3. 2 CHAPTER 12 GASES

4. The Gas Laws u Describe HOW gases behave. u Can be predicted by the theory. u Amount of change can be calculated with mathematical equations.

5. The effect of adding gas. u When we blow up a balloon we are adding gas molecules. u Doubling the the number of gas particles doubles the pressure. (of the same volume at the same temperature).

6. Pressure and the number of molecules are directly related u More molecules means more collisions. u Fewer molecules means fewer collisions. u Gases naturally move from areas of high pressure to low pressure because there is empty space to move in.

7. 1 atm u If you double the number of molecules

8. u You double the pressure. 2 atm

9. u As you remove molecules from a container 4 atm

10. u As you remove molecules from a container the pressure decreases 2 atm

11. u As you remove molecules from a container the pressure decreases u Until the pressure inside equals the pressure outside u Molecules naturally move from high to low pressure 1 atm

12. Changing the size of the container u In a smaller container molecules have less room to move. u Hit the sides of the container more often. u As volume decreases pressure increases.

13. 1 atm 4 Liters u As the pressure on a gas increases

14. 2 atm 2 Liters u As the pressure on a gas increases the volume decreases u Pressure and volume are inversely related

15. Temperature u Raising the temperature of a gas increases the pressure if the volume is held constant. u The molecules hit the walls harder. u The only way to increase the temperature at constant pressure is to increase the volume.

16. u If you start with 1 liter of gas at 1 atm pressure and 300 K u and heat it to 600 K one of 2 things happens 300 K

17. u Either the volume will increase to 2 liters at 1 atm 300 K 600 K

18. 300 K 600 K Or the pressure will increase to 2 atm. Or someplace in between

19. 18 Kinetic Molecular Theory u Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight-line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature.

20. 19 Real Gases u Particles in a REAL gas… have their own volume attract each other u Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

21. 20 Characteristics of Gases u Gases expand to fill any container. random motion, no attraction u Gases are fluids (like liquids). no attraction u Gases have very low densities. no volume = lots of empty space

22. 21 Characteristics of Gases u Gases can be compressed. no volume = lots of empty space u Gases undergo diffusion & effusion. random motion

23. 22 Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 u Always use absolute temperature (Kelvin) when working with gases.

24. 23 Pressure Which shoes create the most pressure?

25. 24 Pressure u Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

26. 25 Pressure u Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

27. 26 Pressure u KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

28. 27 STP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

29. 28 Boyle’s Law P V PV = k

30. 29 Boyle’s Law u The pressure and volume of a gas are inversely related at constant mass & temp P V PV = k

31. 30 V T Charles’ Law

32. 31 V T Charles’ Law u The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

33. 32 P T Gay-Lussac’s Law

34. 33 P T Gay-Lussac’s Law u The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

35. 34 = kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

36. 35 GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gas Law Problems u A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

37. 36 GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gas Law Problems u A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

38. 37 GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 Gas Law Problems u A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

39. 38 GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Gas Law Problems u A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

40. u A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is change to 1.5 atm what is the new volume? u A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change to volume to 43 L? Your Turn

41. u What is the temperature of a gas that is expanded from 2.5 L at 25ºC to 4.1L at constant pressure. u What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC?

42. Your Turn u What is the pressure inside a 0.250 L can of deodorant that starts at 25ºC and 1.2 atm if the temperature is raised to 100ºC? u At what temperature will the can above have a pressure of 2.2 atm?

43. Your Turn u A 15 L cylinder of gas at 4.8 atm pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume? u If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas?

44. 43 V n Avogadro’s Principle u Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

45. 44 PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R You don’t need to memorize these values! Merge the Combined Gas Law with Avogadro’s Principle:

46. 45 Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT You don’t need to memorize these values!

47. The Ideal Gas Law u Pressure times Volume equals the number of moles times the Ideal Gas Constant (R) times the temperature in Kelvin. u R does not depend on anything, it is really constant u R = 0.0821 (L atm)/(mol K) u R = 62.4 (L mm Hg)/(K mol)

48. 47 GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law Problems u Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

49. 48 GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K Ideal Gas Law Problems u Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3

50. Examples u How many moles of air are there in a 2.0 L bottle at 19ºC and 747 mm Hg? u What is the pressure exerted by 1.8 g of H 2 gas exert in a 4.3 L balloon at 27ºC?

51. 50 Dalton’s Law u The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

52. 51 GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa Dalton’s Law u Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

53. 52 GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr u A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? Look up water-vapor pressure for 35.0°C. Sig Figs: Round to least number of decimal places. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

54. Density u The Molar mass of a gas can be determined by the density of the gas. u D= mass = m Volume V u Molar mass = mass = m Moles n u n = PV RT

55. u Molar Mass = m (PV/RT) u Molar mass = m RT V P u Molar mass = DRT P

56. You might need to review Chapter 8

57. At STP u At STP determining the amount of gas required or produced is easy. u 22.4 L = 1 mole u For example How many liters of O 2 at STP are required to produce 20.3 g of H 2 O?

58. Not At STP u Chemical reactions happen in MOLES. u If you know how much gas - change it to moles u Use the Ideal Gas Law n = PV/RT u If you want to find how much gas - use moles to figure out volume V = nRT/P

59. Example #1 u HCl(g) can be formed by the following reaction u 2NaCl(aq) + H 2 SO 4 (aq) 2HCl(g) + Na 2 SO 4 (aq) u What mass of NaCl is needed to produce 340 mL of HCl at 1.51 atm at 20ºC?

60. Example #2 u 2NaCl(aq) + H 2 SO 4 (aq) 2HCl(g) + Na 2 SO 4 (aq) u What volume of HCl gas at 25ºC and 715 mm Hg will be generated if 10.2 g of NaCl react?

62. Ideal Gases don’t exist u Molecules do take up space u There are attractive forces u otherwise there would be no liquids

63. Real Gases behave like Ideal Gases u When the molecules are far apart u The molecules do not take up as big a percentage of the space u We can ignore their volume. u This is at low pressure

64. Real Gases behave like Ideal gases when u When molecules are moving fast. u Collisions are harder and faster. u Molecules are not next to each other very long. u Attractive forces can’t play a role.

65. Diffusion u Effusion Gas escaping through a tiny hole in a container. u Depends on the speed of the molecule. u Molecules moving from areas of high concentration to low concentration. u Perfume molecules spreading across the room.

66. Graham’s Law u The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules. u Kinetic energy = 1/2 mv 2 u m is the mass v is the velocity. Chem Express

67. u bigger molecules move slower at the same temp. (by Square root) u Bigger molecules effuse and diffuse slower u Helium effuses and diffuses faster than air - escapes from balloon. Graham’s Law

68. 67 Graham’s Law u Diffusion Spreading of gas molecules throughout a container until evenly distributed. u Effusion Passing of gas molecules through a tiny opening in a container

69. 68 Graham’s Law KE = ½mv 2 u Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. –Larger m  smaller v

70. 69 Graham’s Law u Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed.

71. 70 u Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

72. 71 u 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 Put the gas with the unknown speed as “Gas A”.

73. 72 u An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.

74. 73 Gas Stoichiometry u Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law u Non- STP Given liters of gas? –start with ideal gas law Looking for liters of gas? –start with stoichiometry conv.

75. 74 1 mol CaCO 3 100.09g CaCO 3 Gas Stoichiometry Problem u What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.26 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

76. 75 WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 Gas Stoichiometry Problem u What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm 3  kPa/mol  K

77. 76 WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = 0.597 mol O 2 Gas Stoichiometry Problem u How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm 3  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O 3 15.0 L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT 

78. 77 2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem u How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.