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Chapter 5 Gases Chemical Principles 7th Edition Steven S. Zumdahl 5.1 Early Experiments 5.2 The Gas Laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stoichiometry 5.5 Dalton’s Law of Partial Pressures 5.6 The Kinetic Molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions 5.10 Real Gases 5.11 Characteristics of Real Gases 5.12 Chemistry in the Atmosphere Hot air balloon taking off from the ski resort of Chateau d’Oex in the Swiss Alps. – Matter exists in three distinct physical states: gas, liquid, and solid. Of these, the gaseous state is the easiest to describe both experiment and theoretically. 5.1 Early Experiments – In 1643 an Italian physicist named Evangelista Torricelli (16081647), who had been a student of Galileo, performed experiments that showed that the air in the atmosphere exerts pressure. (In fact, as we will see all gases exert pressure.) Torricelli designed the first barometer by filling a tube that was closed at one end with mercury and then inverting it in a dish of mercury (see Fig. 5.1). FIGURE 5.1 A torricellian barometer. The tube, completely filled with mercury, is inverted in a dish of mercury. Mercury flows out of the tube until the pressure of the column of mercury (shown by black arrow) “standing on the surface” of the mercury in the dish is equal to the pressure of the air (shown by green arrows) on the rest of the surface of the mercury in the dish. Units of Pressure – Because instruments used for measuring pressure, such as the manometer (see Fig. 5.2), often use columns of mercury because of its high density, the most commonly used units for pressure are based on the height of the mercury column (in millimeters) the gas pressure can support. – The unit millimeters of mercury (mm Hg) is called the torr in honor of Torricelli. – A related unit for pressure is the standard atmosphere: 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr – 1 atm: 760 mm Hg, 760 torr, 101.325 Pa (N/m2), 1.01325 bar; 29.92 in Hg, 14.7 lb/in2 (Psi). FIGURE 5.2 A simple manometer, a device for measuring the pressure of a gas in a container. The pressure of the gas is given by h (the difference in mercury levels) in units of torr (equivalent to mm Hg). (a) Gas pressure = atmospheric pressure – h. (b) Gas pressure = atmospheric pressure + h. – However, since pressure is defined as force per unit area, Pressure = force/area – In the SI system the unit of force is the Newton (N) and the unit of area is meters squared (m²). – Thus the unit if pressure in the SI system is newtons per meter squared (N/m²), called the pascal (Pa). In terms of pascals the standard atmosphere is 1 standard atmosphere = 101,325 Pa – The International Union of Pure and Applied Chemist (IUPAC) has adopted 1 bar (100,000 Pa) as the standard pressure instead of 1 atm (101,325 Pa). Both standards are now widely used. 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle’s Law – The first quantitative experiments on gases were performed by an Irish chemist, Robert Boyle (1627-1691). Using a Jshaped tube closed at one end (Fig. 5.3), which he reportedly set up in the multistory entryway of his house, Boyle studied the relationship between the pressure of the trapped gas and its volume. FIGURE 5.3 A J-tube similar to the one used by Boyle. – Respective values from Boyle’s experiments are given in Table 5.1. – These data show that the product of the pressure and volume for the trapped air sample is constant within the accuracies of Boyle’s measurements (note the third column in Table 5.1). – This behavior can be represented by the equation PV = k – which is called Boyle’s law, where k is a constant at a specific temperature for a given sample of air. – Figure 5.4(a) shows a plot of P versus V, which produces a hyperbola. Notice that as the pressure drops by half, the volume doubles. Thus there is an inverse relationship between pressure and volume. FIGURE 5.4 Plotting Boyle’s data from Table 5.1. (a) A plot of P versus V shows that the volume doubles as the pressure is halved. (b) A plot of V versus 1/P gives a straight line. The slope of this line equals the value of the constant k. – The second type of plot can be obtained by rearranging Boyle’s law to give V = k/p – which is the equation for a straight line of the type y = mx + b – where m represents the slope and b is the intercept of the straight line. In this case, y = V, x = 1/P, m = k, and b = 0. Thus a plot of V versus 1/P using Boyle’s data gives a straight line with an intercept of zero, as shown in Fig. 5.4(b). – Highly accurate measurements in various gases at a constant temperature have shown that the product PV is not quite constant but changes with pressure. – Results for several gases are shown in Fig. 5.5. FIGURE 5.5 A plot of PV versus P for several gases. An ideal gas is expected to have a constant value of PV, as shown by the dashed line. Carbon dioxide shows the largest change in PV, and this change is actually quite small: PV changes from approximately 22.39 L atm at 0.25 atm to 22.26 L atm at 1.00 atm. Thus Boyle’s law is a good approximation at these relatively low pressures. – Note the small changes that occur in the product PV as the pressure is varied. – Boyle’s law: V ∝ 1/P at constant temperature. – A gas that obeys Boyle’s law is called an ideal gas. Example 5.1 – In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measurements were made at various pressures, using 1.0 mol of NH3 gas at a temperature of 0°C. Using the results listed blow, calculate the Boyle’s law constant for NH3 at the various pressure. Experiment 1 2 3 4 5 6 Pressure (atm) 0.1300 0.2500 0.3000 0.5000 0.7500 1.000 Volume (L) 172.1 89.28 74.35 44.49 29.55 22.08 Solution – To determine how closely NH3 gas follows Boyle’s law under these conditions, we calculate the value of k (in L atm) for each set of values: Experiment k = PV 1 22.37 2 22.32 3 22.31 4 22.25 5 22.16 6 22.08 – Although the deviations from true Boyle’s law behavior are quite small at these low pressures, the value of k changes regularly in one direction as the pressure is increased. Thus, to calculate the “ideal” value of k for NH3, plot PV versus P, as shown in Fig. 5.6, and extrapolate (extend the line beyond the experimental points) back to zero pressure, where, for reasons we will discuss later, a gas behaves most ideally. The value of k obtained by this extrapolation is 22.41 L atm. This is the same value obtained from similar plots for the gases CO2, O2, and Ne at 0°C, as shown in Fig. 5.5. FIGURE 5.6 A plot of PV versus P for 1 mol of ammonia. The dashed line shows the extrapolation of the data to zero pressure to give the “ideal” value of PV of 22.41 L atm. Charles’s Law – Charles found in 1787 that the volume of a gas at constant pressure increases linearly with the temperature of the gas. That is, a plot of the volume of a gas (at constant pressure) versus its temperature (°C) gives a straight line. – This behavior is shown for several gases in Fig. 5.7. – As with Boyle’s law, Charles’s law is obeyed exactly only at relatively low pressures. – One very interesting feature of these plots is that the volumes of all the gases extrapolate