2. Chapter 10 Gases

3. Review Quiz Net Ionic Equations (includes questions about the reactions).

4. Teacher Note The solutions to many of the calculations are worked out in a packet in the folder for Chapter 10.

5. Barometers and Standard Atmospheric Pressure

6. Standard atmospheric pressure defined as the pressure sufficient to support a mercury column of 760mm (units of mmHg, or torr).

7. Barometers and Standard Atmospheric Pressure Standard atmospheric pressure defined as the pressure sufficient to support a mercury column of 760mm (units of mmHg, or torr). Another unit was introduced to simplify things, the atmosphere (1 atm = 760 mmHg).

8. Barometers and Standard Atmospheric Pressure Standard atmospheric pressure defined as the pressure sufficient to support a mercury column of 760mm (units of mmHg, or torr). Another unit was introduced to simplify things, the atmosphere (1 atm = 760 mmHg). 1 atm = 760 mmHg = 760 torr = 101.325 kPa (page 262).

9. STP standard temperature and pressure Standard temperature 0°C or 273 K Standard pressure 1 atm (or equivalent)

10. Boyle’s Law

11. –Pressure varies inversely with the volume of a sample of gas if temperature remains constant. ”

12. Boyle’s Law

13. Boyle’s Law: Pressure & Volume

14. Boyle’s Law: Pressure-Volume Relationships A sample of air occupies 73.3 mL at 98.7 atm and 0 ºC. What volume will the air occupy at 4.02 atm and 0 ºC? 1800 mL

15. Boyle’s Law: Pressure-Volume Relationships A sample of helium occupies 535 mL at 988 mmHg and 25 °C. If the sample is transferred to a 1.05-L flask at 25 °C, what will be the gas pressure in the flask? 503 mm Hg

16. Effects of temperature on a gas Volume varies directly with Temperature “The volume of a quantity of gas, held at constant pressure, varies directly with the Kelvin temperature.”

17. Charles’s Law a

18. Charles Law: Volume and Temperature (Figure 10.8 Page 266)

19. Charles’ Law and Absolute Zero Extrapolation to zero volume gives a temperature of -273°C or 0 K

20. A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy at 150 °C? (Assume no change in pressure.) Charles’s Law: Temperature-Volume Relationships 2.98 L

21. A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. At what Celsius temperature will the volume of oxygen occupy 0.750 L? (Assume no change in pressure.) Charles’s Law: Temperature-Volume Relationships -167°C

22. Pressure vs. Temperature Pressure varies directly with Temperature If the temperature of a fixed volume of gas doubles its pressure doubles.

23. Pressure vs. Temperature The pressure exerted by a gas is directly related to the Kelvin temperature. V is constant.

24. Pressure vs. Temperature

25. Example A gas has a pressure of 645 torr at 128°C. What is the temperature in Celsius if the pressure increases to 1.50 atm? P i = 645 torr P f = 1.50 atm 760 torr = 1140 torr 1 atm T i = 128°C + 273 = 401 KT f = ?K

26. Solution T 2 = 401 K x 1140 torr = 709K 645 torr 709K - 273 = 436°C

27. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

28. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? x 3.20 atm x 90.0 mL 0.800 atm 180.0 mL 604 K - 273 = 331 °C 302 K= 604 K

29. Combined Gas Law A 10.0 cm 3 volume of gas measured 75.6 kPa and 60.0  C is to be corrected to correspond to the volume it would occupy at STP. 6.12 cm 3

30. Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers. Gay-Lussac’s Law

31. How many liters of steam can be formed from 8.60L of oxygen gas? 17.2 L

32. How many liters of hydrogen gas will react with 1L of nitrogen gas to form ammonia gas? 3L H 2

33. A

34. How many mL of hydrogen are needed to produce 13.98 mL of ammonia? A 20.97 ml NH 3

35. Avogadro’s Law: Equal volumes of gases at the same temperature and pressure contain the same number of particles.

36. The molar volume of a gas at STP = 22.4L 22.4 L

37. Ideal Gas An ideal gas is defined as one for which both the volume of molecules and forces of attraction between the molecules are so small that they have no effect on the behavior of the gas.

38. Ideal Gas Equation PV=nRT

39. R values a I R values for atm and kPa on Page 272 in book. A

40. Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37 atm and a temperature of 315 K. 15.9 L

41. Find the pressure in millimeters of mercury of a 0.154 g sample of helium gas at 32°C and contained in a 648 mL container. 1130 mm Hg

42. Boyle’s Law and Diving What is more dangerous? A deep dive or a shallow dive.

43. Boyle’s Law and Diving We can answer this by first noting that for every 10 meters you descend in the water, the pressure increases by about 1 atm.

45. Boyle’s Law and Diving When you hold your breath, you create a closed system with your lungs and Boyle's law applies. If you are down at 90 meters (at 10 atm) and you rise 10 meters to 80 meters (at 9 atm), the pressure has decreased by about 10%, and since PV is a constant your lungs expand by about 10% (probably not too bad).

46. Why is PV is constant? PV=nRT The right side of the equation has not changed (we have not changed temperature or the amount of gas in the lungs). Therefore the left side of the equation, (PV) must remain constant as well.

47. Boyle’s Law and Diving Now if you are at 10 meters (at 2 atm), and you rise 10 meters to the surface (at 1 atm) the pressure had decreased by 50% and expanding your lungs by this factor could cause significant damage, maybe death!

48. An experiment shows that a 113 mL gas sample has a mass of 0.171 g at a pressure of 721 mm Hg and a temperature of 32°C. What is the molar mass (molecular weight) of the gas? 40.0 g/mol

49. Can the ideal “gas” equation be used to determine the molar mass of a liquid?

51. Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 o C and 736 mm Hg pressure. If the mass of the flask before and after the experiment was 148.375g and 149.457 g, what is the molar mass of the liquid? 57.9 g/mol

52. Homework  See AP sample problem (1999 Test question #5) in notebook (include as part of your homework). You may find this problem easier after doing the lab for this chapter: “Molar Mass of a Volatile Liquid”

53. Homework Attempt the pre-lab for “The Molecular Mass of a Volatile Liquid”. –Problem #1 is due ______. –The rest of the pre-lab is due ______. Do the lab summary for “The Molecular Mass of a Volatile Liquid”. It is due ____.

54. What is the density of methane gas (natural gas), CH 4, at 125 o C and 3.50 atm? 1.71 g/L

55. Calculate the density in g/L of O 2 gas at STP. 1.43 g/L

56. Dalton’s Law of Partial Pressure The total pressure in a container is the sum of the partial pressures of all the gases in the container. In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container. P total = P 1 + P 2 + P 3 P total = 100 KPa + 250 KPa + 200 KPa = 550 KPa

57. AB Total = 6.0 atm PVV mixture P A2.0 atm1.0 L2.0 atm B4.0 atm1.0 L4.0 atm Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container B. Find the total pressure of the gas mixture in B. 1.0 L

58. Dalton’s Law Problem Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen at standard conditions if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 KPa, 0.04 KPa, and 0.94 KPa respectively? P total = P O 2 + P N 2 + P CO 2 + P Other gases 101.3 KPa = P O 2 + 79.1 KPa + 0.04 KPa + 0.94KPa P O 2 = 101.3 KPa – (79.1 KPa + 0.04 KPa + 0.94KPa) P O 2 = 21.2 KPa

59. ABZ Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container Z (vol. 2.0 L). Find the total pressure of mixture in Z.

60. ABZ Total = 3.0 atm Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container Z (vol. 2.0 L). Find the total pressure of mixture in Z. PXPX VXVX VZVZ P X,Z A2.0 atm1.0 L 2.0 L 1.0 atm B4.0 atm1.0 L2.0 atm

61. AB ZC Find total pressure of the gas mixture in Container Z. 1.3 L 2.6 L 3.8 L 2.3 L 3.2 atm 1.4 atm 2.7 atm X atm

62. AB ZC Total = 7.9 atm Find total pressure of the gas mixture in Container Z. 1.3 L 2.6 L 3.8 L 2.3 L 3.2 atm 1.4 atm 2.7 atm X atm PXPX VXVX VZVZ P X,Z A3.2 atm1.3 L 2.3 L 1.8 atm B1.4 atm2.6 L1.6 atm C2.7 atm3.8 L4.5 atm

63. Dalton’s Law Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 422

64. Dalton’s Partial Pressures Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 421

65. Dalton’s Law of Partial Pressures The mole ratio in a mixture of gases determines each gas’s partial pressure. Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa. Find partial pressure of each gas

66. Dalton’s Law of Partial Pressures Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa. Find partial pressure of each gas.

67. 80.0 g each of He, Ne, and Ar are in a container. The total pressure is 780 mm Hg. Find each gas’s partial pressure.

68. Total: 26 mol gas P He = 20 / 26 of total P Ne = 4 / 26 of total P Ar = 2 / 26 of total

71. Example: A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758.0 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? (P water = 23.8 torr at 25 o C) P O 2 = P T - P water = 758.0 torr - 23.8 torr = 734.2 torr Now What?

72. Find the molar mass of an unknown gas if a 0.16 g sample of the gas is collected over water and equalized to a pressure of 781.7 torr and a volume of 90.0 mL at a temperature of 28°C.

74. 44 g/mol

75. Homework Do the AP sample problem (1999 Test question #5) in notebook. It will be included as part of your homework. Don’t forget the pre-lab and lab summary for “The Molecular Mass of a Volatile Liquid”.

76. Gas Diffusion and Effusion Graham's Law: governs the rate of effusion and diffusion of gas molecules.

77. “Stink” or “Die” a

78. The Root Mean Square Speed Fig. 10.17 Page 285

79. NET NET MOVEMENT To use Graham’s Law, both gases must be at same temperature. diffusion diffusion: particle movement from high to low concentration effusion effusion: diffusion of gas particles through an opening For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast

80. Gas Diffusion and Effusion Graham's Law: governs the rate of effusion and diffusion of gas molecules. Rate of diffusion/effusion is inversely proportional to its molar mass. Rate of diffusion/effusion is inversely proportional to its molar mass.

81. Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. Graham’s Law The lightest gas is “Gas A” and the heavier gas is “Gas B”. Relative rate means find the ratio “v A /v B ”. Kr 83.80 36 Br 79.904 35

82. A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Graham’s Law O 15.9994 8 H 1.00794 1

83. An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. Graham’s Law The lightest gas is “Gas A” and the heavier gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign. O 15.9994 8 H2H2 2.0 1

84. Theory developed to explain gas behavior. Theory of moving molecules. Assumptions: –Gases consist of a large number of molecules in constant random motion. –Volume of individual molecules negligible compared to volume of container. –Intermolecular forces (forces between gas molecules) negligible. –Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature. –Average kinetic energy of molecules is proportional to temperature. Kinetic Molecular Theory

85. Kinetic molecular theory gives us an understanding of pressure and temperature on the molecular level. Pressure of a gas results from the number of collisions per unit time on the walls of container. Magnitude of pressure is determined by how often and with what force the gas molecules strike. Gas molecules have an average kinetic energy. Each molecule has a different energy. Kinetic Molecular Theory

86. The Maxwell – Boltzman Distribution shows us that gas molecules have an average kinetic energy however each molecule can have a different energy. Maxwell – Boltzman Distribution The peak of the curve represents the most probable energy (E mp ). The average energy (E bar) is further to the right.

87. Since temperature affects the energy of molecules, the temperature of the system has an effect on the size and shape of the Maxwell-Boltzmann distribution. Maxwell – Boltzman Distribution

88. As kinetic energy increases, the velocity of the gas molecules increases. Therefore the Maxwell – Boltzman Distribution can be interpreted on the y-axis not only as energy but also as velocity (speed). Kinetic Molecular Theory a

89. The Maxwell – Boltzman Distribution shows that the distribution of kinetic energies becomes greater (more disperse) as temperature increases. Kinetic Molecular Theory a

90. As the volume of a container of gas increases at constant temperature, the gas molecules have to travel further to hit the walls of the container. There are fewer collisions by the gas molecules with the walls of the container. Therefore, pressure decreases. If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases. How does this theory explain Boyles Law?

91. If temperature increases at constant volume, the average kinetic energy of the gas molecules increases and they speed up. Therefore, there are more frequent and more forceful collisions with the container walls by the gas molecules and the pressure increases. How does this theory explain Charles Law?

92. Ideal Gases vs. Real Gases An ideal gas is an “imaginary gas” made up of particles with negligible particle volume and negligible attractive forces.

93. Ideal Gases vs. Real Gases In a “Real Gas” the molecules of a gas do have volume and the molecules do attract each other. Therefore anything that makes gas particles more likely to stick together or stay close to one another make them behave less ideally.

94. As the volume becomes smaller, the molecules get closer together, and a greater fraction of the occupied space is actually taken up by gas molecules. Therefore, the higher the pressure, the less the gas resembles an ideal gas. Real Gases: Deviations from Ideal Behavior As the pressure on a gas increases, the molecules are forced into a smaller volume.

95. The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules. As temperature increases, the gas molecules move faster and are further apart. Also, higher temperatures mean more energy available to break intermolecular forces. Therefore, the higher the temperature, the more ideal the gas. Real Gases: Deviations from Ideal Behavior

96. A real gas typically exhibits behavior closest to “ideal gas” behavior at low pressures and high temperatures. At the opposite conditions gases behave less ideally and high pressures and low temperatures can cause to the condensation of gases because of the increased effectiveness of intermolecular forces. Real Gases and Ideal Behavior

97. Heat vs. Work The transfer of energy between objects can occur as a result of heat or work. Work includes all types of energy transfer other than heat and is a concept covered more fully in classical physics. An example of energy transfer through work is the expansion of a gas in a piston within an engine.

98. Work and Gas Expansion/Compression When the volume of a gas in the piston increases, work is done by the gas. When the volume of a gas in the piston decreases, work is done on the gas by an external force.

99. Work = P x ΔV The magnitude of the work done on or by a gas is therefore equal to the product of the pressure of the gas times the change in the volume of the gas.