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1. Properties of Gases
2. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following.
3. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure.
4. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law
5. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law
6. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law 4. Avogadro’s Hypothesis
7. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law 4. Avogadro’s Hypothesis 5. Ideal gas law
8. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law 4. Avogadro’s Hypothesis 5. Ideal gas law 6. Dalton’s law of partial pressures
9. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law 4. Avogadro’s Hypothesis 5. Ideal gas law 6. Dalton’s law of partial pressures 7. Graham’s law of effusion
10. Key objectives: Be able to state (i.e. define) and use in practical calculations, the following. 1. Concept of force and pressure. 2. Boyle’s Law 3. Charles’ Law 4. Avogadro’s Hypothesis 5. Ideal gas law 6. Dalton’s law of partial pressures 7. Graham’s law of effusion 8. Kinetic theory and real gases
11. Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures.
12. Gas: Refers to a substance that is entirely gaseous at ordinary temperatures and pressures. The 4 important measurable properties of gases are:
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
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
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
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)
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.
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
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
20. Units of pressure
21. Units of pressure distance velocity = time
22. Units of pressure distance velocity = time velocity acceleration = time
23. Units of pressure distance velocity = time velocity acceleration = time i.e. distance acceleration = time2
24. To find the SI units of force:
25. To find the SI units of force: force = mass x acceleration
26. To find the SI units of force: force = mass x acceleration distance force = mass x time2
27. To find the SI units of force: force = mass x acceleration distance force = mass x time2 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.
28. To find the SI units of force: force = mass x acceleration distance force = mass x time2 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
29. The SI units of pressure are: Nm-2
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.
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
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.
33. 1 atm = 76 cm Hg
34. 1 atm = 76 cm Hg = 760 mm Hg
35. 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli)
36. 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition)
37. 1 atm = 76 cm Hg = 760 mm Hg = 760 torr (named after Torricelli) = 101325 Pa (by definition) 1 bar = 10 5 Pa
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
39. Atmospheric pressure is measured by a barometer.
40. The difference between a gas and a vapor:
41. The difference between a gas and a vapor: A gas is a substance normally in the gaseous state at ordinary temperatures and pressures.
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.
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.
44. Temperature: For the gas equations, the temperature is measured on the Kelvinscale – the units are K (kelvin).
45. Temperature: For the gas equations, the temperature is measured on the Kelvinscale – the units are K (kelvin). The absolute lowest temperature is zero degrees kelvin.
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 oC by: degrees K = degrees oC + 273.15
47. The Gas Laws
48. The Gas Laws The first systematic and quantitative study of gas behavior was carried out by Boyle.
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.
50. (const. T and n) This is a statement of Boyle’s Law. (V is the volume, P is the pressure, T is the temperature, and n is the moles of gas).
51. (const. T and n) This is a statement of Boyle’s 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. (const. T and n) This is a statement of Boyle’s 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. (const. T and n) This is a statement of Boyle’s 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 Boyle’s Law.
54. We can write Boyle’s law as: PV = C (const. T and n)
55. We can write Boyle’s law as: PV = C (const. T and n) If the initial values of pressure and volume are Pi and Viand if the conditions are changed to a final pressure Pf and final volume Vf then we can write:
56. We can write Boyle’s law as: PV = C (const. T and n) If the initial values of pressure and volume are Pi and Viand if the conditions are changed to a final pressure Pf and final volume Vf then we can write: initial conditions PiVi = C
57. We can write Boyle’s law as: PV = C (const. T and n) If the initial values of pressure and volume are Pi and Viand if the conditions are changed to a final pressure Pf and final volume Vf then we can write: initial conditions PiVi = C final conditions PfVf = C