to zero at the same temperature, – 273.2°C. – On the Kelvin temperature scale this point is defined as 0 K, which leads to the following relationship between the Kelvin and Celsius scales: Temperature (K) = 0°C + 273 FIGURE 5.7 Plots of V versus T (°C) for several gases. The solid lines represent experimental measurements on gases. The dashed lines represent extrapolation of the data into regions where these gases would become liquids or solids. Note that the samples of the various gases contain different numbers of moles. – When the volumes of the gases shown in Fig. 5.7 are plotted versus temperature on the Kelvin scale, the plots in Fig. 5.8 result. – In this case the volume of each gas is directly proportional to temperature and extrapolates to zero when the temperature is 0 K. – This behavior is represented by the equation known as Charles’s law, V = bT – where T is the temperature (in Kelvins) and b is a proportionality constant. – 0 K is called absolute zero, and there is much evidence to suggest that this temperature cannot be attained. – Temperature of approximately 10–6 K have been produced in laboratories, but 0 K has never been reached. FIGURE 5.8 Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature. Avogadro’s Law – Equal volumes of gases at the same temperature and pressure contain the same number of “particles”. – This observation is called Avogadro’s law, which can be stated mathematically as V = an – where V is the volume of the gas, n is the number of moles, and a is a proportionality constant. – This equation states that for a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas. – This relationship is obeyed closely by gases at low pressures. Cold Atoms The latest low-temperature of Colorado in Boulder when a team of scientists led by Carl Wieman reported that they had cooled a sample containing 2 × 107 cesium atoms to 1.1 × 10–6 K, about one-millionth of a degree above absolute zero. This record-low temperature was achieved by a technique know as laser cooling, in which a laser beam is directed against a beam individual atoms, dramatically slowing the movement of the atoms. The atoms are then further cooled in an “optical molasses” produced by intersection of field, the Colorado scientists were able to hold the supercold cesium atoms for about 1 s, and the possibility exists that the cold atoms could be trapped for much longer periods of time with improvements in the apparatus. 5.3 The Ideal Gas Law – We have considered three laws that describe the behavior of gases as revealed by experimental observations: Boyle’s law: Charles’s law: Avogadro’s law: V = k/P V = bT V = an (at constant T and n) (at constant P and n) (at constant T and P) – Number of moles of gas present can be combined as follows: V = R(Tn/P) – where R is the combined proportionality constant called the universal gas constant. – R has the value of 0.08206 L atm K–1 mol–1. The preceding equation can be rearranged to the more familiar form of the ideal gas law: PV = nRT – The ideal gas law is an equation of state for a gas, where the state of the gas in its condition at a given time. A particular state of a gas is described by its pressure, volume, temperature, and number of moles. – It is important to recognize that the ideal gas law is an empirical equation – it is based on the experimental measurements of the properties of gases. 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 gasses approach at low pressure and high temperature. – The idea gas law applied best at pressure below 1 atm. Example 5.2 – A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H2 present in this gas sample. Solution – Solving the ideal gas law for n gives n = PV/RT – In this case P = 1.5 atm, V = 8.56 L, T = 0°C + 273 = 273 K, and R = 0.08206 L atm K–1 mol–1. Thus n = [(1.5 atm)(8.56 L)]/[(0.08206 L atm/K mol)(273 K)] = 0.57 mol Example 5.3 – Suppose we have a sample of ammonia gas with a volume of 3.5 L at a pressure of 1.68 atm. The gas is compressed to a volume of 1.35 L at a constant temperature. Use the ideal gas law to calculate the final pressure. Solution – The basic assumption we make when using the ideal gas law to describe a change in state for a gas is that the equation applies equally well to both the initial and the final states. In dealing with a change in state, we always place the variables on one side of the equals sign and the constants on the other. In this case the pressure and volume change, while the temperature and the number of moles remain constant (as does R, by definition). Thus we write the ideal gas law as PV = nRT Change Remain constant – Since n and T remain the same in this case, we can write P1V1 = nRT and P2V2 = nRT. Combining these equations gives P1V1 = nRT = P2V2 or P1V1 = P2V2 – We are giver P1 = 1.68 atm, V1 = 3.5 L, V2 = 1.35 L. Solving for P2 gives P2 = (V1/V2)P1 = (3.5 L/1.35 L) 1.68 atm = 4.4 atm – Check: Does this answer make sense? The volume decreased (at constant temperature), so the pressure should increase, as the result of the calculation indicates. Note that the calculated final pressure is 4.4 atm. Because most gases do not behave ideally above 1 atm, we might find that if we measured the pressure of this gas sample, the observed pressure would differ slightly from 4.4 atm. Example 5.4 – 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. Solution – To solve this problem, we take the ideal gas law and segregate the changing variables and the constants by placing them on opposite sides of the equation. In this case volume and temperature change, and number of moles and pressure (and of course R) remain constant. Thus PV = nRT becomes V/T = nR/P – which leads to V1/T1 = nR/P and V2/T2 = nR/P – Combining these equations gives V1/T1 = nR/P = V2/T2 or V1/T1 = V2/T2 – We are given T1 = 5°C + 273 = 278 K V1 = 3.8 L T2 = 86°C + 273 = 359 K V2 = ? – Thus V2 = T2V1/T1 = (359 K)(3.8 L)/278 K = 4.9 L – Check: Is the answer sensible? In this case the temperature was increased (at constant pressure), so the volume should increase. Thus the answer makes sense. Example 5.5 – A sample of diborane gas (B2H6), a substance that bursts into flames 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? Solution – Since, for this sample, pressure, temperature, and volume all change while the number of moles remains constant, we use the ideal gas law in the form PV/T = nR – which leads to P1V1/T1 = nR = P2V2/T2 or P1V1/T1 = P2V2/T2 – Then V2 = (T2P1V1)/(T1P2) – We have P1 = 345 torr T1 = –15°C + 273 = 258 K V1 = 3.48 L P2 = 468 torr T2 = 36°C + 273 = 309 K V2 = ? – Thus V2 = [(309 K)(345 torr)(3.48 L)]/[(258 K)(468 torr)] = 3.07 L – Always convert temperature to the Kelvin scale when applying the ideal gas law. 5.4 Gas Stoichiometry – Suppose we have 1 mole of an ideal gas at 0°C (273.2 K) and 1 atm. From the ideal gas law, the volume of the gas is given by V = nRT/P = [(1.000 mol)(0.08206 L atm K–1 mol–1)(273.2 K)]/1.000 atm = 22.42 L – This volume of 22.42 liters is called the molar volume of an ideal gas. – The measured molar volumes of several gases are listed in Table 5.2. – The conditions 0°C and 1 atm, called standard temperature and pressure (abbreviated STP) are common reference