1. Chapter 6

2. Fig. 6-CO, p. 208

3. p. 242

4. What do SCUBA and airbags have to do with gases? Let’s find out…..

5. 6.1 Properties and Measurements of Gases Objectives  To describe the characteristics of the three states of matter – solid, liquid, and gas  To define the pressure of a gas and the units in which it is measured

6.  A solid has fixed shape and volume. Solid Phase Solid Br 2 at low temperature

7.  A liquid has fixed volume but no definite shape. The density of a solid or a liquid is given in g/mL. Liquid Phase Liquid Br 2

8.  A gas has no fixed volume or definite shape. The density of a gas is given in g/L whereas liquids and solids are in g/mL. Can expand or be compressed depending on the pressure exerted on a system Gas Phase Gaseous Br 2

9. 6.1 Properties and Measurements of Gases cont’d  Condensed phases Includes solids & liquids because they are resistant to changes in volume caused by changes in pressure The density of gases can increase 1000-fold or more when compressed

10. 6.1 Properties and Measurements of Gases cont’d  Pressure of a Gas = the force per unit area exerted on a surface Ex: the atmosphere exerts a pressure due to the weight of gas molecules in the air ○ Air pressure at higher elevations is lower due to fewer gas molecules above you than at sea level A barometer measures the pressure of the atmosphere A manometer measures pressure differences

11.  The pressure of the atmosphere is measured with a barometer. Pressure of a Gas

12.  Both open and closed end manometers measure pressure differences. Manometers P gas < P atm P gas > P atm h = P gas

13. 6.1 Properties and Measurements of Gases cont’d  Units of Pressure Measurement SI unit is the pascal (Pa) 1 Pa = 1N/m 2 or 1 kg/ms 2 (N = newton; m = meter, s = seconds) This unit of pressure is very small for experiments typically carried out by chemists Typically see other units used as shown in Table 6.1 (p. 212 in TB) 1 atm of pressure = the normal pressure at sea level which is defined as the pressure exerted on a column of mercury 760 mm high

14. 6.1 Properties and Measurements of Gases cont’d Relationships Between Pressure Units 1 atm = 760 mm Hg 1 atm = 760 mm Hg 1 torr = 133.3 Pa 1 torr = 133.3 Pa 1 atm = 760 torr 1 atm = 760 torr 1 atm = 14.7 psi 1 atm = 14.7 psi 1 atm = 101.325 kPa 1 atm = 101.325 kPa 1 atm = 1.01325 bar 1 atm = 1.01325 bar 1 atm = 29.92 in Hg 1 atm = 29.92 in Hg

15. 6.1 Properties and Measurements of Gases cont’d Express 433 torr in atmospheres Answer: 0.570 atm Express 0.450 atm in kPa Answer: 45.6 kPa

16. 6.2 Gas Laws Objectives  Determine how a gas sample responds to changes in volume, pressure, moles, and temperature

17. Why are we interested in determining how a gas sample responds to changes in one or more physical properties?  What if you lived near a gas storage tank? … Would you want to know how a change in temperature would affect the pressure in the metal tank that stores gaseous materials? Would you want to know if the tank is designed to withstand such an increase in pressure?  So let’s learn how to assess how gases behave….

18. 6.2 Gas Laws  The physical properties of all gases behave in the same general manner, regardless of the identity of the gas  Four independent properties define the physical state of a gas Pressure Volume Temperature # of moles

19.  Boyle’s Law is the relationship between volume & pressure (at a constant temperature and amount of gas)  Increasing the pressure on a gas sample, by addition of mercury to an open ended manometer, causes the volume of the gas to decrease. The Effect of Pressure on Gas Volume

20.  A plot of volume versus 1/P is a straight line.  V = k 1 x 1 P can be rewritten as PV = constant Boyle’s Law

21. 6.2 Gas Laws cont’d  Boyle’s Law cont’d Can predict what will happen to the pressure or volume of a gas if its volume or pressure changes (when T and n are held constant) Since P 1 V 1 = k 1 and P 2 V 2 = k 1 then P 1 V 1 = P 2 V 2

