas Law Summary so far: 1 Boyles law: V (const. T and n) P Charles law V T (const. P and n) Avogadros law V n (const. P and T) 103

104 Math Aside: 104

105 Math Aside: If f X 105

106 Math Aside: If f X and f Y 106

107 Math Aside: If f X and f Y and f Z -1 107

108 Math Aside: If f X and f Y and f Z -1 Then f X Y Z -1 108

109 Using the result from the math aside, we can put the expressions for Boyles law, Charles law, and Avogadros law together, so that: nT V P 109

110 Using the result from the math aside, we can put the expressions for Boyles law, Charles law, and Avogadros law together, so that: nT V P The proportionality can be replaced by an equation by inserting a proportionality constant. In this case the constant is represented by the symbol R, and is called the gas constant. Hence: 110

111 Using the result from the math aside, we can put the expressions for Boyles law, Charles law, and Avogadros law together, so that: nT V P The proportionality can be replaced by an equation by inserting a proportionality constant. In this case the constant is represented by the symbol R, and is called the gas constant. Hence: PV = n R T Ideal gas equation 111

112 Ideal gas: Is defined to be a gas that satisfies the ideal gas equation. 112

113 Ideal gas: Is defined to be a gas that satisfies the ideal gas equation. Evaluation of the gas constant: Experiments show that 1 mol of any ideal gas at 0 O C and 1 atm pressure occupies 22.414 liters. The conditions 0 O C and 1 atm are called standard temperature and pressure, abbreviated STP. 113

114 Ideal gas: Is defined to be a gas that satisfies the ideal gas equation. Evaluation of the gas constant: Experiments show that 1 mol of any ideal gas at 0 O C and 1 atm pressure occupies 22.414 liters. The conditions 0 O C and 1 atm are called standard temperature and pressure, abbreviated STP. (1 atm) (22.414 l) R = ( 1 mol) (273.15 K) = 0.082057 l atm K -1 mol -1 114

115 In the SI unit system, R = 8.3145 J K -1 mol -1 115

116 In the SI unit system, R = 8.3145 J K -1 mol -1 Here, J stands for joule, the SI unit of energy. 116

117 In the SI unit system, R = 8.3145 J K -1 mol -1 Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m 2 s -2 117

118 In the SI unit system, R = 8.3145 J K -1 mol -1 Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m 2 s -2 Exercise: Try to convert R = 0.082057 l atm K -1 mol -1 to the value given in SI units. (Its a factor-label exercise). 118

119 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO 2 at STP. Assume CO 2 can be treated as an ideal gas. 119

120 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO 2 at STP. Assume CO 2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 o C = 273.15 K (exact value) 120

121 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO 2 at STP. Assume CO 2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 o C = 273.15 K (exact value) n R T V = P 121

122 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO 2 at STP. Assume CO 2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 o C = 273.15 K (exact value) n R T V = P (0.168 mol) (0.08206 l atm mol -1 K -1 ) (273 K) V = (1 atm) = 3.76 l 122

123 Combined Gas Law 123

124 Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are P i, V i, T i, and n i, and if the conditions are changed to a final pressure, volume, temperature, and moles P f, V f, T f, and n f respectively, then we can write: 124

125 Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are P i, V i, T i, and n i, and if the conditions are changed to a final pressure, volume, temperature, and moles P f, V f, T f, and n f respectively, then we can write: P i V i initial conditions = R n i T i 125

126 Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are P i, V i, T i, and n i, and if the conditions are changed to a final pressure, volume, temperature, and moles P f, V f, T f, and n f respectively, then we can write: P i V i initial conditions = R n i T i P f V f final conditions = R n f T f 126

127 Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are P i, V i, T i, and n i, and if the conditions are changed to a final pressure, volume, temperature, and moles P f, V f, T f, and n f respectively, then we can write: P i V i initial conditions = R n i T i P f V f final conditions = R n f T f Hence: P i V i P f V f = n i T i n f T f 127

128 Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are P i, V i, T i, and n i, and if the conditions are changed to a final pressure, volume, temperature, and moles P f, V f, T f, and n f respectively, then we can write: P i V i initial conditions = R n i T i P f V f final conditions = R n f T f Hence: P i V i P f V f = n i T i n f T f This is the Combined Gas Law. 128

