Chapter 5  The Gas Phase
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Chapter 5 The Gas Phase

Course Number: CHEM 425, Spring 2010

College/University: Chicago State

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Chapter 5: The Gas Phase CH1010-1040 homepage Section 5-1: Macroscopic Properties of Gases Section 5-2: The Ideal Gas Law Section 5-3: Kinetic Molecular Theory Section 5-4: The Historical Role of Gas Phase Chemical Reactions Section 5-5: Gas Production via Double Displacement Reactions Applications Major Concept Area: Electrical Forces and Bonding Specific Concepts in this Chapter: In the gas phase, molecules are...

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5: Chapter The Gas Phase CH1010-1040 homepage Section 5-1: Macroscopic Properties of Gases Section 5-2: The Ideal Gas Law Section 5-3: Kinetic Molecular Theory Section 5-4: The Historical Role of Gas Phase Chemical Reactions Section 5-5: Gas Production via Double Displacement Reactions Applications Major Concept Area: Electrical Forces and Bonding Specific Concepts in this Chapter: In the gas phase, molecules are far apart. The behavior of all gases is described by a simple equation called the ideal gas law, PV = nRT. The kinetic energies of gas molecules are distributed according to the Maxwell-Boltzmann equation. The average kinetic energy of a gas molecule, KEmolecule, is 3kT/2, where T is the absolute temperature and k is a constant of nature. In this chapter we begin an examination of the phases (physical states) in which a pure substance may exist. For most substances there are three phases -- solid (s), liquid (l), and gas (g). The gas phase is the simplest of the three to deal with theoretically, primarily because the molecules behave completely chaotically and therefore give rise to simply-formulated average properties. The solid phase is also relatively easily dealt with, because it is so highly organized. The liquid phase, intermediate in its degree of order, is the most difficult to model. This chapter deals with the gas phase. A subsequent chapter deals with the solid and liquid phases. We follow the chapter on the liquid phase with chapters on the energetics of conversions among phases, and equilibria between two or more phases. Of interest throughout will be the question, "Why do some substances exist as solids, some as liquids, and still others as gases at room temperature and subjected to 1 atmosphere of pressure?" Gases and their properties play an ubiquitous and critical role in our daily lives. Life is supported by the oxygen of the air, which reacts with glucose in biochemical combustion. Living systems have evolved complex molecular architecture to absorb oxygen from the air; carry it to the cells; deliver it to the cells; transport it within the cell; and supervise its multistep, controlled reaction with glucose. It is the rapid expansion of gases produced in the explosive combustion of gasoline that performs the work of the internal combustion engine. Expansion and contraction of gases in response to temperature and pressure changes is at the basis of continually changing weather patterns. Clearly an understanding of the properties, behavior, and chemical reactivity of gases is an important foundation for developing an appreciation of complex biological and technological systems. These properties, behaviors, and reactivities are the focus of this chapter. Following a discussion of the ideal gas law, a simple mathematical model for gas behavior, we discuss the Kinetic Molecular Theory of gases, which establishes a link between molecular motion and temperature. We finish the chapter with a look at the historical significance of gas phase chemical reactions in chemistry, and a treatment of double displacement reactions that produce gases as products. 