1. 1 Chapter 8 Gases, Liquids, and Solids

2. 2  Solids have  A definite shape.  A definite volume.  Particles that are close together in a fixed arrangement.  Particles that move very slowly. Solids 8.1 State of Matter and Their Changes

3. 3  Liquids have  An indefinite shape, but a definite volume.  The same shape as their container.  Particles that are close together, but mobile.  Particles that move slowly. Liquids

4. 4  Gases have  An indefinite shape.  An indefinite volume.  The same shape and volume as their container.  Particles that are far apart.  Particles that move fast. Gases

5. 5 Identify each as: 1) solid 2) liquid or 3) gas. ___ A. It has a definite volume, but takes the shape of the container. __ B. Its particles are moving rapidly. __ C. It fills the volume of a container. __ D. It has particles in a fixed arrangement. __ E. It has particles close together that are mobile. Learning Check

6. 6 Identify each as: 1) solid 2) liquid or 3) gas. 2 A. It has a definite volume, but takes the shape of the container. 3 B. Its particles are moving rapidly. 3 C. It fills the volume of a container. 1 D. It has particles in a fixed arrangement. 2 E. It has particles close together that are mobile. Solution

7. 7 8.2 Gases and the Kinetic- Molecular Theory

8. 8 Particles of a gas  Move rapidly in straight lines.  Have kinetic energy that increases with an increase in temperature.  Are very far apart.  Have essentially no attractive (or repulsive) forces.  Have very small individual volume compared to the volume of the container they occupy. Kinetic Theory of Gases

9. 9 Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Properties of Gases

10. 10 A barometer measures the pressure exerted by the gases in the atmosphere. The atmospheric pressure is measured as the height in mm of the mercury column. Barometer

11. Chapter 09Slide 11 Gas Pressure02 Units of pressure: atmosphere (atm) Pa (N/m 2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in 2 (14.696 lb/in 2 = 1 atm) in Hg (29.921 in Hg = 1 atm)

12. 12 A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense 2) H 2 O is heavier 3) air is more dense than H 2 O Learning Check

13. 13 A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense Solution

14. 14  A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area  One atmosphere (1 atm) is 760 mm Hg.  1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr Pressure 8.3 Pressure

15. 15 In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa). Units of Pressure

16. 16 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3)22,300 mm Hg Learning Check

17. 17 A. What is 475 mm Hg expressed in atm? 2) 0.625 atm 475 mm Hg x 1 atm = 0.625atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm Solution

18. 18 8.4 Boyle’s Law: The Relation Between Volume and Pressure

19. 19  The pressure of a gas is inversely related to its volume when T and n are constant.  If volume decreases, the pressure increases. Boyle’s Law

20. 20 The product P x V remains constant as long as T and n do not change. P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyle’s Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyle’s Law

21. 21 Temperature and Volume (Charles’ Law)

22. 22 The Kelvin temperature of a gas is directly related to the volume (P and n are constant). When the temperature of a gas increases, its volume increases. Charles’ Law

23. 23 For two conditions, Charles’ Law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles’ Law to solve for V 2 V 2 = V 1 T 2 T 1 Charles’ Law V and T

24. 24 Learning Check Solve Charles’ Law expression for T 2. V 1 = V 2 T 1 T 2

25. 25 Solution V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 =V 2 T 1 Isolate T 2 by dividing through by V 1 V 1 T 2 =V 2 T 1 V 1 V 1 T 2 =V 2 T 1 V 1

26. 26 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443°C 2) 170°C 3) – 82°C Learning Check

27. 27 2) 170°C T 2 = T 1 V 2 V 1 T 2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 K = 170°C Solution

28. 28 The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. P 1 = P 2 T 1 T 2 8.6 Gay-Lussac’s Law: The Relation Between Pressure and Temperature

29. 29 A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) 1. Set up a data table. Conditions 1Conditions 2 P 1 = 2.0 atmP 2 = T 1 = 18°C + 273 T 2 = 62°C + 273 = 291 K = 335 K Calculation with Gay-Lussac’s Law ?

