1. GASES Chapter 5

2. A Gas -Uniformly fills any container. -Mixes completely with any other gas -Exerts pressure on its surroundings.

3. Simple barometer invented by Evangelista Torricelli

4. Barometer The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. 1 atm = 760 mm Hg 1 atm Pressure 760 mm Hg Vacuum

5. Manometer Gas h Column of mercury to measure pressure. h is how much lower the pressure is than outside.

6. Manometer h is how much higher the gas pressure is than the atmosphere. h Gas

7. Simple Manometer

8. Units of pressure 1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr 1 atm = 101,235 Pascals = 101.325 kPa 14.69 psi=1Atm Can make conversion factors from these. What is 724 mm Hg in atm in torr? in? kPa?

9. Pressure -is equal to force/unit area -SI units = Newton/meter 2 = 1 Pascal (Pa) -1 standard atmosphere = 101,325 Pa -1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

10. Pressure Unit Conversions The pressure of a tire is measured to be 28 psi. What would the pressure in atmospheres, torr, and pascals. (28 psi)(1.000 atm/14.69 psi) = 1.9 atm (28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 10 3 torr (28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 10 5 Pa

11. Volume of a gas decreases as pressure increases at constant temperature

12. Boyle’s Law * P 1 V 1 = P 2 V 2 (T = constant) ( * Holds precisely only at very low pressures.) Boyle’s Law Pressure and volume are inversely related at constant temperature.

13. V P (at constant T)

14. Boyle’s Law Calculations A 1.5-L sample of gaseous CCl 2 F 2 has a pressure of 56 torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be? Decrease P 1 = 56 torr P 2 = 150 torr V 1 = 1.5 L V 2 = ? V 1 P 1 = V 2 P 2 V 2 = V 1 P 1 /P 2 V 2 = (1.5 L)(56 torr)/(150 torr) V 2 = 0.56 L

15. Boyle’s Law Calculations In an automobile engine the initial cylinder volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure? P 1 = 1.00 atm P 2 = ? V 1 = 0.725 L V 2 = 0.075 L V 1 P 1 = V 2 P 2 P 2 = V 1 P 1 /V 2 P 2 = (0.725 L)(1.00 atm)/(0.075 L) P 2 = 9.7 atm Is this answer reasonable?

16. Volume of a gas increases as heat is added when pressure is held constant.

17. Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

18. Charles’s Law

19. Examples What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC, if the pressure is held constant?

20. Combined Gas Law If the moles of gas remains constant, use this formula and cancel out the other things that don’t change. P 1 V 1 = P 2 V 2. T 1 T 2

21. Examples A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC?

22. Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

23. At constant temperature and pressure, increasing the moles of a gas increases its volume.

24. AVOGADRO’S LAW V 1 /n 1 = V 2 /n 2

25. Avogadro simple A 5.20L sample at 18.0 C and 2.00 atm pressure contains 0.436 moles of gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be ?

26. AVOGADRO’S LAW dif A 12.2 L sample containing 0.50 mol of oxygen gas, O 2, at a pressure of 1.00 atm and a temperature of 25 o C is converted to ozone, O 3, at the same temperature and pressure, what will be the volume of the ozone? 3 O 2(g) ---> 2 O 3(g) (0.50 mol O 2 )(2 mol O 3 /3 mol O 2 ) = 0.33 mol O 3 V 1 = 12.2 L V 2 = ? n 1 = 0.50 mol n 2 = 0.33 mol V 2 = (12.2 L)(0.33 mol)/(0.50 mol) V 2 = 8.1 L

27. Ideal Gas Law -An equation of state for a gas. -“state” is the condition of the gas at a given time. PV = nRT

28. IDEAL GAS 1. Molecules are infinitely far apart. 2. Zero attractive forces exist between the molecules. 3. Molecules are infinitely small--zero molecular volume.

29. REAL GAS 1. Molecules are relatively far apart compared to their size. 2. Very small attractive forces exist between molecules. 3. The volume of the molecule is small compared to the distance between molecules.

30. Ideal Gas Law PV = nRT R = proportionality constant = 0.0821 L atm   mol  P = pressure in atm V = volume in liters n = moles T = temperature in Kelvins Holds closely at P < 1 atm

31. Ideal Gas Law Calculations A 1.5 mol sample of radon gas has a volume of 21.0 L at 33 o C. What is the pressure of the gas? p = ? V = 21.0 L n = 1.5 mol T = 33 o C + 273 T = 306 K R = 0.0821 Latm/molK pV = nRT p = nRT/V p = (1.5mol)(0.0821Latm/molK)(306K) (21.0L) p = 1.8 atm

32. Ideal Gas Law Calculations A sample of hydrogen gas, H 2, has a volume of 8.56 L at a temperature of O o C and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present. p = 1.5 atm V = 8.56 L R = 0.0821 Latm/molK n = ? T = O o C + 273 T = 273K pV = nRT n = pV/RT n = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K) n = 0.57 mol

33. Standard Temperature and Pressure “STP” P = 1 atmosphere T =  C or 273K The molar volume of an ideal gas is 22.42 liters at STP

34. Molar Volume pV = nRT V = nRT/p V = (1.00 mol)(0.08206 Latm/molK)(273K) (1.00 atm) V = 22.4 L

35. Gases at STP A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present? (1.75L N 2 )(1.000 mol/22.4 L) = 7.81 x 10 -2 mol N 2