129 Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 10 2 atm at 25.0 o C. What volume of air does this correspond to at STP? 129

130 Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 10 2 atm at 25.0 o C. What volume of air does this correspond to at STP? Given initial data: P i = 1.50 x 10 2 atm V i = 21. 0 l T i = 298.2 K (25.0 + 273.15) 130

131 Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 10 2 atm at 25.0 o C. What volume of air does this correspond to at STP? Given initial data: P i = 1.50 x 10 2 atm V i = 21. 0 l T i = 298.2 K (25.0 + 273.15) Given final data: P f = 1 atm T f = 273.15 K 131

132 P i V i P f V f = n i T i n f T f Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: P i V i P f V f = T i T f 132

133 P i V i P f V f = n i T i n f T f Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: P i V i P f V f = T i T f The equation can be rearranged, so that P i T f V f = V i P f T i 133

134 (1.50 x 10 2 atm) (273.15 K) V f = (21.0 l) (1 atm) (298.2 K) = 2.89 x 10 3 l at STP 134

135 Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 O C and 492 torr has a mass 0f 0.116 g. Two steps: first find moles of gas, then determine the molar mass. 135

136 Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 O C and 492 torr has a mass 0f 0.116 g. Two steps: first find moles of gas, then determine the molar mass. From PV = nRT, PV n = RT 136

137 Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 O C and 492 torr has a mass 0f 0.116 g. Two steps: first find moles of gas, then determine the molar mass. From PV = nRT, PV n = RT 1 atm 1 l (492 torr)( )( 279 ml)( ) 760 torr 1000 ml = (0.08206 l atm mol -1 K -1 )(304.5 K) = 7.23 x 10 -3 mol 137

138 The molar mass is given by: 138

139 The molar mass is given by: 0.116 g = 7.23 x 10 -3 mol 139

140 The molar mass is given by: 0.116 g = 7.23 x 10 -3 mol = 16.0 g mol -1 140

141 Daltons Law of Partial Pressures 141

142 Daltons Law of Partial Pressures Daltons Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume. 142

143 Daltons Law of Partial Pressures Daltons Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume. Consider a simple case: A mixture of two gases A and B in a container of volume V and temperature T. 143

144 The pressure exerted by gas A – called the partial pressure of gas A – is given by: 144

145 The pressure exerted by gas A – called the partial pressure of gas A – is given by: n A RT P A = V 145

146 The pressure exerted by gas A – called the partial pressure of gas A – is given by: n A RT P A = V Similarly for gas B, n B RT P B = V 146

147 Now the total pressure P Total is n Total RT P Total = V 147

148 Now the total pressure P Total is n Total RT P Total = V Now n Total = n A + n B so that: 148

149 Now the total pressure P Total is n Total RT P Total = V Now n Total = n A + n B so that: (n A + n B )RT n A RT n B RT P Total = = + V V V 149

150 Now the total pressure P Total is n Total RT P Total = V Now n Total = n A + n B so that: (n A + n B )RT n A RT n B RT P Total = = + V V V Hence, P Total = P A + P B 150

151 Now the total pressure P Total is n Total RT P Total = V Now n Total = n A + n B so that: (n A + n B )RT n A RT n B RT P Total = = + V V V Hence, P Total = P A + P B This is Daltons Law of partial pressures: The total pressure is the sum of the partial pressures. 151

152 The preceding result can be generalized to any number of components: P Total = P A + P B + P C + P D + … where P A, P B, P C, P D, etc. are the partial pressures of the individual gases. 152

153 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm. 153

154 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm. 154

155 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm. 155

156 156

157 157

158 But 158

159 Similar calculations give and 159

160 Daltons law has a practical application when calculating the volume of gases collected over water. For a gas collected over water, the measured pressure is given by where denotes the vapor pressure of water. 160

161 161

162 (p. 217) 162

163 Problem example: O 2 generated in the decomposition of KClO 3 is collected over water. The volume of the gas collected at 24 o C and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O 2 obtained. The vapor pressure of H 2 O at 24 o C is 22.4 torr. First step: Calculate the partial pressure of O 2. (Daltons law) 163

