2. 1 Bellringer In the three boxes, draw what the molecules look like in a solid, liquid, and gas. Solid Liquid Gas

3. 2 GASES

4. 3 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 THREE STATES OF MATTER

6. 5 Have a definite shape Have a definite volume Molecules are held close together and there is very little movement between them. Kinetic Molecular Theory

7. 6 Have an indefinite shape Have a definite volume Kinetic Molecular Theory: Atoms and molecules have more space between them than a solid does, but less than a gas (ie. It is more “fluid”.)

8. 7 Have an indefinite shape Have an indefinite volume Kinetic Molecular Theory: Molecules are moving in random patterns with varying amounts of distance between the particles.

9. 8 At 100°C, water becomes water vapor, a gas. Molecules can move randomly over large distances. Below 0°C, water solidifies to become ice. In the solid state, water molecules are held together in a rigid structure. Between 0°C and 100 °C, water is a liquid. In the liquid state, water molecules are close together, but can move about freely.

10. 9 Changing states requires energy in either the form of heat. Changing states may also be due to the change in pressure in a system.

11. 10 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 fill containers uniformly and completely.Gases fill containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.

12. 11 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) –ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions! n = amount (moles)n = amount (moles) P = pressure (atmospheres)P = pressure (atmospheres)

13. 12 Temperature T = temperature (K)T = temperature (K) –ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions! Converting is simple! K = C + 273 C = K - 273 Converting is simple! K = C + 273 C = K - 273 Convert the temperature of this room (23 C) to Kelvin

14. 13 Pause for a Cause Convert the following to Kelvin: Water freezes at 0 o C Water boils at 100 o C Human Body Temperature is 37 o C Convert the following to Celsius: Absolute Zero is 0 K Standard Pressure is 273.15 K

15. 14 Force Pressure = Area

16. 15 Pressure Is caused by the collisions of molecules with the walls of a container is equal to force/unit area. SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere (atm) = 101,325 Pa 1 standard atmosphere (atm) = 760 mm Hg 1 standard atmosphere (atm) = 760 torrs 1 standard atmosphere (atm) = 14.7 psi 1 standard atmosphere (atm) = 101.325 kPa

17. 16Pressure Column height measures Pressure of atmosphere 1 standard atmosphere (atm) *1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = about 34 feet of water!

18. 17 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm 5.2

19. 18 Pressure Conversions What is 475 mm Hg expressed in atm? 1 atm 760 mm Hg = 0.625 atm 475 mm Hg x

20. 19 Pressure Conversions A. What is 2 atm expressed in torr?

21. 20 Pause for a Cause Convert a pressure of 1.75 atm to kPa and to mm Hg. Convert a pressure of 570. torr to atmospheres and to kPa. Convert a pressure of 550 mm of Hg into atm and Pa Convert a pressure of 150. KPa to mm of Hg and atm

22. 21 Bellringer – answer in complete sentences 1. If you want to reduce the pressure of gas in a tire, what could you do? 2. As you let the gases in a hair spray can escape, what happens to the volume and pressure of the gas? 3. If you want air to enter your lungs, what must you do to the space inside your chest cavity?

23. 22 Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P 1 V 1 = P 2 V 2 Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

24. 23 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.

25. 24 Boyle’s Law Problem A sample of oxygen gas has a volume of 150 mL when its pressure is 0.923 atm. If the pressure is increased to 0.987 atm and the temperature remains constant, what will the new volume be?

26. 25 You try! A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height?

27. 26 Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V 1 T 2 = V 2 T 1 If the temperature goes up, the volume goes up!If the temperature goes up, the volume goes up! Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

28. 27 Charles’s original balloon Modern long-distance balloon

29. 28 Charles’s Law

30. 29 Practice A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

31. 30 You try! A helium-filled balloon has a volume of 2.75 L at 20°C. The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K and in °C?

32. 31 Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 T 2 = P 2 T 1 If one temperature goes up, the pressure goes up!If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850)

33. 32 Practice The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C?

34. 33 Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20°C. At the end of the trip, the pressure gauge reads 1.9 atms. What is the new Celsius temperature of the air inside the tire?

35. 34 Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 T 2 = P 2 V 2 T 1 No, it’s not related to R2D2

36. 35 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? Set up Data Table P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

37. 36 Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

38. 37 Practice The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C?

39. 38 You try! The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm?

40. 39 One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

41. 40 And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

42. 41 Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

43. 42 Avogadro’s Law V  number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 5.3 Constant temperature Constant pressure

44. 43 Dalton’s Law of Partial Pressures

45. 44 Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... This is particularly useful in calculating the pressure of gases collected over water. P atm = P gas + P water

46. 45 Dalton’s Law of Partial Pressure Problem Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing only a mixture of these gases at exactly 1.00 atm., the partial pressure of the carbon dioxide and nitrogen are given as P CO 2 = 0.285 torr, P N 2 = 593.525 torr. What is the partial pressure of oxygen?

47. 46 Dalton’s Law- You Try! Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing only a mixture of these gases. The partial pressure of the carbon dioxide, nitrogen, and oxygen are given as P CO2 = 0.350 atm, P N2 = 75.2 kPa, and P O2 = 410. torr. What is the partial pressure of oxygen in atm? What is the total pressure?

48. 47 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

49. 48 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6

50. 49 2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O 22 5.6

51. 50 IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION! P V = n R T

52. 51 Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases are most ideal at high temperatures, low pressures, with few molecules. Under these conditions there are less intermolecular forces.

