1. Chapter 1 Gases Dr. Hisham E Abdellatef Professor of pharmaceutical analytical chemistry 2011 - 2012

2. Characteristics of Gases Expand to fill and assume the shape of their container Compressible Next

3. Diffuse into one another and mix in all proportions. homogeneous mixtures Particles move from an area of high concentration to an area of low concentration. Next

4. PropertiesProperties that determine physical behavior of a gas 1. VOLUME 2. PRESSURE 3. TEMPERATURE

5. 1. Volume l x w x h π r²h capacity of the container enclosing it. (m 3 ), (dm 3 ), or liter. For smaller volumes (cm 3 ), (ml).

6. 2. Temperature Temperature: Three temperature scales Fahrenheit (ºF) Celsius (ºC) Kelvin K (no degree symbol) K performing calculations with the gas law equations.

7. gas Expand when heated

8. Temperature K = (ºC) + 273.15 Example: ºC = 20º K = 293.15 Absolute or Kelvin Scale: =-273.15 (ºC as its zero). (T when V= 0 ) 1 K = 1 ºC

9. 3. Pressure The molecules in a gas are in constant motion. gaseous atoms that collide with each other and the walls of the container. "Pressure" is a measure of the collisions of the atoms with the container.

10. Pressure Force per unit area Equation: P = F/A F = force A = area Next

11. Barometer -Atmospheric pressure is measured with a barometer. Height of mercury varies with atmospheric conditions and with altitude.

12. Mercury Barometer Measurement of Gas Pressure

14. -Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column. -There are several units used for pressure: -Pascal (Pa), N/m 2 -Millimeters of Mercury (mmHg) -Atmospheres (atm)

15. Manometers Used to compare the gas pressure with the barometric pressure. Next

17. Types of Manometers Closed-end manometer The gas pressure is equal to the difference in height (Dh) of the mercury column in the two arms of the manometer

18. Open-end Manometer The difference in mercury levels (D h ) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure

19. Three Possible Relationships 1.Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal. P gas = P bar

21. 2.Gas pressure is greater than the barometric pressure. ∆P > 0 P gas = P bar + ∆P

22. 3.Gas pressure is less than the barometric pressure. ∆P < 0 P gas = P bar + ∆P

23. The Simple Gas Laws 1. Boyle’s Law 2. Charles’ Law 3. Gay-Lussac’s Law 4. Combined Gas Law

24. variables required to describe a gas Amount of substance: moles Volume of substance: volume Pressures of substance: pressure Temperature of substance: temperature

25. 25 The Pressure-Volume Relationship: Boyle’s Law Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

27. The Pressure-Volume Relationship: Boyle’s Law

28. 28 Example An ideal gas is enclosed in a Boyle's-law apparatus. Its volume is 247 ml at a pressure of 625 mmHg. If the pressure is increased to 825 mmHg, what will be the new volume occupied by the gas if the temperature is held constant?

29. Solution Method 1: P 1 V 1 = P 2 V 2 or, solving for V 2 the final volume

30. Solution Method 2: The pressure of the gas increases by a factor 825 / 625, the volume must decrease by a factor of 625 / 825 V 2 = V 1 X (ratio of pressures)

31. Charles’ Law The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. V 1 = V 2 T 1 T 2 or V 1 T 2 = V 2 T 1

34. Example. Charles’ Law A 4.50-L sample of gas is warmed at constant pressure from 300 K to 350 K. What will its final volume be? Given: V 1 = 4.50 L T 1 = 300. K T 2 = 350. K V 2 = ? Equation: V 1 = V 2 T 1 T 2 or V 1 T 2 = V 2 T 1 (4.50 L)(350. K) = V 2 (300. K) V 2 = 5.25 L

35. Gay-Lussac’s Law The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. P 1 = P 2 T 1 T 2 or P 1 T 2 = P 2 T 1

36. On the next slide The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.

38. Combined Gas Law Pressure and volume are inversely proportional to each other and directly proportional to temperature. P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1

39. Example. Combined Gas Law A sample of gas is pumped from a 12.0 L vessel at 27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure? Given: P 1 = 760 Torr P 2 = ? V 1 = 12.0 L V 2 = 3.5 L T 1 = 300 K T 2 = 325 K Equation: P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1 (760 Torr)(12.0 L)(325 K) = ( P 2) (3.5 L)(300 K) P 2 = 2.8 x 10³ Torr

40. Avogadro’s Law Volume & Moles

41. Avogadro’s Law At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V = c · n V = volume c = constant n= # of moles Doubling the number of moles will cause the volume to double if T and P are constant.

42. Ideal Gas Law

43. Equation Includes all four gas variables: Volume Pressure Temperature Amount of gas Next

44. PV = nRT Gas that obeys this equation if said to be an ideal gas (or perfect gas). No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures. Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K R = 0.082058 L·atm/mol· K

45. Example Suppose 0.176 mol of an ideal gas occupies 8.64 liters at a pressure of 0.432 atm. What is the temperature of the gas in degrees Celsius?