58. Hence: PiVi = PfVf (const. T and n) This is also a statement of Boyle’s Law.
59. Hence: PiVi = PfVf (const. T and n) This is also a statement of Boyle’s 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. No mention is made about the moles of gas – so make the assumption that n remains constant. Since temperature is constant, we can use Boyle’s Law. PiVi = PfVf
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 Boyle’s Law. PiVi = PfVf This can be rearranged to read: Pi Vf = Vi Pf
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 Boyle’s Law. PiVi = PfVf This can be rearranged to read: Pi Vf = Vi Pf 760.0 torr Vf = x 0.100 l 800.0 torr Vf = 0.0950 l
63. Charles and Gay-Lussac’s Law
64. Charles and Gay-Lussac’s 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. Charles and Gay-Lussac’s 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. The proportionality can be turned into an equality: V = C’ T (const. P and n)
67. 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 Boyle’s law).
68. If the initial values of temperature and volume are Ti and Vi and if the conditions are changed to a final temperature Tf and final volume Vf then we can write:
69. If the initial values of temperature and volume are Ti and Vi and if the conditions are changed to a final temperature Tf and final volume Vf then we can write: Vi initial conditions = C’ Ti
70. If the initial values of temperature and volume are Ti and Vi and if the conditions are changed to a final temperature Tf and final volume Vf then we can write: Vi initial conditions = C’ Ti Vf final conditions = C’ Tf
71. If the initial values of temperature and volume are Ti and Vi and if the conditions are changed to a final temperature Tf and final volume Vf then we can write: Vi initial conditions = C’ Ti Vf final conditions = C’ Tf Hence: ViVf = (const. P, n) Ti Tf This is a statement of Charles’ Law.
72. Problem Example: A gas occupying 4.50 x 102 ml is heated from 22.0 oC to 187 oC at constant pressure. What is the final volume of the gas?
73. Problem Example: A gas occupying 4.50 x 102 ml is heated from 22.0 oC to 187 oC 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:
74. Problem Example: A gas occupying 4.50 x 102 ml is heated from 22.0 oC to 187 oC 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: ViVf = Ti Tf
75. Problem Example: A gas occupying 4.50 x 102 ml is heated from 22.0 oC to 187 oC 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: ViVf = Ti Tf Note that it is necessary to convert the temperatures given to units of K.
76. Ti = 22.0 + 273.15 = 295.2 K Tf = 187 + 273.15 = 4.60 x 102 K
77. Ti = 22.0 + 273.15 = 295.2 K Tf = 187 + 273.15 = 4.60 x 102 K Tf Vf = Vi Ti 4.60 x 102 K = 4.50 x 102 ml 2.952 x 102 K = 701 ml
78. Avogadro’s Law
79. Avogadro’s 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).
80. Avogadro’s 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.
81. It is more convenient to work in terms of moles. A mole contains 6.02214 x 1023 items.
82. It is more convenient to work in terms of moles. A mole contains 6.02214 x 1023 items. For example, one mole of argon contains 6.02214 x 1023 atoms of argon, a mole of dinitrogen contains 6.02214 x 1023 molecules of N2.
83. It is more convenient to work in terms of moles. A mole contains 6.02214 x 1023 items. For example, one mole of argon contains 6.02214 x 1023 atoms of argon, a mole of dinitrogen contains 6.02214 x 1023 molecules of N2. A mole of apples is really, really, ...really big!
84. It is more convenient to work in terms of moles. A mole contains 6.02214 x 1023 items. For example, one mole of argon contains 6.02214 x 1023 atoms of argon, a mole of dinitrogen contains 6.02214 x 1023 molecules of N2. A mole of apples is really, really, ...really big! The mole is a convenient “lab-sized” unit for amount of substance.
85. Avogadro’s law can be written as: (const. P and T)
86. Avogadro’s law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = C’’n (const. P and T)
87. Avogadro’s law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = C’’n (const. P and T) This is a statement of Avogadro’s Law.
