2. 1

3. 2 Properties of Gases

4. 3 May be compressed Expand to fill container Low density May be mixed Constant, uniform pressure on container walls

5. 4 The Kinetic- Molecular Theory

6. 5 KMT is based on the motions of gas particles. A gas that behaves exactly as outlined by KMT is known as an ideal gas. While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure (high temperature & low pressure).

7. 6 Principle Assumptions of the KMT 1.Gases consist of tiny subatomic particles. 2.The distance between particles is large compared with the size of the particles themselves. 3.Gas particles have no attraction for one another.

8. 7 4.Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container. Principle Assumptions of the KMT

9. 8 5.No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic. 6.The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature. Principle Assumptions of the KMT

10. 9 Kinetic Energy

11. 10 All gases have the same kinetic energy at the same temperature. As a result, lighter molecules move faster than heavier molecules. mHmH 2 = 2 mOmO 2 = 32 vHvH 2 vOvO 2 = 1 4 Kinetic Energy

12. 11 Parameters for Describing Gases A measurable factor forming one of a set that defines a system or sets the conditions of its operation.

13. 12 ParameterSymbolDescriptionUnits P Pressure Force per unit area mmHg (torr) or atm Volume V No.of moles n of gas mole Volume of container liter Temperature T Absolute temp of gas K

14. 13 Measurement of Pressure of Gases

15. 14 The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.

16. 15 Pressure equals force per unit area.

18. 17 Evangelista Torricelli 1644

19. Measuring Pressure Attempts to pump water out of flooded mines often failed because H 2 O can’t be lifted more than 34 feet.

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

21. Pet watering dish Chicken watering dish

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

23. 22

24. 34’ column of water 1 Atm The atmosphere would support a column of H 2 O > 34 feet high. Measuring Pressure

25. Torricelli Barometer Pressure of the atmosphere supports a column of Hg 760 mm high. 1 atm 1 atm 1 atm = 760 mm Hg 760 torr 29.92 in Hg 14.7 lb/in 2 101,325 Pa vacuum Mercury used because it’s so dense.

26. 25 Boyle’s Law 1662

27. 26 Robert Boyle

28. 27 At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

29. 28 Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

30. 29 The effect of pressure on the volume of a gas.

31. 30 V 1 = 8.00 LP 1 = 500 torr V 2 = 3.00 L P 2 = ? An 8.00 L sample of N 2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant). Step 1. Organize the given information:

32. 31 Step 2. Write and solve the equation for the unknown. An 8.00 L sample of N 2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

33. 32 Step 3. Put the given information into the equation and calculate. An 8.00 L sample of N 2 is at a pressure of 500. torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

34. 33 A tank of O 2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr? Step 1. Organize the given information: V 1 = 1500. ml P 1 = 350. torr P 2 = 1000. torr V 2 = ?

35. 34 A tank of O 2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr? Step 2. Write and solve the equation for the unknown.

36. 35 A tank of O 2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr? Step 3. Put the given information into the equation and calculate. = 525 ml

37. 36 Charles’ Law 1787

38. 37 Jacques Charles

39. December 1783 38

40. 39 Volume-temperature relationship of methane (CH 4 ). O2O2 N2N2

41. 40 Charles’ Law At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

42. 41 Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

43. 42 Step 1. Organize the information (remember to make units the same): V 1 = 255 mLT 1 = 75 o C = 348 K V 2 = ?T 2 = 250 o C = 523 K A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

44. 43 Step 2. Write and solve the equation for the unknown: A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

45. 44 Step 3. Put the given information into the equation and calculate : V 1 = 255 mLT 1 = 75 o C = 348 K V 2 = ?T 2 = 250 o C = 523 K A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

46. 45 A tank of O 2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure). Step 1. Organize the given information: V 1 = 16.0 L T 1 = 500.K V 2 = 20.0 L T 2 = ?

47. 46 A tank of O 2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure). Step 2. Write and solve the equation for the unknown.

48. 47 A tank of O 2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure). Step 3. Put the given information into the equation and calculate: = 625 K

49. 48 Gay-Lussac’s Law 1802

50. 49 Joseph Gay-Lussac

51. 50 The pressure of a gas in a fixed volume increases with increasing temperature. Lower T Lower P Higher T Higher P Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature.

52. 51 The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

53. 52 Step 1. Organize the information (remember to make units the same): P 1 = 21.5 atmT 1 = 40 o C = 313 K P 2 = ?T 2 = 100 o C = 373 K At a temperature of 40. o C an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100. o C what will be the pressure of the oxygen?

54. 53 Step 2. Write and solve the equation for the unknown: At a temperature of 40. o C an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100. o C what will be the pressure of the oxygen?

