1. 1 Chapter 12 Gases

2. 2 Overview  Gas Laws –Gas pressure and its measurement –Empirical gas laws –Ideal gas laws –Stoichiometry and gases –Gas Mixtures; Law of Partial Pressures  Kinetic and Molecular Theory –Kinetic Theory of an Ideal Gas –Molecular speeds: diffusion and effusion –Real gases

3. 3 Measurements on Gases  The most readily measured properties of a gas are:  Temperature  Volume  Pressure

4. 4 Measuring Pressure  Pressure (P) is the force (F) that acts on a given area (A)  One of the most important of the measured quantities for gases  Pressure has traditionally been measured in units relating to the height of the Hg and is thus expressed as mm Hg = 1 Torr.

5. 5 Atmospheric Pressure and the Barometer  Due to gravity, the atmosphere exerts a downward force and therefore a pressure upon the earth's surface  Force = (mass*acceleration) or F=ma  The earth's gravity exerts an acceleration of 9.8 m/s 2  A column of air 1 m 2 in cross section, extending through the atmosphere, has a mass of roughly 10,000 kg

6. 6  Atmospheric pressure can be measured by using a barometer  A glass tube with a length somewhat longer than 760 mm is closed at one end and filled with mercury  The filled tube is inverted over a dish of mercury such that no air enters the tube  Some of the mercury flows out of the tube, but a column of mercury remains in the tube. The space at the top of the tube is essentially a vacuum  The dish is open to the atmosphere, and the fluctuating pressure of the atmosphere will change the height of the mercury in the tube

7. 7 The mercury is pushed up the tube until the pressure due to the mass of the mercury in the column balances the atmospheric pressure

8. 8 Standard Atmospheric Pressure  Standard atmospheric pressure corresponds to typical atmospheric pressure at sea level  It is the pressure needed to support a column of mercury 760 mm in height  In SI units it equals 1.01325 x 10 5 Pa

9. 9  Relationship to other common units of pressure:  (Note that 1 torr = 1 mm Hg)

10. 10  A manometer is used to measure the pressure of an enclosed gas. Their operation is similar to the barometer, and they usually contain mercury.  It consists of a tube of liquid connected to enclosed container which makes it possible to measure pressure inside the container.

11. 11  A closed tube manometer is used to measure pressures below atmospheric  An open tube manometer is used to measure pressures slightly above or below atmospheric  In a closed tube manometer the pressure is just the difference between the two levels (in mm of mercury)

12. 12  In an open tube manometer the difference in mercury levels indicates the pressure difference in reference to atmospheric pressure

13. 13 Manometer

14. 14

15. 15  Other liquids can be employed in a manometer besides mercury.  The difference in height of the liquid levels is inversely proportional to the density of the liquid i.e. the greater the density of the liquid, the smaller the difference in height of the liquid  The high density of mercury (13.6 g/ml) allows relatively small manometers to be built

16. 16 The Gas Laws  Boyle's Law: For a fixed amount of gas and constant temperature, PV = constant.

17. 17  The volume of some amount of a gas was 1.00 L when the pressure was 10.0 atm; what would the volume be if the pressure decreased to 1.00 atm?

18. 18 The Gas Laws  Charles's Law: at constant pressure the volume is linearly proportional to temperature. V/T = constant

19. 19 A gas occupied a volume of 6.54 L at 25°C what would its volume be at 100°C?

20. 20 The Gas Laws  Avagadro’s law for a fixed pressure and temperature, the volume of a gas is directly proportional to the number of moles of that gas. V/n = k = constant.

21. 21 The volume of 0.555 mol of some gas was 100.0 L; what would be the volume of 15.0 mol of the same gas at the same T and P?

22. 22  The three historically important gas laws derived relationships between two physical properties of a gas, while keeping other properties constant:

23. 23  These different relationships can be combined into a single relationship to make a more general gas law:

24. 24  If the proportionality constant is called "R", then we have:

25. 25  Rearranging to a more familiar form:  This equation is known as the ideal-gas equation

26. 26  Values for R are determined by the values used for volume and pressure. The value that we will use is 0.0821 l atm/mole K

27. 27  When any of the other three quantities in the ideal gas law have been determined the last one can be calculated.

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29. 29  Calculate the pressure inside a TV picture tube, if it's volume is 5.00 liters, it's temperature is 23.0  C and it contains 0.0100 mg of nitrogen.

30. 30 Further Applications of Ideal-Gas Equation  The density of a gas the density of a gas can be related to the pressure from the ideal gas law using the definition of density: d = mass/vol.

31. 31 Estimate the density of air at 20.0  C and 1.00 atm by supposing that air is predominantly N 2.

32. 32  Rearrangement permits the determination of molecular mass of a gas from a measure of the density at a known temperature and pressure.

33. 33 A certain gas was found to have a density of 0.480 g/L at 260  C and 103 Torr. Determine the MM of the compound.

