1. Chapter 11 - Gases Properties of Gases Gas Pressure Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. Chapter 11 Slide 2 of 89 Kinetic Theory of Gases Gases consist of small particles that Move rapidly in straight lines. Have essentially no attractive (or repulsive) forces. Are very far apart. Have very small volumes compared to the volume of the container they occupy. Have kinetic energies that increase with an increase in temperature. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

3. Chapter 11 Slide 3 of 89 Properties of Gases Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n, number of moles). Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 11.1

4. Chapter 11 Slide 4 of 89 Units of Pressure Gas pressure Is described as a force acting on a specific area. Pressure (P) = Force Area Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 11.1

5. Chapter 11 Slide 5 of 89 What is 475 mm Hg expressed in atm? 475 mm Hg x 1 atm = 0.625 atm 760 mm Hg The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm Learning Check

6. Chapter 11 Slide 6 of 89 Atmospheric Pressure The atmospheric pressure Is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Is about 1 atmosphere or a little less at sea level. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. Chapter 11 Slide 7 of 89 Altitude and Atmospheric Pressure Atmospheric pressure Depends on the altitude and the weather. Is lower at high altitudes where the density of air is less. Is higher on a rainy day than on a sunny day. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

8. Chapter 11 Slide 8 of 89 Barometer A barometer Measures the pressure exerted by the gases in the atmosphere. Indicates atmospheric pressure as the height in mm of the mercury column. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

9. Chapter 11 Slide 9 of 89 Chapter 11Gases Pressure and Volume Boyle’s Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

10. Chapter 11 Slide 10 of 89 Boyle’s Law Boyle’s Law states that The pressure of a gas is inversely related to its volume when T and n are constant. If volume decreases, the pressure increases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

11. Chapter 11 Slide 11 of 89 In Boyle’s Law The product P x V is 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

12. Chapter 11 Slide 12 of 89 Solving for a Gas Law Factor The equation for Boyle’s Law can be rearranged to solve for any factor. P 1 V 1 = P 2 V 2 Boyle’s Law To obtain V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 P 2 P 2 P 1 V 1 = V 2 P 2

13. Chapter 11 Slide 13 of 89 PV in Breathing In inhalation, The lungs expand. The pressure in the lungs decreases. Air flows towards the lower pressure in the lungs. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

14. Chapter 11 Slide 14 of 89 PV in Breathing In exhalation Lung volume decreases. Pressure within the lungs increases. Air flows from the higher pressure in the lungs to the outside. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

15. Chapter 11 Slide 15 of 89 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

16. Chapter 11 Slide 16 of 89 Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T? 1. Set up a data table Conditions 1Conditions 2 P 1 = 550 mm HgP 2 = 2200 mm Hg V 1 = 8.0 LV 2 = (predict smaller V 2 ) Calculation with Boyle’s Law ?

17. Chapter 11 Slide 17 of 89 2. Because pressure increases, we predict that the volume will decrease. Solve Boyle’s Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 V 2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume Calculation with Boyle’s Law (Continued)

18. Chapter 11 Slide 18 of 89 Learning Check For a cylinder containing helium indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases B 2) Pressure increases A Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

19. Chapter 11 Slide 19 of 89 Learning Check If the helium in a cylinder has a volume of 120 mL and a pressure of 850 mm Hg, what is the new volume if the pressure is changed to 425 mm Hg inside the cylinder? A) 60 mL B) 120 mLC) 240 mL P 1 = 850 mm Hg P 2 = 425 mm Hg V 1 = 120 mL V 2 = ?? V 2 = P 1 V 1 = 120 mL x 850 mm Hg = 240 mL P 2 425 mm Hg P-1

20. Chapter 11 Slide 20 of 89 Learning Check A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T constant), is the new volume represented by A, B, or C? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings P-2

21. Chapter 11 Slide 21 of 89 If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 LB) 6.4 LC) 12.8 L V 2 = V 1 P 1 P 2 V 2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant.) Learning Check

22. Chapter 11 Slide 22 of 89 A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg Learning Check Practice At Home

23. Chapter 11 Slide 23 of 89 Chapter 11Gases Temperature and Volume (Charles’ Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

24. Chapter 11 Slide 24 of 89 Charles’ Law In Charles’ Law 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. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

25. Chapter 11 Slide 25 of 89 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 T 2 xV 1 = V 2 x T 2 T 1 T 2 V 2 = V 1 T 2 T 1 Charles’ Law: V and T