5-1 Macroscopic Properties of Gases. All gases have observable properties in common: Gases flow readily from one space to another and occupy all available space. Gases assume the shapes of their containers. Gases are readily compressible. By exerting force, we can cause a gas to occupy a smaller volume. Two or more gases form homogeneous mixtures (solutions) in all proportions. An example is air, a mixture of primarily nitrogen and oxygen with small amounts of other gases. Gases diffuse rapidly. This means that a gas can move across a large space in a relatively short time. Examples are the odors of perfume and skunk. The state of a gas is described by the values of 4 macroscopic variables: the volume (V) occupied by the gas; the temperature (T) of the gas; the pressure (P) that the gas exerts on the walls of its container; and the amount of gas in moles (n). The values of these quantities are related by a very simple equation called the ideal gas law. We will discuss this equation after a brief word on pressure. Pressure. Pressure is defined as force per unit area. We illustrate the concept of pressure with an example. Example 5-1. The bottom of each foot of a 150-pound man has a surface area of about 24 square inches. When the man stands on a bathroom scale, what pressure is exerted by his feet? Solution. The force exerted by the feet is the man's weight, 150 lb. This is distributed over the area of the bottoms of both feet, or 48 square inches. The pressure is then P = F/A = 150 lb/ 48 in2 = 3.1 lb/in2 If the man stands on the scale with only one foot, the force is the same, but the pressure is doubled because the area is cut in half. Gases exert pressure on the walls of their containers. Since a gas uniformly occupies a container, it exerts the same pressure on all the walls. The gases that make up the atmosphere exert pressure on the surface of the earth. This pressure, called the atmospheric pressure, can be measured using a barometer, pictured in Figure 5-1. The left side of the figure shows a beaker or cup containing mercury. A glass tube of length 1 m and cross-sectional area, A, fitted with a stopcock on top, is inserted into the beaker so that the end of the tube is below the level of mercury in the beaker. Initially, the stopcock is open and the levels of mercury inside and outside the glass tube are the same. A vacuum pump is then attached to the top of the glass tube, and the air is pumped out. As this occurs, the mercury level rises in the tube until it reaches a height, h. Despite further pumping, the mercury level rises no higher in the tube. When all air has been pumped out of the tube, the stopcock is closed, and the mercury maintains its position in the tube. This behavior is understood as follows. Initially, before pumping, the atmosphere exerts the same pressure on the surfaces of mercury inside and outside the tube. Since Pout = Pin = atmospheric pressure, the mercury levels inside and outside the tube are the same. As air is pumped out of the tube, its pressure on the mercury inside the tube decreases, and mercury rises in the tube to a height, h, such that the force due to the mass of mercury in the tube just counterbalances the pressure of the atmosphere on the mercury outside the tube. The height h is found to vary from day to day because the pressure of the atmosphere varies with weather conditions. However, its value is always in the neighborhood of 0.76 meters. For this reason, a unit of pressure called the atmosphere (atm) is defined as the pressure that will support a column of mercury 76 cm, or 760 mm, in height. 1 atm = 760 mm Hg The pressure exerted by a column of mercury 1 mm in height is called a torr after Evangelista Torricelli, inventor of the barometer. Thus 1 atm = 760 mm Hg = 760 torr In the English and SI systems, the units of pressure are respectively the pound per square inch and the pascal (1 Pascal = 1 Newton per square meter, where a Newton is the force required to accelerate a 1 kg body at 1 m/s2). The pressure in Pa exerted by a column of mercury 760 mm in height can be readily calculated and will provide a conversion factor between SI and traditional non-SI units. The force exerted by a mass, m, of mercury is mg, where g is the acceleration due to gravity. This force is distributed over the cross sectional area, A, of the tube. The mercury therefore exerts the pressure, P = mg/A on the surface of mercury at the bottom of the tube. The mass of mercury can be calculated from its density, , and its volume, which is Ah. Substituting for m in the expression for pressure gives equation 5-1-1: (5-1-1): P = *(Ah)g/A = *gh The pressure exerted by a column of mercury (or any liquid) of