22. 6.2 Gas Laws, cont’d Boyle’s Law cont’d Example: A balloon containing 575 mL nitrogen gas at a pressure of 1.03 atm is compressed to a final volume of 355 mL. What is the resulting pressure of the nitrogen? A: 1.67 atm

23. 6.2 Gas Laws, cont’d Boyle’s Law cont’d Example : In the lungs of a deep-sea diver (V= 6.0 L) the pressure of the air is 7400 torr. At a constant temperature of 37°C, to what volume would the air expand if the diver were immediately brought to the surface (1.0 atm) A: 58 L

24. Fig. 6-7, p. 216 The Effect of Temperature on Gas Volume

25.  A plot of volume versus temperature is a straight line.  Extrapolation to zero volume yields absolute zero in temperature: -273 o C.  V = k 2 x T, where T is given in units of kelvin. *** Charles’s Law

26. 6.2 Gas Laws cont’d  Charles’ Law V = k 2 x T ○ Note temperatures must be in degrees Kelvin!!!! ○ can be rewritten as V/T = constant Can predict what will happen to the temperature or volume of a gas if its volume or temperature changes (when P and n are held constant) Use V 1 /T 1 = constant = V 2 /T 2

27. Problem Solving Using Charles’ Law The volume of a sample of nitrogen gas increases from 0.440 L at 27°C to 1.01 L as it is heated to a new temperature. Calculate the new temperature of the nitrogen gas sample. A: 416°C

28. 6.2 Gas Laws cont’d  Avagadro’s Law At a constant temperature and pressure, equal volumes of gases contain the same number of particles (independent of the type of gas) Avagadro’s Law:V = k 3 x n

29. Avogadro’s Hypothesis Both containers are of equal volume, contain the same # of moles of gas, are at the same pressure, but the mass of the gases is different.

30.  A plot of the volume of all gas samples, at constant T and P, vs. the number of moles (n) of gas is a straight line.  V = k 3 x n Avogadro’s Law

31. 6.2 Gas Laws cont’d  The Combined Gas Law For a given amount of gas, n, the three remaining properties of a gas can be related by the combined gas law P 1 V 1 = P 2 V 2 T 1 T 2 *** Remember, temperatures must be in degrees Kelvin!

32. Practice Problem Using the Combined Gas Law 1. The pressure of a sample of gas is 2.6 atm in a 1.54 L container at a temperature of 0°C. Calculate the pressure exerted by this sample if the volume changes to 1.00 L and the temperature changes to 27°C. A1: 4.40 atm 2. A sample of a gas occupies 4.0 L at 25 ° C and 2.0 atm of pressure. Calculate the volume at STP (T = 0 ° C, P = 1 atm). A2: 7.3 L

33. If you aren’t given gas law relationships, what can you do…..  If you ever forget any of the gas law relationships, including the combined gas law, you can use the ideal gas law to help you out.  Remember P 1 V 1 = n 1 RT 1 & P 2 V 2 = n 2 RT 2  To solve problems, the ideal gas law is rearranged so variable quantities are on the left side and constant quantities are listed on the right. Then the two left hand sides of the resulting equations are set equal to one another and you solve for the unknown quantity  Let’s practice in class……

34. 6.3 The Ideal Gas Law Objectives  Write the ideal gas law  Calculate the pressure, volume, amount, or temperature of a gas given the other three properties  Calculate the molar mass and the density of gas samples by using the ideal gas law

35.  The ideal gas law combines the three gas laws into a single equation: PV = nRT where: R = 0.08206 L. atm/mol. K (R determined experimentally)  All common gases follow the ideal gas law at normal temperatures and pressures  FYI: The volume of one mole of an ideal gas at STP is 22.4 L. This is a commonly used conversion factor that might come in handy. STP = 1 atm, 0°C Ideal Gas Law

