2. Revision history: 10/5/01 2/20/03 5/21/03 6/24/04 12/27/06 12/29/06 01/05/10 01/09/10 050212 Gases Pisgah High School M. Jones

3. The Properties of Gases Part I

4. Properties of Gases 1. Gases expand to fill the container. 2. Gases take on the shape of the container. 3. Gases are highly compressible. (Can be liquefied at high pressures). 4. Gases have low densities. 5. Gases mix uniformly.

5. The Kinetic Molecular Theory The kinetic molecular theory describes the behavior of ideal gases. An ideal gas is one that conforms to the KMT.

6. 1. Molecules are in constant random motion Temperature is proportional to the average kinetic energy of the molecules. KE = ½ mv 2 KE = ½ mass times speed squared The speed is proportional to the absolute temperature (Kelvin).

7. 2. A gas is mostly empty space Molecules are far apart from each other. This accounts for the low density and high compressibility. The volume of the individual molecules is negligible compared to the volume of the gas.

8. 3. No intermolecular forces There are no attractive or repulsive forces between gas molecules. Adjacent molecules do not attract or repel each other.

9. 4. Collisions are elastic When gas molecules collide with each other they may speed up or slow down, BUT … The net (total) energy of the gas molecules does not change.

10. Kinetic Molecular Theory 1. Gases in constant motion, speed depends on temperature. 2. Molecules have negligible volume. 3. No intermolecular forces. 4. Elastic collisions. No change in energy.

11. Temperature reminder When doing calculations, temperature must always be in an absolute temperature scale … … where the lowest possible temperature is zero degrees. Use Kelvin degrees!

12. Temperature conversion K = C + 273

13. 1. Pressure is the force per unit area exerted by the gas molecules. 2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container. Pressure

14. Pressure 1. Pressure is a measure of the force per unit area. P = force area Pressure can be in pounds per square inch (PSI), or … … newtons per square meter (N/m 2 ) pascal (Pa) = N/m 2

15. Pressure 1. Pressure is a measure of the force per unit area. P = force area Glass tube with Hg Bowl of Hg At sea level, air pressure holds up a column of mercury 760 mm high. Torricelli

16. Pressure Measurements Standard sea level pressure is… 1.00 atmospheres (atm) 760 mm Hg 760 torr (from Torricelli) 101.3 kilopascals (kPa) 14.7 lb/in 2

17. Pressure Measurements Standard sea level pressure is… 1.00 atmospheres (atm) 760 mm Hg 760 torr (from Torricelli) 101.3 kilopascals (kPa) 14.7 lb/in 2 Exact

18. Pressure 2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container. If you change the number of collisions, you change the pressure.

19. The Gas Laws Part II

20. The Gas Laws Boyle’s Law Boyle’s Law Amonton’s Law Amonton’s Law Charles’s Law Charles’s Law Combined Gas Law Combined Gas Law Gay-Lussac’s Law Avogadro’s Law Dalton’s Law

21. Boyle’s Law At a constant temperature, pressure is inversely proportional to volume. Pressure Volume

22. Boyle’s Law At a constant temperature, pressure is inversely proportional to volume. 1/Pressure Volume

23. Boyle’s Law At a constant temperature, pressure is inversely proportional to volume. P 1 V 1 = P 2 V 2 PV = k P  1V Year: 1662

24. Amonton’s Law “Air thermometer”, 1695. This is not Gay-Lussac’s law. P  T P1P1P1P1 T1T1T1T1 = P2P2P2P2 T2T2T2T2 Diagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

25. Amonton’s Law P  T P1P1P1P1 T1T1T1T1 = P2P2P2P2 T2T2T2T2 This is why you measure your tire pressure when the tire is cold. Tire pressures vary with temperature.