height h is independent of the cross sectional area of the column. The diameter of the barometer tube is therefore unimportant. Substituting the density of mercury (13.595 g/cm3), the acceleration constant (g = 9.80665 m/s2), and h = 0.760 m in 5-1-1 gives P = 1.0132 x 105 Nm-2 (1 Nm-2 = 1 Pascal). The following conversion relations are thus obtained: 1 atm = 14.7 lb/in2 = 1.013 * 105 Pa (1 Pa = 1N/m2) = 1.013 bar (1 bar = 105 Pa) Although the bar is official, the atmosphere and the torr are still the most commonly used pressure units in chemistry. We will therefore use the atm and the torr. Note that atmospheric pressure is 14.7/3.1 = 4.7 times greater than the pressure exerted by the feet of the man in Example 1-1. The magnitude of the pressure exerted by the atmosphere is equivalent to having that 150-lb man, holding a 550-lb weight, standing on your stomach. The pressure of a gas confined in a container cannot be measured with a barometer. For this a variety of devices may be used, the simplest being the manometer. A manometer is a U-shaped glass tube containing mercury or some less toxic fluid. The gas container is attached to one side of the manometer, and the other side is either open to the atmosphere (an open-end manometer) or evacuated (a closed-end manometer). An open-end manometer is shown in Figure 5-2. The difference in the levels of fluid in the two arms of the manometer is a measure of the pressure of gas in the bulb through the relationship, P1 = P2 + P. Here P1 is the pressure exerted by the gas on arm 1 of the manometer, P2 is the pressure exerted on the arm of the manometer not connected to the gas container (arm 2), and P is the height of fluid in arm 2 minus the height in arm 1. P is positive if fluid is higher in arm 2 and negative if it is higher in arm 1. For an open-end manometer, P2 is atmospheric pressure; for a closed-end manometer, P2 is zero. If the manometer fluid is mercury, the pressure is in torr and may be converted to atm using the conversion factor given above. If the manometer contains some other fluid, the height of fluid may be converted to an equivalent height of mercury using the densities of the fluid and mercury. Example 5-2. An open-end manometer containing water as the working fluid is used to measure the pressure of gas in a bulb. The level of water in the side attached to the bulb is 8.65 cm higher than that in the side open to the atmosphere. Atmospheric pressure is 693 torr. What is the pressure of gas in the bulb? Solution. Use the relation P = *g*h. To find the height of mercury equivalent to 8.65 cm of water, we equate *gh for water to *gh for mercury. The gravitational acceleration cancels, giving (1.00 g/cm3)(86.5 mm) = (13.60 g/cm3)(h) h(Hg) = 6.4 mm The pressure of gas is this amount less than atmospheric pressure: Pgas = 693 - 6.4 = 687 torr. 5-2 The Ideal Gas Law. The ideal gas law is a relationship between the pressure (P), the volume (V), the amount (n), and the temperature (T) of a gas. It is presented in Equation 5-2-1. The ideal gas law is called an equation of state, because it represents all possible states -- combinations of P,V and T -- in which the amount of gas, n, can be found. (5-2-1): PV = nRT In this equation, P is the pressure in atm; V is the volume in L; n is the amount of gas in moles, and T is the Kelvin temperature. R is a proportionality constant, remarkably the same for all gases, which relates the 4 variables. Equation 5-2-1 is of simple form, which belies the effort expended in developing it. Over 200 years elapsed between the first quantitative experiments on gases and the final enunciation of equation 5-2-1. The simple form of the equation hides a remarkable sophistication in understanding. We now examine what the equation says regarding gas behavior. Volume-Pressure relation. If we rearrange the equation to solve for the volume, the equation states that volume decreases as the pressure exerted on (also by) the gas increases. The inverse relation of pressure and volume is a fact of every day experience. For example, when you compress the air in the chamber of a tire pump, you feel the increase in pressure of the air within. In 1662, Robert Boyle developed the quantitative relationship between pressure and volume