1. Chem 106, Prof. T. L. Heise 1 CHE 106: General Chemistry  CHAPTER TEN Copyright © Tyna L. Heise 2001 All Rights Reserved

2. Chem 106, Prof. T. L. Heise 2 Do Now  Turn in VSEPR Lab  Make sure you have included bond angles and the DIPOLE MOMENTS for any of the polar molecules  Any questions about the upcoming test?

3. Chem 106, Prof. T. L. Heise 3  How is matter encountered? –NOT on the atomic or molecular scale –As a large collection of atoms or molecules –Recognized as solids liquids and gases Gases Chapt. 10.1

4. Chem 106, Prof. T. L. Heise 4  Characteristics of Gases –Expands to fill container completely –Highly compressible –Form homogeneous mixtures regardless of identities or proportions Gases Chapt. 10.1

5. Chem 106, Prof. T. L. Heise 5  Why exhibit these characteristics? –Individual particles are far apart –Act as if they are only molecule present Gases Chapt. 10.1

6. Chem 106, Prof. T. L. Heise 6 Properties of Gases Chapt. 10.2 When measuring gases, easiest properties are 1) Temperature - thermochemistry 2) Volume - solution chemistry 3) Pressure

7. Chem 106, Prof. T. L. Heise 7 Pressure Chapt. 10.2 - conveys idea of force P = FP = pressure AF = force A = area - pressure is exerted on any surface a gas contacts

8. Chem 106, Prof. T. L. Heise 8 Atmospheric pressure Gases of our atmosphere exert a force on surface of the earth due to gravity. - force exerted is equal to mass times acceleration due to gravity F = ma since P = F then P = maa = 9.8 m A A s 2 Chapt. 10.2

9. Chem 106, Prof. T. L. Heise 9 Atmospheric pressure Chapt. 10.2 Atmospheric pressure is measured using a mercury barometer Tube is filled with mercury Small portion falls back into dish when inverted, vacuum exists above liquid in column Column moves up or down depending on atmospheric force on surface of mercury in dish

10. Chem 106, Prof. T. L. Heise 10 Atmospheric pressure Chapt. 10.2 Units of Measurement: 1 atm 760 mmHg 760 torr 1.01325 x 10 5 Pa 101.325 kPa Convert 13.33 kPa into atm 13.33 kPa 1 atm= 0.1316 atm 101.325 kPa

11. Chem 106, Prof. T. L. Heise 11 Pressures of Enclosed gases Chapt. 10.2 Manometers Similar in operation to barometer Two types, closed tube and open tube Closed tube - measures pressures below atmospheric Difference in tube heights = pressure

12. Chem 106, Prof. T. L. Heise 12 Pressures of Enclosed gases Chapt. 10.2 Open tube - measures pressures near atmospheric Difference in tube heights relates pressure of gas to atmospheric pressure

13. Chem 106, Prof. T. L. Heise 13 Pressures of Enclosed gases Chapt. 10.2

14. Chem 106, Prof. T. L. Heise 14 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference?

15. Chem 106, Prof. T. L. Heise 15 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? a) Convert atm into torr

16. Chem 106, Prof. T. L. Heise 16 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? a) Convert atm into torr 0.835 atm 760 torr= 1 atm

17. Chem 106, Prof. T. L. Heise 17 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? a) Convert atm into torr 0.835 atm 760 torr= 634 torr 1 atm* which is less than atmospheric pressure

18. Chem 106, Prof. T. L. Heise 18 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? a) Convert atm into torr atmospheric pressure makes arm open to air lower and arm attached to gas container higher

19. Chem 106, Prof. T. L. Heise 19 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? b) P gas = P atm + difference in heights

20. Chem 106, Prof. T. L. Heise 20 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? b) P gas = P atm + difference in heights P gas = 634 torr P atm = 755 torr 634 = 755 + difference

21. Chem 106, Prof. T. L. Heise 21 Pressures of Enclosed gases Chapt. 10.2 Sample Problem: A vessel connected to an open end manometer is filled with gas to a pressure of 0.835 atm. The atmospheric pressure is 755 torr. a) which arm of manometer is higher? b) what is the height difference? b) P gas + difference in heights = P atm P gas = 634 torr P atm = 755 torr 634 + X = 755 X = 121 torr

22. Chem 106, Prof. T. L. Heise 22 Now  Open the floor to questions about the upcoming test

23. Chem 106, Prof. T. L. Heise 23 The Gas Laws Chapt. 10.3 Four variables needed to adequately describe a gas Temperature (T) Pressure (P) Volume (V) Number of moles (n) Equations that express relationships between these variables are the gas laws.