26. Measure the pressures of a gas at various temperatures at a constant volume. Amonton’s air thermometer was used to find the value of absolute zero. Amonton’s Law

27. Finding Absolute Zero Pressure Temperature (C) -300 -150 0 100 200 300 -273 C Extrapolate to the x-axis

28. Charles’s Law At constant pressure, volume is directly proportional to temp. Volume Temperature

29. Charles’s Law At constant pressure, volume is directly proportional to temperature. V  T VT = k V1V1V1V1 T1T1T1T1 = V2V2V2V2 T2T2T2T2

30. Charles’s Law Studied gases during 1780’s.Studied gases during 1780’s. Hydrogen balloon assents, 3000 m, in 1783.Hydrogen balloon assents, 3000 m, in 1783. Collaborated with the Montgolfier brothers on hot air balloons, 1783.Collaborated with the Montgolfier brothers on hot air balloons, 1783. Charles’s gas studies published by Gay-Lussac in1802.Charles’s gas studies published by Gay-Lussac in1802.

31. Charles’s Law Hydrogen balloon assent, 3000 m, 1783.

32. Combined Gas Law gives P  TV 1V T

33. Combined Gas Law P  TV Next we convert equation by adding a constant. to an to an P  kTV

34. Combined Gas Law Rearranging the equation gives: = k PVT P  kTV

35. Combined Gas Law Combining the laws of Boyle, Amonton and Charles produces the combined gas law. = k PVT

36. Combined Gas Law P1V1P1V1P1V1P1V1 T1T1T1T1 = k P2V2P2V2P2V2P2V2 T2T2T2T2 = k Consider a confined gas at two sets of conditions. Since the number of molecules is the same and the values of k are the same, then we can combine the two equations.

37. Combined Gas Law P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 P1V1P1V1P1V1P1V1 T1T1T1T1 = k P2V2P2V2P2V2P2V2 T2T2T2T2 = k

38. P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 Use the combined gas law whenever you are asked to find a new P, V or T after changes to a confined gas.

39. Sample Combined Gas Law Problem Consider a confined gas in a cylinder with a movable piston. The pressure is 0.950 atm. Find the new pressure when the volume is reduced from 100.0 mL to 65.0 mL, while the temperature remains constant? 100 mL 65 mL

40. Sample Combined Gas Law Problem Start with the equation for the combined gas law. 100 mL 65 mL

41. Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 Since the temperature is constant, we can cancel out T 1 and T 2.

42. Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1= P2V2P2V2P2V2P2V2 Next, solve for P 2. This becomes Boyle’s Law P1V1P1V1P1V1P1V1= V2V2V2V2 P2P2P2P2

43. Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1= V2V2V2V2 P2P2P2P2 (0.950 atm)(100.0 mL) = 65.0 mL P2P2P2P2 = P2P2P2P2 1.46 atm

44. Combined Gas Law Problems 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and 0.900 atm. What is the new pressure when the volume is reduced to 0.750 L and the temperature increases to 24.0 C? The answer is 2.43 atm

45. Combined Gas Law Problems 2. Some “left over” propane gas in a rigid steel cylinder has a pressure of 24.6 atm at a temperature of 20.C. When thrown into a campfire the temperature in the cylinder rises to 313C. What will be the pressure of the propane? The answer is 49.2 atm

46. Combined Gas Law Problems 3. Consider the fuel mixture in the cylinder of a diesel engine. At its maximum, the volume is 816 cc. The mixture comes in at 0.988 atm and 31 C. What will be the temperature (in C) when the gas is compressed to 132 cc and 42.4 atm? The answer is 1837 C

47. Gay-Lussac’s Law Year: 1802 At a given temperature and pressure, the volumes of reacting gases are in a ratio of small, whole numbers. Also known as the law of combining volumes.

48. Gay-Lussac’s Law Gay-Lussac found that the volumes of gases in a reaction were in ratios of small, whole numbers. 2 H 2 (g) + O 2 (g)  2 H 2 O(g) 200 mL of hydrogen reacts with 100 mL of oxygen. 2:1

49. Gay-Lussac’s Law 2 H 2 (g) + O 2 (g)  2 H 2 O(g) The ratio of volumes of gases come from the ratios of the coefficients in the balanced equation.

