1. Physical Properties Gases

2. Kinetic Molecular Theory b 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 average kinetic energy is directly proportional to the absolute temperature b To change pressure you can: Change temperature Change volume Change amount

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

4. Characteristics of Gases b Gases can be compressed no volume = lots of empty space b Gases undergo diffusion & effusion random motion State Changes

5. Describing Gases b Gases can be described by their: Temperature Pressure Volume Number of molecules/moles K atm L #

6. TemperatureTemperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 b Temperature: Every time temperature is used in a gas law equation it must be stated in Kelvin (an absolute temperature)

7. PressurePressure Which shoes create the most pressure? Pressure of a gas is the force of its particles exerted over a unit area (number of collisions against a container’s walls)

8. PressurePressure b Pressure may have different units: torr, millimeters of mercury (mm Hg), atmospheres (atm), Pascals (Pa), or kilopascals (kPa). b KEY EQUIVALENT UNITS 760 torr 760 mm Hg 1 atm 101, 325 Pa 101. 325 kPa (kilopascal) 14.7 psi

9. STPSTP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa The volume of 1 mole of gas at STP is 22.4 L -OR- STP

10. Pressure Problem 1 b The average pressure in Denver, Colorado, is 0.830 atm. Express this in kPa. 0.830 atm 1 atm 101.325 kPa = 84.1 kPa

11. Pressure Problem 2 b Convert a pressure of 1.75 atm to psi. 1.75 atm 1 atm 14.7 psi = 25.7 psi

12. Pressure Problem 3 b Convert a pressure of 570. torr to atmospheres and kPa. (a) 570 torr 760 torr 1 atm =.750 atm (b) 570 torr 760 torr 101.325 kPa = 76.0 kPa

13. The Gas Laws The Behavior of Gases The Behavior of Gases

14. Boyle’s Law P V b Investigated the relationship between pressure and volume on a gas. b He found that pressure and volume of a gas are inversely related at constant mass & temp

15. Boyle’s Law b As pressure increases b Volume decreases Boyles LAW: P V P 1 V 1 = k

16. Boyle’s Law

17. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 = P 2 V 2 Gas Law Problems b 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

18. V T Charles’ Law Jacques Charles investigated the relationship of temperature & volume in a gas. He found that the volume of a gas and the Kelvin temperature of a gas are directly related.

19. V T Charles’ Law b As volume increases, temperature increases b Charles Law:

20. Charles’ Law

21. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: V 1 T 2 = V 2 T 1 Gas Law Problems b 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

22. GIVEN: V 1 = 43.5 L T 1 = 0°C = 273K V 2 = ? T 2 = 35.4°C= 3 08.4K WORK: V 1 T 2 = V 2 T 1 Gas Law Problems b Example: The volume of a sample of gas is 43.5 L at STP. What will the volume of the gas be if the temperature is increased to 35.4°C? CHARLES’ LAW TT VV (43.5 L)(308.4 K)=V 2 (273 K) V 2 = 49.1 L

23. P T Gay-Lussac’s Law Since temperature and volume are related, it would make sense that temperature and pressure are related. The equation for this will resemble Charle’s law:

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

25. GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 T 2 = P 2 T 1 Gas Law Problems b 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)(296K) T 2 = 217 K = -56°C

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

27. 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 b 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

28. The First 4 Gas Laws b The Gas Laws Table

29. Gases II. Ideal Gas Law Ideal Gas Law and Gas Stoichiometry

30. Part 1 Ideal Gas Law

31. Avogadro’s Law b The volume of a gas increases or reduces, as its number of moles is being increased or decreased. b The volume of an enclosed gas is directly proportional to its number of moles.

32. V n Avogadro’s Principle Avogadro’s Principle b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas n = number of moles

33. PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.08206 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

34. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.08206 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

35. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.08206 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 Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0.412 mol of Heat 16°C & occupying 3.25 L.

36. GIVEN: V = ?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 b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol O 2 = 2.7 mol 32.00 gO 2 PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3

37. Ideal Gas Law and Molar Mass b The Ideal Gas Law can be used to calculate the mass or the molar mass of a gas sample if the mass of the sample is known. b number of moles gas = mass gas or n gas = m gas_ Molar Mass gas MM gas b substituting this for n gas in PV = nRT yields b P V = m gas__ R T b MM gas

38. GIVEN: P = 1.00 atm T = 22°C = 295. K V =.600 L R = 0.08206 L  atm/mol  K MM = 28.01 g m N 2 = ? WORK: MM P V = m gas R T (28.01)(1.0)(.60)=m (.0821) (295) g/mol atm L L  atm/mol  K K m N 2 =.69 g of N 2 Applications of Ideal Gas Law Applications of Ideal Gas Law b Calculate the grams of N 2 present in a 0.60 L sample kept at 1.00 atm pressure and a temperature of 22.0 o C.

39. Applications of Ideal Gas Law b Can be used to calculate the molar mass of a gas and the density b Substitute this into ideal gas law b And m/V = d in g/L, so

40. GIVEN: P = 1.50 atm T = 27°C = 300. K d = 1.95 g/L R = 0.08206 L  atm/mol  K MM = ? WORK: MM = dRT/P MM=(1.95)(0.08206)(300.)/1.50 g/L L  atm/mol  K K atm MM = 32.0 g/mol Applications of Ideal Gas Law Applications of Ideal Gas Law b The density of a gas was measured at 1.50 atm and 27 ° C and found to be 1.95 g/L. Calculate the molar mass of the gas.

41. GIVEN: d = ? g/L CO 2 T = 25°C = 298 K P = 750. torr R = 0.08206 L  atm/mol  K MM = 44.01 g/mol MM = dRT/P →d = MM P/RT d=(44.01 g/mol )(.987 atm ) (0.08206 L  atm/mol  K )(298 K ) d = 1.78 g/L CO 2 Applications of Ideal Gas Law Applications of Ideal Gas Law b Calculate the density of carbon dioxide gas at 25 °C and 750. torr. WORK: 750 torr 1 atm =.987 atm 760 torr =.987 atm

42. Part 2 Gas Stoichiometry

43. * Stoichiometry Steps Review * 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Molarity - moles  liters soln Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

44. 1 mol of a gas=22.4 L at STP Molar Volume at STP Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

45. Molar Volume at STP Molar Volume at STP Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Volume (22.4 L/mol) LITERS OF GAS AT STP

46. Gas Stoichiometry Gas Stoichiometry b Liters of one Gas  Liters of another Gas: Avogadro’s Principle Coefficients give mole ratios and volume ratios b Moles (or grams) of A  Liters of B: STP – use 22.4 L/mol Non-STP – use ideal gas law & stoich b Non- STP Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conversion

47. Gas Stoichiometry - Volume b What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C 3 H 8 )? C 3 H 8 + O 2  CO 2 + H 2 O 5 3 4 4.00 L ?L 4.00 L C 3 H 8 5 L O 2 1 L C 3 H 8 = 20.0 L O 2

48. b How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L Gas Stoichiometry Problem – STP 2KClO 3  2KCl + 3O 2

49. 1 mol CaCO 3 100.09g CaCO 3 Gas Stoichiometry Problem – Non-STP b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 0.0525 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 for n to find liters NEXT  P = 103 kPa V = ? n = ? R = 8.315 dm 3 kPa/molK T = 298K

50. WORK: PV = nRT (103 kPa)V =(0.0525mol)(8.315 dm 3  kPa/mol  K ) (298K) V = 1.26 dm 3 CO 2 b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 0.0525 mol T = 25°C = 298 K R = 8.315 dm 3  kPa/mol  K Gas Stoichiometry Problem – Non-STP

51. 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 b 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 = ?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 

52. 2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem b 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

53. Ch. 14 - Gases III. Two More Laws Dalton’s Law & Graham’s Law

54. A. Dalton’s Law b 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 +...P n

55. GIVEN: P O2 = ? P total = 1.00 atm P CO2 = 0.12 atm P N2 = 0.70 atm WORK: P total = P CO2 + P N2 + P O2 1.00 atm = 0.12 atm +.70 atm + P O2 P O2 = 0.18 atm A. Dalton’s Law b A sample of air has a total pressure of 1.00 atm. The mixture contains only CO 2, N 2, and O 2. If P CO2 = 0.12 atm and P N2 = 0.70, what is the partial pressure of O 2 ?

56. A. Dalton’s Law b Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture, represented by chi ( ) b Because n is directly proportional to P according to the ideal gas law, we can derive

57. GIVEN: χ N2 = 0.7808 P total = 760. torr P N2 = ? WORK: χ N2 = P N2 / P TOTAL 0.7808 = (P N2 )/(760. torr) P N2 = 593 torr A. Dalton’s Law b The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N 2 in air when the atmospheric pressure is 760. torr.

58. A. Dalton’s Law When H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor Water exerts a pressure known as water-vapor pressure So, to determine total pressure of gas and water vapor inside a container, make total pressure inside bottle = atmosphere and use this formula: P total = P gas + P H 2 O

59. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 20.4 mm Hg = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa A. Dalton’s Law b 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, convert to kPa Sig Figs: Round to lowest 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.

60. 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 b 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. A. 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.

61. B. Graham’s Law b Diffusion Spreading of gas molecules throughout a container until evenly distributed b Effusion Passing of gas molecules through a tiny opening in a container

62. B. Graham’s Law KE = ½mv 2 b 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

63. B. Graham’s Law b 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.

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

65. b 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? B. Graham’s Law Put the gas with the unknown speed as “Gas A”.

66. b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. B. 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.