conditions for the properties of gasses. For example the molar volume of an ideal gasses is 22.42 liter at STP. Example 5.6 – Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 produced at STP from the decomposition of 152 g of CaCO3 according to the reaction. Solution – We use the same strategy we used in the stoichiometry problems earlier in the book. That is, we compute the number of moles of CaCO3 consumed and the number of moles of CO2 produced. The moles of CO2 can then be converted to volume by using the molar volume of an ideal gas. – Using the molar of CaCO3, we can calculate the number of moles of CaCO3: 152 g CaCO3 × (1 mole CaCO3/100.1 g CaCO3) = 1.52 mol CaCO3 – Since each mole of CaCO3 produces 1 mol of CO2, 1.52 mol of CO2 will be formed. We can compute the volume of CO2 at STP by using the molar volume: 1.52 mol CaCO3 × 22.42 L CO2/1 mol CO2 = 34.1 L CO2 – Thus the decomposition of 152 g of CaCO3 will produce 34.1 L of CO2 at STP. – Remember that the molar volume of an ideal gas is 22.42 L at STP. Molar Mass – One very important use of the ideal gas law is in the calculation of the molar mass (molecular weight) of a gas from its measured density. To understand the relationship between gas density and molar mass, note that the number of moles of gas n can be expressed as n = (grams of gas/molar mass) = (mass/molar mass) = (m/molar mass) – Substitution into the ideal gas equation gives P = nRT/V = [(m/molar mass)RT]/V = [m(RT))/(V(molar mass)] – But m/V is the gas density, d, in units of grams per liter. Thus or P = dRT/molar mass Molar mass = dRT/P – Thus, if the density of a gas at a given temperature and pressure is known, its molar mass can be calculated. 5.5 Dalton’s Law of Partial Pressures – In 1803 Dalton summarized his observations as follows: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. – This statement, known as Dalton’s law of partial pressures, can be expressed as follows: PTotal = P1 + P2 + P3 + ⋅⋅⋅ – The pressures P1, P2, P3, and so on, are called partial pressures. – Assuming that each gas behaves ideally, the partial pressure of each gas can be calculated from the ideal gas law: P1 = n1RT/V, P2 = n2RT/V, P3 = n3RT/V, ⋅⋅⋅ – The total pressure of the mixture, PTotal, can be represented as PTotal = P1 + P2 + P3 + ⋅⋅⋅ = n1RT/V + n2RT/V + n3RT/V = (n1 + n2 + n3 +⋅⋅⋅)(RT/V) = nTotal(RT/V) – where nTotal is the sum of numbers of moles of the various gases. – Thus, 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 individual gas particles. – This idea is illustrated in Fig. 5.9. – At this point we need to define the mole fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. – The Greek letter chi (χ) is used to symbolize the mole fraction. For a given component in a mixture, the mole fraction χ1 is χ1 = n1/nTotal = n1/(n1 + n2 + n3 +⋅⋅⋅) FIGURE 5.9 The partial pressure of each gas in a mixture of gases depends on the number of moles of that gas. The total pressure is the sum of the partial pressures and depends on the total moles of gas particles present, no matter what their identities. + = – Since n = P(V/RT) – That is, for each component in the mixture, n1 = P1(V/RT), n2 = P2(V/RT), ⋅⋅⋅ – Therefore, we can represent the mole fraction in terms of pressures: χ1 = n1/nTotal = P1(V/RT)/[P1(V/RT) + P2(V/RT) + P3(V/RT) + ⋅⋅⋅] n1 n1 n2 n3 = [(V/RT)P1]/[(V/RT)(P1 + P2 + P3 + ⋅⋅⋅)] = P1/(P1 + P2 + P3 + ⋅⋅⋅) = P1/PTotal – Similarly, χ2 = n2/nTotal = P2/PTotal – and so on. Thus the mole fraction of a particular component in a mixture of ideal gases is directly related to its partial pressure. – The expression for the mole fraction, χ1 = P1 / PTotal – can be rearranged: P1 = χ1 × PTotal – That is, the partial pressure of a particular component of a gaseous mixture is equal to the mole fraction of that component times the total pressure. – Fig. 5.10 shows the collection of oxygen gas produced by the decomposition of solid potassium chlorate. FIGURE 5.10 The production of oxygen by thermal decomposition of KClO3. The MnO2 catalyst is mixed with the KClO3 to make the reaction faster. The Chemistry of Air Bags Most experts agree that air bags represent a very important advance in automobile safety. Air bag is activated when a serve deceleration (an impact) causes a steel ball to compress a spring and electrically ignite detonator cap, which, in turn, causes sodium azide (NaN3) to decompose explosively, forming sodium and nitrogen gas: 2NaN3(s) → 2Na(s) + 3N2(g) This system works very well and required only a relatively small amount of sodium azide [100 g yield 56 L of N2(g) at 25°C and 1 atm]. Sodium azide, besides being explosive, has a toxicity roughly equal to that of sodium cyanide. It also forms hydrazoic acid (HN3), a toxic and explosive liquid, when treated with acid. NaN3 Example 5.7 – The mole fraction of nitrogen in air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760 torr. Solution – The partial pressure of N2 can be calculated as follows: PN = χN × PTotal = 0.7808 × 760 torr = 593 torr 2 2 Example 5.8 – A sample of solid potassium chlorate (KClO3) was heated in a test tube (see Fig. 5.10) and decomposed according to the following reaction: 2KClO3(s) → 2KCl(s) + 3O2(g) – The oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure of water at 22°C is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed. Solution – First, we find the partial pressure of O2 from Dalton’s law of partial pressure: PTotal = PO + PH O = PO + 21 torr = 754 torr 2 2 2 – Thus PO = 754 torr – 21 torr = 733 torr 2 – Now we use the ideal gas law to find the number of moles of O2: nO = (PO V)/RT 2 2 – In this case PO = 733 torr = 733 torr/(760 torr/atm) = 0.964 atm 2 V = 0.650 L T = 22°C + 273 = 295 K R = 0.08206 L atm K–1 mol–1 – Thus nO = [(0.964 atm)(0.650 L)]/[(0.08206 L atm K–1 mol–1)(295 K)] 2 = 2.59 × 10–2 mole – Next, we calculate the moles of KClO3 needed to produce this quantity of O2, using the mole ratio from the balanced equation for the decomposition of KClO3: 2.59 × 10–2 mole O2 × (2 mol KClO3/3 mol KClO3) = 1.73 × 10–2 mole KClO3 – Using the molar mass of KClO3, we calculate the grams of KClO3: 1.73 × 10–2 mole KClO3 × (122.6 g KClO3/1 mol KClO3) = 2.12 g KClO3 – Thus the original sample contained 2.12 g of KClO3. 