164 = 762 torr – 22.4 torr = 739.6 torr (extra sig. fig) = 0.973 atm 164

165 From the ideal gas equation PV = nRT, PV n = RT (0.973 atm) (0.128 l) = (0.08206 l atm mol -1 K -1 )(297 K) = 0.00511 mols 165

166 Grahams Law of Effusion 166

167 Grahams Law of Effusion Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent. 167

168 Grahams Law of Effusion Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent. Effusion: The process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening. 168

169 Grahams law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate: abbreviated R (dont get this confused with the gas constant). 169

170 Grahams law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate: abbreviated R (dont get this confused with the gas constant). Gas density: abbreviated d. 170

171 Grahams law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate: abbreviated R (dont get this confused with the gas constant). Gas density: abbreviated d. 1 R (const. P, T) 171

172 Grahams law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate: abbreviated R (dont get this confused with the gas constant). Gas density: abbreviated d. 1 R (const. P, T) This is Grahams law of effusion. 172

173 The proportionality can be turned into an equality: R = C (const. P and T) This is a statement of Grahams law of effusion. (The constant is unrelated to any of the previous constants in Boyles law, etc.). 173

174 For two gases A and B, the relative rates of effusion can be evaluated as follows: 174

175 For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: R A = C 175

176 For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: R A = C For gas B: R B = C 176

177 For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: R A = C For gas B: R B = C Hence: R A = R B 177

178 For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: R A = C For gas B: R B = C Hence: R A = R B R A = (const. P, T) R B 178

179 For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: R A = C For gas B: R B = C Hence: R A = R B R A = (const. P, T) R B This is a statement of Grahams law of effusion Law. 179

180 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? 180

181 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N 2 and B be CO 2. 181

182 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N 2 and B be CO 2. 182

183 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N 2 and B be CO 2. 183

184 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N 2 and B be CO 2. = 1.25 184

185 Problem example, Graham s law: Under conditions for which the density of carbon dioxide is 1.96 g/l and that of dinitrogen is 1.25 g/l, which gas will effuse more rapidly? What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Let A be N 2 and B be CO 2. = 1.25 So = 1.25 x, that is, N 2 effuses 1.25 times faster than CO 2. 185

186 An alternative form can be developed for Grahams law of effusion, where the ratio of the rates is determined from the molar masses of the two gases. 186

187 An alternative form can be developed for Grahams law of effusion, where the ratio of the rates is determined from the molar masses of the two gases. m m d = M = V n 187

188 An alternative form can be developed for Grahams law of effusion, where the ratio of the rates is determined from the molar masses of the two gases. m m d = M = V n so nM d = V 188

189 An alternative form can be developed for Grahams law of effusion, where the ratio of the rates is determined from the molar masses of the two gases. m m d = M = V n so nM d = V and PV = n RT, so (n/V) = P/(RT) 189

190 An alternative form can be developed for Grahams law of effusion, where the ratio of the rates is determined from the molar masses of the two gases. m m d = M = V n so nM d = V and PV = n RT, so (n/V) = P/(RT) PM d = RT 190

191 Now plug this result for d into the expression for Grahams law. 191

192 Now plug this result for d into the expression for Grahams law. That is (const. P, T) 192

193 Now plug this result for d into the expression for Grahams law. That is (const. P, T) This is a statement of Grahams law of effusion Law. 193

194 Exercise, Graham s law: What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? 194

195 Exercise, Graham s law: What is the ratio of the rates of effusion of dinitrogen to carbon dioxide? Note: In this example, no density information is given, so use the formula involving the molar masses. 195

196 Kinetic Theory of Gases 196

197 The kinetic theory of gases provides a working model of a gas. It is an attempt to interpret experimental observations at the molecular level. 197

198 The kinetic theory of gases provides a working model of a gas. It is an attempt to interpret experimental observations at the molecular level. Basic Postulates of Kinetic Theory: 198

199 The kinetic theory of gases provides a working model of a gas. It is an attempt to interpret experimental observations at the molecular level. Basic Postulates of Kinetic Theory: 1. A gas consists of an extremely large number of tiny particles that are in constant random motion. 199