30. 30 Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P 2 P 1 = P 2 T 1 T 2 P 2 = P 1 T 2 T 1 P 2 = 2.0 atm x 335 K = 2.3 atm 291 K

31. 31 Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12.0 L to 24.0 L. D. Volume _______when T changes from 15.0 °C to 45.0°C. Learning Check

32. 32 Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure 1) Increases, when V decreases. B. When T decreases, V 2) Decreases. C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C Solution

33. 33 8.7 The Combined gas Law

34. 34 Combined Gas Law Charles’ law: V  T  (at constant n and P) Boyle’s law: V  (at constant n and T) 1 P V V  T P V = constant x = K T P T P K is a constant PV / T= K

35. 35 The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P 1 V 1 =P 2 V 2 T 1 T 2 Combined Gas Law

36. 36 A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1Conditions 2 P 1 = 0.800 atm P 2 = 3.20 atm V 1 = 0.180 L (180 mL) V 2 = 90.0 mL T 1 = 29°C + 273 = 302 KT 2 = ?? Combined Gas Law Calculation

37. 37 2. Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 2 = T 1 P 2 V 2 P 1 V 1 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T 2 = 604 K – 273 = 331 °C Combined Gas Law Calculation (continued)

38. 38 A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)? Learning Check

39. 39 Data Table T 1 = 308 K T 2 = -95°C + 273 = 178K V 1 = 675 mL V 2 = ??? P 1 = 646 mm Hg P 2 = 802 mm Hg Solve for V 2 V 2 =V 1 P 1 T 2 P 2 T 1 V 2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K Solution

40. 40 The volume of a gas is directly related to the number of moles of gas when T and P are constant. V 1 = V 2 n 1 n 2 Avogadro's Law: Volume and Moles 8.8 Avogadro’s law: The Relation Between Volume and Molar Amount

41. 41 Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L

42. 42 Solution 3) 2.4 L Conditions 1Conditions 2 V 1 = 1.5 LV 2 = ??? n 1 = 0.75 mole Hen 2 = 1.2 moles He V 2 = V 1 n 2 n 1 V 2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

43. 43 The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg) STP

44. 44 At STP, 1 mole of a gas occupies a volume of 22.4 L. The volume of one mole of a gas is called the molar volume. Molar Volume

45. 45 The molar volume at STP can be used to form conversion factors. 22.4 L and 1 mole 1 mole 22.4 L Molar Volume as a Conversion Factor

46. 46 A. What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g2) 0.357 g3) 1.43 g Learning Check

47. 47 A. 1) 5.60 L 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He Solution

48. 48 8.9 The Ideal Gas Law _McG 3/4

49. 49 Ideal Gas Equation Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

50. 50 The universal gas constant, R, can be calculated using the molar volume of a gas at STP. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = 0.0821 L atm mole K Note there are four units associated with R. Universal Gas Constant, R PV = nRT

51. 51 A cylinder contains 5.0 L of O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Learning Check

52. 52 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mole O 2 x 32.0 g O 2 = 5.8 g O 2 1 mole O 2 Solution

53. 53 What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT (0.0821 L atm/mole K)(303K) = 0.00703 mole 2. Set up the molar mass relationship. Molar mass = g = 0.250 g = 35.6 g/mole mole 0.00703 mole Molar Mass of a Gas

54. 54 Gases in Equations The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. Problem: What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s)

55. 55 Gases in Equations (continued) 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl 2 = 0.083 mole Cl 2 27.0 g Al 2 moles Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT = (0.083 mole Cl 2 )(0.0821 Latm/moleK)(300K) P 1.2 atm = 1.7 L Cl 2

56. 56 What volume (L) of O 2 at 24°C and 0.950 atm are needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