36. MOLAR MASS OF A GAS n = m/M n = number of moles m = mass M = molar mass

37. MOLAR MASS OF A GAS P = mRT/VM or P = DRT/M therefore: M = DRT/P

38. A gas at 34.0C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass.

39. What is the density of 2L of O2 gas at STP?

40. GAS STOICHIOMETRY 1. Mass-Volume 2. Volume-Volume

41. Gas Stoichiometry at STP Quicklime, CaO, is produced by heating calcium carbonate, CaCO 3. Calculate the volume of CO 2 produced at STP from the decomposition of 152 g of CaCO 3. CaCO 3(s) ---> CaO (s) + CO 2(g) (152g CaCO 3 )(1 mol/100.1g)(1mol CO 2 /1mol CaCO 3 ) (22.4L/1mol) = 34.1L CO 2 Note: This method only works when the gas is at STP!!!!!

42. Volume-Volume If 25.0 L of hydrogen reacts with an excess of nitrogen gas, how much ammonia gas will be produced? All gases are measured at the standard temperature and pressure. 2N 2(g) + 3H 2(g) ----> 2NH 3(g)

43. 16.7 L NH3

44. Gas Stoichiometry Not at STP (Continued) p = 1.00 atm V = ? n = 1.28 x 10 -1 mol R = 0.08206 Latm/molK T = 25 o C + 273 = 298 K pV = nRT V = nRT/p V = (1.28 x 10 -1 mol)(0.08206Latm/molK)(298K) (1.00 atm) V = 3.13 L O 2

45. Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +...

46. Dalton’s Law of Partial Pressures Calculations A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at.33 atm would have what total pressure? P Total = P 1 + P 2 + P 3 P Total = 1.25 atm + 2.55 atm +.33 atm P total = 4.13 atm

47. Water Vapor Pressure 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) When a sample of potassium chlorate is decomposed and the oxygen produced is collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 o C. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 o C is 21 torr). How many moles of oxygen are produced? P ox = P total - P HOH P ox = 754 torr - 21 torr p ox = 733 torr

48. Dalton’s Law Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25ºC.

49. A Volume of 2.0 L of He at 46° C and 1.2 ATM pressure was added to a vessel that contained 4.5L of N 2 at STP. What is the total pressure of each gas at STP after the He is added. RB P 128

50. MOLE FRACTION -- the ratio of the number of moles of a given component in a mixture to the total number of moles of the mixture.  1 = n 1 / n total  1 = V 1 / V total  1 = P 1 / P total (volume & temperature constant)

51. The partial pressure of Oxygen gas was observed to be 156 torr in the air with a total atmosphere pressure of 743 torr. Calculate the mole fraction of O2 present.

52. Mole Fraction The mole Fraction of nitrogen in the air is.7808. Calculate the partial pressure of N2 in the air when the atmospheric pressure is 760.

53. Root mean square velociyy Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

54. Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C. and100°C.

55. Example Calculate the root mean square velocity of carbon dioxide at 25ºC. Calculate the root mean square velocity of chlorine at 25ºC.

56. Effusion: describes the passage of gas into an evacuated chamber. Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

57. Effusion of a gas into an evacuated chamber

58. Effusion: Diffusion:

59. Grams Law Calculate the ratio of effusion rates of hydrogen gas and uranium hexafluoride.

60. A compound effuses through a porous cylinder 3.20 time faster than helium. What is it’s molar mass?

61. Kinetic Molecular Theory 1.Volume of individual particles is  zero. 2.Collisions of particles with container walls cause pressure exerted by gas. 3.Particles exert no forces on each other. 4.Average kinetic energy  Kelvin temperature of a gas.

62. Real Gases Real molecules do take up space and they do interact with each other (especially polar molecules). Need to add correction factors to the ideal gas law to account for these.

63. Volume Correction The actual volume free to move in is less because of particle size. More molecules will have more effect. Corrected volume V’ = V - nb b is a constant that differs for each gas. P’ = nRT (V-nb)

64. Pressure correction Because the molecules are attracted to each other, the pressure on the container will be less than ideal depends on the number of molecules per liter. since two molecules interact, the effect must be squared.

65. Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared. P observed = P’ - a 2 () V n

66. Altogether P obs = nRT - a n 2 V-nb V Called the Van der Wall’s equation if rearranged Corrected Corrected Pressure Volume ()

67. Where does it come from a and b are determined by experiment. Different for each gas. Bigger molecules have larger b. a depends on both size and polarity. once given, plug and chug.

68. Example Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC Using the ideal gas law. Van der Waal’s equation –a = 6.49 atm L 2 /mol 2 –b = 0.0562 L/mol

69. Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

70. Plots of PV/nRT vs. P for several gases at 200 K. Note the significant deviation from ideal behavior.

71. Real Gases  corrected pressure corrected volume P ideal V ideal

72. Concentration for some smog components vs. time of day

73. NO 2(g)  NO (g) + O (g) O (g) + O 2(g)  O 3(g) NO (g) + 1/2 O 2(g)  NO 2(g) __________________________ 3/2 O 2(g)  O 3(g) What substances represent intermediates? Which substance represents the catalyst?

74. Chemistry is a gas!!