1. Chapter 14 (page 440)

2. Why is it important to know the volume - mass relationship of gases, the Idea Gas law, and the stoichiometry of gases??

3.  Kinetic molecular theory  Brownian motion  Pressure  Pascal  Barometer  Diffusion  Boltzmann’s constant

4. Gases consist of widely separated atoms or molecules in constant, random motion No interaction between atoms or molecules, except during collisions Mostly empty space Straight trajectory until a collision occurs Properties of gases

5. Gases have a unique set of physical properties: 1.Gases are translucent or transparent. 2.Gases have very low densities when compared to liquids or solids. 3.Gases are highly compressible compared to liquids and solids. 4.Gases can expand or contract to fill any container. Mostly empty space Properties of gases These can be explained by the kinetic molecular theory

6. kinetic molecular theory: the theory that explains the observed thermal and physical properties of matter in terms of the average behavior of a collection of atoms and molecules. Gases consist of atoms or molecules with a lot of space in between, that are in constant, random motion

7. If the liquid and gas are both made from the same molecules (H 2 O), you can explain the “disappearance” by assuming that the molecules are much more spread out in the gas phase. Properties of gases Evidence for the atomic / molecular nature of matter:

8. Brownian motion can be seen by magnifying diluted milk and observing tiny fat globules getting knocked around by the surrounding water molecules Brownian motion What Brownian motion tells us: 1.Matter consists of discrete particles (molecules or atoms) 2.Molecules (or atoms) are in constant, vigorous motion as a result of temperature

9. Brownian motion Brownian motion provides a peek into the microscopic world of atoms to see details that are normally hidden by the law of averages, and the enormous number of incredibly small atoms.

10.  In the early 1800’s Joseph Gay-Lussac studied gas volume relationships involving a chemical reaction between hydrogen and oxygen and observed that 2 L of hydrogen would react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure  Hydrogen gas + oxygen gas ----  water vapor 2 L 1 L ----  2 L 2 volumes 1 volume ----  2 volumes

11. Assignment  Take a new sheet of paper and fold it into three sections  Write your name, the title of the chapter and the number  On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

12.  In Gay-Lussac experiment it was found that the reaction took place in a 2:1:2 relationship between hydrogen, oxygen and water vapor  If you had 600 L H 2, 300 L O 2 you would get 600 L H 2 O formed  With hydrogen and chlorine combining then:  Hydrogen gas + chlorine gas ----  hydrogen chloride gas 1 L 1 L ----  2 L 1 volumes 1 volume ----  2 volumes

13.  The relationship that he found between gas volumes is now known to be a law  Gay-Lussac’s law of combining volumes of gases states – at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

14.  Remember from Dalton’s atomic theory that atoms are indivisible  Also remember that Dalton believed that one atom of one reactant always combined with one atom of the other reactant (which caused questions when forming water vapor)  Gay-Lussac disproved the second theory of Dalton, but a scientist by the name of Avogadro formulated an explanation of the problem  Avogadro reasoned that molecules contained more than one atom and formulated his theory which later became a law  Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

16.  Remember that one mole is equal to the atomic mass of that atom  Well, one mole of a molecule or a compound is equal to the combined molecular masses of the atoms  Avogadro suggested that one mole of an element, molecule or an compound contained the same number of particles which is 6.022 x 10 23 particles  So 1 mole of H 2 (2.015g) contained the same number of particles as 1 mole O 2 (32g) at standard pressure (1 atm) and temperature (273 K) STP  Standard volume at STP is 22.414 L

17.  Write a detailed three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)  Turn to page 468 and complete # 1 – 4 then turn them in  Honors chemistry homework  Page 468 # 7 - 16

18.  Molar Volume  Ideal gas

19. Gases consist of widely separated atoms or molecules in constant, random motion Mostly empty space No interaction between gas atoms or molecules except in collisions Straight trajectories until collision occurs

20. Gas pressure is increased by more frequent and/or harder collisions

21. 1. the density More molecules means more impacts and a higher pressure. 2. the volume of the container With less space to move around, there are more collisions and a higher pressure. You can affect the gas pressure by changing:

22. 1. the density More molecules means more impacts and a higher pressure. 2. the volume of the container With less space to move around, there are more collisions and a higher pressure. 3. the temperature With more kinetic energy, the molecules move faster. The collisions are harder and more frequent. You can affect the gas pressure by changing:

23. Boyle’s law: V versus P Robert Boyle’s experiment: - Mercury (Hg) was poured down a tube shaped like the letter “J.” - The tube was closed on the lower end. - The gas inside the tube took up space (volume). - The temperature and number of gas molecules inside the tube stayed constant. - Boyle observed the change in pressure in mmHg as a function of volume. He then graphed the relationship between pressure and volume.

24. Boyle’s law: V versus P Pressure versus volume Temperature Number of moles constant

25. Boyle’s law: V versus P Pressure versus volume Temperature Number of moles constant

26. Boyle’s law: V versus P

27. If a weather balloon is released on the ground with a volume of 3.0 m 3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm? Boyle’s law: V versus P

28. Assignment  On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

29. Boyle’s law: V versus P If a weather balloon is released on the ground with a volume of 3.0 m 3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm? Asked:Volume of the balloon when it reaches 100,000 ft Given: Relationships: Solve: Answer: The balloon will have a volume of 300 m 3.

30. Charles’s law: V versus T The volume increases as the temperature increases

31. Charles’s law: V versus T Volume versus temperature Pressure Number of moles constant

32. Charles’s law: V versus T Doubling the Kelvin temperature will double the volume of a gas Kelvin temperatures simplify the V versus T relationship

33. Charles’s law: V versus T

34. If you inflate a balloon to a size of 8.0 L inside where the temperature is 23 o C, what will be the new size of the balloon when you go outside where it is 3 o C? Charles’s law: V vs. T

35. If you inflate a balloon to a size of 8.0 L inside where the temperature is 23 o C, what will be the new size of the balloon when you go outside where it is 3 o C? Charles’s law: V vs. T Asked:Volume of the balloon when the temperature drops to 3oC Given: Relationships: Solve:

36. Combined gas law

37. Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20 o C, you’ll need 18 m 3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50 o C?

38. Asked:Volume of the balloon when it reaches 100,000 ft Given: Relationships: Solve:

39. Assignment  On the third section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

40. Avogadro’s law: V versus moles Two equal volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor. Gas volumes act like moles because the same size container has the same number of molecules (at the same temperature and pressure).

41. Avogadro’s law: V versus moles Moles versus volume Temperature Pressure constant

42. Avogadro’s law: V versus moles Moles versus volume Temperature Pressure constant

43. Twice the volume, twice the number of molecules assuming the temperature and pressure are constant Moles versus volume Temperature Pressure constant Avogadro’s law: V versus moles

44. Restrictions

45. The ideal gas law Combining the previous gas laws, we obtain the ideal gas law In reality, the ideal gas law is an approximation which is accurate for many gases over a wide range of conditions. The ideal gas law is not accurate at very high density or at very low temperature.

46. The ideal gas law The universal gas constant R is the only constant

47. Calculating the universal gas constant using various units Watch out for the units! The ideal gas law

48. Assignment  On the first section of back side on that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

49. The ideal gas law Limitations of the ideal gas law In an ideal gas, we assume that: 1. individual gas molecules take up no space 2. gas molecules do not interact with each other For very small volumes or very low temperatures, gas atoms and molecules are very close together, and van der Waals attractions are no longer negligible

50. PV = nRT

51. What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20 o C? PV = nRT

52. What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20 o C? PV = nRT Asked:Tank pressure Given: Relationships: Solve:

53. What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20 o C? PV = nRT Asked:Tank pressure Given: Relationships: Solve:

54. What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20 o C) volume of 0.050 cm 3 to a 191 o C oven?

55. Asked:Find the volume of a bubble under changing temperature conditions Given: Relationships: Solve:Pressure and moles stay constant, so they cancel out.

56. What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20 o C) volume of 0.050 cm 3 to a 191 o C oven? Asked:Find the volume of a bubble under changing temperature conditions Given: Relationships: Solve:Pressure and moles stay constant, so they cancel out.

57. Boyle’s law Charles’s law Avogadro’s law Combined gas law Ideal gas law R = universal gas law

58.  Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)  Turn to page 468 and complete # 5 – 6 then turn them in  Honors chemistry Homework:  Page 468 # 17 - 26

59.  Molar Volume  Ideal gas

60.  N/A

61. Ideal gas law R = universal gas law We can now solve stoichiometry problems involving gases

62. and solids solutions other gases

63. Steps for solving stoichiometry problems

65. 2C 20 H 42 + 61O 2 → 40CO 2 + 42H 2 O If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400 o C. Assuming the carbon dioxide produced is at that temperature, what volume of CO 2 is produced at a pressure of 790 mmHg? The combustion reaction is:

66. Asked:Volume of CO 2 produced Given:Paraffin: mass of 125 g CO 2 : P = 790 mmHg, T = 1,400 o C Relationships:Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C 20 H 42 ~ 40 moles CO 2 PV = nRT If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400 o C. Assuming the carbon dioxide produced is at that temperature, what volume of CO 2 is produced at a pressure of 790 mmHg? The combustion reaction is: 2C 20 H 42 + 61O 2 → 40CO 2 + 42H 2 O

67. Asked:Volume of CO 2 produced Given:Paraffin: mass of 125 g; CO 2 : P = 790 mmHg, T = 1,400 o C Relationships:Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C 20 H 42 ~ 40 moles CO 2 PV = nRT Solve: 125 g C 20 H 42 0.442 moles C 20 H 42

68. Asked:Volume of CO 2 produced Given:Paraffin: mass of 125 g; CO 2 : P = 790 mmHg, T = 1,400 o C Relationships:Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C 20 H 42 ~ 40 moles CO 2 PV = nRT Solve: 0.442 moles C 20 H 42 8.85 moles CO 2

69. Asked:Volume of CO 2 produced Given:Paraffin: mass of 125 g; CO 2 : P = 790 mmHg, T = 1,400 o C Relationships:Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C 20 H 42 ~ 40 moles CO 2 PV = nRT Solve: 8.85 moles CO 2

70. Asked:Volume of CO 2 produced Given:Paraffin: mass of 125 g; CO 2 : P = 790 mmHg, T = 1,400 o C Relationships:Molar mass of paraffin = 282.6 g/mole Mole ratio: 2 moles C 20 H 42 ~ 40 moles CO 2 PV = nRT Solve: 8.85 moles CO 2 1,170 L CO 2

71. A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium: If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22 o C? Mg(s) + 2HCl(aq) → H 2 (g) + MgCl 2 (aq)

72. Asked:Pressure of H 2 produced Given:HCl: V = 0.050 L, Molarity = 1.25 moles/L H 2 : V = 1.00 L, T = 22 o C Relationships:Mole ratio: 2 moles HCl ~ 1 mole H 2 PV = nRT Mg(s) + 2HCl(aq) → H 2 (g) + MgCl 2 (aq) A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium: If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22 o C?

73. Asked:Pressure of H 2 produced Given:HCl: V = 0.050 L, Molarity = 1.25 moles/L H 2 : V = 1.00 L, T = 22 o C Relationships:Mole ratio: 2 moles HCl ~ 1 mole H 2 PV = nRT Solve: Volume 0.050 L HCl 1.25 M HCl 0.0625 moles HCl

74. Asked:Pressure of H 2 produced Given:HCl: V = 0.050 L, Molarity = 1.25 moles/L H 2 : V = 1.00 L, T = 22 o C Relationships:Mole ratio: 2 moles HCl ~ 1 mole H 2 PV = nRT Solve: 0.0625 moles HCl 0.0313 moles H 2

75. Asked:Pressure of H 2 produced Given:HCl: V = 0.050 L, Molarity = 1.25 moles/L H 2 : V = 1.00 L, T = 22 o C Relationships:Mole ratio: 2 moles HCl ~ 1 mole H 2 PV = nRT Solve: 0.0313 moles H 2 0.756 atm H 2

76. What volume of butane gas is needed at room temperature (23 o C) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825 o C and a pressure of 1.04 atm? The reaction is: 2C 4 H 10 (g) + 13O 2 (g) → 8CO 2 (g) + 10H 2 O(g)

77. What volume of butane gas is needed at room temperature (23 o C) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825 o C and a pressure of 1.04 atm? The reaction is: Asked:Volume of C 4 H 10 needed to produce 85.0 L of CO 2 Given:C 4 H 10 : P = 0.984 atm, T = 23 o C CO 2 : P = 1.04 atm, T = 825 o C, V = 85.0 L Relationships:Mole ratio: 2 moles C 4 H 10 ~ 8 moles CO 2

78. Asked:Volume of C 4 H 10 needed to produce 85.0 L of CO 2 Given:C 4 H 10 : P = 0.984 atm, T = 23 o C CO 2 : P = 1.04 atm, T = 825 o C, V = 85.0 L Relationships:Mole ratio: 2 moles C 4 H 10 ~ 8 moles CO 2 Solve: C 4 H 10 CO 2

79. Asked:Volume of C 4 H 10 needed to produce 85.0 L of CO 2 Given:C 4 H 10 : P = 0.984 atm, T = 23 o C CO 2 : P = 1.04 atm, T = 825 o C, V = 85.0 L Relationships:Mole ratio: 2 moles C 4 H 10 ~ 8 moles CO 2 Solve: C 4 H 10 CO 2 Use the mole ratio to substitute for n CO2

80. Asked:Volume of C 4 H 10 needed to produce 85.0 L of CO 2 Given:C 4 H 10 : P = 0.984 atm, T = 23 o C CO 2 : P = 1.04 atm, T = 825 o C, V = 85.0 L Relationships:Mole ratio: 2 moles C 4 H 10 ~ 8 moles CO 2 Solve:

81. Asked:Volume of C 4 H 10 needed to produce 85.0 L of CO 2 Given:C 4 H 10 : P = 0.984 atm, T = 23 o C CO 2 : P = 1.04 atm, T = 825 o C, V = 85.0 L Relationships:Mole ratio: 2 moles C 4 H 10 ~ 8 moles CO 2 Solve:

82. M = m R T PV m = mass M = molar mass D = M P RT D = density

83.  Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)  Turn to page 469 and complete # 32 then turn it in.  Honors chemistry Homework:  Page 471 # 73 - 81

84.  Homework requirement: Learn all terms and concepts covered on this topic.  Make sure you have all assignments between page 224 and 227 completed and turned in by your test date.