57. 57 1. Calculate the moles of O 2 needed. 28.0 g NH 3 x 1 mole NH 3 x 5 mole O 2 17.0 g NH 3 4 mole NH 3 = 2.06 mole O 2 2. Place the moles O 2 in the ideal gas equation. V = nRT = (2.06 moles)(0.0821 L atm/moleK)(297K) P 0.950 atm = 52.9 L O 2 Solution

58. 58 Mixture of Gases

59. 59 Gases We Breathe

60. 60 8.10 Partial Pressure and Dalton’s law

61. 61 In a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container. Partial Pressure

62. 62 The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 +..... Dalton’s Law of Partial Pressures

63. 63 The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. Partial Pressures

64. 64 For example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L. V = 22.4 L Total Pressure 0.5 mole O 2 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 mole N 2 0.4 mole O 2 0.6 mole He 1.0 mole 1.0 atm

65. 65 A scuba tank contains O 2 with a pressure of 0.450 atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank? Learning Check

66. 66 1. Convert the pressure in atm to mm Hg 0.450 atm x 760 mm Hg = 342 mm Hg = P O 1 atm 2 2. Calculate the sum of the partial pressures. P total = P O + P He 2 P total = 342 mm Hg + 855 mm Hg = 1197 mm Hg Solution

67. 67 For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium? 1) 520 mm Hg 2) 2040 mm Hg 3) 4800 mm Hg Learning Check

68. 68 3) 4800 mm Hg P Total = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm P Total = P O + P He 2 P He = P Total - P O 2 P He = 6080 mm Hg - 1280 mm Hg = 4800 mm Hg Solution

69. 69 Henry’s Law According to Henry’s Law, the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.

70. 70 Blood Gases In the lungs, O 2 enters the blood, while CO 2 from the blood is released. In the tissues, O 2 enters the cells, which release CO 2 into the blood.

71. 71 Blood Gases In the body, cells use up O 2 and give off CO 2. O 2 flows into the tissues because the partial pressure of O 2 is higher (100 mm Hg) in oxygenated blood, and lower (<30 mm Hg) in the tissues. CO 2 flows out of the tissues because the partial pressure of CO 2 is higher (>50 mm Hg) in the tissues and lower (40 mm Hg) in the blood.

72. 72 In ionic compounds, ionic bonds are strong attractive forces that hold positive and negative ions together. 8.11 Intermolecular Forces

73. 73 Attractive Forces between Covalent Compounds In covalent compounds, polar molecules exert attractive forces called dipole-dipole attractions. Hydrogen bonds are strong dipole attractions between hydrogen atoms and atoms of F, O, or N, which are very electronegative.

74. 74 Dipole–Dipole Force: Molecule containing polar covalent bond may have a net molecular polarity. In such cases, the positive and negative ends of different molecules are attracted to each other what is called a dipole–dipole force. Fig 8.14 Attraction between dipoles in polar molecules

75. 75 Hydrogen bonds: A hydrogen bond is an attractive interaction between an unshared electron pair on an electronegative O, N, and F and a positively polarized hydrogen atom bonded to another O, N, or F. Hydrogen bonding in water and ammonia

76. 76 London dispersion force: The short-lived attractive force due to constant motion of electrons within molecules. Fig 8.15 London dispersion forces and the electron distribution in bromine

77. 77 Melting Points and Attractive Forces Ionic compounds require large amounts of energy to break apart ionic bonds. Thus, they have high melting points. Hydrogen bonds are the strongest type of dipole- dipole attractions. They require more energy to break than other dipole attractions. Dispersion forces are weak interactions and very little energy is needed to change state.

78. 78 Melting Points and Attractive Forces of Some Typical Substances

79. 79 Learning Check Identify the type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion A. NCl 3 B. H 2 O C. Br-Br D. KCl E. NH 3

80. 80 Solution Identify the type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds4) Only dispersion 2 A. NCl 3 3 B. H 2 O 4 C. Br-Br 1 D. KCl 3 E. NH 3

81. 81 Energy and States of Matter Heating and Cooling Curves

82. 82 8.15 Changes of States Heat of fusion: The quantity of heat required to completely melt one gram of a substance once it has reached its melting point. Heat of vaporization: The quantity of heat required to completely vaporize one gram of a substance once it has reached its boiling point.