5.6 The Kinetic Molecular Theory of Gases – On the basis of observations from different types of experiments, we know that at pressure less than 1 atm most gases closely approach the behavior described by the ideal gas law. Now we want to construct a model to explain this behavior. – However, although laws summarize observed behavior, they do not tell us why nature behaves in the observed fashion. – The model in chemistry consist of speculations about what individual atoms or molecules might be doing to cause the observed behavior of the macroscopic system. – A model is consisted successful if it explains the observed behavior in question and predicts correctly the results of future experiments. Note that a model can never be proved to be absolutely true. – In fact, any model is an approximation by its very nature and is bound to fail at some point. – An example of this type of model is the kinetic molecular theory, 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). Kinetic Molecular Theory – 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). – 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. – The particles are assumed to exert no forces on each other; they are assumed to neither attract nor repel each other. – The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. – Of course, real gas particle do have a finite volume and do exert forces on each other. Thus they do not conform exactly to these assumptions. But we will see that these postulates do indeed explain ideal gas behavior. – The true test of a model is how well its predictions fit the experimental observations. The postulates of the kinetic molecular model picture an ideal gas as consisting of particles have no volume and no attraction for each other, and the model assumes that the gas produces pressure on its container by collisions with the walls. Separating Gases Assume you work for an oil company that owns a huge natural gas reservoir containing a mixture of methane and nitrogen gases. Your job is to separate the nitrogen (N2 size ≈ 430 pm) from the methane (CH4 size ≈ 410 pm). How might you accomplish this task? They produced a “molecular sieve” in which the pore (passage) sizes can be adjusted precisely enough to separate N2 molecules from CH4 molecules. The material involved is a special hydrate trianosilicate (contains H2O, Ti, Si, O, and Sr) compounds and known as ETS-4. The researchers has shown that the material can be used to separate N2 (≈ 410 pm) from O2 (≈ 390 pm). They have also shown that it is possible to reduce the N2 content of natural gas from 18% to less than 5% with a 90% recovery of CH4. Molecular sieve framework of titanium (blue), silicon (green), and oxygen (red) atoms contract on heating–at room temperature (left) d = 4.27 Å; at 250°C (right) d = 3.94 Å. The Quantitative Kinetic Molecular Model – Suppose there are n moles of an ideal gas in a cubical container with sides each of length L in meters Assume each gas particle has a mass m and that it is in rapid, random, straight-line motion colliding with the walls, as shown in Fig. 5.11. – The collisions will be assumed to be elastic – no loss of kinetic energy occur. – Each particle in the gas has a particular velocity u, which can be broken into components ux, uy, and uz, as shown in Fig. 5.12. uxy2 = ux2 + uy2 Hypotenuse of right triangle Sides of right triangle FIGURE 5.11 An ideal gas particle in a cube whose sides are of length L (in meters). The particle collides elastically with the walls in a random, straightline motion. FIGURE 5.12 (a) The Cartesian coordinate axes. This can be represented as a rectangular solid with sides ux, uy, and uz and body diagonal u. (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, ux, uy, and uz. (c) In the xy plane, ux2 + uy2 = uxy2 by the Pythagorean theorem. Since uxy and uz are also perpendicular, u2 = uxy2 + uz2= ux2 + uy2 + uz2. – Then, constructing another triangle as shown in Fig. 5.12, we find u2 = uxy2 + uz2 u2 = ux2 + uy2 + uz2 – Note that only the x component of the velocity affects the particle’s impacts on these two walls, as shown in Fig. 5.13. – The larger the x component of the velocity, the faster the particle travels between these two walls, thus producing more impacts per unit of time on these walls. – Remember that the pressure of the gas is caused by these collisions with the walls. FIGURE 5.13 (a) Only the x component of the gas particle’s velocity affects the frequency of impacts on the shaded walls, the walls that are perpendicular to the x axis. (b) For an elastic collision, there is an exact reversal of the x component of the velocity and of the total velocity. The change in momentum (final – initial) is then – mux – mux= –2mux – The collision frequency (collisions per unit of time) with the two walls that are perpendicular to the x axis is given by (Collision frequency)x = velocity in the x direction/distance between the walls = ux/L – Next , what is the force of a collision? Force is defined as mass times acceleration: F = ma = m(Δu/Δt) – where F represents forces, a represents the acceleration, Δu represents a change in velocity, Δt represents a given length of time. – Since we assume that the particle has constant mass, we can write F = (mΔu)/Δt = Δ(mu)/Δt – Remember that an elastic collision means that there is no change in the magnitude of the velocity. The change in momentum in the x direction is Change in momentum = Δ(mux) = final momentum – initial momentum = –mux – mux Final momentum in x diretion Iinal momentum in x diretion = –2mux – Recall that since force is the change in momentum per unit of time, then Forcex = Δ(mux)/Δt – This expression can be obtained by multiplying the change in momentum per impact by the number of impacts per unit of time: Forcex = (2mux)(ux/L) = change in momentum per unit of time Change in Momentum per impact Impacts per unit of time – That is, Forcex = 2mux2/L Forcey = 2muy2/L Forcez = 2muz2/L – Since we have shown that u2 = ux2 + y2 + uz2 – The total force on the box is ForceTotal = forcex + forcey + forcez = 2mux2/L + 2muy2/L + 2muz2/L = (2m/L)(ux2 + y2 + uz2) = (2m/L)(u2) – We use the average of the square of the velocity u2 to obtain ForceTotal = (2m/L)(u2) – Next, we need to calculate the pressure: Pressure caused by “average” particle = forceTotal / areaTotal = (2mu2/L)/6L2 = mu2/3L3 The 6 sides of the cube Area of each side – Since the volume V of the cube is equal to L3, we can write Pressure = P = mu2/3V – The number of particles in a given gas sample can be expressed as follows: Number of gas particles = nNA – The total pressure on the box caused by n moles of a gas is therefore P = nNA(mu2/3V) – We have P = (2/3)[nNA(1/2mu2)/V)] – KE = 1/2mu2, the energy caused by the motion of particle. – We get the average kinetic energy for a mole of gas particles: (KE)avg = NA(1/2mu2) – Use this definition, we can rewrite the expression for pressure as P = (2/3)[n(KE)avg/V)] or PV/n = (2/3)(KE)avg – The fourth postulate of the kinetic molecular theory is that the average kinetic energy of the particles in the gas sample is directly proportional to the temperature in Kelvins. Thus, since (KE)avg ∝ T, we can write PV/n = (2/3)(KE)avg ∝ T or PV/n ∝ T – Compare the ideal gas law, PV/n = RT From experiment – With the result from the kinetic molecular theory, PV/n ∝ T From theory – These expression have exactly the same form of R, the universe gas constant, is considered the proportionally constant in the second case. (Left) A balloon filled with air at room temperature. (Center) The balloon is dipped into liquid nitrogen at 77 K. (Right) The balloon collapses at the molecules inside slow down because of the decrease temperature. Slower molecules produce a lower pressure. Room temperature 77 K The Meaning of Temperature – The exact relationship between temperature and average kinetic energy can be obtained by combining the equations PV/n = RT = (2/3)(KE)avg – which yields the expression (KE)avg = (3/2)RT – The Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperature meaning greater motion. Root Mean Square Velocity – The square root of u² is called the root mean square velocity and is