34. 34 Partial Pressure and Dalton’s Law  Dalton's Law = the sum of the partial pressures of the gases in a mixture = the total pressure or P = P A + P B + P C +...where P i = the partial pressure of component i.

35. 35  Dalton found that gases obeying the ideal gas law in the pure form will continue to act ideally when mixed together with other ideal gases.  The individual partial pressures are used to determine the amount of that gas in the mixture, not the total pressure, P A = n A RT/V.  Since they are in the same container T and V will be the same for all gases.

36. 36 1.00 g of air consists of approximately 0.76 g nitrogen and 0.24 g oxygen. Calculate the partial pressures and the total pressure when this sample occupies a 1.00 L vessel at 20.0  C. 1.00 g of air consists of approximately 0.76 g nitrogen and 0.24 g oxygen. Calculate the partial pressures and the total pressure when this sample occupies a 1.00 L vessel at 20.0  C.

37. 37 Partial Pressure and Dalton’s Law2  Mole fraction another quantity commonly determined for gas mixtures. It is defined the number of moles of one substance relative to the total number of moles in the mixture or  X can be calculated from –moles of each gas in the mixture or –the pressures of each gas

38. 38 Gas Collection by Water Displacement  Certain experiments involve the determination of the number of moles of a gas produced in a chemical reaction  Sometimes the gas can be collected over water  Potassium chlorate when heated gives off oxygen:  2KClO 3 (s) -> 2KCl(s) + 3O 2 (g)  The oxygen can be collected in a bottle that is initially filled with water

39. 39

40. 40  The volume of gas collected is measured by first adjusting the beaker so that the water level in the beaker is the same as in the pan.  When the levels are the same, the pressure inside the beaker is the same as on the water in the pan (i.e. 1 atm of pressure)  The total pressure inside the beaker is equal to the sum of the pressure of gas collected and the pressure of water vapor in equilibrium with liquid water Pt = PO 2 + PH 2 O

41. 41 Suppose KClO 3 was decomposed according to 2 KClO 3 (s)+   2KCl(s) + 3O 2 (g). P T = 755.2 Torr and 370.0 mL of gas was collected over water at 20.0  C. Determine the number of moles of O 2 if the vapor pressure of water is 17.5 torr at this temperature.

42. 42

43. 43 Stoichiometric Relationships with Gases  When gases are involved in a reaction, gas properties must be combined with stoichiometric relationships.  Two types exist –Volume of gas and volume of gas –Condensed phase and volume of gas

44. 44  Determine the volume of oxygen gas needed to react with 1.00 L of hydrogen gas at the same temperature and pressure to produce water.

45. 45

46. 46 Determine the volume of gas produced at 273.15 K and 1.00 atm if 1.00 kg of calcium oxide reacts with a sufficent amount of carbon. Assume complete reaction (i.e. 100% yield) CaO(s) + 3C(s)  CaC 2 (s) + CO(g).

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48. 48 The Behavior of Real Gases  The molar volume is not constant as is expected for ideal gases.  These deviations due to an attraction between some molecules.  Applicable at high pressures and low temperatures.

49. 49  For compounds that deviate from ideality the van der Waals equation is used: where a and b are constants that are characteristic of the gas.

50. 50 The Kinetic Molecular Theory of Gases  Microscopic view of gases assumes that – A gas is a collection of molecules (atoms) in continuous random motion. – The molecules are infinitely small point- like particles that move in straight lines until they collide with something.

51. 51 KMT cont. –Gas molecules do not influence each other except during collision. –All collisions are elastic; the total kinetic energy is constant at constant T. –Average kinetic energy is proportional to T.

52. 52 The Kinetic Theory – Molecular Theory of Gases  Theory leads to a description of bulk properties i.e. observable properties.  The average kinetic energy of the molecule is where N A = Avagadro’s number.

53. 53  Average kinetic energy of moving particles can also be obtained from where u = average velocity

54. 54 Combine 1 & 2 to get a relationship between the velocity, temperature and molecular mass.

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56. 56 Determine average velocity of He at 300 K.

57. 57 Predict the ratio of the speeds of a gas if the temperature is increased from 300 K to 450 K.

58. 58 Graham’s Law: Diffusion and Effusion of Gases  Diffusion the process whereby a gas spreads out through another gas to occupy the space with uniform partial pressure.  Effusion the process in which a gas flows through a small hole in a container.

59. 59  Graham’s Law of Effusion the rate of effusion of gas molecules through a hole is inversely proportional to the square root of the molecular mass of the gas at constant temperature and pressure.

60. 60 Graham’s Law for Two Gases

61. 61 Determine the molecular mass of an unknown compound if it effused through a small orifice 3.55 times slower than CH 4.

62. 62 A compound with a molecular mass of 32.0 g/mol effused through a small opening in 35 s; determine the effusion time for the same amount of a compound with a molecular mass of 16.0.

63. 63