that we now know as Boyle's Law, which states that the volume occupied by a fixed amount of gas at a fixed temperature varies inversely with the pressure exerted on the gas. Boyle's Law was the first-discovered component of equation 5-2-1. Note that the ideal gas law predicts that the volume should fall to 0 as pressure gets very large. For real gases this does not happen. At ordinary temperatures, most real gases will liquefy if pressure is made large enough. Once liquefication has occured, volume does not decrease much with increasing pressure. Real gases obey equation 5-2-1 closely only when P is not too large. This is why equation 5-2-1 is called the ideal gas law -- it applies strictly only to an idealized gas, which does not liquefy but instead shrinks to zero volume when pressure is made very large. Volume-Temperature relation. The ideal gas law states that the volume occupied by an amount of gas, n, varies directly with the absolute temperature of the gas. A particular sample of gas, then, should double in volume if heated from 300 to 600 K. This behavior, too, is quite well known, at least qualitatively, in every day experience, as in the expansion (contraction) of a heated (cooled) balloon. The relationship between gas volume and Celsius temperature was first established by Jacques Charles in 1800. He clearly established the linear relationship between the two quantities. Similar experiments relating gas pressure and temperature were carried out at about the same time by Gay-Lussac, and similar proportionality was found. In both cases, the relationships obtained were not so simple as the ideal gas law, because the absolute scale of temperature was not then recognized. The relation between volume and Celsius temperature is in equation 5-2-2. (5-2-2): V = *t(oC) + Plots of V versus t(oC) for gas samples containing different amounts of gas give different slopes and different intercepts on the V axis. However, they all give the same intercept on the t axis, -273.16 oC. This suggests that the zero of temperature be shifted 273.16 Celsius degrees to the left. The resulting scale of temperature is, of course, the Kelvin scale. Equation 5-2-2 takes the much simpler form, V = *T. A profound implication of equation 5-2-2 is that there exists a temperature at which the volume of gas should become zero. This temperature is called absolute zero. The Kelvin scale is set up so that 0 K corresponds to the temperature at which the volume of an ideal gas sample would become zero. This corresponds to -273.16 on the Celsius scale. Any temperature lower than this value would cause the gas to have negative volume. Since this is physically meaningless, the conclusion is that temperatures lower than absolute zero do not exist. That there is a lower limit on attainable temperature is not obvious from our everyday experience and from the manner in which we normally measure and think about temperature; the conclusion is nonetheless correct. Relation Between Volume and Amount of Gas. The ideal gas law states that the volume occupied should increase in direct proportion with the amount (moles) of gas, when the gas is kept at the same pressure and temperature throughout. This is an extension of the hypothesis by Amadeo Avogadro in 1811 that equal volumes of different gases contain the same number of molecules. Avogadro made this statement in explanation of many observations about the relative volumes of gaseous elements that react to form compounds. It was in fact this hypothesis that finally allowed correct molecular formulas to be deduced and the atomic mass scale to be put on a firm basis. The Value of the Gas Constant R. The gas constant R turns out to be an ubiquitous and extremely important quantity in physical science. We begin by obtaining a value for it based on the experimentallymeasured volume of 1.0 mole of gas at 1.00 atm pressure and a temperature of 273.16 K. It is found that under these conditions the gas occupies a volume of 22.414 L. Rearranging equation 5-2-1 to solve for R, we can calculate its value: R = PV/nT = (1.000 atm)(22.414 L)/(1.000 mole)(273.16K) = 0.08206 L-atm/K-mole The numerical value of R depends upon the units in which pressure and volume are expressed. P is commonly expressed in