symbolized urms: urms = √ u² – We can obtain an expression for urms from the equations (KE)avg = NA[(1/2)mu2)] and (KE)avg = (3/2)RT – Combination of these equations gives NA[(1/2)mu2)] = (3/2)RT or u2 = (3RT)/(NAm) – Taking the square root of both sides of the last equation produce √ u² = urms = √(3RT)/(NAm) – In these expression m represents the mass in kilograms of a single gas particle. – When NA, the number of particles in a mole, is multiplied by m, the product is the mass of a mole of gas particles in kilograms. We will call this quantity M. Substituting M for NAm in the equation for urms, we obtain urms = √(3RT)/(M) – Before we can use this equation, we need to consider the units for R. So far we have used 0.08206 L atm K–1 mol–1 as the value for R. – A joule (J) is defined as a kilogram meter squared per second squared (kg m2/s2). When R is converted from liter atmospheres to joules, it has the value 8.3145 J K–1 mol–1. When R with these units is used in the expression √(3RT)/(M), urms has units of meters per second, as described. – If the path of a particular gas particle could be monitored, it would probably look very erratic, something like that shown in Fig. 5.14. FIGURE 5.14 Path of particle in a gas. Any given particle will continuously change its course as a result of collisions with other particles, as well as with the walls of the container. – The average distance a particle travels between collisions in a particular gas sample is called the mean free path. – It is typically a very small distance (1 × 10–7 m for O2 at STP). – The actual distribution of molecular velocities for oxygen gas at STP is shown in Fig. 5.15. – Figure 5.16 shows the velocity distribution for nitrogen gas at three temperature. – The distribution of velocities of the particles in an ideal gas is described by the Maxwell–Boltzmann distribution law: f(u) = 4π(m/2πkBT)(3/2)u2e(–mu – – – – 2/2k T) B where u = velocity in m/s m = mass of a gas particle in kg kB = Boltzmann’s constant = 1.38066 × 10–23 J/K T = temperature in K FIGURE 5.15 A plot of the relative number of O2 molecules that have a given velocity at STP. FIGURE 5.16 A plot of the relative number of N2 molecules that have a given velocity at three temperatures. Note that as the temperature increases, both the average velocity (reflected by the curve’s peak) and the spreads of velocities increase. 273 K 1273 K 2273 K – Analysis of the expression for f(u) yields the following equation for the most probable velocity ump (the velocity possessed by the greatest number of gas particles): ump = √(2kBT)/(m) = √(2RT)/(M) – – – – where M = molar mass of the gas particles in kg = 6.022 × 1023 × m R = gas constant = 6.022 × 1023 × kB Another type of velocity that can be obtained from f(u) is the average velocity uavg (sometimes written u), which is given by equation uavg = u = √(8kBT)/(πm) = √(8RT)/(πM) – ump: uavg: urms stand in the ratios: ump: uavg: urms = 1.000 : 1.128 : 1.225 – This relationship is shown for nitrogen gas at 0°C in Fig. 5.17. FIGURE 5.17 A velocity distribution for nitrogen gas at 273 K, with the values of most probable velocity (ump, the velocity at the curve maximum), the average velocity (uavg), and the root mean square velocity (urms) indicated. 5.7 Effusion and Diffusion – Diffusion is the term used to describe the mixing of gases. The rate of diffusion is the rate of the mixing of gases. – Effusion is the term used to describe the passage of a gas through a tiny orifice into an evacuated chamber, as shown in Fig. 5.18. The rate of effusion measures the rate at which the gas is transferred into the chamber. FIGURE 5.18 The effusion of a gas into an evacuated chamber. The rate of effusion (the rate at which the gas is transferred across the barrier through the pin hole) is inversely proportional to the square root of the mass of the gas molecules. Effusion – The relative rates of effusion of two gases at the same temperature and pressure are given by the inverse ratio of the square roots of the masses of the gas particles: Rate of effusion for gas 1/Rate of effusion for gas 2 = √M2/√M1 – where M1 and M2 represent the molar masses of the gases. This equation is called Graham’s law of effusion. – In Graham’s law the units for molar mass can be g/mol or kg/mol, since the units cancel in the ratio √M2/√M1. – This reasoning leads to the following prediction for two gases at the same temperature T: Effusion rate for gas 1/Effusion rate for gas 2 = (uavg for gas 1)/(uavg for gas 2) = √(8RT)/(πM1)/√(8RT)/(πM2) = √M2/√M1 Diffusion – Diffusion is frequently illustrated by the lecture demonstration represented in Fig. 5.19, in which two cotton plugs, one soaked in ammonia and the other hydrochloric acid, are simultaneously placed at the ends of a long tube. A white ring of ammonium chloride (NH4Cl) forms where the NH3 and HCl molecules meet several minutes later: NH3(g) + HCl(g) → NH4Cl(s) White solid – The progress of the gases through the tube is surprisingly slow in light of the fact that the velocities of the HCl and NH3 molecules at 25°C are approximately 450 and 660 meter per second, respectively. Why does it take several minutes for the NH3 and HCl molecules to meet? FIGURE 5.19 (a) A demonstration of the relative diffusion rates of NH3 and HCl molecules through air. Two cotton plugs, one dipped in HCl(aq) and one dipped in NH3(aq), are simultaneously inserted into the ends of the tube. Gaseous NH3 and HCl vaporizing from the cotton plugs diffuse toward each other and, where they meet, react to form NH4Cl(s). (b) When HCl(g) and NH3(g) meet in the tube, a white ring of NH4Cl(s) forms. – The answer is that the tube contains air and thus the NH3 and HCl molecules undergo many collision with O2 and N2 molecules as they travel through the tube. dNH /dHCl = distance traveled by NH3/distance traveled HCl 3 = uavg(NH )/uavg(HCl) = √MHCl/√MNH = √36.5/√17 = 1.5 3 3 – However, careful experiments show that this prediction is not borne out-the observed ratio of distances is 1.3, not 1.5 as predicted by Graham’s law. – If no air were present in the tube, the ratio of distances would be 1.5 as predicted from Graham’s law. – Natural uranium is mostly 92238U, which cannot be fissioned to produce energy. It contains only about 0.7% of the fissionable nuclide 92235U must be increased to about 3%. In the gas diffusion enrichment process, the natural uranium (containing 238U and a small amount of 235U) react with fluorine to 92 92 form a mixture of 238UF6 and 235UF6. – Because these molecules have slightly different masses, which allows them to be separated by a multistage diffusion process. (238UF6 99.3% v.s 235UF6 0.7%) – Although the process is called gaseous diffusion, because the chambers are separated by barriers that effectively allow only individual UF6 molecules to pass through, it behaves like an effusion process. – Thus we can find the actual ratio of the two types of UF6 in chamber 2 from Graham’s law: Diffusion rate for 235UF6/Diffusion rate for 238UF6 = √Mass(238UF6)/Mass(235UF6) = √(352.05 g/mol)/(349.03 g/mol) = 1.0043 – We can use this factor to calculate the ratio of 235UF6 / 238UF6 in chamber 2: 235UF 6/ 238UF 6 = 1.0043 × (235UF6/238UF6) = 1.0043(7/993) Chamber 2 Chamber 1 = 1.0043(7.0493 × 10–3) = 7.0797 × 10–3 – This enrichment process (in 235UF6) continues as the slight enriched gas in chamber 2 diffuses into chamber 3 and is again enriched by a factor of 1.0043. – To calculate the number of steps required to enrich from 0.700% 235U to 3.00% 235U, we have the following equation: (0.700 235UF6/99.3 238UF6) × (1.0043)N = 3.00 235UF6/97.0 238UF6 Original ratio – where N represents the number of states. Desired ratio – Thus Origin ratio × 1.0043 × 1.0043 × 1.0043 × ⋅⋅⋅ = final ratio First stage Second stage Third stage – Solving this equation for N yields 345. – A photo of actual diffusion cells is shown in Fig. 5.20. FIGURE 5.20 Uranium-enrichment converters from the Paducah gaseous diffusion plant in Kentucky. 5.8 Collisions of Gas Particles with the Container Walls – We will define the quantity we are looking for as ZA, the collision rate (per second) of the gas particles with a section of wall that has an area A (in m²). – We expect ZA to depend on the following factors: 1. The average velocity of the gas particles 2. The size of the area being considered 3. The number of particles in the container – How is the ZA expected to depend on the average velocity of the gas particles? For example, if we double the average velocity, we double the number of wall impacts, so ZA should double. Thus, ZA depends directly on uavg: ZA ∝ uavg – Similarly, ZA depends directly on A, the area of the wall under consideration. Thus ZA ∝ A – Thus ZA is expected to depend directly on N/V. That is, ZA ∝ N/V. – In summary, ZA should be directly proportional to uavg, A, and N/V: ZA ∝ uavg × A × (N/V) – Note that the units for ZA expected from this relationship are (m/s) × m2 × (particles)/m3 → (particles)/s or (collisions)/s – The correct units for ZA are 1/s, or s–1. – Substituting the expression for uavg gives ZA ∝ (N/V)A [√(8RT)/(πM)] – A more detail analysis of this situation shows that proportionality constant is 1/4. Thus the exact equation for ZA is ZA = (1/4) (N/V)A(√(8RT)/(πM)) = A(N/V)(√(RT)/(2πM)) Example 5.9 – Calculate the impact rate in a 1.00-cm² section of a vessel containing oxygen gas at a pressure of 1.00 atm and 27°C. Solution – To calculate ZA, we must identify the values of the variables in the equation ZA = A(N/V)[√(RT)/(2πM)] – In this case A is given as 1.00 cm². However, to be inserted into the expression for ZA, A must have the unit m². The appropriate conversion gives A = 1.00 × 10–4 m². – The quantity N/V can be obtained from the ideal gas law by solving for n/V and then converting to the appropriate units: n/V = P/RT = 1.00 atm/[(0.08206 L atm/K mol)(300 K)] = 4.06 × 10–2 mol/L – To obtain N/V, which has the units (molecules)/m³, from n/V, we make the following conversion: N/V = (4.06 × 10–2 mol/L) × [6.022 × 1023 (molecules/mole)] × (1000 L/m3) = 2.44 × 1025 molecules/m3 – The quantity M represents the molar mass of O2 in kg. Thus M = (32.0 g/mol) × (1 kg/1000 g) = 3.20 × 10–2 kg/mol – Next, we insert these quantities into the expression for ZA: ZA = A(N/V)[√(RT)/(2πM)] = (1.00 × 10–4 m2)(2.44 × 1025 m–3) × √[(8.3145 J/K mol)(300 K)]/[(2)(3.14)(3.20 × 10–2 kg/mol)] = 2.72 × 1023 s–1 – That is, in this gas 2.72 × 10²³ collisions per second occur on each 1.00-cm² area of the container. 5.9 Intermolecular Collisions – In this section we will consider the collision frequency of the particles in a gas. We will start by considering a single spherical gas particle with diameter d (in meters) that moving with velocity uavg. As this particle moves through the gas in a straight line, it will collide with another particle only if the other particle has its center in a cylinder with radius d, as shown in Fig. 5.21. – Thus our particle “sweeps out” a cylinder of radius d and length uavg × 1 second during every second of it flight. Therefore, the volume of the cylinder swept out per second is V = volume = (πd)2(uavg)(1 s) Area of Length of cylinder cylinder slice FIGURE 5.21 The cylinder swept out by a gas particle of diameter d. – To specify the number of gas particles, we use the number density of the gas N/V, which indicates the number of gas particles per unit volume. Thus we can write Number of collisions = (volume swept out) × N/V = πd2(uavg)(N/V) per second = πd2[√(8RT)/(πM)](N/V) = (N/V)d2[√(8πRT)/(M)] – What about the motions of the other particles? – They are moving in various directions with various velocities, – When the motions of the particles are accounted for, the relative velocity of the primary particle becomes √2uavg rather than the value uavg that we have been using. Thus the expression for the collision rate becomes Collision rate (per second) = Z = √2 (N/V) d2 [√(8πRT)/(M)] = 4 (N/V) d2 [√(πRT)/(M)] Example 5.10 – Calculate the collision frequency for an oxygen molecule in a sample of pure oxygen gas at 27°C and 1.0 atm. Assume that the diameter of an O2 molecule is 300 pm. Solution – To obtain the collision frequency, we must identify the quantities in the expression Z = 4(N/V)d2√(πRT)/(M) – that are appropriate to this case. We can obtain the value of N/V for this sample of oxygen by assuming ideal behavior. From the ideal gas law n/V = P/RT = 1.0 atm/[(0.08206 L atm/K mol)(300 K)] = 4.1 × 10–2 mol/L – Thus N/V = (4.1 × 10–2 mol/L)(6.022 × 1023 molecules/mol)(1000 L/m3) = 2.5 × 1025 molecules/m3 – From the given information we know that d = 300 pm = 300 × 10–12 m or 3 × 10–10 m – Also, for O2, M = 3.20 × 10-² kg/mol. Thus Z = 4(2.5 × 1025 m–3)(3 × 10–10 m) × √[π(8.314 J K–1 mol–1)(300 K)]/(3.20 × 10–2 kg/mol) = 4 × 109 (collisions)/s = 4 × 109 s–1 – Notice how large this number is. Each O2 molecule undergoes approximately 4 billion collisions per second in this gas sample. Mean Free Path – If we multiply 1/Z by the average velocity, we obtain the mean free path λ: λ = (1/Z) × uavg = distance between collisions Time between Distance traveled collisions (s) per second – Substituting the expressions for 1/Z and uavg gives λ = [1/(4(N/V)d2√(πRT)/(M)][√(8RT)/(πM)] = 1/(√2(N/V)πd2) Example 5.11 – Calculate the mean free path in a sample of oxygen gas at 27°C and 1.0 atm. Solution – Using data from the previous example, we have λ = 1/[√2(2.5 × 1025 m–3)(π)(2(3 × 10–10 m)2] = 1 × 10–7 m – Note that an O2 molecule travels only a very short distance before it collides with another O2 molecule. This produces a path for a given O2 molecule like the one represented in Fig. 5.14, where the length of each straight line is ∼10–7 m. 5.10 Real Gases – 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. Thus ideal gas behavior can best be thought of as the behavior approached by real gases under certain conditions. – Plots of PV/nRT versus P are shown for several gases in Fig. 5.22. For an ideal gas PV/nRT equals 1 under all conditions, but notice that for real gases PV/nRT approaches 1 only at low pressures (typically at 1 atm). – To illustrate the effect of temperature, we have plotted PV/nRT versus P for nitrogen gas at several temperatures in Fig. 5.23. FIGURE 5.22 Plots of PV/nRT versus P for several gases (200 K). Note the significant deviations from ideal behavior (PV/nRT = 1). The behavior is close to ideal only at low pressure (less than 1 atm). FIGURE 5.23 Plots of PV/nRT versus P for nitrogen gas at three temperatures. Note that, although nonideal behavior is evident in each case, the deviations are smaller at the higher temperatures. – The most important conclusion to be drawn from these plots is that a real gas typically exhibits behavior that is closet to ideal behavior at low pressures and high temperatures. – To follow van der Waals analyses, we start with the ideal gas law, P = nRT/V – Remember that this equation describes the behavior of a hypothetical gas consisting of volumeless entities that do not interact with each other. – P = nRT/V is also 1 at high pressures for many gases because of a canceling of nonideal effects. – P’ is corrected for the finite volume of the particles. The attractive forces have not yet been taken into account. – In contrast, a real gas consists of atoms or molecules that have finite volumes. Thus the volume available to a given particle in a real gas is less than the volume of container because the gas particle themselves take up some of the space. – Pobs is usually called just P. – We have now corrected for both the finite volume and the attractive forces of the particles. – The volume actually available to a given gas molecule is given by the difference V – nb. – This modification of the ideal gas equation leads to the expression P’ = nRT/(V – nb) – The volume of the gas particles has now been taken into account. – The next step is to account for the attractions that occur among the particles in a real gas. The effect of these attractions is to make the observed pressure Pobs smaller that it would be if the gas particles did not interact: Pobs = (P’ – correction factor) = (nRT/(V – nb) – correction factor) – When gas particles come close together, attractive force occur, which cause the particles to hit the wall slightly less often than they would in the absence of these interactions (see Fig. 5.24). – The size of the correction factor depends on the concentration of gas molecules defined in terms of moles of gas particles per liter (n/V). – In a gas sample containing N particles, there are N – 1 partners available for each particle, as shown in Fig. 5.25. FIGURE 5.24 (a) Gas at low concentration-relatively few interactions between particles. The indicated gas particle exerts a pressure on the wall close to that predicted for an ideal gas. (b) Gas at high concentrationmany more interactions between particles. Because of these interactions the collision frequency with the walls is lowered, thus causing the observed pressure to be smaller than if the gas were behaving ideally. FIGURE 5.25 Illustration of pairwise interactions among gas particles. In a sample with 10 particles, each particle has 9 possible partners, to give 10(9)/2 = 45 distinct pairs. The factor of ½ arises, because when particle 1 is the particle of interest, we count the ①…② pair; and when particle ② is the particle of interest, we count the ②…① pair. However, ①…② and ②…① are the same pair, which we thus have counted twice. Therefore, we must divide by 2 to get the correct number of pairs. – Since the 1 · · · 2 pair is the same as the 2 · · · 1 pair, this analysis counts each pair twice. Thus for N particles there are N(N – 1)/2 pairs. If N is a very large number, N – 1 approximately equals N, given N2/2 possible pairs. – Thus the correction to ideal pressure for the attractions of the particles has the form Pobs = P’ – a(n/V)2 – where a is a proportionality constant (which includes the factor of 1/2 from N2/2) – Inserting the corrections for both the volume and of the particle and the attractions of the particles give the equation Pobs = Observed Pressure [nRT/( V – nb)] – a(n/V)2 Volume Volume of the correction container Pressure correction – This equation can be rearranged to give the van der Waals equation: [Pobs + a(n/V)2] (V – nb) = nRT Corrected pressure Corrected volume Pideal Videal – The values of a and b for various gases are given in Table 5.3. – These observations are illustrated in Fig. 5.26. – The fact that a real gas tends to behavior more ideally at high temperatures can also be explained in terms of the van der Waals model. At high pressure the particles are moving so rapidly that the effects of interparticle interactions are not very important. a H2 < N2 < CH4 < CO2 FIGURE 5.26 The volume occupied by the gas particles themselves is less important at (a) large container volumes (low pressure) than at (b) small container volumes (high pressure). Cold Sounds In view of the 1996 prohibition on the production of the ozonedamaging chloroflurocarbons (CFCs) formerly used in many air conditioners and refrigerators, innovative methods for cooling without CFCs are being considered. Because of the forces that exist among its molecules, a real gas becomes hot when it is compressed and cools when it expands. A novel idea is to use sound waves to power a refrigerator. The apparatus is fairly simple: Loudspeakers are mounted at each end of a U-shaped tube filled with an inert gas. The sound waves produced by the speakers cause pressure variations in the tube. In low-pressure areas the gas cools; in high-pressure areas the gas become hot. The hot areas are cooled by a heat exchanger and over time the gas in the tube becomes very cold. Acoustic refrigeration units has some real advantages: They have no moving parts (except the speaker drivers) and they are environmentally safe. Presently the “sound-fridge” is neither as cheap nor as efficient as conventional cooling devices, but Garrett believes that the unit can be made competitive within several years. 5.11 Characteristic of Several Real Gases – The plot for H2(g) (Fig. 5.22) never drops below the ideal value (1.0) in contrast to all the other gases. What is special about H2 compared to these other gases? – Recall from Section 5.8 that the reason that the compressibility of a real gas falls below 1.0 is that the actual (observed) pressure is lower than occur in real gases. This must mean that H2 molecules have very low attractive forces for each other. – Note that H2 has the lowest value of a among the gases H2, N2, CH4, and CO2 (Table 5.3). – Remember that the value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure: Pideal = Pobs + a(n/V)2 – A low value of a reflects weak intermolecular forces among the gas molecules. – Also notice that although the compressibility for N2 dips below 1.0, it dose not show as much deviation as that for CH4, which in turn does not show as much deviation as the compressibility for CO2. – Based on this behavior we can surmise that the importance of intermolecular interactions increases in this order: H2 < N2 < CH4 < CO2 – This order is reflected by the relative a values for these gases in Table 5.3. 