26. Chapter 11 Slide 26 of 89 Learning Check Solve Charles’ Law expression for T 2. V 1 = V 2 T 1 T 2

27. Chapter 11 Slide 27 of 89 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 T 2 =V 2 T 1 V 1

28. Chapter 11 Slide 28 of 89 A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0°C, what is the new volume of the balloon (P constant)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? (decrease) T 1 = 21°C = 294 KT 2 = 0°C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations. Calculations Using Charles’ Law

29. Chapter 11 Slide 29 of 89 Calculations Using Charles’ Law (continued) 2. Solve Charles’ law for V 2 V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K When temperature decreased, the volume decreased as predicted.

30. Chapter 11 Slide 30 of 89 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)? A) 443°C B) 170°C C) - 82°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 Learning Check 18 o C + 273 = 291 K

31. Chapter 11 Slide 31 of 89 Chapter 11Gases Temperature and Pressure (Gay-Lussac’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

32. Chapter 11 Slide 32 of 89 Gay-Lussac’s Law: P and T In Gay-Lussac’s Law The pressure exerted by a gas is directly related to the Kelvin temperature. V and n are constant. P 1 = P 2 T 1 T 2 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

33. Chapter 11 Slide 33 of 89 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 = (increase) T 1 = 18°C + 273 T 2 = 62°C + 273 = 291 K = 335 K Calculation with Gay- Lussac’s Law ?

34. Chapter 11 Slide 34 of 89 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

35. Chapter 11 Slide 35 of 89 Use the gas laws to complete with B) Increases D) Decreases 1. Pressure _______, when V decreases. 2. When T decreases, V _______. 3. Pressure _______ when V changes from 12 L to 24 L 4. Volume _______when T changes from 15 °C to 45°C 5. Pressure ________when n changes from 1mole to 3 moles. Quiz 11-1

36. Chapter 11 Slide 36 of 89 Vapor Pressure and Boiling Point Vapor pressure is the Pressure of gas molecules above the surface of a liquid. At the boiling point Gas bubbles within the liquid have a pressure equal to the vapor pressure. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

37. Chapter 11 Slide 37 of 89 Boiling Point of Water The boiling point of water Depends on the vapor pressure. Is lower at higher altitudes. Is increased by using an autoclave to increase external pressure. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 11.4

38. Chapter 11 Slide 38 of 89 Chapter 11 Gases The Combined Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

39. Chapter 11 Slide 39 of 89 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

40. Chapter 11 Slide 40 of 89 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

41. Chapter 11 Slide 41 of 89 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)

42. Chapter 11 Slide 42 of 89 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

43. Chapter 11 Slide 43 of 89 1. Data Table Conditions 1Conditions 2 T 1 = 308 KT 2 = -95°C + 273 = 178K V 1 = 675 mL V 2 = ??? P 1 = 646 mm Hg P 2 = 802 mm Hg 2. 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 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)? P 1 V 1 = P 2 V 2 T 1 T 2

44. Chapter 11 Slide 44 of 89 Chapter 11 Gases Volume and Moles (Avogadro’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

45. Chapter 11 Slide 45 of 89 Avogadro's Law: Volume and Moles In Avogadro’s Law The volume of a gas is directly related to the number of moles of gas. T and P are constant. V 1 = V 2 n 1 n 2 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

46. Chapter 11 Slide 46 of 89 Learning Check If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

47. Chapter 11 Slide 47 of 89 If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L Conditions 1Conditions 2 V 1 = 1.5 LV 2 = ??? n 1 = 0.75 mol Hen 2 = 1.2 mol He V 2 = V 1 n 2 n 1 V 2 = 1.5 L x 1.2 mol He = 2.4 L 0.75 mol He V 1 = V 2 n 1 n 2

48. Chapter 11 Slide 48 of 89 The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP, Standard Temperature and Pressure). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg) STP

49. Chapter 11 Slide 49 of 89 Molar Volume At standard temperature and pressure (STP). 1 mol of a gas occupies a volume of 22.4 L, which is called its molar volume. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

50. Chapter 11 Slide 50 of 89 The molar volume At STP Can be used to form conversion factors. 22.4 L and 1 mol 1 mol 22.4 L Molar Volume as a Conversion Factor