atmospheres or torr (both non-SI) or Pascal (SI), and V is commonly expressed in mL. Values of R in common P-V unit combinations are presented in Table 5-1. Table 5-1: R in Common Pressure-Volume Unit Systems Pressure Unit Volume Unit R atm bar torr Pa L L mL m3 0.08206 L-atm/K-mole 0.08314 L-bar/K-mole 6.237 x 104mL-torr/K-mole 8.314 J/K-mole From the Pascal-m3 value, we see that R has units of energy/K-mole. It follows that the combined units Lbar, L-atm, and mL-torr are also energy units. We can express R in terms of any desired energy unit by using an appropriate conversion factor (see Appendix C). In using the ideal gas law, equation 5-2-1, the units of P, V, and R must be self-consistent, and the temperature must be expressed in Kelvins. A few examples of the use of the ideal gas law are now presented. Example 5-3. An ideal gas occupies 326 mL at 730 torr of pressure and a temperature of 19oC. What is its volume at 1.00 atm and 0oC? Solution. We take two approaches to the solution of this problem. In the first approach, we use the ideal gas law twice. First, we calculate the moles of gas under the original V,T,P conditions; second, we calculate the volume occupied by this number of moles at 1 atm and 0oC. Convert to the necessary units of T, P: Initial T(K) = 273 + 19 = 292 K Final T(K) = 273 K P = 730 torr = 730/760 atm = 0.961 atm P = 1.00 atm V = 326 mL = 0.326 L V is unknown Calculate the number of moles of gas under the initial conditions: n = PV/RT = (0.961 atm)(0.326 L)/(0.08206 L-atm/K-mole)(292K) = 1.307 * 10-2 moles Calculate volume at final T and P: V = nRT/P = (1.307 * 10-2 moles)(0.08206 L-atm/K-mole)(273K)/(1.00 atm) = 0.29 L (2 significant figures) To shortcut this long process we take advantage of the constancy of the amount (moles) of gas. Since n is constant, PV/RT must also be constant, and its initial and final values are the same: PiVi/Ti = PfVf/Tf Solving for Vf and substituting numbers gives Vf = 0.29 L. Example 5-4. 3.24 * 10-4 moles of ideal gas occupies 5.31 mL at 25oC. Calculate the gas pressure. Solution. Convert units, and apply the ideal gas law to the problem. Convert units: T(K) = t(oC) + 273 = 25 + 273 = 298K V(L) = 5.31 mL/(1000 mL/L) = 5.31 * 10-3 L Calculate P: P = nRT/V = (3.24 * 10-4)(0.08206)(298)/(0.00531) = 1.49 atm Example 5-5. Consider the apparatus in Figure 5-3. Initially, a quantity of gas is confined in the left box at a pressure of 3.40 atm and a temperature of 286 K. 4 sequential processes, a, b, c, and d, are carried out. Work out the value of each indicated quantity after the completion of each process. 1. Valve 1 is opened with valve 2 closed. T is kept constant. What are the final volume and pressure of the gas? 2. T is raised to 350 K. What are the final V and P of the gas? 3. Valve 1 is closed, valve 2 is opened. What are the final V and P of the gas to the right of valve 1? 4. T is lowered to 300 K. What are the final V and P of the gas to the right of valve 1? Solution. We apply the ideal gas law to each process in turn. 1. When valve 1 is opened, the gas will expand to occupy its original box and the 3.0-L box at uniform pressure. T is constant, and the amount of gas is too. The gas volume goes up by a factor of 4, so the pressure goes down by the same factor. V occupied by gas = 4.0 L P of gas = 3.4/4 = 0.85 atm 2. When T is raised to 350 K, the gas pressure will increase in direct proportion, because V will remain constant. Therefore V occupied by gas = 4.0 L P of gas = TfPi/Ti = (350)(0.85)/(286) = 1.04 atm 3. When valve 1 is closed, the gas in the 1.0-L box will no longer communicate with the gas in the middle box. Only the gas in the middle box will be involved when valve 2 is opened. This gas will expand from its initial pressure of 1.04 atm until its final pressure is 0.25 atm. At this point, the piston in the cylinder will stop moving. V occupied by gas to right of valve 1 = PiVi/Pf = (1.04)(3.0)/(0.25) = 12.48 L. P of gas to the right of valve 1 = 0.25 atm 4. When T is lowered to 300 K, the volume occupied by the gas will decrease, as the piston moves in to maintain pressure at 0.25 atm. V occupied by gas to the right of valve 1 = ViTf/Ti = (12.48 L)(300)/(350) = 10.70 L. P of gas to the right of valve 1 = 0.25 atm Example 5-6. 