55. 54 Step 3. Put the given information into the equation and calculate: At a temperature of 40. o C an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100. o C what will be the pressure of the oxygen? P 1 = 21.5 atmT 1 = 40 o C = 313 K P 2 = ?T 2 = 100 o C = 373 K

56. 55 Combined Gas Laws

57. 56 A combination of Boyle’s and Charles’ Laws or Charles’ and Gay-Lussac’s Laws. Used when pressure and temperature change at the same time. Solve the equation for any one of the 6 variables

58. 57 A sample of hydrogen occupies 465 ml at at 0 o C and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume? o C + 273 = K 0 o C + 273 = 273 K -15 o C + 273 = 258 K Step 1. Organize the given information, putting temperature in Kelvins:

59. 58 Step 1. Organize the given information: P 1 = 760 torrP 2 = 950 torr V 1 = 465 mLV 2 = ? T 1 = 273 KT 2 = 258 K A sample of hydrogen occupies 465 ml at 0 o C and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?

60. 59 A sample of hydrogen occupies 465 ml at 0 o C and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume? Step 2. Write and solve the equation for the unknown V 2.

61. 60 Step 3 Put the given information into the equation and calculate. A sample of hydrogen occupies 465 ml at 0 o C and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?

62. 61 Dalton’s Law of Partial Pressures

63. 62 John Dalton

64. 63 Each gas in a mixture exerts a pressure that is independent of the other gases present. The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. P total = P a + P b + P c + P d + ….

65. 64 A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container? P total = P He + P Ne + P Ar P total = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm

66. 65 The pressure in the collection container is equal to the atmospheric pressure. The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure. Collecting a Gas Sample Over Water

67. 66 Oxygen collected over water.

68. 67 A sample of O 2 was collected over water in a bottle at a temperature of 25 o C when the atmospheric pressure was 760 torr. What is the pressure of the O 2 alone? The vapor pressure of water at 25 o C is 23.8 torr.

69. 68 Temp (ºC)Press (torr)Temp (ºC)Press (torr) 04.62625.2 14.92726.7 25.32828.3 35.72930.0 46.13031.8 56.53133.7 67.03235.7 77.53337.7 88.03439.9 98.63542.2 109.23644.6 119.83747.1 1210.53849.7 1311.23952.4 1412.04055.3 1512.84158.3 1613.64261.5 1714.54364.8 1815.54468.3 1916.54571.9 2017.54675.7 2118.74779.6 2219.84883.7 2321.04988.0 2422.45092.5 2523.85197.2

70. 69 A sample of nitrogen was collected over water and occupies 300. ml at 23 o C and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0 o C, what would be the new volume of the dry nitrogen? Vapor pressure of H 2 0 @ 23 o C is 21.0 torr. 0 o C + 273 = 273 K 23 o C + 273 = 296 K Step 1. Organize the given information, putting temperature in Kelvin and correcting for water vapor pressure:

71. 70 A sample of nitrogen was collected over water and occupies 300. ml at 23 o C and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0 o C, what would be the new volume of the dry nitrogen? Vapor pressure of H 2 0 @ 23 o C is 21.0 torr. Step 1. Organize the given information, P 1 = 729 torrP 2 = 760 torr V 1 = 300. mLV 2 = ? T 1 = 296 KT 2 = 273 K

72. 71 A sample of nitrogen was collected over water and occupies 300. ml at 23 o C and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0 o C, what would be the new volume of the dry nitrogen? Vapor pressure of H 2 0 @ 23 o C is 21.0 torr. Step 2. Write and solve the equation for the unknown V 2.

73. 72 A sample of nitrogen was collected over water and occupies 300. ml at 23 o C and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0 o C, what would be the new volume of the dry nitrogen? Vapor pressure of H 2 0 @ 23 o C is 21.0 torr. Step 3. Put the given information into the equation and calculate.

74. 73 Avogadro’s Law 1811

75. 74 Amadeo (Amedeo) Avogadro

76. 75 Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

77. 76 Standard Temperature and Pressure

78. 77 Standard Temperature and Pressure Standard Conditions Standard Temperature and Pressure STP 273.15 K or 0.00 o C 1 atm or 760 torr or 760 mm Hg Selected common reference points of temperature and pressure.

79. 78 Volume of one mole of any gas at STP = 22.4 L.STP 22.4 L at STP is known as the molar volume of any gas.

80. 79

81. 80 Ultimate Combined Gas Law

82. 81 A combination of Boyle’s, Charles’ and Avogadro’s Laws. Used when pressure, temperature, and moles change at the same time. Solve the equation for any one of the 8 variables

83. 82 A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume? Step 1. Organize the given information: P 1 = 9.7 atm V 1 = 8.74 L T 1 = 300 K n 1 = 3.45 moles P 2 = 15.0 atm V 2 = ? T 2 = 310 K n 2 = 3.45 + 1.27 = 4.72

84. 83 A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume? Step 2. Write and solve the equation for the unknown V 2.