5.12 Chemistry in the Atmosphere – The gases that are most important to us are located in the atmosphere that surrounds the earth’s surface. – The average composition of the earth’s atmosphere near sea level, with the water vapor removed, is shown in Table 5.4. – The atmosphere is a highly complex and dynamic system, but for convenience we divide it into several layers based on the way the temperature changes with altitude. (The lowest layer, called the troposphere, is shown in Fig. 5.27.) – The chemistry occurring in the troposphere is strongly influenced by human activities. Millions of tons of gases and particulates are released into the troposphere by our highly industrial civilization. FIGURE 5.27 The variation of temperature and pressure with altitude. Note that the pressure steadily decreases with increasing altitude but that the temperature does not change monotonically. – Actually, it is amazing that the atmosphere can absorb so much material with relatively small permanent changes. – Significant changes, however, are occurring. Severe air pollution is found around many large cities, and it is probable that long-range changes in the planet’s weather are taking place. – The two main sources of pollution are transportation and the production of electricity. The combustion of petroleum in vehicles produced CO, CO2, NO, and NO2, along with unburned molecules from petroleum. – The complex chemistry of polluted air appears to center on ozone and the nitrogen oxides (NOx). – At the high temperatures in the gasoline and diesel engines of cars and trucks, N2 and O2 react to form a small quantity of NO, which is emitted into the air with the exhaust gases (see Fig. 5.28). FIGURE 5.28 Concentration (in molecules per million molecules of “air”) of some smog components versus time of day. After P.A. Leighton’s classic experiment, “Photochemistry of Air Pollution,” in Physical Chemistry: A Series of Monographs, ed. Eric Hutchinson and P. Van Rysselberghe, Vol. IX, New York: Academic Press, 1961. – This NO is oxidized in air to NO2, which in turn absorbs energy from sunlight and breaks up into nitric oxide and free oxygen atoms: Radiant energy NO2(g) ⎯⎯→ NO(g) + O(g) – Oxygen atoms are very reactive and can be combine with O2 to form ozones: O(g) + O2(g) → O3(g) – Although represented here as O2, the actual oxidant is an organic peroxide such as CH3COO, formed by reaction of O2 with organic pollutants. – Ozone in also very reactive. It can react with the unburned hydrocarbons in the polluted air to produce chemicals that cause the eyes to water and burn and are harmful to the respiratory system. – 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. NO2(g) → NO(g) + O(g) O(g) + O2(g) → O3(g) NO(g) + 1/2O2(g) → NO2(g) Net reaction: 3/2O2(g) → O3(g) – We can observe this process by analyzing polluted air at various times during a day (see Fig. 5.28). – Current efforts to combat the formation of photochemical smog are focused on cutting down the amounts of molecules from unburned fuel in automobile exhaust and designing engines that produce less nitric oxide (see Fig. 5.29). FIGURE 5.29 Our various modes of transportation produce large amounts of nitrogen oxides, which facilitate the formation of photochemical smog. – The other major source of pollution results from burning coal to produce electricity. Much of the coal found in Midwest contains significant quantities of sulfur, which, when burned, produces sulfur oxide: S(In coal) + O2(g) → SO2(g) – A further oxidation reaction occurs when sulfur dioxide is changed to sulfur trioxide in the air: 2SO2(g) + O2(g) → 2SO3(g) – The production of sulfur trioxide is significant because it can combine with droplet of water in the air to form sulfuric acid: SO3(g) + H2O(l) → H2SO4(aq) – Sulfuric acid is very corrosive to both living things and building materials. Another result of this type of pollution is acid rain (see Fig. 5.30). FIGURE 5.30 A helicopter dropping lime in a lake in Sweden to neutralize excess acid from acid rain. – One way to use high-sulfur coal without further harming the air quality is to remove the sulfur oxide from the exhaust gas by means of a system called a scrubber before it is emitted from the power plant stock. A common method called scrubbing involves blowing powdered limestone (CaCO3) into the combustion camber, where it is decomposed to lime and carbon dioxide: CaCO3(s) → CaO(s) + CO2(g) – The lime then combines with the sulfur dioxide to form calcium sulfite: CaO(s) + SO2(g) → CaSO3(s) – The calcium sulfite and any remaining unreacted sulfur dioxide are removed by injecting an aqueous suspension of lime into the combustion chamber and the stack, producing a slurry (a thick suspension), as shown in Fig. 5.31. FIGURE 5.31 Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants. – Unfortunately, there are many problems associated with scrubbing. The systems are complicated and expensive and consume a great deal of energy. The large quantities of calcium sulfite produced in the process present a disposal problem. The Important of Oxygen Oxygen has been present only for the last half of the earth’s history – appearing about 2.5 billion years ago when organisms began to use chlorophyll to convert sunlight to stored energy. The burial of plants to begin the formation of coal and petroleum also released oxygen into the atmosphere. In contrast, oxygen was removed from the atmosphere by mountain building and erosion as freshly exposed rocks combined with oxygen to form oxygen-containing minerals. Atmospheric Concentration of Oxygen Acid Rain: An Expensive Problem Rainwater, even in pristine wildness areas, is slightly acidic because some of the carbon dioxide present in the atmosphere dissolves in the raindrops to produce H+ ions by the following reaction: H2O(l) + CO2(g) → H+(aq) + HCO3–(aq) This process produces only very small concentrations of H+ ions in the rainwater. However, gases such as NO2 and SO2, which are by-products of energy use, can produce significantly higher H+ concentrations. Nitrogen dioxide reacts with water to give a mixture of nitrous acid and nitric acid: 2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq) Sulfur dioxide is oxidized to sulfur trioxide, which then reacts with water to form sulfuric acid: 2SO2(aq) + O2(g) → 2SO3(g) SO3(g) + H2O(l) → H2SO4(aq) The damage cause by the acid formed in polluted air is a growing worldwide problem. Lakes are dying in Norway, the forests are sick in Germany, and buildings and status are deteriorating all over the world. Both marble and limestone react with sulfuric acid to form calcium sulfate. The process can be represented most simply as CaCO3(s) + H2SO4(aq) → Ca2+(aq) + SO42–(aq) + H2O(l) + CO2(g) What can be done to protect limestone and marble structures from this kind of damage? Of course, one approach is to lower sulfur dioxide and nitrogen oxide emissions from power plant. In addition, scientists are experimenting with coatings to protect marble from the acidic atmosphere. However, a coating can do more harm than good unless it “breathes”. If moisture trapped beneath the coating freeze, the expanding ice can fracture the marble. Needless to say, it is difficult to find a coating that will allow water to pass but not allow acid to pass, so the search continues. The damaging effects of acid rain can be seen by comparing these photos of a decorative statue on the Field Museum in Chicago. The photo on the left was taken c. 1920; the photo on the right was taken in 1990. Recent renovation has since replaced the deteriorating marble. Homework – 25, 39, 45, 67, 69, 81, 97, 109, 113, 125. 146