24. Chem 106, Prof. T. L. Heise 24 Boyle’s Law Chapt. 10.3 Boyle investigated the relationship between volume and pressure. As pressure increased, the volume decreased; proves an inverse relationship PV = constant

25. Chem 106, Prof. T. L. Heise 25 Charles’s Law Chapt. 10.3 Charles investigated the relationship between volume and temperature. As temp decreased, the volume decreased; proves a direct relationship V = constant T

26. Chem 106, Prof. T. L. Heise 26 Avogadro’s Law Chapt. 10.3 Avogadro investigated the relationship between volume and amount of substance. As number of molecules doubles, the volume as doubles - proves a direct relationship Avogadro’s Hypothesis - equal volumes of gases at same temperature and pressure contain equal numbers of molecules 22.4 L

27. Chem 106, Prof. T. L. Heise 27 Check it out!  P vs V  T vs P  T vs V

28. Chem 106, Prof. T. L. Heise 28 The Ideal Gas Equation Chapt. 10.4 Combining the previous three laws, allows for a better mathematical look at gases. The term R in the gas equation is called the gas constant. The conditions of 0°C and 1 atm are referred to as standard temperature and pressure!

29. Chem 106, Prof. T. L. Heise 29 The Ideal Gas Equation Chapt. 10.4 The ideal gas law allows us to isolate all variables that are to be held constant and set up proportionalities. P 1 P 2 T 1 = T 2 V 1 V 2 T 1 = T 2 P 1 V 1 = P 2 V 2

30. Chem 106, Prof. T. L. Heise 30 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: Tennis balls are usually filled with air or N 2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24°C? 1) PV = nRT

31. Chem 106, Prof. T. L. Heise 31 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: Tennis balls are usually filled with air or N 2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24°C? 2) PV = nRT P = x R = 0.08206 L-atm/mol-K V = 144 cm 3 T = 24°C n = 0.33 g N 2

32. Chem 106, Prof. T. L. Heise 32 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: Tennis balls are usually filled with air or N 2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24°C? 3) PV = nRT P = x R = 0.08206 L-atm/mol-K V = 144 cm 3 = 144 mL = 0.144 L T = 24°C = 297 K n = 0.33 g N 2 1 mol N 2 = 0.012 mol N 2 28 g N 2

33. Chem 106, Prof. T. L. Heise 33 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: Tennis balls are usually filled with air or N 2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24°C? 4) P = nRT = 0.012 mol N 2 (0.08206 L-atm/mol-K) (297 K) V0.144 L

34. Chem 106, Prof. T. L. Heise 34 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: Tennis balls are usually filled with air or N 2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24°C? 2) P = nRT = 0.012 mol N 2 (0.08206 L-atm/mol-K) (297 K) V0.144 L P = 2.0 atm

35. Chem 106, Prof. T. L. Heise 35 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?

36. Chem 106, Prof. T. L. Heise 36 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 1) V 1 = V 2 T 1 T 2

37. Chem 106, Prof. T. L. Heise 37 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 1) V 1 = V 2 V 1 = 28,500 ft 3 T 1 T 2 T 1 = -15°C V 2 = x T 2 = 31°C

38. Chem 106, Prof. T. L. Heise 38 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 2) V 1 = V 2 T 1 T 2 V 1 = 28,500 ft 3 (12) 3 in 3 16.4 cm 3 1 L = 8.08 x 10 5 L 1 ft 3 1 in 3 10 3 cm 3

39. Chem 106, Prof. T. L. Heise 39 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 3) V 1 = V 2 V 1 = 8.08 x 10 5 L T 1 T 2 T 1 = -15°C = 258 K V 2 = x T 2 = 31°C = 304 K

40. Chem 106, Prof. T. L. Heise 40 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 4) V 1 = V 2 8.08 x 10 5 L = X T 1 T 2 258 K 304 K