200 2. The particles are separated by distances far greater than their own dimensions. The molecules can be considered as point-like, that is, they posses mass but have negligible volume (compared with the volume of the container). 200

201 2. The particles are separated by distances far greater than their own dimensions. The molecules can be considered as point-like, that is, they posses mass but have negligible volume (compared with the volume of the container). 3. All molecular collisions are elastic, that is, the sum of the kinetic energy of the colliding molecules remains unchanged before and after the collision. 201

202 It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v 2 T The bar over KE means average, and over v 2 means the average of v 2. 202

203 It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v 2 T The bar over KE means average, and over v 2 means the average of v 2. v 2 is called the mean square velocity. 203

204 It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v 2 T The bar over KE means average, and over v 2 means the average of v 2. v 2 is called the mean square velocity. If there are N molecules v 1 2 + v 2 2 + … + v N 2 v 2 = N 204

205 Molecules exert neither attractive nor repulsive forces on one another. This ties in directly with postulate 3. 205

206 Molecules exert neither attractive nor repulsive forces on one another. This ties in directly with postulate 3. Compressibility: Since gas molecules are separated by large distances (relative to the molecular size), they can be compressed easily to occupy smaller volumes. 206

207 Pressure – volume connection (p. 215): 207

208 Pressure – volume connection: Increase the external pressure on the piston at constant temperature, then more molecules strike the container walls, so the force per unit area increases, and hence the pressure of the gas increases. 208

209 Pressure – volume connection: Increase the external pressure on the piston at constant temperature, then more molecules strike the container walls, so the force per unit area increases, and hence the pressure of the gas increases. That is, a smaller volume implies a larger pressure. Detailed arguments lead to P V -1. 209

210 Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules. 210

211 Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. 211

212 Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. If the volume of the container is kept constant, the area being struck is the same, so the force per unit area, that is the pressure, increases. 212

213 Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. If the volume of the container is kept constant, the area being struck is the same, so the force per unit area, that is the pressure, increases. Expect P T (const. V, n). 213

214 Temperature – volume connection: Increase the temperature at constant pressure. 214

215 Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. 215

216 Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. 216

217 Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. This would increase the pressure, but if this is held constant, it would be necessary for the volume of the container to increase to maintain this constant pressure. 217

218 Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules. At higher velocities, the gas molecules strike the container walls more frequently and with greater force. This would increase the pressure, but if this is held constant, it would be necessary for the volume of the container to increase to maintain this constant pressure. Expect V T (const. P, n). 218

219 Grahams law of effusion: Consider two gases (call them A and B) at the same temperature. 219

220 Grahams law of effusion: Consider two gases (call them A and B) at the same temperature. Kinetic theory indicates that the average kinetic energy for both gases must be the same. 220

221 Grahams law of effusion: Consider two gases (call them A and B) at the same temperature. Kinetic theory indicates that the average kinetic energy for both gases must be the same. gas A gas B KE A = ½ m A v A 2 KE B = ½ m B v B 2 221

222 Grahams law of effusion: Consider two gases (call them A and B) at the same temperature. Kinetic theory indicates that the average kinetic energy for both gases must be the same. gas A gas B KE A = ½ m A v A 2 KE B = ½ m B v B 2 If gas B molecules have higher mass, then the only way the average KE can be the same is 222

223 that the average velocity of the B molecules is smaller than gas A. 223

224 that the average velocity of the B molecules is smaller than gas A. The molecules moving more quickly are more likely to hit the opening, and hence the gas with the molecules having the lower mass will effuse faster. 224

225 Maxwell Distribution of Molecular Speeds 225

226 Maxwell Distribution of Molecular Speeds At a given instant, how many molecules are moving at a particular speed? 226

227 Maxwell Distribution of Molecular Speeds At a given instant, how many molecules are moving at a particular speed? The answer to this question is provided by Maxwells distribution of speeds curve. 227

228 Maxwell Distribution of Molecular Speeds (p. 205) 228

229 One important point to note is that although there are always some slow-moving molecules, there are fewer slow-moving molecules at higher temperatures. 229

230 One important point to note is that although there are always some slow-moving molecules, there are fewer slow-moving molecules at higher temperatures. This is important for understanding the rates (i.e. speeds) of gas phase chemical reactions. 230


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