51. Chapter 11 Slide 51 of 89 Guide to Using Molar Volume Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

52. Chapter 11 Slide 52 of 89 1. What is the volume at STP of 4.00 g CH 4 ? A) 5.59 LB) 11.2 LC) 44.8 L 4.00 g CH 4 x 1 mol CH 4 x 22.4 L (STP) = 16.04 g CH 4 1 mol CH 4 Learning Check 5.59 L

53. Chapter 11 Slide 53 of 89 2. How many grams of He are present in 8.00 L at STP? A) 25.6 gB) 0.357 gC) 1.43 g 8.00 L x 1 mol He x 4.003 g He = 22.4 L 1 mol He Learning Check 1.43 g He

54. Chapter 11 Slide 54 of 89 AT STP, the density of gas can be calculated using the mass of the gas and its volume. Density = Molar mass Molar volume Density of a Gas at STP

55. Chapter 11 Slide 55 of 89 Density of a Gas Calculate the density in g/L of O 2 gas at STP. At STP, we know that 1 mol O 2 (32.00 g) occupies a volume of 22.4 L. Density O 2 at STP = 32.00 g O 2 = 1.43 g/L 22.4 L (STP)

56. Chapter 11 Slide 56 of 89 Chapter 11 Gases The Ideal Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

57. Chapter 11 Slide 57 of 89 The ideal gas law Provides a relationship between the four properties (P, V, n, and T) of gases that can be written equal to a constant R. PV = R nT Rearranges these properties to give the ideal gas law expression. PV = nRT Ideal Gas Law

58. Chapter 11 Slide 58 of 89 The universal gas constant, R Can be calculated using the molar volume of a gas at STP. Calculated at STP uses 273 K,1.00 atm, 1 mol of a gas, and a molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mol) (273K) n T = 0.0821 L atm mol K Universal Gas Constant, R

59. Chapter 11 Slide 59 of 89 Another value for the universal gas constant is obtained using mm Hg for the STP pressure. The units of R are dependent on the units of the parameters that are used to calculate R. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? Learning Check

60. Chapter 11 Slide 60 of 89 What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? R = PV = (760 mm Hg) (22.4 L) nT (1 mol) (273 K) = 62.4 L mm Hg mol K Solution

61. Chapter 11 Slide 61 of 89 Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.86 mol N 2 O at 23°C, what is the pressure (mm Hg) in the tank? Learning Check Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

62. Chapter 11 Slide 62 of 89 1. Adjust the units of the given properties to match the units of R. V = 20.0 L20.0 L T = 23°C + 273 296 K n = 2.86 mol2.86 mol P = ? ? Solution

63. Chapter 11 Slide 63 of 89 2. Rearrange the ideal gas law for P. P = nRT V 3. Substitute quantities and solve. P = (2.86 mol)(62.4 L mm Hg)(296 K) (20.0 L) (mol K) = 2.64 x 10 3 mm Hg Solution

64. Chapter 11 Slide 64 of 89 Learning Check A cylinder contains 5.0 L O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

65. Chapter 11 Slide 65 of 89 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)(mol K) = 0.18 mol O 2 (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = 0. 18 mol O 2 x 32.0 g O 2 = 5.8 g O 2 1 mol O 2 Solution

66. Chapter 11 Slide 66 of 89 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) molK = 0.00703 mol RT (0.0821 L atm)(303K) 2. Set up the molar mass relationship. Molar mass = g = 0.250 g mol 0.00703 mol = 35.6 g/mol Molar Mass of a Gas

67. Chapter 11 Slide 67 of 89 Chapter 11 Gases Gas Laws and Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

68. Chapter 11 Slide 68 of 89 Gases in Equations The volume or amount of a gas in a chemical reaction can be calculated from STP conditions or the ideal gas law. Mole factors from the balanced equation.

69. Chapter 11 Slide 69 of 89 STP and Gas Equations What volume (L) of O 2 gas is needed to completely react with 15.0 g of aluminum at STP? 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Plan: g Al mol Al mol O 2 L O 2 (STP) 15.0 g Al x 1 mol Al x 3 mol O 2 x 22.4 L (STP) 26.98 g Al 4 mol Al 1 mol O 2 = 9.34 L O 2 at STP

70. Chapter 11 Slide 70 of 89 Ideal Gas Equation and Reactions What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) Plan? g Al moles Al moles Cl 2 L Cl 2

71. Chapter 11 Slide 71 of 89 Ideal Gas Equation and Reactions 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 mol Al x 3 mol Cl 2 = 0.083 mol Cl 2 26.98 g Al 2 mol Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT =(0.083 mol Cl 2 )(0.0821 L atm/mol K)(300 K) P 1.2 atm = 1.7 L Cl 2