50. Avogadro’s Law Equal volumes of gases at the same temperature and pressure have equal numbers of molecules. Vn = k Avogadro’s law followed Dalton’s atomic theory and Gay-Lussac’s law. Year: 1811. V  n V1V1V1V1 n1n1n1n1 = V2V2V2V2 n2n2n2n2

51. Dalton’s Law Dalton’s law of partial pressures deals with mixtures of gases. P total = P 1 + P 2 + P 3 … Use when dealing with the pressure of H 2 O(g) when collecting a gas over water. The total pressure is the sum of the partial pressures:

52. Dalton’s Law Inverted gas collecting bottle Flask with metal and HC l Rubber tubing carrying H 2 gas Water in trough “Collecting a gas over water”

53. Dalton’s Law Inverted gas collecting bottle Flask with metal and HC l Rubber tubing carrying H 2 gas Water in trough “Collecting a gas over water”

54. Dalton’s Law Flask with metal and HC l Water in trough Water is displaced through the mouth of the bottle as H 2 gas bubbles in. H 2 gas

55. Dalton’s Law Flask with metal and HC l Water in trough A mixture of hydrogen gas and water vapor is in the collection bottle. H 2 gas

56. Dalton’s Law Flask with metal and HC l Water in trough The pressure of the mixture is the sum of the pressures of H 2 O gas and H 2 gas. H 2 gas

57. Dalton’s Law Flask with metal and HC l Water in trough P total = P H 2 O + P H 2 H 2 gas

58. Sample Problem Flask with metal and HC l Water in trough The total pressure in the bottle is 712.7 torr. The temperature is 19 C. What is the H 2 pressure? H 2 gas

59. Flask with metal and HC l Water in trough At 19C, the vapor pressure of water is 16.5 torr. The H 2 pressure is … H 2 gas Sample Problem

60. Flask with metal and HC l Water in trough P H 2 = P total - P H 2 O H 2 gas P H 2 = 712.7 torr – 16.5 torr = 696.2 torr Sample Problem

61. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

62. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P total = P 1 + P 2

63. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P total = P air + P Hg

64. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = P total - P air

65. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = 693 torr – 684 torr

66. Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = 9 torr

67. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html More on the gas laws:

68. Gases and Moles The Ideal Gas Equation Part III

69. What factors affect the pressure of a confined gas? 1. Number of molecules 2. Temperature 3. Volume of the container Think in terms of the number of collisions.

70. Number of molecules Increasing the number of molecules increases the number of collisions … P  n Where n is the number of moles of molecules … which increases the pressure.

71. Temperature Increasing the temperature makes the molecules move faster, increasing the number of collisions … P  T Where T is the absolute temperature … which increases the pressure.

72. Volume Increasing the volume of the container decreases the number of collisions … P  Where V is the volume 1V … which decreases the pressure.

73. Sooooo… P  n P  T P  1V n V T

74. nTR Make it into an equation P  n V T P  V

75. The Ideal Gas Equation n TR P  V n TR PV  … is usually written as …

76. The Ideal Gas Equation nTR P  V R = 0.0821 L atm mol K R is the “gas constant”

77. The Ideal Gas Equation nTR P  V Can R be in units other than L atm mol K ?

78. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

79. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

80. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

81. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

82. Ideal Gas Equation The ideal gas equation relates pressure, volume, temperature and the number of moles of a quantity of gas. PV = nRT

83. Ideal Gas Equation Use the ideal gas equation whenever the problem gives you mass or moles, or asks for a mass or a number of moles. PV = nRT

84. Ideal gas equation problem: Some ammonia gas (NH 3 ) is contained in a 2.50 L flask at a temperature of 20.0 C. If there are 0.0931 moles of the gas, what is its pressure?