88. Avogadro’s law can be written as: (const. P and T) The proportionality can be replaced by an equality: V = C’’n (const. P and T) This is a statement of Avogadro’s Law. (The double prime is used to signify that this constant is different to the constants appearing in Boyle’s law and Charles’ law).
89. If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write:
90. If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write: Vi initial conditions = C’’ ni
91. If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write: Vi initial conditions = C’’ ni Vf final conditions = C’’ nf
92. If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write: Vi initial conditions = C’’ ni Vf final conditions = C’’ nf Hence: ViVf = (const. P, T) ninf
93. If the initial values of volume and moles are Vi and ni and if the conditions are changed to a final volume Vf and final moles nf then we can write: Vi initial conditions = C’’ ni Vf final conditions = C’’ nf Hence: ViVf = (const. P, T) ninf This is a statement of Avogadro’s Law.
94. Avogadro’s 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.
95. Avogadro’s 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 H2(g) + N2(g) 2 NH3(g)
96. Avogadro’s 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 H2(g) + N2(g) 2 NH3(g) 3 volumes 1 volume 2 volumes
97. Ideal Gas Law
98. Ideal Gas Law Summary so far:
99. Ideal Gas Law Summary so far: 1 Boyle’s law: V (const. T and n) P
100. Ideal Gas Law Summary so far: 1 Boyle’s law: V (const. T and n) P Charles’ law V T (const. P and n)
101. Ideal Gas Law Summary so far: 1 Boyle’s law: V (const. T and n) P Charles’ law V T (const. P and n) Avogadro’s law V n (const. P and T)
102. Math Aside:
103. Math Aside: If f X
104. Math Aside: If f X and f Y
105. Math Aside: If f X and f Y and f Z-1
106. Math Aside: If f X and f Y and f Z-1 Then f X Y Z-1
107. Using the result from the math aside, we can put the expressions for Boyle’s law, Charles’ law, and Avogadro’s law together, so that: nT V P
108. Using the result from the math aside, we can put the expressions for Boyle’s law, Charles’ law, and Avogadro’s 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:
109. Using the result from the math aside, we can put the expressions for Boyle’s law, Charles’ law, and Avogadro’s 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
110. Ideal gas: Is defined to be a gas that satisfies the ideal gas equation.
111. 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 OC and 1 atm pressure occupies 22.414 liters. The conditions 0 OC and 1 atm are called standard temperature and pressure, abbreviated STP.
112. 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 OC and 1 atm pressure occupies 22.414 liters. The conditions 0 OC 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
113. In the SI unit system, R = 8.3145 J K-1 mol-1
114. In the SI unit system, R = 8.3145 J K-1 mol-1 Here, J stands for joule, the SI unit of energy.
115. 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 m2 s-2
116. 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 m2 s-2 Exercise: Try to convert R = 0.082057 l atm K-1 mol-1 to the value given in SI units. (It’s a factor-label exercise).
117. Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas.
118. Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value)
119. Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value) n R T V = P
120. Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 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
121. Combined Gas Law
122. Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write:
123. Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti
124. Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nfTf
125. Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nfTf Hence: Pi Vi Pf Vf = niTi nfTf
126. Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nfTf Hence: Pi Vi Pf Vf = niTi nfTf This is the Combined Gas Law.
127. 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 102atm at 25.0 oC. What volume of air does this correspond to at STP?
128. 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 102atm at 25.0 oC. What volume of air does this correspond to at STP? Given initial data: Pi = 1.50 x 102atm Vi = 21. 0 l Ti = 298.2 K (25.0 + 273.15)
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 102atm at 25.0 oC. What volume of air does this correspond to at STP? Given initial data: Pi = 1.50 x 102atm Vi = 21. 0 l Ti = 298.2 K (25.0 + 273.15) Given final data: Pf = 1 atm Tf = 273.15 K
130. Pi Vi Pf Vf = niTi nfTf Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: Pi Vi Pf Vf = Ti Tf
131. Pi Vi Pf Vf = niTi nfTf Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: Pi Vi Pf Vf = Ti Tf The equation can be rearranged, so that Pi Tf Vf = Vi Pf Ti
132. (1.50 x 102atm) (273.15 K) Vf = (21.0 l) (1 atm) (298.2 K) = 2.89 x 103 l at STP
133. Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC and 492 torr has a mass 0f 0.116 g. Two steps: first find moles of gas, then determine the molar mass.
134. Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC 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
135. Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC 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
136. The molar mass is given by:
137. The molar mass is given by: 0.116 g = 7.23 x 10-3 mol
138. The molar mass is given by: 0.116 g = 7.23 x 10-3 mol = 16.0 g mol-1
139. Dalton’s Law of Partial Pressures
140. Dalton’s Law of Partial Pressures Dalton’s 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.
141. Dalton’s Law of Partial Pressures Dalton’s 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.
142. The pressure exerted by gas A – called the partial pressureof gas A – is given by:
143. The pressure exerted by gas A – called the partial pressureof gas A – is given by: nART PA = V
144. The pressure exerted by gas A – called the partial pressureof gas A – is given by: nART PA = V Similarly for gas B, nBRT PB = V
145. Now the total pressure PTotalis nTotal RT PTotal = V
146. Now the total pressure PTotalis nTotal RT PTotal = V Now nTotal = nA + nBso that:
147. Now the total pressure PTotalis nTotal RT PTotal = V Now nTotal= nA + nBso that: (nA + nB)RT nART nBRT PTotal = = + V VV
148. Now the total pressure PTotalis nTotal RT PTotal = V Now nTotal= nA + nBso that: (nA + nB)RT nART nBRT PTotal = = + V VV Hence, PTotal = PA + PB
149. Now the total pressure PTotalis nTotal RT PTotal = V Now nTotal= nA + nBso that: (nA + nB)RT nART nBRT PTotal = = + V VV Hence, PTotal = PA + PB This is Dalton’s Law of partial pressures: The total pressure is the sum of the partial pressures.
150. The preceding result can be generalized to any number of components: PTotal = PA + PB + PC + PD + … where PA, PB, PC, PD, etc. are the partial pressures of the individual gases.
151. 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.
152. 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. 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. But
155. Similar calculations give and
156. Dalton’s law has a practical application when calculating the volume of gases collected overwater. For a gas collected over water, the measured pressure is given by where denotes the vapor pressure of water.
157. (p. 217)
158. Problem example: O2 generated in the decomposition of KClO3 is collected over water. The volume of the gas collected at 24 oC and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O2 obtained. The vapor pressure of H2O at 24 oC is 22.4 torr. First step: Calculate the partial pressure of O2. (Dalton’s law)
159. = 762 torr – 22.4 torr = 739.6 torr (extra sig. fig) = 0.973 atm
160. From the ideal gas equation PV = nRT, PV n = RT (0.973 atm) (0.128 l) = (0.08206 l atm mol-1K-1)(297 K) = 0.00511 mols
161. Graham’s Law of Effusion
162. Graham’s 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.
163. Graham’s 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.
164. Graham’s 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 (don’t get this confused with the gas constant).
165. Graham’s 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 (don’t get this confused with the gas constant). Gas density: abbreviated d.
166. Graham’s 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 (don’t get this confused with the gas constant). Gas density: abbreviated d. 1 R (const. P, T)
167. Graham’s 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 (don’t get this confused with the gas constant). Gas density: abbreviated d. 1 R (const. P, T) This is Graham’s law of effusion.
168. The proportionality can be turned into an equality: R= C (const. P and T) This is a statement of Graham’s law of effusion. (The constant is unrelated to any of the previous constants in Boyle’s law, etc.).