36. Table 6-2, p. 219

37. 6.3 The Influences of Changing Conditions on Gases  Q: In A balloon filled with oxygen gas at 25°C occupies a volume of 2.1 L. Assuming that the pressure remains constant, what is the volume at 100°C?  A: 2.6 L  Q: The volume of a sample of nitrogen gas at 27°C increases from 0.440 L to 1.101 L when the sample is heated. What is the second temperature of the nitrogen sample?  A: 416°C

38. Practice Problems Using the Ideal Gas Law  Calculate the number of moles of argon gas in a 30.0 L container at a pressure of 10.0 atm and temperature of 298 K.  A: 12.3 mol

39. Using the Ideal Gas Law to Calculate Molar Mass and Density  The ideal gas law (PV=nRT) can be used to calculate the density (mass/volume) and molar mass (mass/moles) of a gas.  Since the ideal gas law does not contain a variable for mass, moles must be calculated first to help determine a molar mass or n/V must be calculated first to determine a density.  Let’s see how …..

40. Calculating Density from the Ideal Gas Law  Calculate the density of CO 2 (g) at STP. Strategies to consider: ○ Remember that the density of gases is in units of grams/L ○ Given PV = nRT, you can calculate n/V in terms of mol/L. ○ Then you can use the molar mass of the given gas to calculate density: mol/L x g/mol = g/L  A: 1.96 g/L  Please Note: At constant pressure and temperature the density of a gas is proportional to its molar mass, so the higher the molar mass, the greater the density of the gas.

41. Calculating Molar Mass from the Ideal Gas Law  Calculate the molar mass of a gas if a 1.02 g sample occupies 220 mL at 95.0 ° C and a pressure of 750 torr. Strategies to consider: ○ Remember that molar mass = grams/mol ○ You are given grams, so if you calculate moles from PV=nRT you have everything you need to calculate molar mass  A: 142 g/mol

42. Using the Ideal Gas Law to Calculate Molar Mass  A 4.25 g sample of an ideal gas occupies a volume of 1250 mL at 100.0°C and a pressure of 707 torr. What is the molar mass of the gas?  A: 112 g/mol  A gas has a density of 0.725 g/L and a pressure of 1.033 atm. What is the molar mass of the gas?  A: 17.0 g/mol

43. 6.4 Stoichiometry Calculations Involving Gases Objectives  Perform stoichiometric calculations for reactions in which some or all of the reactants or products are gases  Use relative volumes of gases directly in equation stoichiometry problems

44. Gases and Chemical Equations  The ideal gas law can be used to determine the number of moles, n, for use in problems involving reactions.  The ideal gas law relates n to the volume of gas just as molar mass is used with masses of solids and molarity is used with volumes of solutions.

45. Fig. 6-11, p. 224

46. Example: Gases with Equations  Calculate the volume of O 2 gas formed in the decomposition of 2.21 g of KClO 3 at STP. 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

50. 6.5 Dalton’s Law of Partial Pressure Objectives  Use Dalton’s law of partial pressure in calculations involving mixtures of gases  Calculate the partial pressure of a gas in a mixture from its mole fractions

51.  The pressure exerted by each gas in a mixture is called its partial pressure.  For a mixture of two gases A and B, the total pressure, P T, is  P T = P A + P B Dalton’s Law of Partial Pressure

52. Pressure of a Mixture of Gases

53. Solving Partial Pressures Problems  Calculate the pressure in a container that contains O 2 gas at a pressure of 3.22 atm and N 2 gas at a pressure of 1.29 atm.  A: 4.51 atm

54. 6.5 Dalton’s Law of Partial Pressure Mole Fraction  Mole fraction ( , chi) is the number of moles of one component of a mixture divided by the total number of moles of all substances present in the mixture.   A  +  B +  C = 1  The partial pressure of any gas, A, in a mixture is given by: P A =  A x P T