41. Chem 106, Prof. T. L. Heise 41 The Ideal Gas Equation Chapt. 10.4 Sample Exercise: A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft 3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C? 4) V 1 = V 2 8.08 x 10 5 L = X T 1 T 2 258 K 304 K X = 9.52 x 10 5 L

42. Chem 106, Prof. T. L. Heise 42 Further Applications of Law Chapt. 10.5 Ideal gas equation can be used to determine - density of gas - molar mass of gas - volumes of gases formed or consumed

43. Chem 106, Prof. T. L. Heise 43 Further Applications of Law Chapt. 10.5 Gas Densities and Molar Mass n = moles V L * multiply both sides by molar mass, M n M = P M V R T

44. Chem 106, Prof. T. L. Heise 44 Further Applications of Law Chapt. 10.5 Gas Densities and Molar Mass nM = moles g V L moles * multiplying both sides by molar mass, M, will give us g, so that L Density = PM RT

45. Chem 106, Prof. T. L. Heise 45 Further Applications of Law Chapt. 10.5 Gas Densities and Molar Mass this equation can algebraically be rearranged to solve for molar mass Density = PM RT Molar Mass = d R T P

46. Chem 106, Prof. T. L. Heise 46 Further Applications of Law Chapt. 10.4 Sample Exercise: Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr 1) M = d R T P

47. Chem 106, Prof. T. L. Heise 47 Further Applications of Law Chapt. 10.4 Sample Exercise: Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr 2) M = d R Td = 1.17 g/L PR = 62.36 L-torr/mol-K T = 21°C = 294 K P = 740.0 torr

48. Chem 106, Prof. T. L. Heise 48 Further Applications of Law Chapt. 10.4 Sample Exercise: Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr 3) M = d R Td = 1.17 g/L PR = 62.36 L-torr/mol-K T = 21°C = 294 K P = 740.0 torr = 1.17 g/L ( 62.36 L-torr/mol-K) (294 K) 740.0 torr

49. Chem 106, Prof. T. L. Heise 49 Further Applications of Law Chapt. 10.4 Sample Exercise: Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr 4) M = d R Td = 1.17 g/L PR = 62.36 L-torr/mol-K T = 21°C = 294 K P = 740.0 torr = 1.17 g/L ( 62.36 L-torr/mol-K) (294 K) 740.0 torr = 29.0 g/L

50. Chem 106, Prof. T. L. Heise 50 Further Applications of Law Chapt. 10.4 Sample Exercise: The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere. 1) d = P M R T

51. Chem 106, Prof. T. L. Heise 51 Further Applications of Law Chapt. 10.4 Sample Exercise: The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere. 2) d = P MPressure = 1.6 atm R TM = 28.6 g/mol R = 0.08206 L-atm/mol-K T = 95 K

52. Chem 106, Prof. T. L. Heise 52 Further Applications of Law Chapt. 10.4 Sample Exercise: The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere. 3) d = P M = 1.6 atm(28.6 g/mol) R T 0.08206 L-atm/mol-K ( 95 K)

53. Chem 106, Prof. T. L. Heise 53 Further Applications of Law Chapt. 10.5 Sample Exercise: The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere. 4) d = P M = 1.6 atm(28.6 g/mol) R T 0.08206 L-atm/mol-K ( 95 K) = 5.9 g/L

54. Chem 106, Prof. T. L. Heise 54 Further Applications of Law Chapt. 10.5 Volumes of Gases in Chemical Reactions A knowledge of gases is often important because gases are often reactants or products in chemical reactions - coefficients in balanced equations is again going to be very important n = PV RT Gas Data Moles of Gas A Moles of Gas B g of Gas B

55. Chem 106, Prof. T. L. Heise 55 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction?

56. Chem 106, Prof. T. L. Heise 56 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction? 1) No need to convert into moles of O 2, we were given that information

57. Chem 106, Prof. T. L. Heise 57 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction? 2) Convert from mol O 2 into mol NH 3 1.00 mol O 2 4 mol NH 3 = 0.800 mol NH 3 5 mol O 2

