Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

9,043 views

8,757 views

8,757 views

Published on

No Downloads

Total views

9,043

On SlideShare

0

From Embeds

0

Number of Embeds

1,169

Shares

0

Downloads

106

Comments

0

Likes

3

No embeds

No notes for slide

- 1. Gases <ul><li>AP Chemistry Unit 5: Chapter 5 </li></ul>
- 2. Gases <ul><li>How can we explain the behavior of Gases? </li></ul>
- 3. Intro <ul><li>Earth ’s atmosphere is a gaseous solution that consists mainly of nitrogen (N 2 ) and oxygen (O 2 ). This atmosphere supports life and acts as a waste receptacle for many industrial processes. The chemical reactions that follow often lead to various types of pollution, including smog and acid rain. </li></ul>
- 4. Intro <ul><li>The gases in the atmosphere also shield us from harmful radiation from the sun and keep the earth warm by reflecting heat radiation back toward the earth. In fact, there is now great concern that an increase in atmospheric carbon dioxide, a product of the combustion of fossil fuels , is causing a dangerous warming of the earth. </li></ul>
- 5. Pressure
- 6. Pressure <ul><li>Gas uniformly fills a container, is easily compressed, and mixes completely with any other gas. One of the most important properties is that it exerts pressure on its surroundings equally. </li></ul>
- 7. Pressure <ul><li>Incredible Tank Implosion </li></ul>
- 8. Pressure
- 9. Pressure
- 10. Pressure
- 11. Pressure <ul><li>Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). </li></ul><ul><ul><li>Torricelli ’s barometer is constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury. </li></ul></ul>
- 12. Pressure <ul><li>Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). </li></ul><ul><ul><li>Notice that a large quantity of mercury stays in the tube. In fact, at sea level the height of this column of mercury averages 760 mm. </li></ul></ul>
- 13. Pressure <ul><li>Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). </li></ul><ul><ul><ul><li>Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity. </li></ul></ul></ul>
- 14. Pressure <ul><li>Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). </li></ul><ul><ul><ul><li>Atmospheric pressure varies with weather changes and altitude. </li></ul></ul></ul>
- 15. Pressure <ul><li>Manometer – an instrument for measuring pressure often below that of atmospheric pressure. </li></ul>
- 16. Units of Pressure <ul><li>Pressure = force/area </li></ul><ul><ul><li>torr – in honor of Torricelli is equal to a mm Hg. </li></ul></ul><ul><ul><ul><li>760 mm Hg = 760 torr </li></ul></ul></ul><ul><ul><li>1 atm = 760 torr </li></ul></ul><ul><ul><li>Pascal = N/m 2 </li></ul></ul><ul><ul><ul><li>1atm = 101,325 Pa </li></ul></ul></ul>
- 17. Practice Problems <ul><li>Page 225 #35, 37, 39 </li></ul>
- 18. The Gas Laws of Boyle, Charles and Avogadro
- 19. Boyles Law <ul><li>Boyle (1627-1691) – performed the first quantitative experiments on gases. Used a J tube to measure pressures. </li></ul><ul><ul><ul><li>PV = k; P 1 V 1 =P 2 V 2 </li></ul></ul></ul><ul><ul><ul><li>k is a constant for a given sample of air at a specific temperature. </li></ul></ul></ul>
- 20. Boyles Law <ul><li>Pressure and volume are often plotted. </li></ul><ul><ul><li>P vs V – gives a hyperbola and an inverse relationship. </li></ul></ul><ul><ul><li>Boyles law rearranged is </li></ul></ul><ul><ul><ul><li>V=k/P=k1/P; when plotted as V vs 1/P – gives a straight line with the intercept of zero </li></ul></ul></ul>
- 21. Boyles Law
- 22. Boyles Law <ul><ul><ul><li>Boyles ’s law holds precisely at very low temperatures, but varies at higher pressures. PV will vary as pressure is varied. </li></ul></ul></ul><ul><ul><ul><li>An ideal gas is a gas that strictly obeys Boyles ’ law. </li></ul></ul></ul>
- 23. Boyles Law
