1. Chapter 12 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena The Gaseous State of Matter The air in a hot air balloon expands When it is heated. Some of the air escapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.

2. Chapter Outline Copyright 2012 John Wiley & Sons, Inc 12.1 General PropertiesGeneral Properties 12.2 The Kinetic-Molecular TheoryThe Kinetic-Molecular Theory 12.3 Measurement of PressureMeasurement of Pressure 12.4 Dependence of Pressure on Number of Molecules and TemperatureDependence of Pressure on Number of Molecules and Temperature 12.5 Boyle’s LawBoyle’s Law 12.6 Charles’ LawCharles’ Law 12.7 Gay-Lussac’s LawGay-Lussac’s Law 12.8 Combined Gas LawsCombined Gas Laws 12.9 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures 12.10 Avogadro’s LawAvogadro’s Law 12.11 Mole-Mass-Volume Relationships of GasesMole-Mass-Volume Relationships of Gases 12.12 Density of GasesDensity of Gases 12.13 Ideal Gas LawIdeal Gas Law 12.14 Gas StoichiometryGas Stoichiometry 12.15 Real GasesReal Gases

3. Objectives for Today  Kinetic Molecular Theory of Gases  Gas Measurements  Boyle’s, Charles’ and Gay-Lussac’s Laws Copyright 2012 John Wiley & Sons, Inc

4. GASES AND KINETIC MOLECULAR THEORY Copyright 2012 John Wiley & Sons, Inc

5. General Properties Gases Have an indefinite volume Expand to fill a container Have an indefinite shape Take the shape of a container Have low densities Have high kinetic energies Copyright 2012 John Wiley & Sons, Inc

6. Kinetic Molecular Theory (KMT) Assumptions of the KMT and ideal gases include: 1.Gases consist of tiny particles 2.The distance between particles is large compared with the size of the particles. 3.Gas particles have no attraction for each other 4.Gas particles move in straight lines in all directions, colliding frequently with each other and with the walls of the container. Copyright 2012 John Wiley & Sons, Inc

7. Kinetic Molecular Theory Assumptions of the KMT (continued): 5.Collisions are perfectly elastic (no energy is lost in the collision). 6.The average kinetic energy for particles is the same for all gases at the same temperature. 7. The average kinetic energy is directly proportional to the Kelvin temperature. Copyright 2012 John Wiley & Sons, Inc

8. Diffusion Copyright 2012 John Wiley & Sons, Inc

9. Effusion Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure. Graham’s law of effusion: Copyright 2012 John Wiley & Sons, Inc

10. Your Turn! Which gas will diffuse most rapidly? a.He b.Ne c.Ar d.Kr Copyright 2012 John Wiley & Sons, Inc

11. Measurement of Pressure Copyright 2012 John Wiley & Sons, Inc Pressure depends on the Number of gas molecules Temperature of the gas Volume the gas occupies

12. Atmospheric Pressure Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface. Copyright 2012 John Wiley & Sons, Inc

13. Barometer Copyright 2012 John Wiley & Sons, Inc

14. Pressure Conversions Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm Copyright 2012 John Wiley & Sons, Inc Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr.

15. Your Turn! A pressure of 3.00 atm is equal to a.819 torr b.3000 torr c.2280 torr d.253 torr Copyright 2012 John Wiley & Sons, Inc

16. Dependence of Pressure on Number of Molecules Copyright 2012 John Wiley & Sons, Inc P is proportional to n (number of molecules) at T c (constant T) and V c (constant V). The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.

17. Dependence of Pressure on Temperature Copyright 2012 John Wiley & Sons, Inc P is proportional to T at n c (constant number of moles) and V c. The increased pressure is due to more frequent collisions higher energy collisions

18. Your Turn! If you change the temperature of a sample of gas from 80°C to 25°C at constant volume, the pressure of the gas a.will increase. b.will decrease. c.will not change Copyright 2012 John Wiley & Sons, Inc

19. Boyle’s Law Copyright 2012 John Wiley & Sons, Inc What happens to V if you double P? V decreases by half! What happens to P if you double V? P decreases by half!