83. 83 Heating Curve A heating curve illustrates the changes of state as a solid is heated. Sloped lines indicate an increase in temperature. Plateaus (flat lines) indicate a change of state.

84. 84 A. A flat line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state Learning Check

85. 85 A. A flat line on a heating curve represents 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change Solution

86. 86 Cooling Curve  A cooling curve illustrates the changes of state as a gas is cooled.  Sloped lines indicate a decrease in temperature.  This cooling curve for water begins at 140°C and ends at -30°C.

87. 87 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0°C2) 50°C3) 100°C B. At a temperature of 0°C, water 1) freezes2) melts3) changes to a gas C. At 40 °C, water is a 1) solid 2) liquid3) gas D. When water freezes, heat is 1) removed2) added Learning Check

88. 88 Use the cooling curve for water to answer each. A. Steam condenses at a temperature of 3) 100°C B. At a temperature of 0°C, water 1) freezes C. At 40 °C, water is a 2) liquid D. When water freezes, heat is 1) removed Solution

89. 89 8.14 Solids Amorphous solid: One whose constituent particles are randomly scattered and has no long range structure. Crystalline solid: One whose particles – whether atoms, ions, or molecules- have an ordered arrangement extending over a long range. Crystalline solids can be categorized as: Ionic such as Na + Cl - whose constituent particles are ions. Covalent such as diamond or quartz where units are held together by covalent bonds. Metallic Crystalline solid

90. 90 Covalent Crystalline solid such as sucrose or ice whose constituent particles are molecules held together by the intermolecular forces.

91. Metallic Crystalline solid 91 Metallic such as silver or iron – three dimensional array of metal ions immersed in electrons that are free to move about.

92. 92 Chapter Summary According to the kinetic molecular theory: The physical behavior of gases can be explained by assuming that they consist of particles moving rapidly at random, separated from each other by great distances. Gas pressure is the result of molecular collisions with a surface. Boyle’s law: Volume of a fixed amount of gas is inversely proportional to its pressure. Charle’s law: Volume of a fixed amount of gas is directly proportional to its temperature.

93. 93 Gay-Lussac’s law: The pressure of a fixed amount of gas at constant volume is directly proportional to its Kelvin temperature. Avogadro’s law: Equal volume of gases at the same temperature and pressure contain the same number of moles. Combined gas law: Boyle’s law, Charle’s law, and Gay-Lussac’s law together is known as combined gas law. Ideal gas law: Relates the effects of temperature, pressure, volume, and molar amount. Chapter Summary

94. 94 At 0 o C and 1 atm pressure, Standard temperature and pressure (STP), 1 mole of any gas occupies a volume of 22.4 liters. Partial pressure: The amount of pressure exerted by an individual gas in a mixture. Intermolecular forces: Forces that act to hold molecules close to each other. Three major types of intermolecular forces are: Dipole-dipole forces, London dispersion forces, and Hydrogen bonds Solids: Crystalline – Constituent particles are ordered: ionic, molecular, metallic, and covalent. Amorphous - Constituent particles are not ordered. Chapter Summary Contd.

95. 95 When a solid is heated, particles begin to move around freely at the melting point, and the substance becomes liquid. The amount of heat that is necessary to melt a solid at its melting point is the heat of fusion of that solid. As a liquid is heated, molecules escape from the surface of the liquid until a equilibrium is reached between liquid and gas, resulting in a vapor pressure of the liquid. At a liquid’s boiling point, its vapor pressure equal atmospheric pressure. The amount of heat necessary to vaporize a given amount of liquid at its boiling point is called its heat of vaporization. Chapter Summary Contd.

96. 96 End of Chapter 8