46. Solution PV = nRT To degrees Celsius we need only subtract 273 from the above result: =258 - 273 = -15 O C

47. Example Suppose 5.00 g of oxygen gas, O 2, at 35 °C is enclosed in a container having a capacity of 6.00 liters. Assuming ideal-gas behavior, calculate the pressure of the oxygen in millimeters of mercury. (Atomic weight: 0 = 16.0)

48. Solution One mole of O 2 weighs 2(16.0) = 32.0 g. 5.00 g of O 2 is, therefore, 5.00 g/32.0 g mol- 1, or 0.156 mol. 35 °C is 35 + 273 = 308 K PV = nRT

49. Molar volume of an ideal gas at STP The volume occupied by one mole, or molar volume, of an ideal gas at STP is

50. Applications of the Ideal Gas Law

51. Dalton’s Law of Partial Pressure Mixture of Gases

52. Total Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture. P total = P A + P B + ……

53. Total Pressure: Mixture of Gases

54. Example: Gas Mixtures & Partial Pressure A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel? Step 1: n O2 = 6.00 g O 2 x 1 mol O 2 32 g O 2 = 0.188 mol O 2 Next ---- >

55. n CH4 = 9.00 g CH 4 x 1 mol CH 4 16.0 g CH 4 = 0.563 mol CH 4 Step 2: Calculate pressure exerted by each P O2 = nRT V = (0.188 mol O 2 )(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.281 atm

56. P CH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.841 atm Step 3: Add pressures P total = P O2 + P CH4 = 0.281 atm + 0.841 atm P total = 1.122 atm

57. Try ? Q1, Q2, Q3 page 42(lecturer note )

58. Graham’s Law Molecular Effusion and Diffusion

59. Graham’s Law of Effusion Effusion – The escape of gas through a small opening. Diffusion – The spreading of one substance through another.

60. Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Graham’s Law of Effusion

61. Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Graham’s Law of Effusion

62. Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Gas escaping from a balloon is a good example. Graham’s Law of Effusion

63. Chapter 1063 Molecular Effusion and Diffusion Graham’s Law of Effusion

64. Example 1. Rate of Effusion The rate of effusion of an unknown gas (X) through a pinhole is found to be only 0.279 times the rate of effusion of hydrogen (H 2 ) gas through the same pinhole, if both gases are at STP. What is the molecular weight of the unknown gas? (Atomic weight; H = 1.01.) Next

65. Solution M x = 26.0

66. Pv = nRt n=m mass M d Pv= m RT PM = m RT M V PM = dRT or d = PM RT

67. Example2. Rate of Effusion Calculate the ratio of the effusion rates of hydrogen gas (H 2 ) and uranium hexafluoride (UF 6 ), a gas used in the enrichment process fuel for nuclear reactors. Known: Molar Masses H 2 = 2.016 g/mol UF 6 = 352.02 g/mol (Rate of effusion)² = M U compound M H gas = 352.02 2.016 Rate of effusion = 13.21

68. Quize 1. what is the relative rates of diffusion of H 2 and CO 2 under the same condition ? 2. What is the density of gas which it’s diffusion is 1.414 times of the rate of diffusion of CO 2 at STP ???.

69. Problem Calcium hydride, CaH 2, reacts with water to form hydrogen gas: CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g) How many grams of CaH 2 are needed to generate 10.0L of H 2 gas if the pressure of H 2 is 740 torr at 23 o C?

70. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g) -Calculate moles of H 2 formed -Calculate moles of CaH 2 needed -Convert moles CaH 2 to grams

71. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

72. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

73. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

74. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

75. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

76. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

77. Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

78. Kinetic-Molecular Theory of Gases

79. Kinetic-Molecular Theory -Theory developed to explain gas behavior -To describe the behavior of a gas, we must first describe what a gas is: –Gases consist of a large number of molecules in constant random motion. –Volume of individual molecules negligible compared to volume of container. –Intermolecular forces (forces between gas molecules) negligible. –Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature. –Average kinetic energy of molecules is proportional to temperature.

80. Nonideal (Real) Gases

82. Gases may be 1. Ideal gas on obeys the gas law PV = constant (at constant T) 2.Real gas Deviate (not obey to gas law PV ≠ Constant)

83. Real gas deviate from ideal gases when (T is very low and P is very high). two factors : Real gases posse’s attractive forces between molecules. Every molecule in a real has a real volume.

85. True volume (Real) = V container – non copressiable volume (b) = V- b for 1 mole Actual volume = V – nb ( for n mole)

86. force of attraction between the molecules of gases: (pressure) 1 mole n moles

87. van der Waals Equation

88. Equation corrects for volume and intermolecular forces (P + n²a/V²)(V-nb) = nRT n²a/V² = related to intermolecular forces of attraction n²a/V² is added to P = measured pressure is lower than expected a & b have specific values for particular gases V - nb = free volume within the gas

89. EXAMPLE Calculate the pressure exerted by 10.0 g of methane, CH 4, when enclosed in a 1.00-liter container at 25 °C by using (a) the ideal-gas law and (b) the van der Waals equation. Gasa (L 2.atm/mol 2 )b (L/mol) CO 2 3.6580.04286 Ethane C 2 H 6 5.5700.06499 Methane CH 4 2.250.0428 Helium He0.03460.0238 Hydrogen H 2 0.24530.02651 Oxygen O 2 1.3820.03186 Sulfur dioxide SO 2 6.8650.05679

90. The molecular weight of CH 4 is 16.0; so n, the number of moles of methane, is 10.0 g/16.0 g mol -1, or 0.625 mol. (a) Considering the gas to be ideal and solving for P, we obtain (b) Treating the gas as a Van der Waals gas and solving for P, we have

91. Try ? Q1, Q2, Q3 page 50 (lecturer note )