- 24. Charles Law <ul><li>Charles (1746-1823) – the first person to fill a balloon with hydrogen gas and who made the first solo balloon flight. </li></ul><ul><li>Charles found in 1787 that the volume of a gas at constant pressure increases linearly with the temperature of the gas. </li></ul><ul><ul><li>V = bT ; V 1 /T 1 = V 2 /T 2 </li></ul></ul><ul><ul><li>T is in Kelvin, b is a proportionality constant </li></ul></ul>
- 25. Charles Law <ul><li>Temperature vs Volume plots a straight line. </li></ul><ul><ul><li>Slope will vary will type of gas. </li></ul></ul><ul><ul><li>All gas plots of T vs V will extrapolate to zero at the same temperature. </li></ul></ul><ul><ul><ul><li>-273° or 0 K </li></ul></ul></ul>
- 26. Charles Law
- 27. Avogadro ’s Law <ul><li>Avogadro (1811) – postulated that equal volumes of gases at the same temperature and pressure contain the same number of particles (moles). </li></ul><ul><ul><li>V = an; V 1 /n 1 =V 2 /n 2 </li></ul></ul><ul><ul><li>n is number of moles; a is a proportionality constant. </li></ul></ul>
- 28. Ideal Gas Law
- 29. Ideal Gas Law <ul><li>The relationships that Boyle, Charles and Avogadro presented can be combined to show how the volume of a gas depends on pressure, temperature, and number of moles of gas present. </li></ul><ul><ul><li>V = R(Tn/P) </li></ul></ul><ul><ul><li>R is the universal gas constant. </li></ul></ul><ul><ul><ul><li>A combination of proportionality constants </li></ul></ul></ul>
- 30. Ideal Gas Law <ul><li>The equation is often rearranged to form the more common: </li></ul><ul><ul><li>PV=nRT </li></ul></ul><ul><ul><ul><li>R=.08206 L atm/K mol </li></ul></ul></ul>
- 31. Ideal Gas Law <ul><li>Limitations </li></ul><ul><ul><li>A gas that obeys this equation is said to behave ideally. The ideal gas equation is best regarded as a limiting law, it expresses behavior that real gases approach at low pressures and high temperatures. </li></ul></ul><ul><ul><li>Most gases behave ideally at pressures below 1 atm. </li></ul></ul>
- 32. Example Problem <ul><li>A sample of hydrogen gas (H 2 ) has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H 2 molecules present in this gas sample. </li></ul>
- 33. Example Problem <ul><li>V = 8.56 L at a temperature of 0°C </li></ul><ul><li>P =1.5 atm </li></ul><ul><li>Calculate the moles </li></ul>PV=nRT; R = .08206 L atm/K mol .57 mol
- 34. Example Problem 2 <ul><li>You have a sample of ammonia gas with a a volume of 7.0ml at a pressure of 1.68 atm. The gas is compressed to a volume of 2.7 ml at a constant temperature. Use the ideal gas law to calculate the final pressure. </li></ul>
- 35. Example Problem 2 <ul><li>V 1 = 7.0 ml at a pressure of 1.68 atm </li></ul><ul><li>V 2 =2.7 ml at a constant temperature. </li></ul><ul><li>Calculate the final pressure. </li></ul>PV=nRT; but nRT are constant: PV=PV 4.4 atm
- 36. Example Problem 3 <ul><li>A sample of methane gas that has a volume of 3.8 L at 5°C is heated to 86°C at constant pressure. Calculate its new volume. </li></ul>
- 37. Example Problem 3 <ul><li>V 1 = 3.8 L and T 1 5°C </li></ul><ul><li>T 2 =86°C at constant pressure. </li></ul><ul><li>Calculate its new volume. </li></ul>PV=nRT but n, R and P are constant: V 1 /T 1 = V 2 /T 2 4.9 L
- 38. Example Problem 4 <ul><li>A sample of diborane gas (B 2 H 6 ), a substance that burst into flame when exposed to air, has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48 L. If conditions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample. </li></ul>
- 39. Example Problem 4 <ul><li>P 1 =345 torr, T 1 =-15°C and V 1 = 3.48L </li></ul><ul><li>T 2 =36°C and P 2 =468 torr, </li></ul><ul><li>What is the volume? </li></ul>PV = nRT; nR are constant: P 1 V 1 /T 1 = P 2 V 2 /T 2 3.1 L
- 40. Example Problem 5 <ul><li>A sample containing 0.35 mol argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr. Calculate the change in volume that occurs. </li></ul>