20. Boyle’s Law A sample of argon gas occupies 500.0 mL at 920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature. Copyright 2012 John Wiley & Sons, Inc Knowns V 1 = 500 mL P 1 = 920. torr V 2 = 937 mL Calculate Set-Up

21. Boyle’s Law Another approach to the same problem: Since volume increased from 500. mL to 937 ml, the pressure of 920. torr must decrease. Multiply the pressure by a volume ratio that decreases the pressure: Copyright 2012 John Wiley & Sons, Inc

22. Your Turn! A 6.00 L sample of a gas at a pressure of 8.00 atm is compressed to 4.00 L at a constant temperature. What is the pressure of the gas? a.4.00 atm b.12.0 atm c.24.0 atm d.48.0 atm Copyright 2012 John Wiley & Sons, Inc

23. Your Turn! A 400. mL sample of a gas is at a pressure of 760. torr. If the temperature remains constant, what will be its volume at 190. torr? A. 100. mL B. 400. mL C. 25.0 mL D. 1.60x10 2 mL Copyright 2012 John Wiley & Sons, Inc

24. Charles’ Law Copyright 2012 John Wiley & Sons, Inc The volume of an ideal gas at absolute zero (-273°C) is zero. Real gases condense at their boiling point so it is not possible to have a gas with zero volume. The gas laws are based on Kelvin temperature. All gas law problems must be worked in Kelvin!

25. Charles’ Law A 2.0 L He balloon at 25°C is taken outside on a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant? Copyright 2012 John Wiley & Sons, Inc Knowns V 1 = 2.0 L T 1 = 25°C= 298 K T 2 = -15°C = 258 K Calculate Set-Up

26. Charles’ Law Another approach to the same problem: Since T decreased from 25°C to -15°C, the volume of the 2.0L balloon must decrease. Multiply the volume by a Kelvin temperature ratio that decreases the volume: Copyright 2012 John Wiley & Sons, Inc

27. Your Turn The volume of a gas always increases when a.Temperature increases and pressure decreases b.Temperature increases and pressure increases c.Temperature decreases and pressure increases d.Temperature decreases and pressure decreases Copyright 2012 John Wiley & Sons, Inc

28. Your Turn! A sample of CO 2 has a volume of 200. mL at 20.0 ° C. What will be its volume at 40.0 ° C, assuming that the pressure remains constant? a.18.8 mL b.100. mL c.213 mL d.400. mL Copyright 2012 John Wiley & Sons, Inc

29. Your Turn! A sample of gas has a volume of 3.00 L at 10.0 ° C. What will be its temperature in °C if the gas expands to 6.00 L at constant pressure? a.20.0°C b.293°C c.566°C d.142°C Copyright 2012 John Wiley & Sons, Inc

30. Gay-Lussac’s Law Copyright 2012 John Wiley & Sons, Inc

31. 31 40 o C + 273 = 313 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? Method A. Conversion Factors Step 1. Change o C to K: o C + 273 = K 100 o C + 273 = 373 K temperature increases  pressure increases Determine whether temperature is being increased or decreased.

32. 32 Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure: 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?

33. 33 A temperature ratio greater than 1 will increase the pressure 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?

34. 34 Method B. Algebraic Equation 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?

35. 35 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?

36. 36 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 atm T 1 = 40 o C = 313 K P 2 = ?T 2 = 100 o C = 373 K

37. Objectives for Today Kinetic Molecular Theory of Gases Gas Measurements Boyle’s, Charles’ and Gay-Lussac’s Laws Copyright 2012 John Wiley & Sons, Inc

38. Objectives for Today  Combined Gas Law  Dalton’s Law of Partial Pressure  Avogadro’s Law  Density of Gases Copyright 2012 John Wiley & Sons, Inc

39. Combined Gas Laws Copyright 2012 John Wiley & Sons, Inc Used for calculating the results of changes in gas conditions. Boyle’s Law where T c Charles’ Law where P c Gay Lussacs’ Law where V c P 1 and P 2, V 1 and V 2 can be any units as long as they are the same. T 1 and T 2 must be in Kelvin.

40. Combined Gas Law Copyright 2012 John Wiley & Sons, Inc If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr? STP: Standard Temperature 273K or 0°C Standard Pressure 1 atm or 760 torr Knowns V 1 = 500. mL T 1 =273K P 1 = 760 torr T 2 = 85°C = 358K P 2 = 560 torr Set-Up Calculate

41. Combined Gas Law A sample of oxygen gas occupies 500.0 mL at 722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg. Copyright 2012 John Wiley & Sons, Inc Knowns V 1 = 500. mLT 1 = -25°C = 248K P 1 = 722 torr V 2 = 2.53 L = 2530 mLP 2 = 560 torr Set-Up Calculate

42. Your Turn! A sample of gas has a volume of 8.00 L at 20.0 ° C and 700. torr. What will be its volume at STP? a.1.20 L b.9.32 L c.53.2 L d.6.87 L Copyright 2012 John Wiley & Sons, Inc