85. Solution PV = nRT P = (nRT)/V = (0.0931 mol ) P = 0.896 atm 2.50 L )(293 K ) (0.0821 L atm mol K

86. Here’s another one Find the volume of 1.00 mole of nitrogen gas (N 2 ) at 0.0 C and 1.00 atm of pressure.

87. Solution V = (1.00 mol ) V = (1.00 mol ) 1.00 atm )(273 K ) (0.0821 L atm mol K PV = nRT V = (nRT)/P V = 22.4 L

88. Ideal gas equation problem: How many grams of sulfur trioxide are in an 855 mL container at a pressure of 1585 torr and a temperature of 434 C? The answer is 2.46 g SO 3

89. The Ideal Gas Equation can be used to derive the Combined Gas Law

90. The Combined Gas Law Start with the ideal gas equation: PV = nRT

91. The Combined Gas Law P 1 V 1 = nRT 1 P 2 V 2 = nRT 2 and Suppose the volume, pressure and temperature change to give a new pressure, volume and temperature.

92. The Combined Gas Law Now, solve for what doesn’t change, the constants n and R: P 1 V 1 = nRT 1 P 2 V 2 = nRT 2 and

93. The Combined Gas Law P1V1P1V1P1V1P1V1 = nR T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2 Now, solve for what doesn’t change, the constants n and R: and

94. The Combined Gas Law Since both are equal to nR, we can make a new equation. and P1V1P1V1P1V1P1V1 = nR T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

95. The Combined Gas Law Since both are equal to nR, we can make a new equation. = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

96. The Combined Gas Law This is the Combined Gas Law = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

97. The Combined Gas Law It can be derived from the laws of Boyle, Amonton and Charles, or the Ideal Gas Equation the Ideal Gas Equation = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

98. The Ideal Gas Equation and Density

99. Density calculations And an equation for “moles”: Start with the equation for density: D = m V n = mM Where m = mass and M = molar mass

100. Density calculations into the ideal gas equation … PV = nRT Now substitute n = mM PV = mRT M and get and get

101. Density calculations to get to get Now rearrange PV = mRTM P = mRT VMVMVMVM

102. Density calculations Recall that Recall that D = mV P = DRT M P = mRT VMVMVMVM

103. Density calculations Solving for density, Solving for density, P = DRT M becomes: becomes: D =D =D =D = PMPMPMPMRT

104. Density calculations The density of a gas depends on the molar mass and the pressure and temperature. D = PMPMPMPMRT

105. Density Problem (a) at STP (b) at a pressure of 695 torr and a temperature of 40.0 C. 1. Determine the density of nitrogen, N 2, gas The answers are 1.25 g/L, and 0.996 g/L. and 0.996 g/L.

106. 2. Determine the molar mass of a gas which has a density of 8.53 g/L at a pressure of 2.50 atm and a temperature of 500.0 K? Another Problem The answer is 140. g/mol

107. Solutions of Gases

108. Many gases are soluble in water. Were it not for the solubility of oxygen, fish would have wear scuba tanks. Besides the characteristics of the gas itself, what are the two external factors which affect the solubility of a gas?

109. Solutions of Gases The temperature of the solution. The pressure of the gas above the solution. See “Henry’s Law”. Two factors affect the solubility of a gas:

110. Solutions of Gases The temperature of the solution. Why is the champagne chilled?

111. Solutions of Gases More CO 2 is released from the champagne as its temperature rises.

112. Solutions of Gases The solubility of gases decreases with increasing temperature.

113. http://www.switchstudio.com/waterkeeper/issues/Winter%2007/fight_power_plants.html Thermal pollution This 1988 thermal image of the Hudson River highlights temperature changes caused by discharge of 2.5 billion gallons of water each day from the Indian Point power plant. The plant sits in the upper right of the photo — hot water in the discharge canal is visible in yellow and red, spreading and cooling across the entire width of the river. Two additional outflows from the Lovett coal- fired power plant are also clearly visible against the natural temperature of the water, in green and blue. Solutions of Gases

114. Thermal pollution The result is a decrease in dissolved oxygen which is needed by the plants and animals living in the heated water. Solutions of Gases

115. Why do we find trout in cold mountain streams, often near riffles and waterfalls? High dissolved oxygen.

116. Solutions of Gases CO 2 is dissolved in a can of Pepsi. Above the liquid in the can is CO 2 gas. The high pressure of the CO 2 in the can keeps the can rigid and the CO 2 in solution. The second factor is the pressure of the dissolved gas above the solution.