- 41. Example Problem 5 <ul><li>n=0.35, T 1 =13°C, P 1 =568 torr </li></ul><ul><li>T 2 =56°C, P 2 =897 torr </li></ul><ul><li>Calculate the change in volume that occurs. </li></ul>-3 L PV=nRT; R = .08206 L atm/K mol
- 42. Practice Problems <ul><li>Page 226 #41, 43, 45, 47, 49, 51, 53, 57, 59, 61 </li></ul>
- 43. Gas Stoichiometry
- 44. Molar Volume <ul><li>One mole of an ideal gas at: </li></ul><ul><ul><li>0°C (273K) </li></ul></ul><ul><ul><li>1atm </li></ul></ul><ul><ul><li>V=nRT/P = 22.42L </li></ul></ul>
- 45. Gas Stoichiometry <ul><li>We use STP or standard temperature and pressure of an ideal gas to make calculations with a gas. </li></ul><ul><ul><li>1 atm </li></ul></ul><ul><ul><li>0°C (273K) </li></ul></ul><ul><li>1 mole = 22.42 L becomes a conversion factor for dimensional analysis. </li></ul>
- 46. Gas Stoichiometry Example <ul><li>A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N 2 are present? </li></ul>7.81 x 10 -2 mol N 2
- 47. Gas Stoichiometry Example 2 <ul><li>Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO 3 ). Calculate the volume of CO 2 at STP produced from the decomposition of 152g CaCO 3 by the reaction </li></ul><ul><li>CaCO 3(s) CaO (s) + CO 2(g) </li></ul>
- 48. Gas Stoichiometry Example 2 <ul><li>152g CaCO 3 </li></ul><ul><li>22.42 L = 1 mol of gas at STP </li></ul><ul><li>Calculate the volume of CO 2 </li></ul><ul><li>CaCO 3(s) CaO (s) + CO 2(g) </li></ul>34.1 L CO 2 at STP
- 49. Gas Stoichiometry Example 3 <ul><li>A sample of methane gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31°C and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO 2 formed at a pressure of 2.50 atm and a temperature of 125°. </li></ul>
- 50. Gas Stoichiometry Example 3 <ul><li>CH 4 V=2.80 L at 25°C and 1.65 atm </li></ul><ul><li>Oxygen V=35.0 L at 31°C and 1.25 atm. </li></ul><ul><li>Calculate the volume of CO 2 at 2.50 atm and 125°C. </li></ul><ul><li>The mixture was then ignited to form carbon dioxide and water. </li></ul>
- 51. Gas Stoichiometry Example 3 <ul><li>CH 4 V=2.80 L at 25°C and 1.65 atm </li></ul><ul><li>Oxygen V=35.0 L at 31°C and 1.25 atm. </li></ul><ul><li>Calculate the volume of CO 2 at 2.50 atm and 125°C. </li></ul><ul><li>CH 4(g) + O 2(g) CO 2(g) + H 2 O (g) </li></ul>2.47 L
- 52. Practice Problems <ul><li>Page 227 #65, 69 </li></ul>
- 53. Molar Mass of a Gas <ul><li>One use of the ideal gas law is in the calculations of the molar mass of a gas from its measured density. </li></ul><ul><ul><ul><li>n = </li></ul></ul></ul><ul><ul><ul><li>P </li></ul></ul></ul><ul><ul><ul><li>D </li></ul></ul></ul><ul><ul><ul><li>P = </li></ul></ul></ul>
- 54. Gas Density/Molar Mass Example <ul><li>The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. </li></ul>32.0 g/mol
- 55. Dalton ’s Law of Partial Pressures
- 56. John Dalton <ul><li>John Dalton formed his atomic theory from his experiments and studies of the mixture of gases. </li></ul><ul><li>His observations car be summarized as follows: </li></ul><ul><ul><li>For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each as would exert if it were alone. </li></ul></ul>
- 57. John Dalton <ul><li>P total =P 1 +P 2 +P 3 +…. </li></ul><ul><ul><li>Subscripts refer to the individual gases and P x refers to partial pressure that a particular gas would exert if it were alone in the container. </li></ul></ul><ul><ul><li>Each Partial pressure can be derived from the ideal gas law and added together to determine the total. </li></ul></ul>
- 58. John Dalton <ul><li>P total =P 1 +P 2 +P 3 +…. </li></ul><ul><ul><li>Since each partial pressure can be broken down into ; the P total can be represented by: </li></ul></ul><ul><ul><ul><li>P total = </li></ul></ul></ul><ul><ul><ul><li>P total = </li></ul></ul></ul>