53. 52 Deviation of Real Gases from Ideal Behavior  Real Gases  A gas that does not behave completely according to the assumptions of the kinetic- molecular theory  Kinetic-molecular theory holds true for noble gases.

54. 53 Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 5.8 Repulsive Forces Attractive Forces

55. 54 Effect of intermolecular forces on the pressure exerted by a gas. 5.8

56. 55 Using PV = nRT P = Pressure V = Volume n = number of moles T = Temperature R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. R = 0.0821 R = 0.0821 L atm Mol K

57. 56 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 745 mm Hg at 25 o C? Solution Solution 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 And we always know R, 0.0821 L atm / mol K

58. 57 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? Solution Solution 2. Now plug in those values and solve for the unknown. PV = nRT n = 1.1 x 10 3 mol (or about 30 kg of gas) RT RT

59. 58 Learning Check Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

60. 59 Learning Check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

61. 60 Ideal Gas Law Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0L container at 298 K?

62. 61 Ideal Gas Law Problem #2 How many moles of helium gas would it take to fill a balloon with a volume of 1000.0 cm 3 (1 cm 3 = 1mL) when the temperature is 32 0 C and the atmospheric pressure is 752 mm of Hg?

63. 62 A tank with a volume of 658 mL contains 1.50 grams of neon gas. The maximum safe pressure that the tank can withstand is 4.50 X 10 2 kPa. At what temperature will the tank have that pressure?

64. 63 Deviations from Ideal Gas Law Real molecules have volume. The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves. There are intermolecular forces. An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions. –Otherwise a gas could not condense to become a liquid.

65. 64 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents.

66. 65 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4 Solve for n, which = mass/molar mass Or,

67. 66

68. 67 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? 5.3

69. 68 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L 5.5

70. 69 Stoichiometry of gases Steps in Solving Mass-Volume Stoichiometry Problems – 1. Write a balanced equation. – 2. Find the number of moles of a given substance. – 3. Use the ratio of moles of a given substance to moles of a required substance to find the moles of a gas. --4. Express moles of a gas in terms of volume of a gas.

71. 70 Sample Mass- Volume Problem Pg. 900 #315. Sodium reacts vigorously with water to produce hydrogen and sodium hydroxide according to the following equation: 2Na(s) + 2H 2 O(l)  2 NaOH(aq) + H 2 (g) If 0.027 g of sodium reacts with excess water, what volume of hydrogen at STP is formed?

72. 71 Pg. 900 #321. Urea, (NH 2 ) 2 CO is an important fertilizer that is manufactured by the following reaction: 2NH 3 (g) + CO 2 (g)  (NH 2 ) 2 CO(s) + H 2 O(g) What volume of NH 3 at STP will be needed to produce 8.50 X 10 4 kg of urea? Mass-Volume Problem- You Try!

73. 72 Sample Volume Mass Problem Pg. 900 #318. Dinitrogen monoxide can be prepared by heating ammonium nitrate, which decomposes according to the following equation: NH 4 NO 3 (s)  N 2 O (g) + 2H 2 O(l) What mass of ammonium nitrate should be decomposed in order to produce 250. mL of N 2 O measured at STP?

74. 73 Pg. 900 #325. Oxygen can be generated in the laboratory by heating potassium chlorate. The reaction is represented by the following equation: 2KClO 3 (s)  2KCl(s) + 3O 2 (g) What mass of KClO 3 must be used in order to generate 5.00 L of O 2, measured at STP Volume- Mass Problem- You Try!

75. 74 The principal source of sulfur is in the form of deposits of free sulfur occurring in volcanically active regions. The sulfur was initially formed by the reaction between the two volcanic vapors SO 2 and H 2 S to form H 2 O( l ) and S ( s ). What volume of each gas, at 730 mm of Hg and 22 o C, was needed to form a sulfur deposit of 4.50 X 10 5 kg on the slopes of a volcano in Hawaii? Practice Problem on Using Stoichiometry Under Nonstandard Conditions

76. 75 Practice Problem on Using Stoichiometry Under Nonstandard Conditions- You Try! A 3.25-g sample of solid calcium carbide (CaC 2 ) reacts with water to produce acetylene gas (C 2 H 2 ) and aqueous calcium hydroxide. If the acetlylene is at 17 o C and 740 mm of Hg, how many mL of acetylene were produced? CaC 2 + H 2 O  C 2 H 2 + Ca(OH) 2

77. 76 Practice Problem on Using Stoichiometry Under Nonstandard Conditions What volume of oxygen gas in liters can be collected at 0.987 atm pressure and 25 o C when 30.6 g of KClO 3 decompose by heating according to the following equation? 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

78. 77 Practice Problem on Using Stoichiometry Under Nonstandard Conditions- You Try! How many liters of gaseous carbon monoxide at 27 o C and 0.247 atm can be produced from the burning of 65.5 g of carbon according to the following equation: 2C(s) + O 2 (g)  2CO 2 (g)

79. 78 The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 u rms = 3RT M  Velocity of a Gas

80. 79 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl

81. 80 GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases.diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.effusion is the movement of molecules through a small hole into an empty container.

82. 81 GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv 2 Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.

83. 82 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

84. 83 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.

85. 84 Graham’s Law Problem 1 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees celsius according to the equation: 4 NH 3 (g) + 5 O 2 (g) --> 4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)?

86. 85 Graham’s Law Problem 2 What is the rate of effusion for H 2 if 15.00 ml of CO 2 takes 4.55 sec to effuse out of a container?

87. 86 Graham’s Law Problem 3 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)?

88. 87