72. Chapter 11 Slide 72 of 89 What volume (L) of O 2 at 24°C and 0.950 atm is needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

73. Chapter 11 Slide 73 of 89 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) 1. Calculate the moles of O 2 needed. 28.0 g NH 3 x 1 mol NH 3 x 5 mol O 2 17.03 g NH 3 4 mol NH 3 = 2.06 mol O 2 2. Place the moles of O 2 in the ideal gas equation. V = nRT =(2.06 mol)(0.0821 L atm/mol K)(297 K) P 0.950 atm = 52.9 L O 2 Solution

74. Chapter 11 Slide 74 of 89 Chapter 11 Gases Partial Pressure (Dalton’s Law) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

75. Chapter 11 Slide 75 of 89 The partial pressure of a gas Is the pressure of each gas in a mixture. Is the pressure that gas would exert if it were by itself in the container. Partial Pressure

76. Chapter 11 Slide 76 of 89 Dalton’s Law of Partial Pressures states that the total pressure Depends on the total number of gas particles, not on the types of particles. Exerted by a gas mixture is the sum of the partial pressures of those gases. P T = P 1 + P 2 +..... Dalton’s Law of Partial Pressures

77. Chapter 11 Slide 77 of 89 Illustrating Partial Pressures Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

78. Chapter 11 Slide 78 of 89 At STP, One mole of a pure gas in a volume of 22.4 L will exert the same pressure as one mole of a gas mixture in 22.4 L. V STP = 22.4 L Gas mixtures Total Pressure 0.5 mol O 2 0.3 mol He 0.2 mol Ar 1.0 mol 1.0 mol N 2 0.4 mol O 2 0.6 mol He 1.0 mol 1.0 atm

79. Chapter 11 Slide 79 of 89 Scuba Diving When a scuba diver makes a deep dive, the increased pressure causes N 2 (g) to dissolve in the blood. If the rise is too fast, the dissolved N 2 forms bubbles in the blood, a dangerous condition called "the bends". Helium is mixed with O 2 to prepare breathing mixtures for deep descents. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

80. Chapter 11 Slide 80 of 89 Learning Check 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? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

81. Chapter 11 Slide 81 of 89 1. Convert the pressure in atm to mm Hg 0.450 atm x 760 mm Hg = 342 mm Hg = P O 2 1 atm 2. Calculate the sum of the partial pressures. P total = PO 2 + P He P total = 342 mm Hg + 855 mm Hg = 1197 mm Hg Solution

82. Chapter 11 Slide 82 of 89 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

83. Chapter 11 Slide 83 of 89 A) 520 mm Hg B) 2040 mm Hg C) 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

84. Chapter 11 Slide 84 of 89 Gases Collected Over Water A gas produced in the laboratory Usually contains water vapor. P T = P water + P gas Has a partial pressure obtained by subtracting the vapor pressure of water from the P T. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

85. Chapter 11 Slide 85 of 89 Learning Check The decomposition of KClO 3 produces O 2 gas and solid KCl. If 124 mL of O 2 is collected over water at 762 mm Hg and 20.0  C, how many moles of O 2 were produced? 2KClO 3 (s) 2KCl(s) + 3O 2 (g)

86. Chapter 11 Slide 86 of 89 Solution Partial pressure water at 20.0  C = 18 mm Hg P T = P water + P gas 762 mm Hg = 18 mm Hg + PO 2 PO 2 = 762 mm Hg - 18 mm Hg = 744 mm Hg PV = nRT n = PV = (744 mm Hg)(0.124 L) = 0.00505 mol O 2 RT (62.4 L mm Hg)(293 K) mol K

87. Chapter 11 Slide 87 of 89 Gases We Breathe The air we breathe Is a gas mixture. Contains mostly N 2 and O 2 and small amounts of other gases. Table 11.7 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

88. Chapter 11 Slide 88 of 89 1. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? A) 35.6 B) 156 C) 760 P oxygen = ( % oxygen / 100 ) X P T P oxygen = ( 21 / 100 ) X 745 mm = 156 mm Learning Check

89. Chapter 11 Slide 89 of 89 B. At an atmospheric pressure of 714 mm, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 mm 2) 9.14 mm 3) 0.109 mm P nitrogen = (594 mm / 760 mm) X 714 mm or = (0.78) X 714 mm = 557 mm Learning Check