2. 1 © 2006 Brooks/Cole - Thomson Chemistry and Chemical Reactivity 6th Edition John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 12 Gases and Their Properties © 2006 Brooks/Cole Thomson Lectures written by John Kotz

3. 2 © 2006 Brooks/Cole - Thomson BEHAVIOR OF GASES Chapter 12

4. 3 © 2006 Brooks/Cole - Thomson Importance of Gases Airbags fill with N 2 gas in an accident.Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3.Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 22 NaN 3 ---> 2 Na + 3 N 2

5. 4 © 2006 Brooks/Cole - Thomson Hot Air Balloons — How Do They Work?

6. 5 © 2006 Brooks/Cole - Thomson THREE STATES OF MATTER

7. 6 © 2006 Brooks/Cole - Thomson General Properties of Gases There is a lot of “free” space in a gas.There is a lot of “free” space in a gas. Gases can be expanded infinitely.Gases can be expanded infinitely. Gases occupy containers uniformly and completely.Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.

8. 7 © 2006 Brooks/Cole - Thomson Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L)V = volume of the gas (L) T = temperature (K)T = temperature (K) n = amount (moles)n = amount (moles) P = pressure (atmospheres)P = pressure (atmospheres)

9. 8 © 2006 Brooks/Cole - Thomson Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

10. 9 © 2006 Brooks/Cole - Thomson Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to Hg densityHg density column heightcolumn height

11. 10 © 2006 Brooks/Cole - Thomson Pressure Column height measures P of atmosphere 1 standard atm = 760 mm Hg1 standard atm = 760 mm Hg = 29.9 inches Hg = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = 101.325 kPa

12. 11 © 2006 Brooks/Cole - Thomson IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. P V = n R T

13. 12 © 2006 Brooks/Cole - Thomson Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

14. 13 © 2006 Brooks/Cole - Thomson Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

15. 14 © 2006 Brooks/Cole - Thomson Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

16. 15 © 2006 Brooks/Cole - Thomson Charles’s original balloon Modern long-distance balloon

17. 16 © 2006 Brooks/Cole - Thomson Charles’s Law Balloons immersed in liquid N 2 (at -196 ˚C) will shrink as the air cools (and is liquefied).

18. 17 © 2006 Brooks/Cole - Thomson Charles’s Law

19. 18 © 2006 Brooks/Cole - Thomson Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules

20. 19 © 2006 Brooks/Cole - Thomson Avogadro’s Hypothesis The gases in this experiment are all measured at the same T and P. 2 H 2 (g) + O 2 (g) 2 H 2 O(g)

21. 20 © 2006 Brooks/Cole - Thomson Combining the Gas Laws V proportional to 1/PV proportional to 1/P V prop. to TV prop. to T V prop. to nV prop. to n Therefore, V prop. to nT/PTherefore, V prop. to nT/P V = 22.4 L for 1.00 mol whenV = 22.4 L for 1.00 mol when –Standard pressure and temperature (STP) –T = 273 K –P = 1.00 atm

22. 21 © 2006 Brooks/Cole - Thomson

23. 22 GAS DENSITY Screen 12.5 Higher Density air Low density helium

24. 23 © 2006 Brooks/Cole - Thomson What are the units of density? Grams/volume Molar mass / molar volume = Molar mass / 22.4 L = Density of a gas at STP Molar mass / molar volume = Molar mass / 22.4 L = Density of a gas at STP D(gas) = M / 22.4 L

25. 24 © 2006 Brooks/Cole - Thomson Given: d(gas) = M / 22.4 L d Is directly related to …. 1.M 2.22.4 3.Gas 4.Molar mass 5.Both 1 and 4

26. 25 © 2006 Brooks/Cole - Thomson The gas with greatest density at STP is …..... 1.NH 3 2.H 2 O 3.CO 2 4.SO 2

27. 26 © 2006 Brooks/Cole - Thomson USING GAS DENSITY The density of air at 15 o C and 1.00 atm is 1.23 g/L. What is the molar mass of air? 1. Calc. moles of air. V = 1.00 LP = 1.00 atmT = 288 K V = 1.00 LP = 1.00 atmT = 288 K n = PV/RT = 0.0423 mol n = PV/RT = 0.0423 mol 2. Calc. molar mass mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

28. 27 © 2006 Brooks/Cole - Thomson

29. 28 The Ideal Gas Law PV = nRT n is moles of gas The ideal gas equation is used whenever moles (or grams) of a substance is either given or needed. Watch out for the units! Units must match the units of R. R = 0.082057 L atm /mol K

30. 29 © 2006 Brooks/Cole - Thomson Using PV = nRT How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/Kmol

31. 30 © 2006 Brooks/Cole - Thomson Work on this problem while the class registers their clickers.