169. For two gases A and B, the relative rates of effusion can be evaluated as follows:
170. For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: RA = C
171. For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: RA = C For gas B: RB = C
172. For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: RA = C For gas B: RB = C Hence: RA = RB
173. For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: RA = C For gas B: RB = C Hence: RA = RB RA = (const. P, T) RB
174. For two gases A and B, the relative rates of effusion can be evaluated as follows: For gas A: RA = C For gas B: RB = C Hence: RA = RB RA = (const. P, T) RB This is a statement of Graham’s law of effusion Law.
175. 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?
176. 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 N2 and B be CO2.
177. 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 N2 and B be CO2.
178. 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 N2 and B be CO2.
179. 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 N2 and B be CO2. = 1.25
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? Let A be N2 and B be CO2. = 1.25 So = 1.25 x , that is, N2 effuses 1.25 times faster than CO2.
181. An alternative form can be developed for Graham’s law of effusion, where the ratio of the rates is determined from the molar masses of the two gases.
182. An alternative form can be developed for Graham’s 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
183. An alternative form can be developed for Graham’s 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
184. An alternative form can be developed for Graham’s 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)
185. An alternative form can be developed for Graham’s 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
186. Now plug this result for d into the expression for Graham’s law.
187. Now plug this result for d into the expression for Graham’s law. That is (const. P, T)
188. Now plug this result for d into the expression for Graham’s law. That is (const. P, T) This is a statement of Graham’s law of effusion Law.
189. Exercise, Graham ’s law: What is the ratio of the rates of effusion of dinitrogen to carbon dioxide?
190. 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.

191. Kinetic Theory of Gases

192. The kinetic theory of gases provides a “working model” of a gas. It is an attempt to interpret experimental observations at the molecularlevel.
193. The kinetic theory of gases provides a “working model” of a gas. It is an attempt to interpret experimental observations at the molecularlevel. Basic Postulates of Kinetic Theory:
194. The kinetic theory of gases provides a “working model” of a gas. It is an attempt to interpret experimental observations at the molecularlevel. Basic Postulates of Kinetic Theory: 1. A gas consists of an extremely large number of tiny particles that are in constant random motion.
195. 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).
196. 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.
197. It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v2 T The bar over KE means average, and over v2 means the average of v2.
198. It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v2 T The bar over KE means average, and over v2 means the average of v2. v2 is called the mean square velocity.
199. It is found: The average kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature. KE = ½ m v2 T The bar over KE means average, and over v2 means the average of v2. v2 is called the mean square velocity. If there are N molecules v12 + v22 + … + vN2 v2 = N
200. Molecules exert neither attractive nor repulsive forces on one another. This ties in directly with postulate 3.
201. 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.
202. Pressure – volume connection (p. 215):
203. 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.
204. 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 smallervolume implies a larger pressure. Detailed arguments lead to P V-1.
205. Pressure – temperature connection: An increase in temperature increases the average velocity of the gas molecules.
206. 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.
207. 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.
208. 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).
209. Temperature – volume connection: Increase the temperature at constant pressure.
210. Temperature – volume connection: Increase the temperature at constant pressure. An increase in temperature increases the average velocity of the gas molecules.
211. 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.
212. 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.
213. 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).
214. Graham’s law of effusion: Consider two gases (call them A and B) at the same temperature.
215. Graham’s 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.
216. Graham’s 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 KEA = ½ mA vA2 KEB = ½ mB vB2
217. Graham’s 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 KEA = ½ mA vA2 KEB = ½ mB vB2 If gas B molecules have higher mass, then the only way the average KE can be the same is
218. that the average velocity of the B molecules is smaller than gas A.
219. 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.
220. Maxwell Distribution of Molecular Speeds
221. Maxwell Distribution of Molecular Speeds At a given instant, how many molecules are moving at a particular speed?
222. 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 Maxwell’s distribution of speeds curve.
223. Maxwell Distribution of Molecular Speeds (p. 205)
224. One important point to note is that although there are always some slow-moving molecules, there are fewer slow-moving molecules at higher temperatures.
225. 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.