58. Chem 106, Prof. T. L. Heise 58 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction? 3) Use ideal gas equation and given info to solve for final variable PV = nRT P = 5.00 atmn = 0.800 mol NH 3 T = 850°C = 1123 K V = XR = 0.08206 L-atm/mol-K

59. Chem 106, Prof. T. L. Heise 59 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction? 4) Use ideal gas equation and given info to solve for final variable PV = nRTV = nRT = 0.800 mol ( 0.08206) (1123 K) P 5.00 atm

60. Chem 106, Prof. T. L. Heise 60 Further Applications of Law Chapt. 10.5 Sample Exercise: In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH 3 + 5O 2 --> 4NO + 6H 2 0 How many liters of NH 3 at 850°C and 5.00 atm are required to react with 1.00 mol of O 2 in this reaction? 4) Use ideal gas equation and given info to solve for final variable PV = nRTV = nRT = 0.800 mol ( 0.08206) (1123 K) P 5.00 atm V = 14.7 L

61. Chem 106, Prof. T. L. Heise 61 Gas Mixtures and Partial Pressures Chapt. 10.6 John Dalton’s work with air allowed him to make the following observation: total pressure of a mixture equals the sum of the pressures that each gas would exert if it were present alone

62. Chem 106, Prof. T. L. Heise 62 Gas Mixtures and Partial Pressures Chapt. 10.6 This equation implies that each gas behaves independently, so each gas has a unique mole quantity, and total moles equals sums of each individual amount... n t = n 1 + n 2 + n 3 +...

63. Chem 106, Prof. T. L. Heise 63 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 1) Solve for H 2 PV = nRT

64. Chem 106, Prof. T. L. Heise 64 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 2) Solve for H 2 PV = nRT P = X V = 10.0 L n = 2.00 g H 2 1 mol H 2 = + 8.00 g N 2 1 mol N 2 = 2 g H 2 28 g N 2 R = 0.08206 L-atm/mol-K T = 273 K

65. Chem 106, Prof. T. L. Heise 65 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 3) Solve for H 2 PV = nRT P = XP = nRT V = 10.0 L V n = 1.28 moles total = 1.28 mol (0.08206) ( 273 K) 10.0 L R = 0.08206 L-atm/mol-K = 2.88 atm T = 273 K

66. Chem 106, Prof. T. L. Heise 66 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 4) Solve for N 2 PV = nRT

67. Chem 106, Prof. T. L. Heise 67 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 5) Solve for N 2 PV = nRT P = X V = 10.0 L n = 8.00 g N 2 1 mol N 2 28 g N 2 R = 0.08206 L-atm/mol-K T = 273 K

68. Chem 106, Prof. T. L. Heise 68 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 6) Solve for N 2 PV = nRT P = XP = nRT V = 10.0 L V n = 8.00 g N 2 1 mol N 2 = 0.286 mol (0.08206) ( 273 K) 28 g N 2 10.0 L R = 0.08206 L-atm/mol-K = 0.641 atm T = 273 K

69. Chem 106, Prof. T. L. Heise 69 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: What is the total pressure exerted by a mixture of 2.00 g of H 2 and 8.00 g of N 2 at 273 K in a 10.0 L vessel? 7) Sum individual pressures to identify total pressure P t = P(H 2 ) + P(N 2 ) = 2.24 atm + 0.641 atm = 2.88 atm

70. Chem 106, Prof. T. L. Heise 70 Gas Mixtures and Partial Pressures Chapt. 10.6 Partial pressures and Mole Fractions - each gas behaves independently and so it is easy to relate the amount of a given gas to its partial pressure - since P = nRT then P 1 = n 1 RT and P 1 = n 1 V VP t n t P t = n t RT V - the ratio n 1 /n t is denoted the mole fraction of gas 1 - mole fractions are unitless values expressing ratio

71. Chem 106, Prof. T. L. Heise 71 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 1) Solve for N 2

72. Chem 106, Prof. T. L. Heise 72 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 2) Solve for N 2 Total pressure = 1220 torr Mole fraction = 82/100 = 0.82 P N = 0.82 (1220 torr) = 1.0 x 10 3 torr

73. Chem 106, Prof. T. L. Heise 73 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 3) Solve for Ar

74. Chem 106, Prof. T. L. Heise 74 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 4) Solve for Ar Total pressure = 1220 torr Mole fraction = 12/100 = 0.12 P Ar = 0.12 (1220 torr) = 150 torr

75. Chem 106, Prof. T. L. Heise 75 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 5) Solve for CH 4

76. Chem 106, Prof. T. L. Heise 76 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample Exercise: From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N 2, 12 mol % Ar and 6.0 mol % CH 4. Calculate the partial pressure of each gas. 2) Solve for CH 4 Total pressure = 1220 torr Mole fraction = 6.0/100 = 0.060 P CH4 = 0.060(1220 torr) = 73 torr

77. Chem 106, Prof. T. L. Heise 77 Gas Mixtures and Partial Pressures Chapt. 10.6 Collecting gases over water - when determining moles of produced gases, the best way to collect a gas sample is over water

78. Chem 106, Prof. T. L. Heise 78 Gas Mixtures and Partial Pressures Chapt. 10.6 Collecting gases over water - when determining moles of produced gases, the best way to collect a gas sample is over water - the volume of gas is measured by raising and lowering collecting container until water levels are equal inside and out…when this occurs, atmospheric pressures are equal - total pressure inside is equal to sum of pressure of gas collected and pressure of water vapor P total = P gas + P H 2 O

79. Chem 106, Prof. T. L. Heise 79 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 1) Calculate partial pressure of N 2

80. Chem 106, Prof. T. L. Heise 80 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 1) Calculate partial pressure of N 2 P t = P N + P water P t = 745 torr P water = 25.21 torr

81. Chem 106, Prof. T. L. Heise 81 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 1) Calculate partial pressure of N 2 P t = P N + P water P t = 745 torr P water = 25.21 torr 745 torr = X + 25.21 torr

82. Chem 106, Prof. T. L. Heise 82 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 1) Calculate partial pressure of N 2 P t = P N + P water P t = 745 torr P water = 25.21 torr 745 torr = X + 25.21 torr P N = 720. torr

83. Chem 106, Prof. T. L. Heise 83 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 2) Use Ideal Gas Equation to determine moles of N 2

84. Chem 106, Prof. T. L. Heise 84 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 2) Use Ideal Gas Equation to determine moles of N 2 PV = nRT

85. Chem 106, Prof. T. L. Heise 85 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 2) Use Ideal Gas Equation to determine moles of N 2 PV = nRT P = 720. Torr V = 0.511 L n = X R = 62.36 L-torr/mol-K T = 26°C = 299 K

86. Chem 106, Prof. T. L. Heise 86 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 2) Use Ideal Gas Equation to determine moles of N 2 PV = nRT n = PV P = 720. torr RT V = 0.511 L = 720. torr(0.511 L) n = X 62.36 (299 K) R = 62.36 L-torr/mol-K T = 26˚C = 299 K

87. Chem 106, Prof. T. L. Heise 87 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 2) Use Ideal Gas Equation to determine moles of N 2 PV = nRT n = PV P = 720. Torr RT V = 0.511 L = 720. torr(0.511 L) n = X 62.36 (299 K) R = 62.36 L-torr/mol-K = 0.0197 mol N 2 T = 26˚C = 299 K

88. Chem 106, Prof. T. L. Heise 88 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 3) Convert from moles of N 2 to g NH 4 NO 2

89. Chem 106, Prof. T. L. Heise 89 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 3) Convert from moles of N 2 to g NH 4 NO 2 0.0197 mol N 2 1 mol NH 4 NO 2 64.0 g NH 4 NO 2 1 mol N 2 1 mol NH 4 NO 2

90. Chem 106, Prof. T. L. Heise 90 Gas Mixtures and Partial Pressures Chapt. 10.6 Sample exercise: When a sample of NH 4 NO 2 is decomposed in a test tube, 511 mL of N 2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? 3) Convert from moles of N 2 to g NH 4 NO 2 0.0197 mol N 2 1 mol NH 4 NO 2 64.0 g NH 4 NO 2 1 mol N 2 1 mol NH 4 NO 2 = 1.26 g NH 4 NO 2