32. 31 © 2006 Brooks/Cole - Thomson Using PV = nRT How much N 2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C + 273 = 298 K T = 25 o C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

33. 32 © 2006 Brooks/Cole - Thomson Using PV = nRT How much N 2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 2. Now calc. n = PV / RT n = 1.1 x 10 3 mol (or about 30 kg of gas)

34. 33 © 2006 Brooks/Cole - Thomson Given: Ab=XYZ Variables that are directly related are …and… 1.A,Z 2.Y, b 3.X, b 4.All of above

35. 34 © 2006 Brooks/Cole - Thomson Given: aB = DF Inversely related are… and … 1.D, F 2.B, D 3.B, a 4.Both 1 and 3

36. 35 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

37. 36 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Strategy: Can you get moles of anything? What? Can you use moles of a liquid in the ideal gas equation?

38. 37 © 2006 Brooks/Cole - Thomson NO!NO! Gases only It is the ideal GAS law!

39. 38 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Strategy: What moles do you need? How will you get this?

40. 39 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution Now what will you do?

41. 40 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution P of O 2 = 0.16 atm

42. 41 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution How can you find the pressure of water? Do a calculation like we just did but for water? Any other way?

43. 42 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry What is P of H 2 O? Could calculate as above. But recall Avogadro’s hypothesis. V  n at same T and P P  n at same T and V P is proportional to n. There are 2 times as many moles of H 2 O as moles of O 2. Therefore, P of H 2 O is twice that of O 2. P of H 2 O = 0.32 atm 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g)

44. 43 © 2006 Brooks/Cole - Thomson Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P(H 2 O) + P(O 2 ) = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

45. 44 © 2006 Brooks/Cole - Thomson Dalton’s Law John Dalton 1766-1844

46. 45 © 2006 Brooks/Cole - Thomson

47. 46 The Disaster of Lake Nyos Lake Nyos in Cameroon was a volcanic lake. CO 2 built up in the lake and was released explosively on August 21, 1986. 1700 people and hundreds of animals died. See Chapter 14 for more information

48. 47 © 2006 Brooks/Cole - Thomson KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

49. 48 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases — and so does speed.

50. 49 © 2006 Brooks/Cole - Thomson

51. 50 Kinetic Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases — and so does speed.

52. 51 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory where u is the speed and M is the molar mass. speed INCREASES with Tspeed INCREASES with T speed DECREASES with Mspeed DECREASES with M Maxwell’s equation

53. 52 © 2006 Brooks/Cole - Thomson Distribution of Gas Molecule Speeds Boltzmann plotsBoltzmann plots Named for Ludwig Boltzmann doubted the existence of atoms.Named for Ludwig Boltzmann doubted the existence of atoms. This played a role in his suicide in 1906. This played a role in his suicide in 1906.

54. 53 © 2006 Brooks/Cole - Thomson Velocity of Gas Molecules Molecules of a given gas have a range of speeds.

55. 54 © 2006 Brooks/Cole - Thomson Velocity of Gas Molecules Average velocity decreases with increasing mass.

56. 55 © 2006 Brooks/Cole - Thomson GAS DIFFUSION AND EFFUSION DIFFUSION is the gradual mixing of molecules of different gases.

57. 56 © 2006 Brooks/Cole - Thomson GAS EFFUSION Figure 12.19 EFFUSION is the movement of molecules through a small hole into an empty container.

58. 57 © 2006 Brooks/Cole - Thomson GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He

59. 58 © 2006 Brooks/Cole - Thomson GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.

60. 59 © 2006 Brooks/Cole - Thomson Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. Active Figure 12.18

61. 60 © 2006 Brooks/Cole - Thomson Using KMT to Understand Gas Laws Recall that KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

62. 61 © 2006 Brooks/Cole - Thomson Avogadro’s Hypothesis and Kinetic Molecular Theory P proportional to n — when V and T are constant

63. 62 © 2006 Brooks/Cole - Thomson Gas Pressure, Temperature, and Kinetic Molecular Theory P proportional to T — when n and V are constant

64. 63 © 2006 Brooks/Cole - Thomson Boyle’s Law and Kinetic Molecular Theory P proportional to 1/V — when n and T are constant

65. 64 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Real molecules have volume.Real molecules have volume. There are intermolecular forces.There are intermolecular forces. –Otherwise a gas could not become a liquid.

66. 65 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with VAN DER WAALS’s EQUATION. Measured V = V(ideal) Measured P intermol. forces vol. correction J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. nRT V - nb V 2 n 2 a P + ----- )(

67. 66 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Cl 2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm

68. 67 © 2006 Brooks/Cole - Thomson Please make your selection... 1.Choice One 2.Choice Two 3.Choice Three 4.Choice Four Answer Now 0%0% 0%0%100%100%

69. 68 © 2006 Brooks/Cole - Thomson Please make your selection... 1.Choice One 2.Choice Two 3.Choice Three 4.Choice Four Answer Now 0%0% 0%0%100%100%