0.432 g of an ideal gas exerts pressure of 0.46 atm in a volume of 768 mL at 0oC. Calculate the molar mass of the gas. Solution. The ideal gas law can be expressed directly in terms of MM because moles is mass over MM: n = PV/RT = (mass)/MM Rearrange for MM: MM = mRT/PV, where m is mass. Solve: MM = (0.432 g)(0.08206 L-atm/K-mole)(273 K)/(0.46 atm)(0.768 L) = 27.4 g/mole The gas could be nitrogen, N2; carbon monoxide, CO; or ethylene, C2H4. This method for determining the MM of an unknown gas or volatile liquid is called the Dumas method. It is simple, fast, and reasonable accurate. Example 5-7. Calculate the density of oxygen gas at a pressure of 0.940 atm at room temperature (25oC). Solution. Rearrangement of the ideal gas law with the recognition that density is mass per volume gives density directly. Rearrange: PV = nRT n/V = m/(MM)V = /MM = P/RT = P(MM)/RT Solve the equation with the specific variables given: = (0.940 atm)(32.00 g/mole)/(0.08206 L-atm/K-mole)(298K) = 1.23 g/L The value has units g/L, and is about 1000 times less than the density of water at the same temperature. Gases are much less dense than liquids, implying that the gas molecules are much further apart than are liquid molecules. Dalton's Law of Partial Pressure. The ideal gas law can be applied to mixtures of gases as well as to single pure gases. Consider a mixture of two non-reacting gases, A and B, in a container of volume V. According to the ideal gas law, the pressure exerted by the mixture is given by equation 5-2-3. (5-2-3): PT = nTRT/V Here nT, the total moles of gas, is the sum of the moles of A and the moles of B: (5-2-4): nT = nA + nB Substitute equation 5-2-4 into equation 5-2-3 and expand: (5-2-5): PT = (nA + nB)RT/V = nART/V + nBRT/V Suppose that we could remove all of the molecules of B from the container. What pressure would A exert in the container alone? This is easily obtained from the ideal gas law to be PA = nART/V. Similarly, PB = nBRT/ V. Substituting the expressions for PA and PB into equation 5-2-5 gives equation 5-2-6: (5-2-6): PT = PA + PB This is Dalton's Law of partial pressures. In words, it says that the total pressure of a mixture of gases (PT) is the sum of the partial pressures of the gases composing the mixture. The partial pressure of gas A is the pressure that nA moles of A would exert if present alone in the container. Dividing the expression for PA by equation 5-2-5 gives equation 5-2-7: (5-2-7): PA/PT = nA/nT = nA/(nA + nB) The ratio of the moles of A to the total moles of all gases in the system is called the mole fraction of A, and is symbolized XA. Thus XA = nA/nT. Substituting this into 5-2-7 and rearranging, we obtain equation 5-2-8: (5-2-8): PA = XAPT The partial pressure of A in a mixture of A and other gases is the total pressure multiplied by the mole fraction of A. Example 5-8. A gas mixture containing equal numbers of nitrogen and oxygen molecules exerts 1.32 atm pressure. What is the partial pressure of nitrogen? Solution. Since half the molecules are nitrogen, half the moles are also nitrogen, so the mole fraction of nitrogen is 0.5. PN2 = 0.5 (1.32 atm) = 0.66 atm We close this section by emphasizing an implicit assumption of the above treatment. Figure 5-4 shows two gases, A and B, in a box. If the box has volume V, what fraction of this volume is occupied by A, and what fraction by B? A frequent answer might be that since half the molecules in the box are A, A occupies half the volume. This is, however, incorrect. If, when we fill the box with gas, we put A in first, the molecules of A will spread out to occupy the whole box. Now add some molecules of B. Since these molecules are "unaware" of the presence of A, they, too, will spread out in the entire volume. Thus both gases occupy the same volume -- the entire volume of the container. VA = VB = V, the box volume. This is implicit in equations 5-2-3 and 5-2-5. Similarly, we assumed that temperature is the same for both gases: TA = TB = T. In contrast, the total moles, nT, is the sum of the moles of A and B; and the total pressure is the sum of the partial pressures. These relationships are summarized in Table 5-2. Table 5-2 Relationships Among State Variables for a Gas Mixture TA = TB = Tmixture nA + nB = nmixture