- 59. <ul><li>For a mixture of ideal gases, it is the total number of moles of particles that is important, not the identity or composition of the involved gas particle. </li></ul>
- 60. Dalton ’s Law Example <ul><li>Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25 and 1.0 atm and 12 L O 2 at 25° and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25° C. </li></ul>
- 61. Mole Fraction <ul><li>The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. </li></ul><ul><ul><li>χ= n x /n total </li></ul></ul><ul><ul><li>χ= P 1 /P total </li></ul></ul>
- 62. Dalton ’s Law Example <ul><li>The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. </li></ul>
- 63. Dalton ’s Law Example <ul><li>The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N 2 in air when the atmospheric pressure is 760 torr. </li></ul>
- 64. Collecting Gas over Water <ul><li>A mixture of gases results whenever a gas is collected by displacement of water. In this situation, the gas in the bottle is a mixture of water vapor and the oxygen being collected. </li></ul>
- 65. Collecting Gas over Water <ul><li>Water vapor is present because molecules of water escape from the surface of the liquid and collect in the space above the liquid. </li></ul>
- 66. Collecting Gas over Water <ul><li>Molecules of water also return to the liquid. When the rate of escape equals the rate of return, the number of water molecules in the vapor state remain constant. </li></ul>
- 67. Collecting Gas over Water <ul><li>When the number of water molecules in the vapor state remain constant the pressure of the water vapor remains constant. </li></ul>
- 68. Collecting Gas over Water <ul><li>This pressure, which depends on temperature, is called vapor pressure of water. </li></ul>
- 69. Collecting Gas over Water Example <ul><li>A sample of solid potassium chlorate (KClO 3 ) was heated in a test tube and decomposed by the reaction: </li></ul>
- 70. Collecting Gas over Water Example <ul><li>The oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr. The volume of gas collected was .650L, and the vapor pressure of water at 22°C is 21 torr. Calculate the partial pressure of O 2 in the gas collected and the mass of KClO 3 in the sample that was decomposed. </li></ul>
- 71. Collecting Gas over Water Example <ul><li>oxygen T=22°C and V=.650L </li></ul><ul><li>Total pressure = 754 torr. </li></ul><ul><li>vapor pressure at 22°C is 21 torr. </li></ul><ul><li>Calculate the partial pressure of O 2 and the mass of KClO 3 in the sample </li></ul>2.59 x 10 -2 mol O 2 2.12 g KClO 3
- 72. The Kinetic Molecular Theory of Gases
- 73. Kinetic Molecular Theory KMT <ul><li>A simple model that attempts to explain the properties of an ideal gas. This model is based on speculations about the behavior of the individual gas particles (atoms or molecules). </li></ul>
- 74. Kinetic Molecular Theory KMT <ul><li>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). </li></ul>
- 75. Kinetic Molecular Theory KMT <ul><li>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. </li></ul>
- 76. Kinetic Molecular Theory KMT <ul><li>The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. </li></ul>
- 77. Kinetic Molecular Theory KMT <ul><li>The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. </li></ul>
- 78. KMT and Boyle ’s Law <ul><li>Because a decrease in volume, the gas particles will hit the walls more often, thus increasing the pressure </li></ul>
- 79. KMT and Charles Law <ul><li>When the gas is heated to a higher temperature, the speeds of its molecules increase and thus hit the walls more often and with more force. Volume and/or pressure will increase. </li></ul>
- 80. KMT and Advogadro ’s Law <ul><li>An increase in the number of gas particles at the same temperature would cause the pressure to increase if the volume were constant. </li></ul>