85. 84 A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume? Step 3. Put the given information into the equation and calculate. = 7.99 L

86. 85 AVOGADRO'S LAW Explained Gay Lussac's Law of Combining Volumes Provided a method for the determination of mole weights of gases comparing densities of gases of known mole weight Served as a foundation for the devolopment of the Kinetic-Molecular Theory

87. 86 Mole-Mass-Volume Relationships

88. 87 Density of Gases

89. 88

90. 89 Density of Gases liters grams

91. 90 Density of Gases depends on T and P

92. 91 The density of neon at STP is 0.900 g/L. What is the mole weight of neon?

93. 92 The mole weight of SO 2 is 64.07 g/mol. Determine the density of SO 2 at STP. 1 mole of any gas occupies 22.4 L at STP

94. 93 Ideal Gas Equation

95. 94 V  P nT

96. 95 V  P nT atmospheres

97. 96 V  P nT liters

98. 97 V  P nT moles

99. 98 V  P nT Kelvin

100. 99 V  P nT Ideal Gas Constant

101. 100 A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750. torr. What is the balloon’s volume? Step 1. Organize the given information. Convert temperature to kelvins. K = o C + 273 K = 25 o C + 273 = 298K Convert pressure to atmospheres.

102. 101 Step 2. Write and solve the ideal gas equation for the unknown. Step 3. Substitute the given data into the equation and calculate. A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750. torr. What is the balloon’s volume? (25º + 273) (750/760)

103. 102 Mole-Weight Calculations

104. 103 Determination of Molecular Weights Using the Ideal Gas Equation

105. 104 Calculate the mole weight of an unknown gas, if 0.020 g occupies 250 mL at a temperature of 305 K and a pressure of 0.045 atm. V = 250 mL = 0.250 Lg = 0.020 g T = 305 K P = 0.045 atm

106. 105 Gas Stoichiometry

107. 106 Definition Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

108. 107 All calculations are done at STP. Gases are assumed to behave as ideal gases. A gas not at STP is converted to STP.

109. 108 Gas Stoichiometry Primary conversions involved in stoichiometry.

110. 109 Mole-Volume Calculations Mass-Volume Calculations

111. 110 Step 1 Write the balanced equation 2 KClO 3  2 KCl + 3 O 2 Step 2 The starting amount is 0.500 mol KClO 3. The conversion is moles KClO 3  moles O 2  liters O 2 What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?

112. 111 Step 3. Solve the problem in one continuous calculation. What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? 2 KClO 3  2KCl + 3 O 2

113. 112 2 Al (s) + 6 HCl (aq)  2AlCl 3(aq) + 3 H 2(g) Step 1 Calculate moles of H 2. grams Al  moles Al  moles H 2 What volume of hydrogen, collected at 30. o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

114. 113 Convert o C to K: 30. o C + 273 = 303 K Convert torr to atm: 2 Al(s) + 6 HCl(aq)  2AlCl 3 (aq) + 3 H 2 (g) Step 2 Calculate liters of H 2. What volume of hydrogen, collected at 30. o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

115. 114 What volume of hydrogen, collected at 30. o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? PV = nRT Step 3. Solve the ideal gas equation for V

116. 115 Volume-Volume Calculations

117. 116 Gay Lussac’s Law of Combining Volumes 1809 N2N2 1 volume + + 3 H 2 3 volumes → → 2 NH 3 2 volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

118. 117 H 2 (g) + Cl 2 (g)  2HCl(g) 1 mol H 2 1 mol Cl 2 2 mol HCl 22.4 L STP 22.4 L STP 2 x 22.4 L STP 1 volume 2 volumes Y volume 2Y volumes For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.

119. 118 What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed? N 2(g) + 3H 2(g)  2NH 3(g)

120. 119 Molecular Formula Calculations

121. Chapter 7 Review 120

122. 121 A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula. N3O6N3O6

123. 122 An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260  C and 2.84 atm. What is the molecular formula of the unknown liquid? Step 1. Determine the Mole Weight:

124. 123 An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260  C and 2.84 atm. What is the molecular formula of the unknown liquid? mass = 29.4 gramsV = 3.60 L T = 260 + 273 = 533 KP = 2.84 atm

125. 124 An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260  C and 2.84 atm. What is the molecular formula of the unknown liquid? Step 2. Use the mole weight and percentages to find the formula: C 9 H 18

126. 125 Diffusion The ability of two or more gases to mix spontaneously until they form a uniform mixture. Stopcock closed No diffusion occurs Stopcock open Diffusion occurs

127. 126 Effusion A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

128. 127 Thomas Graham

129. 128 Graham’s Law of Effusion/Diffusion The rates of effusion or diffusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their mole weights.

130. 129 What is the ratio of the rate of diffusion of CO to CO 2 ?

131. 130 It takes 30.0 sec for 10. ml of O 2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

132. 131 It takes 30.0 sec for 10. ml of O 2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

133. 132 Real Gases

134. 133 Ideal Gas An ideal gas obeys the gas laws. –The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures. –The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.

135. 134 Real Gases Deviations from the gas laws occur at high pressures and low temperatures. –At high pressures the volumes of the real gas molecules are not negligible compared to the volume of the gas –At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.

136. 135