43. Dalton’s Law of Partial Pressures 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 + …. Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air. Copyright 2012 John Wiley & Sons, Inc

44. Collecting Gas Over Water Gases collected over water contain both the gas and water vapor. The vapor pressure of water is constant at a given temperature Pressure in the bottle is equalized so that the P inside = P atm Copyright 2012 John Wiley & Sons, Inc

45. Your Turn! A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor pressure of water at 22 ° C is 19.8 torr. a.742 torr b.782 torr c.784 torr d.750. torr Copyright 2012 John Wiley & Sons, Inc

46. Avogadro’s Law Equal volumes of different gases at the same T and P contain the same number of molecules. Copyright 2012 John Wiley & Sons, Inc 1 volume 1 molecule 1 mol 1 volume 1 molecule 1 mol 2 volumes 2 molecules 2 mol The ratio is the same:

47. Mole-Mass-Volume Relationships Molar Volume: One mole of any gas occupies 22.4 L at STP. Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP. Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm Set-Up Calculate 3.94 g 3.52 L = 25.1 g/mol

48. Your Turn! What is the molar mass of a gas if 240. mL of the gas at STP has a mass of 0.320 grams? a.8.57 g b.22.4 g c.16.8 g d.29.9 g

49. Density of Gases Calculate the density of nitrogen gas at STP. Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.

50. Your Turn! Which of the following gases is the most dense? a.H 2 b.N 2 c.CO 2 d.O 2 Carbon dioxide fire extinguishers can be used to put out fires because CO 2 is more dense than air and can be used to push oxygen away from the fuel source.

51. Objectives for Today Combined Gas Law Dalton’s Law of Partial Pressure Avogadro’s Law Density of Gases Copyright 2012 John Wiley & Sons, Inc

52. Topics for Today  The Ideal Gas Law  Gas Stoichiometry  Real Gases & Air Pollutants

53. IDEAL GAS LAW

54. P is in units of Atmospheres V is in units of Liters n is in units of moles T is in units of Kelvin

55. Example 1: Calculate the volume of 1 mole of any gas at STP. Knowns n = 1 mole T = 273K P = 1 atm Set-Up Calculate Molar volume!

56. Example 2: How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg? Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm Set-Up Calculate

57. Example 3: Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr. Knowns m =4.12 gV = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm Set-Up Calculate

58. Your Turn! What is the molar mass of a gas if 40.0 L of the gas has a mass of 36.0 g at 740. torr and 30.0 ° C? a.33.1 g b.23.0 g c.56.0 g d.333 g

59. STOICHIOMETRY & GASES

60. Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L. Use the Ideal Gas Law if the conditions are not at STP.

61. Example 4: Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm. P 4(s) + 6H 2(g)  4PH 3(g) Knowns V = 4.0 LT = 273 K P = 1 atm Solution Map L H 2  mol H 2  mol P 4 Calculate

62. Example 5: What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C 2 H 6 ? 2 C 2 H 6(g) + 7 O 2(g)  4 CO 2(g) + 6 H 2 O (l) Solution Map m C 2 H 6  mol C 2 H 6  mol O 2  volume O 2 Knowns m = 3.2 g C 2 H 6 T = 25°C = 298K P = 1 atm Calculate

63. Your Turn! How many moles of oxygen gas are used up during the reaction with 18.0 L of CH 4 gas measured at STP? CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) a.1.61 moles b.2.49 moles c.18.0 moles d.36.0 moles

64. Example 6: Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm. N 2(g) + 3H 2(g)  2NH 3(g) Knowns V = 9.0 LT = 450K P = 5.00 atm Solution Map Assume T and P for both gases are the same. Use volume ratio instead of mole ratio! L H 2  L N 2 Calculate

65. Your Turn! What volume of sulfur dioxide gas will react when 12.0 L of oxygen is consumed at constant temperature and pressure? 2 SO 2 + O 2  2 SO 3 a.6.00 L b.12.0 L c.24.0 L d.60.0 L

66. REAL GASES & AIR POLLUTANTS

67. Most real gases behave like ideal gases under ordinary temperature and pressure conditions. Conditions where real gases don’t behave ideally: At high P because the distance between particles is too small and the molecules are too crowded together. At low T because gas molecules begin to attract each other. High P and low T are used to condense gases.

68. Stratospheric Ozone Tropospheric Ozone Oxides of Nitrogen Acid Rain Greenhouse Gases

69. Topics for Today The Ideal Gas Law Gas Stoichiometry Real Gases & Air Pollutants