- 81. KMT and Advogadro ’s Law <ul><li>The volume of a gas (at constant T and P) depends only on the number of gas particles present. The individual particles are not a factor because the particle volumes are so small compared with the distances between the particles. </li></ul>
- 82. KMT and Dalton ’s Law <ul><li>All gas particles are independent of each other and that the volumes of the individual particles are unimportant. Identities of the gas particles do not matter. </li></ul>
- 83. The Meaning of Temperature <ul><li>Kelvin temperature indicates the average kinetic energy of the gas particles. </li></ul><ul><li>The exact relationship between temperature and average kinetic energy can be expressed: </li></ul><ul><ul><li>(KE) avg =3/2 RT </li></ul></ul>
- 84. The Meaning of Temperature <ul><li>The Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperature meaning greater motion. </li></ul>
- 85. Root Mean Square Velocity <ul><li>u 2 =the average of the squares of the particle velocities. </li></ul><ul><li>The square root of u 2 is called the root mean square velocity and is symbolized with u rms </li></ul><ul><ul><li>u rms = = </li></ul></ul><ul><ul><li>M= mole of gas particles (kg) </li></ul></ul><ul><ul><li>R = ; J = kg m 2 /s 2 </li></ul></ul>
- 86. Root Mean Square Velocity Example <ul><li>Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C. </li></ul>1.36 x 10 3 m/s
- 87. Mean Free Path <ul><li>The average distance a particle travels between collisions in a particular gas sample. </li></ul><ul><ul><li>1 x 10 -7 m for O 2 at STP </li></ul></ul><ul><ul><li>u rms =500 m/s </li></ul></ul>
- 88. Mean Free Path <ul><li>A velocity distribution that show the effect of temperature on the velocity distribution in a gas. </li></ul>
- 89. Effusion and Diffusion
- 90. Diffusion <ul><li>Diffusion describes the mixing of gases. </li></ul><ul><ul><li>The rate of mixing gases, is the same as the rate of diffusion </li></ul></ul><ul><ul><li>Dependent upon u rms </li></ul></ul>
- 91. Effusion <ul><li>Effusion describes the transfer of gas from one chamber to another (usually through a small hole or porous opening). </li></ul><ul><ul><li>The rate of transfer is said to be the rate of effusion. </li></ul></ul>
- 92. Effusion <ul><li>The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. </li></ul>
- 93. Effusion <ul><li>Temperature must be the same for both gases. </li></ul><ul><li>M represents the molar masses of the gases. </li></ul><ul><ul><li>Units can be in g or kg since the units will cancel out. </li></ul></ul><ul><li>This is called Graham ’s law of effusion: </li></ul>
- 94. Effusion Example <ul><li>Calculate the effusion rates of hydrogen gas (H 2 ) and Uranium hexafluoride (UF 6 ), a gas used in the enrichment process to produce fuel for nuclear reactors. </li></ul>13.2 : 1
- 95. Real Gases
- 96. Real Gases <ul><li>An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law, although many gases come very close at low pressures and/or high temperatures. </li></ul>
- 97. Real Gases <ul><li>Thus ideal gas behavior can best be thought of as the behavior approached by real gases under certain conditions. </li></ul>
- 98. Real Gases <ul><li>Plots of PV/nRT vs. P for several gases (200K). Ideal behavior only at low pressures. </li></ul>
- 99. Real Gases <ul><li>Plots of PV/nRT vs. P for N 2 at three temperatures. Ideal behavior at higher temperatures. </li></ul>
- 100. KMT Modifications <ul><li>Johannes van der Walls (1837-1923), a physics professor at the University of Amsterdam started work in the area of ideal vs real gas behavior. He won the nobel prize in 1910 for his work. </li></ul>
- 101. KMT Modifications <ul><li>van der Waals modifications to the ideal gas law accounted for the volume of particle space. Therefore adjusting for the volume actually available to a give gas molecule. </li></ul><ul><ul><li>V-nb </li></ul></ul><ul><ul><ul><li>n is number of moles </li></ul></ul></ul><ul><ul><ul><li>b is an empirical constant </li></ul></ul></ul>
