1. 1 Version 1

2. 2 Properties of Gases

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

4. 4 The Kinetic- Molecular Theory (KMT)

5. 5 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.

6. 6 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

7. 7 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

8. 8 The Kinetic-Molecular Theory KMT is based on the motions of gas particles. A gas that behaves exactly as outlined by KMT is known as an ideal gas. While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure (high temperature & low pressure).

9. 9 Kinetic Energy

10. 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

11. 11 Parameters for Describing Gases

12. 12 Variables that describe a gas: VariableSymbolUnitsConversions Pressure P - mm Hg - torr - atm torr = mm Hg 760 torr = 1 atm Volume VL, mL 1000 mL = 1 L TemperatureTK K = o C + 273 Molesnmolesn = g/mw

13. 13 Measurement of Pressure of Gases

14. 14 Pressure results from the collisions of gas molecules with the walls of the container, whether it is a flexible wall or not.

15. 15 Pressure equals force per unit area.

16. 16 Evangelista Torricelli 1644

17. 17 Mercury Barometer The full tube of mercury is inverted and placed in a dish of mercury. The barometer is used to measure atmospheric pressure.

18. 18 Pressure Units Equivalent to 1 Atmosphere 760 torr 760 mm Hg 76 cm Hg 1013 mbar 29.9 in Hg 14.7 lb/in 2

19. 19 The Gas Laws P total = P a + P b + P c + P d + ….

20. 20 Boyle’s Law 1662

21. 21 Robert Boyle

22. 22 At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

23. 23 Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

24. 24 The effect of pressure on the volume of a gas.

25. 25 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:

26. 26 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).

27. 27 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).

28. 28 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 = ?

29. 29 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.

30. 30 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

31. 31 Charles’ Law 1787

32. 32 Jacques Charles

33. 33 Absolute Zero on the Kelvin Scale If a given volume of any gas at 0 o C is cooled by 1 o C the volume of the gas decreases by. If a given volume of any gas at 0 o C is cooled by 20 o C the volume of the gas decreases by.

34. 34 Absolute Zero on the Kelvin Scale If a given volume of any gas at 0 o C is cooled by 273 o C the volume of the gas decreases by. -273 o C (more precisely –273.15 o C) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.

35. 35 Volume-temperature relationship of methane (CH 4 ).

36. 36 Charles’ Law At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

37. 37 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.

38. 38 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?

39. 39 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?

40. 40 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?

41. 41 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 = ?

42. 42 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.

43. 43 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

44. 44 Gay-Lussac’s Law 1802

45. 45 Joseph Gay-Lussac

46. 46 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.

47. 47 The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

48. 48 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?

49. 49 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?

50. 50 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

51. 51 Combined Gas Laws

52. 52 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

53. 53 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:

54. 54 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?

55. 55 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.

56. 56 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?

57. 57 Dalton’s Law of Partial Pressures

58. 58 John Dalton

59. 59 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 + ….

60. 60 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

61. 61 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

62. 62 Oxygen collected over water.

63. 63 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.

64. 64 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:

65. 65 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

66. 66 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.

67. 67 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.

68. 68 Avogadro’s Law 1811

69. 69 Amadeo (Amedeo) Avogadro

70. 70 Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

71. 71 Standard Temperature and Pressure

72. 72 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.

73. 73 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.

74. 74

75. 75 Ultimate Combined Gas Law

76. 76 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

77. 77 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

78. 78 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.

79. 79 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

80. 80 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

81. 81 Mole-Mass-Volume Relationships

82. 82 Density of Gases

83. 83

84. 84 Density of Gases liters grams

85. 85 Density of Gases depends on T and P

86. 86 The density of neon at STP is 0.900 g/L. What is the mole weight of neon?

87. 87 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

88. 88 Ideal Gas Equation

89. 89 V  P nT

90. 90 V  P nT atmospheres

91. 91 V  P nT liters

92. 92 V  P nT moles

93. 93 V  P nT Kelvin

94. 94 V  P nT Ideal Gas Constant

95. 95 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.

96. 96 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?

97. 97 Mole-Weight Calculations

98. 98 Determination of Molecular Weights Using the Ideal Gas Equation

99. 99 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

100. 100 Gas Stoichiometry

101. 101 Definition Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

102. 102 All calculations are done at STP. Gases are assumed to behave as ideal gases. A gas not at STP is converted to STP.

103. 103 Gas Stoichiometry Primary conversions involved in stoichiometry.

104. 104 Mole-Volume Calculations Mass-Volume Calculations

105. 105 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?

106. 106 Step 3. Calculate the moles of O 2, using the mole-ratio method. What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? Step 4. Convert moles of O 2 to liters of O 2 2 KClO 3  2KCl + 3 O 2

107. 107 The problem can also be solved 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

108. 108 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?

109. 109 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?

110. 110 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 Solve the ideal gas equation for V

111. 111 Volume-Volume Calculations

112. 112 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.

113. 113 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.

114. 114 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)

115. 115 Molecular Formula Calculations

116. 116 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:

117. 117 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

118. 118 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

119. 119 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

120. 120 Effusion A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

121. 121 Thomas Graham

122. 122 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.

123. 123 What is the ratio of the rate of diffusion of CO to CO 2 ?

124. 124 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.

125. 125 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.

126. 126 Real Gases

127. 127 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.

128. 128 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.

129. 129