91. Chem 106, Prof. T. L. Heise 91 Kinetic Molecular Theory Chapt. 10.7 Ideal gas equation describes how gases behave, but not why they do... Kinetic molecular theory explains why - gases consist of large numbers of molecules that are in continuous, random motion - volume of the gas molecules themselves is negligible when compared to the volume of the gas as a whole - attractive and repulsive forces between gases molecules is negligible

92. Chem 106, Prof. T. L. Heise 92 Kinetic Molecular Theory Chapt. 10.7 Ideal gas equation describes how gases behave, but not why they do... Kinetic molecular theory explains why - energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature is remaining constant - i.e. collisions are elastic - average kinetic energy is proportional to the absolute temperature

93. Chem 106, Prof. T. L. Heise 93 Kinetic Molecular Theory Chapt. 10.7 This theory is helpful in explaining the pressure and temperature at a molecular level: pressure of gas is caused by the collisions of the molecules against the wall of a container magnitude of pressure is determined by both how often and how forcefully the molecules strike the wall

94. Chem 106, Prof. T. L. Heise 94 Kinetic Molecular Theory Chapt. 10.7 This theory is helpful in explaining the pressure and temperature at a molecular level: absolute temperature of a gas is the measure of the average kinetic energy of its molecules molecular motion increases with increasing temperature this is on average, individual molecules have individual speeds

95. Chem 106, Prof. T. L. Heise 95 Kinetic Molecular Theory Chapt. 10.7 Applications of the Gas Laws 1. Effect of a volume increase at constant temp - molecules must move a longer distance between collisions - fewer collisions per unit time with wall of container - pressure decreases

96. Chem 106, Prof. T. L. Heise 96 Kinetic Molecular Theory Chapt. 10.7 Applications of the Gas Laws 2. Effect of a temperature increase at constant volume - increase in speed and u - more collisions per unit time with wall of container - pressure increases

97. Chem 106, Prof. T. L. Heise 97 Molecular Effusion and Diffusion Chapt. 10.8 According to the kinetic-molecular theory, the average kinetic energy of any collection of gases, u = 3RTR = gas constant MT = temp M = molar mass

98. Chem 106, Prof. T. L. Heise 98 Molecular Effusion and Diffusion Chapt. 10.8 has a specific value of u at any given temperature -two gases at same temp have same avg. kinetic energy -if masses are different, than the speed of particles will be different because of the inclusion of M in formula

99. Chem 106, Prof. T. L. Heise 99 Molecular Effusion and Diffusion Chapt. 10.8 The dependence of speed on mass, has several implications: Effusion - the escape of a molecule through a tiny hole Effusion rate is inversely proportional to the square root of molar mass Normally rates are compared as a ratio

100. Chem 106, Prof. T. L. Heise 100 Molecular Effusion and Diffusion Chapt. 10.8 The dependence of speed on mass, has several implications: Diffusion - spreading of one substance throughout a space or through another substance Diffusion is also related to size of particle, however, molecule collisions make diffusion much more difficult - average distance traveled by a molecule as it diffuses is called the mean free path, which varies with pressure

101. Chem 106, Prof. T. L. Heise 101 Deviations from an Ideal Gas Chapt. 10.9 Although the ideal gas equation is useful, all real gases fail to obey the relationships to some degree - deviation from an ideal gas occurs most at high pressure and low temperature - to ensure as much compliance as possible to the ideal gas equation, a real gas should be considered when it is at high temperatures and low pressure WHY? Real gases DO- have molecular attractions - lose energy when they collide - have volume

102. Chem 106, Prof. T. L. Heise 102 Deviations from an Ideal Gas Chapt. 10.9

103. Chem 106, Prof. T. L. Heise 103 Deviations from an Ideal Gas Chapt. 10.9 Van der Waals’ Equation: takes into account volume and molecular attraction * constants (a) and (b) are different for each gas and must be identified using a table P = nRT - n 2 a V - nb V 2

104. Chem 106, Prof. T. L. Heise 104  Characteristics of gases  Pressure  Boyle’s, Charles’s, & Avogadro’s Laws  Ideal gas equation  Gas densities and Molar mass  Dalton’s Law of partial pressures  Mole fractions  Kinetic Molecular theory  Effusion, Diffusion, u and mean free path  Deviations from ideal gas behavior  Van der Waals equation Chapter Ten; Review