1. Chapter 5 Gases

2. Chapter 5  One of the most amazing things about gasses is that, despite wide differences in chemical properties, all gases more or less obey the gas law.  The gas law deal with how gases behave with respect pressure, volume, temperature and amount.

3. Section 5.1 Pressure A Gas  Uniformly fills any container.  Easily compressed.  Mixes completely with any other gas.  Exerts pressure on its surroundings. Copyright © Cengage Learning. All rights reserved 3

4. Section 5.1 Pressure  SI units = Newton/meter 2 = 1 Pascal (Pa)  1 standard atmosphere = 101,325 Pa  1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr Copyright © Cengage Learning. All rights reserved 4

5. Section 5.1 Pressure Barometer  Device used to measure atmospheric pressure.  Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish. Copyright © Cengage Learning. All rights reserved 5

6. Chapter 5 Manometer  Device used for measuring the pressure of a gas in a container. Copyright © Cengage Learning. All rights reserved 6

7. Section 5.1 Pressure Pressure Conversions: An Example The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. Copyright © Cengage Learning. All rights reserved 7

8. Section 5.1 Pressure 

9. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  Gas laws can be deduced from observations like these.  Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) can be discovered. Copyright © Cengage Learning. All rights reserved 9

10. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s Law  The volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.  Another way to describing it is that their product are constant.  As one goes up, the other goes down  PV = k

11. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s Law: P 1 V 1 = P 2 V 2

12. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s Law  Pressure and volume are inversely related (constant T, temperature, and n, # of moles of gas).  PV = k (k is a constant for a given sample of air at a specific temperature) Copyright © Cengage Learning. All rights reserved 12

13. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s law Copyright © Cengage Learning. All rights reserved 13 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

14. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? P 1 V 1 = P 2 V 2 = (0.956atm)(12,4l)=(1.20)(V 2 ) (0.956atm)(12.4L)/(1.20) = V 2 9.88 L Copyright © Cengage Learning. All rights reserved 14 EXERCISE!

15. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  A sample of oxygen gas occupies a volume of 250.mL at 740. torr pressure. What volume will it occupy at 800. torr pressure?  (740.torr)(250mL)=(800.torr)(V 2 )  V 2 = 231 mL

16. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters?  (125Kpa)(3,50L)=(p 2 )(2.00L)  P 2 = 219Kpa

17. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  A 2.0 liter container of nitrogen had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm.  (3.2atm)(2,0L) = (1,0atm)(V 2 )  V 2 = 6.4 L

18. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Charles’s Law 

19. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Charles’s Law  Volume and Temperature (in Kelvin) are directly related (constant P and n).  V=bT (b is a proportionality constant)  K = °C + 273  0 K is called absolute zero. Copyright © Cengage Learning. All rights reserved 19

20. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro

21. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Charles’ Law: V 1 /T 1 = V 2 /T 2

22. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Charles’s Law Copyright © Cengage Learning. All rights reserved 22 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

23. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? Remember to convert the temperature to Kelvin (1.30)/(24.7+273)= V 2 /(-78.5+273) 0.849 L Copyright © Cengage Learning. All rights reserved 23 EXERCISE!

24. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro 

25. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  4.40 L of a gas is collected at 50.0 °C. What will be its volume upon cooling to 25.0 °C?  4.06 L

26. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure (as required by Charles' Law)?  127 ◦ C

27. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro . Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00 L.  146.5 K

28. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?  682 mL

29. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Avogadro’s Law  This law states that equal volume of gases at constant temperature and pressure contains the same number of particle  V = an V 1 /n 1 = V 2 /n 2

30. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Avogadro’s Law  Volume and number of moles are directly related (constant T and P).  V = an (a is a proportionality constant) Copyright © Cengage Learning. All rights reserved 30

31. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? (2.45mol)/(89.0l) = (2.10)/ (V 2 ) (89.0)(2.10)/(2.45) 76.3 L Copyright © Cengage Learning. All rights reserved 31 EXERCISE!

32. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Avogadro’s Law  If 0.50 moles of oxygen occupies a volume of 7.20 L at STP, what volume does 1.5 moles of O 2 occupy at STP  21.6 L

33. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  If 3.25 moles of argon gas occupies a volume of 1.00L at STP, what volume does 14.15 mole of argon occupy  4.35 L

34. Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro  If 12.40 moles of CO 2 occupies a volume of 96.8 L, how many moles occupy 72.6 L.  9.30 moles

35. Section 5.3 The Ideal Gas Law  We can bring all of these laws together into one comprehensive law:  V = bT (constant P and n)  V = an (constant T and P)  V = (constant T and n) PV = nRT (where R = 0.08206 L·atm/mol·K, the universal gas constant) Copyright © Cengage Learning. All rights reserved 35

36. Section 5.3 The Ideal Gas Law An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire. PV = nRT (3.18atm)(25.0L) = n(0.08206)(23+273) 3.27 mol Copyright © Cengage Learning. All rights reserved 36 EXERCISE!

37. Section 5.3 The Ideal Gas Law What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25°C? Convert Kg of helium to moles of helium (p)(304.0L) =([5.670×1000]/4.003)(0.0821)(25+273) 114 atm Copyright © Cengage Learning. All rights reserved 37 EXERCISE!