- 102. KMT Modifications <ul><li>van der Waals modifications to the ideal gas law allowed for the attractions that occur among particle in a real gas which is dependent upon the concentration of the particles. </li></ul><ul><li> , pressure correction </li></ul><ul><ul><li>a is proportionality constant. </li></ul></ul>
- 103. van der Waals Equation <ul><li>Insert both corrections and the equation can be written as: </li></ul><ul><li>Rearranged for van der Waals: </li></ul>
- 104. van der Waals Equation <ul><li>a and b values are determined for a given gas by fitting experimental behavior. That is a and b are varied until the best fit of the observed pressure is obtained under all conditions. </li></ul>
- 105. Characteristics of Real Gases <ul><li>A low value for a reflects weak intermolecular forces among gas molecules. </li></ul>
- 106. van der Waals <ul><li>Ideal behavior at low pressure (large volume) makes sense because the small amount of volume that the particles consume are not a factor. </li></ul>
- 107. van der Waals <ul><li>Ideal behavior at high temperatures also makes sense because particles are moving at such a high rate that their interparticle interactions are not very important. </li></ul>
- 108. Chemistry in the Atmosphere
- 109. Chemistry in the Atmosphere <ul><li>The most important gases to us are those in the atmosphere that surround the earth ’s surface. </li></ul><ul><ul><li>The principal components are N 2 and O 2 , but many other important gases, such as H 2 O and CO 2 , are also present. </li></ul></ul>
- 111. Chemistry in the Atmosphere <ul><li>Because of gravitational effects, the composition of the earth ’s atmosphere is not constant; heavier molecules tend to be near the earth’s surface, and light molecules tend to migrate to higher altitudes, with some eventually escaping into space. </li></ul>
- 112. Chemistry in the Atmosphere <ul><li>The chemistry occurring in the higher levels of the atmosphere is mostly determined by the effects of high-energy radiation and particles from the sun and other sources in space. The upper atmosphere serves as a shield to prevent this radiation from reaching earth. </li></ul>
- 113. Chemistry in the Atmosphere <ul><li>The troposphere (closest to earth) is strongly influenced by human activities. Millions of tons of gases and particulates are released into the troposphere by our highly industrial civilization. </li></ul>
- 114. Chemistry in the Atmosphere <ul><li>Severe air pollution is found around many large cities. The two main sources of pollution are transportation and the production of electricity. The combustion of petroleum in vehicles produces CO, CO 2 , NO, NO 2 . </li></ul>
- 115. Chemistry in the Atmosphere <ul><li>The complex chemistry of polluted air appears to center around the nitrogen oxides (NO x ). At high temperatures found in the gasoline and diesel engines of cars and trucks, N 2 and O 2 react to form a small quantity of NO that is emitted into the air with the exhaust gases. NO is immediately oxidized in air to NO 2 . </li></ul>
- 116. Reactions in the Atomsphere Ozone is very reactive and can react directly with other pollutants, or the ozone can absorb light and break up to form an energetically excited O 2 molecule (O 2 *) and excited O (O*).
- 117. Reactions in the Atomsphere The end product of this whole process is often referred to as photochemical smog, so called because light is required to initiate some of the reactions.
- 118. Reactions in the Atomsphere Sulfuric acid is very corrosive to both living things and building materials. Another result of this type of pollution is called acid rain.
- 119. THE END

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment