1. S TATES OF M ATTER

2. State of Matter VolumeShapeDensity Compressibility Motion of Molecules Gas Liquid Solid

3. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. GASES Chapter 5

4. Elements that exist as gases at 25 0 C and 1 atmosphere

6. P HYSICAL C HARACTERISTICS OF G ASES Assume the shape of their container Most compressible state of matter Mix evenly and completely (Diffusion) Very low density compared with solids and liquids. Density usually measured in g/L of gas

7. Pressure = Force Area P RESSURE Pressure (P): the force per unit area on a surface. What causes pressure?  collisions of the gas molecules with each other and with surfaces with which they come into contact.  depends on volume, temperature, and the number of molecules present.

8. C ALCULATING F ORCE The SI Unit for force is the Newton (N) Consider a person with a mass of 51 kg. At Earth’s surface, gravity has an acceleration of 9.8 m/s 2. What is the value of force? Force = mass x acceleration Force = 51 kg × 9.8 m/s 2 = 500 kg m/s 2 = 500 N

9. P = 1.5 N/cm 2 P = 38.5 N/cm 2 P = 77 N/cm 2 The greater area the less pressure on the floor. R ELATIONSHIP B ETWEEN P RESSURE, F ORCE, AND A REA

10. B AROMETER Barometer device used to measure atmospheric pressure

11. Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

13. U NITS OF P RESSURE 1 atm = 101.3 kPa = 760 mmHg = 760 torr

14. G AS VS. V APOR Gas: Substance found normally in the gaseous state at normal temperature and pressure Vapor: Gaseous form of any substance that is a liquid or a solid at normal temperature and pressure

15. GAS LAWS

16. B OYLE ’ S L AW As volume increases, pressure _________________ As volume decreases, pressure _________________ As pressure increases, volume ________________ As pressure decreases, volume _________________ PRESSURE AND VOLUME ARE: __________________ RELATED

17. As P (h) increases V decreases

18. Boyle’s Law

19. B OYLE ’ S L AW E QUATION P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas

20. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

21. C HARLES ’ L AW As temperature increases, volume ________________ As temperature decreases, volume _________________ As volume increases, temperature _________________ As volume decreases, temperature _________________ TEMPERATURE AND VOLUME ARE: __________________ RELATED

22. As T increasesV increases

23. Variation of gas volume with temperature at constant pressure. Charles’ Law

24. C HARLES ’ L AW E QUATION V  TV  T V / T = constant T (K) = t ( 0 C) + 273 Temperature must be in Kelvin Constant pressure Constant amount of gas V 1 = V 2 T 1 T 2

25. A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2

26. A VOGADRO ’ S L AW As amount increases, volume ________________ As amount decreases, volume _________________ As volume increases, amount _________________ As volume decreases, amount _________________ AMOUNT AND VOLUME ARE: __________________ RELATED

27. A VOGADRO ’ S L AW E QUATION V  number of moles (n) V /n = constant V 1 = V 2 n 1 n 2 Constant temperature Constant pressure

28. If 22.3 mol of N 2 gas has a volume of 15 L, how many moles of N 2 gas will have a volume of 12 L? V 1 = 15 L n 1 = 22.3 mol V 2 = 12 L n 2 = ? n 2 = V 2 x n 1 V1V1 12 L x 22.3 mol 15 L = = 18 mol V 1 /n 1 = V 2 /n 2

29. C OMBINED G AS L AW E QUATION Boyle’s law and Charles’s law can be combined into a single equation that can be used for situations in which temperature, pressure, and volume, all vary at the same time. The temperature MUST be in Kelvin

30. A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? V 1 = 50.0 L P 1 = 1.08 atm V 2 = ? P 2 =.855 atm T 1 = 25 o C T 2 = 10 o C

31. I DEAL G AS E QUATION Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

32. 0 0 C and 1 atm PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.0821 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. STP = S TANDARD T EMPERATURE AND P RESSURE

33. N UMERICAL V ALUES OF T HE G AS C ONSTANT “R” ALWAYS MATCH UP YOUR UNITS!!!!

34. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L

35. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

36. 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

37. V and T are constant P1P1 P2P2 P total = P 1 + P 2 D ALTON ’ S L AW OF P ARTIAL P RESSURES

38. Daltons Law of Partial Pressure Gases mix homogeneously (form a solution) in any proportions Each gas in a mixture behaves as if it were the only gas present (assuming no chemical reactions). P T = P gasA + P gasB + P gasC + etc.

39. Examples: 1. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 456 mmHg. If gas M contributes 200 mmHg, and gas L contributes 10 mmHg, what is the pressure exerted by gas N.

40. Examples: 2. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 644 mmHg. If gas M contributes to 21% of the pressure, and gas N contributes 54% what are the pressures exerted by all three gases in mmHg.

41. MOLE FRACTIONS Mole fraction of gas A = Moles of gas A_____ Total number of moles of gas GAS AMOUNT IN MOLES A 0.235 B 1.025 C 2.35 D 0.78 Examples: 1. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.

42. GASAMOUNT IN GRAMS He 23.5 CO 2 45.7 CH 4 32.3 Steam 24.7 2. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.

43. P i = X i P T H OW DO P ARTIAL P RESSURE AND M OLE F RACTION R ELATE ? Where: P i is the pressure i X i is the mole fraction of i P T is the total pressure Examples: 1. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg. GAS AMOUNT IN MOLES O2O2 20 N2N2 79 Ar 1

44. Example: 2. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg. GASGAS AMOUNT IN GRAMS O2O2 45.6 N2N2 32.2 ArAr 100.76

45. W HAT IS VAPOR PRESSURE ? The pressure that exists above the surface of a liquid from particles escaping the surface of the liquid

46. V APOR P RESSURE IS EFFECTED BY HOW VOLATILE THE LIQUID IS

47. Before Evaporation At Equilibrium

48. COLLECTING GASES OVER WATER Dalton ’ s Law can be used to calculate the pressure of gas collected over water Set-up for such a system is shown below

49. The collection flask initially contains all water When a reaction takes place in a reaction chamber Gas travels through tubing attached to the reaction chamber The gas displaces the water in the collection flask When all the water is displaced the flask is stoppered and the gas and some water vapor is collected To find the pressure attributed by the gas collected you need to subtract the pressure due to water vapor at a specific temperature from the total pressure.

50. Examples: 1. A gas is collected by water displacement at 50 o C and barometric pressure of 95 kPa. What is the pressure exerted by the dry gas? 2. Oxygen gas is collected by water displacement from the reaction of Na 2 O 2 and water. The oxygen displaces 318 mL of water at 23 o C and 1.000atm. What is the pressure of dry O 2.

51. 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

52. Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

53. **Gases behave ideally at very high temperatures and low pressures**

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

55. D IFFUSION AND E FFUSION DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion (kinetic properties) EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container

56. E FFUSION R ATE V IDEO

57. G RAHAM ’ S L AW O F E FFUSION Graham’s law of effusion: the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

58. G RAHAM ’ S L AW - V ISUAL P ROBLEM

59. G RAHAM ’ S L AW P ROBLEM Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Given: identities of two gases, H 2 and O 2 Unknown: relative rates of effusion Hydrogen = Compound A Oxygen = Compound B

60. G RAHAM ’ S L AW P ROBLEM HINT: Always put the substance with the larger molar mass on top as compound B. 1. Calculate: 2. Rearrange the equation: rate of effusion of A = 3.98 rate of effusion of B 3. Write a sentence: Hydrogen diffuses 3.98 times faster than Oxygen