38. Section 5.3 The Ideal Gas Law At what temperature (in °C) does 121 mL of CO 2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? P 1 V 1 /T 1 = P 2 V 2 /T 2 (1.05atm)(121mL)/(27+273) = (1.40atm)(293mL)/(T 2 +273) 696°C Copyright © Cengage Learning. All rights reserved 38 EXERCISE!

39. Section 5.3 The Ideal Gas Law A sample of hydrogen gas(H 2 ) has a volume of 8.56L at a temperature of 0 ° C and a pressure of 1.5 atm. Calculate the moles of H 2 molecukes present in this gas sample 

40. Section 5.4 Gas Stoichiometry Molar Volume of an Ideal Gas  For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L.  STP = standard temperature and pressure  0°C and 1 atm  Therefore, the molar volume is 22.42 L at STP. Copyright © Cengage Learning. All rights reserved 40

41. Section 5.4 Gas Stoichiometry What is the volume (in liters) occupied by 49.8 g of HCl at STP? 

42. Section 5.4 Gas Stoichiometry A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O 2 are present? Pv= nRT (1.00atm)(2.40L)= n(0.08206)(273) N= 0.0112mol O 2 × 32.00g/mol 3.57 g Copyright © Cengage Learning. All rights reserved 42 EXERCISE!

43. Section 5.4 Gas Stoichiometry Molar Mass of a Gas  d = density of gas  T = temperature in Kelvin  P = pressure of gas  R = universal gas constant Copyright © Cengage Learning. All rights reserved 43

44. Section 5.4 Gas Stoichiometry What is the density of F 2 at STP (in g/L)? D = (MM)/(R)(T) D = (19.00g/mol×2)(1.00atm)/(0.08206)(273K) 1.70 g/L Copyright © Cengage Learning. All rights reserved 44 EXERCISE!

45. Section 5.4 Gas Stoichiometry A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/l at 36 c and 2.88 atm. Calculate the molar mass of the compound 

46. Section 5.5 Dalton’s Law of Partial Pressures  For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +...  The total pressure exerted by a mixture of gases is the sum of the partial pressure of each individual gas that is present. Copyright © Cengage Learning. All rights reserved 46

47. Section 5.5 Dalton’s Law of Partial Pressures Copyright © Cengage Learning. All rights reserved 47

48. Section 5.5 Dalton’s Law of Partial Pressures Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed.  After the valve is opened, what is the pressure of the helium gas? Copyright © Cengage Learning. All rights reserved 48 3.00 atm 3.00 L 2.00 atm 9.00 L EXERCISE!

49. Section 5.5 Dalton’s Law of Partial Pressures 27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C.  Calculate the new partial pressure of oxygen.P 1 V 1 =P 2 V 2 6.13 atm  Calculate the new partial pressure of helium. 2.93 atm  Calculate the new total pressure of both gases. 6.13 + 2.93 = 9.06 atm Copyright © Cengage Learning. All rights reserved 49 EXERCISE!

50. Section 5.6 The Kinetic Molecular Theory of Gases Mixture of helium and oxygen can be used in scuba diving tanks to help prevent “the bends” for a particular drive, 46 L He and 1.0 atm and 12 L O 2 at 25C and 1.0 atm were pumped into a tank with 5.0 L. calculate the partial pressure of each gas and total pressure in the thank at 25C. 

51. Section 5.6 The Kinetic Molecular Theory of Gases 

52. Section 5.6 The Kinetic Molecular Theory of Gases Postulates of the Kinetic Molecular Theory 1)The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). Copyright © Cengage Learning. All rights reserved 52

53. Section 5.6 The Kinetic Molecular Theory of Gases Postulates of the Kinetic Molecular Theory 2)The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. Copyright © Cengage Learning. All rights reserved 53

54. Section 5.6 The Kinetic Molecular Theory of Gases Postulates of the Kinetic Molecular Theory 3)The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. Copyright © Cengage Learning. All rights reserved 54

55. Section 5.6 The Kinetic Molecular Theory of Gases Postulates of the Kinetic Molecular Theory 4)The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Copyright © Cengage Learning. All rights reserved 55

56. Section 5.6 The Kinetic Molecular Theory of Gases Root-Mean- Square speed  One way to estimate molecular speed it to calculate the root-mean-square speed

57. Section 5.6 The Kinetic Molecular Theory of Gases Root Mean Square Velocity R = 8.3145 J/K·mol (J = joule = kg·m 2 /s 2 ) T = temperature of gas (in K) M = mass of a mole of gas in kg  Final units are in m/s. Copyright © Cengage Learning. All rights reserved 57

58. Section 5.7 Effusion and Diffusion  Diffusion – the mixing of gases.  Effusion – describes the passage of a gas through a tiny orifice into an evacuated chamber.  Rate of effusion measures the speed at which the gas is transferred into the chamber. Copyright © Cengage Learning. All rights reserved 58

59. Section 5.7 Effusion and Diffusion Graham’s Law of Effusion  M 1 and M 2 represent the molar masses of the gases. Copyright © Cengage Learning. All rights reserved 59