117. Solutions of Gases Henry’s Law: At a constant temperature, the amount of a given gas dissolved in a solution is directly proportional to the partial pressure of that gas in equilibrium with the solution.

118. Solutions of Gases Henry’s Law: p = k H c Where p is the partial pressure of the gas, k H is the Henry’s law constant, and c is the concentration of the gas in moles per liter. Some Henry’s law constants at 298K O 2 769.2 Latm/mol CO 2 29.4 Latm/mol H 2 1282.1 Latm/mol

119. Solutions of Gases Henry’s Law: p = k H c The pressure of the CO 2 in the headspace of can of Pepsi is 4.50 atm at 298K. What is the concentration of the CO 2 ? Some Henry’s law constants at 298K O 2 769.2 Latm/mol CO 2 29.4 Latm/mol H 2 1282.1 Latm/mol 0.153 M Ans:

120. Solutions of Gases Henry’s Law: p = k H c How do we predict the relative solubilities of gases based on the values of k H ? Which gas below is the most soluble in water? Some Henry’s law constants at 298K O 2 769.2 Latm/mol CO 2 29.4 Latm/mol H 2 1282.1 Latm/mol CO 2 Answer:

121. Graham’s Law

122. Molecules at the same temperature have the same average kinetic energy. (T  KE) KE is proportional to the speed of the molecules and the mass of the molecules. KE = ½ mv 2

123. Graham’s Law Diffusion – “a gas spreading out” Graham’s Law deals with the effusion of two gases into each other. Effusion – a gas moving through a small hole

124. Graham’s Law Gas A Gas B Consider two gases at the same temperature and pressure in a box with two partitions, separated by a wall with a hole that can be opened.

125. Graham’s Law Gas A Gas B Gas A effuses into Gas B, and … Gas B effuses into Gas A, so that … …eventually both gases will be evenly distributed in the box.

126. Graham’s Law Gas A Gas B The ratio of the rates at which Gas A and Gas B effuse into each other is … … inversely proportional to the square roots of their molar masses.

127. Graham’s Law Gas A Gas B The bottom line: the lighter the gas, the faster it moves!

128. Graham’s Law Gas A Gas B Suppose nitrogen and an unknown gas, both at the same temperature and pressure, are in the box. The rates of effusion are …

129. Graham’s Law Gas A Gas B … 0.0160 moles/L/min for the nitrogen and 0.0205 mol/L/min for the unknown. What is the molecular weight of the unknown?

130. Graham’s Law Gas A Gas B Start with the equation for Graham’s Law

131. Graham’s Law Gas A Gas B Square both sides, then solve for the molecular mass of compound x

132. Graham’s Law Gas A Gas B

133. Graham’s Law Gas A Gas B M x = 17.0g/mol

134. Graham’s Law Gas A Gas B What is the unknown gas? The gas is “smelly”, dissolves in water to make a basic solution, and has a molar mass of 17 g/mol.

135. Graham’s Law Gas A Gas B Ammonia gas, NH 3, has a strong smell, and...

136. OH - Graham’s Law Gas A Gas B … it reacts with water … … it reacts with water … NH 3 (g) + HOH(l)  NH 3 (g) + HOH(l)  … which is a basic solution, OH - NH 4 + +

137. Graham’s Law Gas A Gas B 1 x 14.0 + 3 x 1.0 = 17.0 g/mol … and ammonia, NH 3, has a molar mass of 17.0 g/mol.