55.  Mole fraction of the yellow gas is 3/12 = 0.25 and the mole fraction of the red gas is 9/12 = 0.75 Mole Fraction

56. Solving Partial Pressure Problem s …..  Calculate the partial pressure of Ar gas in a container that contains 2.3 mol of Ar and 1.1 mol of Ne and is at a total pressure of 1.4 atm.  A: 0.95 atm

57.  Now let’s use these skills to solve some practical problems commonly encountered in the chemistry lab….

58.  Water vapor is also present in a sample of O 2 gas collected over water. (Fig 6.14 p. 228) Collecting Gases Over Water

59. 6.5 Dalton’s Law of Partial Pressure  Collecting Gases by Water Displacement Gas samples collected by water displacement are not pure because some water molecules are also present in the gas phase The total pressure is equal to the gas pressure plus the pressure of the water vapor To calculate pressure due to gas alone you must subtract the partial pressure of water vapor from the total pressure Pressures due to water vapor at different temperatures are given in Table 6.3 on p. 228

60. 6.5 Dalton’s Law of Partial Pressure  Q: A sample of KClO 3 is heated and decomposes to produce O 2 gas. The gas is collected by water displacement at 26°C. The total volume of the collected gas is 229 mL at a pressure equal to the measured atmospheric pressure, 754 torr. How many moles of O 2 form?  A: 8.95 x 10 -3 mol O 2  Calculate the number of moles of hydrogen produced by the reaction of sodium with water. In the reaction, 1.3 L of gas are collected by water displacement at 26C. The atmospheric pressure is 756 torr.  A: 0.051 mol H 2

61. Example: Collecting Gases  Sodium metal is added to excess water, and H 2 gas produced in the reaction is collected over water with the gas volume of 1.2 L. If the pressure is 745 torr and the temperature 26 o C, what was the mass of the sodium? The vapor pressure of water at 26 o C is 25 torr. 2Na(s) + 2H 2 O( l )  H 2 (g) + 2NaOH(aq)

62. 6.6 The Kinetic Molecular Theory of Gases Objectives  Show that the predictions of the kinetic molecular theory are consistent with experimental observations  Sketch a Maxwell-Boltzmann distribution curve for the distribution of speeds of gas molecules  Perform calculations using the relationships among molecular speed and the temperature and molar mass of a gas.

63.  How can a single law (the ideal gas law) explain the behavior of all gases, regardless of the nature or size of the gas particles?

64. 6.6 The Kinetic Molecular Theory of Gases  The four postulates of kinetic molecular theory are: 1. A gas consists of small particles that are in constant and random motion 2. Collisions of gas particles with each other and with the walls of the container are elastic – there is no loss in the total kinetic energy when the particles collide – and no attractive or repulsive forces exist between the gases

65. 6.6 The Kinetic Molecular Theory of Gases  The four postulates of kinetic molecular theory are cont’d : 3. Gas particles are very small compared to the average distance that separates them 4. The average kinetic energy of gas particles is proportional to the temperature on the Kelvin scale

66. 6.6 The Kinetic Molecular Theory of Gases  The relationship between kinetic energy and the speed of the particles is given by: KE = ½ mu 2

67.  Gas particles move at different speeds.  Average speed is called the root mean square (rms) speed, u rms, and is the square root of the average squared speed. Average Speed of a Gas Maxwell-Boltzmann distribution curves

68. R = 8.314 J/mol. K; molar mass in kilograms per mole Average Speed of a Gas

70. 6.7 Diffusion & Effusion Objectives  Define diffusion and effusion  Calculate relative rates of effusion of two gases and use the data to calculate molar masses

71. 6.7 Diffusion & Effusion  Diffusion the mixing of particles due to motion the faster the molecular motion, the faster a gas diffuses Rate of diffusion is less than the root mean speed (rms) of the gas because collisions prevent particles from moving in a straight line

72. 6.7 Diffusion & Effusion Effusion Effusion The passage of a gas through a small hole into an evacuated space Graham’s Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass Graham’s Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass

73. 6.7 Diffusion & Effusion  Graham’s Law can be used to determine the molar mass of an unknown gas by measuring the times needed for equal volumes of a known gas and an unknown gas to effuse through the same small hole at constant pressure and temperature  Note that the rate of effusion is inversely related to time so Graham’s Law can be rewritten as: t B = molar mass of B t A √ molar mass of A

74. 6.7 Diffusion & Effusion  Q: Calculate the molar mass of a gas if equal volumes of nitrogen and the unknown gas take 2.2 and 4.1 min, respectively, to effuse through the same hole under conditions of constant temperature and pressure  A: 97 g/mol  Q: Calculate the molar mass of a gas if equal volumes of oxygen gas and the unknown gas take 3.25 and 8.41 min, respectively, to effuse through the same hole under conditions of constant temperature and pressure.  A: 214 g/mol

75. 6.8 Deviations from Ideal Behavior Objectives  Explain why gases deviate from the ideal gas law under certain conditions  Use the van der Waals equation to account for deviations from the ideal gas law

76. 6.8 Deviations from Ideal Behavior  Most gases obey the ideal gas law closely at a pressure of about 1 atm and a temperature well above the boiling point of the substance in other words at low pressures & temperatures 100°C above the BP of the substance, far from conditions under which the gases would condense to a liquid  Gases at high pressures do not behave as predicted by the ideal gas law

77.  Gases deviate from the ideal gas law at high pressures. Deviations from Ideal Behavior

78.  A gas deviates from ideal gas behavior at low temperatures (near the condensation point ) and high pressures.  Behavior of O 2 is shown Deviations from Ideal Behavior

79. 6.8 Deviations from Ideal Behavior  Deviations from the ideal gas law occur under extreme conditions because two assumptions of the kinetic molecular theory are not correct when gas molecules are close together.  These assumptions are: (a) that gas particles are small compared to the distances separating them (b) that there are no attractive forces between gas particles

80.  The assumption that gas particles are small compared to the distances separating them fails at high pressures.  The observed value of PV/nRT will be greater than 1 under these conditions. Deviations from Ideal Behavior

81.  The forces of attraction between closely spaced gas molecules reduce the impact of wall collisions.  These attractive forces cause the observed value of PV/nRT to decrease below the expected value of 1 at moderate pressures. Forces of Attraction in Gases

82. 6.8 Deviations from Ideal Behavior  The van der Waals equation is used to describe gases under non-ideal conditions : (P + an 2 /V 2 )(V-nb) = nRT To correct for the volume occupied by the gas we subtract the term nb from the volume where n is the # of moles of a gas and b is a constant that depends on the gas

83. 6.8 Deviations from Ideal Behavior  The van der Waals equation: (P + an 2 /V 2 )(V-nb) = nRT To correct for attractive forces between gas particles we add the term an 2 /V 2 to the pressure where a is a constant related to the strength of the attractive forces, n= # of moles of the gas, and V = the volume of the gas sample Note that the values of a and b are derived experimentally and are different for each gas

84. 6.9 Deviations from Ideal Behavior van der Waals constants Gas a (atmL 2 /mol 2 b (L/mol) H2H2H2H20.2440.0266 He0.0340.0237 Ne0.2110.0171 H2OH2OH2OH2O5.460.0305 NH 3 4.170.0371 CH 4 2.250.0428 N2N2N2N21.390.0391 O2O2O2O21.360.0318 Ar1.340.0322 CO 2 3.590.0427

85. 6.8 Deviations from Ideal Behavior  Q: Calculate the pressure of 2.01 mol of gaseous H 2 O at 400°C in a 2.55 L container, using the ideal gas law and the van der Waal’s equation.  A: Ideal gas law = 43.6 atm  A: van der Waal’s = 41.2 atm  Q: Calculate the pressure of 0.223 mol of ammonia gas at 30.0°C in a 3.23 L container, using the ideal gas law and the van der Waal’s equation  A: Ideal gas law = 1.72 atm  A: van der Waal’s = 1.70 atm

86. Fig. 6-16, p. 230