PA + PB = Pmixture VA = VB = Vmixture Dalton's Law is particularly useful in experimental situations involving the collection of gases over liquids. In such situations, the space above the liquid contains not only molecules of the gas, but also molecules of the vapor form of the liquid, produced by evaporation of the liquid. The vapor exerts a partial pressure that is part of the total pressure exerted by the gas mixture over the liquid. This partial pressure due to vapor that exists above the corresponding liquid phase is the vapor pressure, Pvap, of the liquid. According to Dalton's Law, PT = Pgas + (Pvap)liquid To calculate Pgas, we must correct the total pressure for the vapor pressure of the liquid, which we can look up in a table. Example 5-9. 0.1140 g of an unknown metal M reacts with excess aqueous HCl according to the reaction M + 2H+ ---> M2+ + H2(g). The H2(g) is collected over water. The volume of the gas above the water is 52.1 mL, the total pressure of the gas is 738 torr, and the temperature is 22.1oC. The vapor pressure of water at 22.1oC is 20.0 torr. Calculate the molar mass of the metal. Solution. Strategy: PT ---> PH2 ---> moles H2 ---> moles M ---> molar mass M. Use Dalton's Law to calculate the pressure of H2 in the gas mixture above the water: Ptotal = PH2 + (Pvap)H2O PH2 = Ptotal - (Pvap)H2O = 738 - 20 = 718 torr Calculate the moles of H2 from the ideal gas law: nH2 = PV/RT = (718/760 atm)(0.0521 L)/(0.08206 L-atm/K-mole)(295.1 K) = 2.033 * 10-3 moles Calculate moles M from stoichiometry: moles M = 2.033 * 10-3 moles H2 * 1mole M/1 mole H2 = 2.033 * 10-3 moles M Calculate molar mass of M from known mass and moles: MM = 0.1140g/2.033 * 10-3 moles = 56.1 g/mole The metal is probably Fe. Example 5-10. Figure 5-5 shows a syringe that has been filled with N2(g) to a volume of 20.0 mL. The end of the syringe has been capped with a rubber septum (a membrane that may be penetrated to add or remove substances from the syringe, but which otherwise seals the syringe off). 2.50 mL of pure liquid ethanol, C2H6O(l), is injected into the syringe. It is observed that the syringe piston moves out until the occupied volume in the syringe is 32.00 mL. Atmospheric pressure is 730.0 torr, and the temperature of the syringe and its contents is 25.00oC. Calculate the pressure exerted by ethanol vapor in the syringe. Solution. See if you can figure this out on your own. 5-3 Kinetic Molecular Theory. The ideal gas law was developed from macroscopic observations, with no knowledge of the behavior of a gas at the molecular level. We now attempt to interpret the ideal gas law in terms of a reasonable molecular model (simple physical picture) of gas behavior. What are the individual molecules doing, and why does their behavior manifest itself in the ideal gas law? We now enter the domain of theory -- a mental interpretation of experimental results. The model of gas behavior currently accepted by scientists is the Kinetic Molecular model. The theory of gas behavior built on this model is called the Kinetic Molecular Theory (KMT). This theory is one of the oldest and most resoundingly successful in science. First, we assume that gases consist of very small molecules. Since a gas occupies the entire volume of its container, and since gases readily diffuse, the molecules must move about in space. Consequently they have kinetic energy -- energy of motion. Since gases are compressible, the molecules must be far apart. Focussing on an individual molecule, it seems reasonable to assume that it is a small particle that moves about at random, occasionally colliding with a wall of the container and with other gas molecules. The molecule sometimes gains and sometimes loses energy in these collisions, so that it frequently changes speed. Sometimes it moves rapidly, sometimes slowly. Thus our picture of the gas is dynamic rather than static. Figure 5-6 is an attempt to portray this simple physical model of the gas phase. These statements about gas molecules are consistent with the properties of gases listed at the beginning of the chapter. None of them have been proven, but they seem reasonable and constitute the postulates of the Kinetic Molecular Theory: 1. 2. 3. 4. 5. Gases consist of molecules that behave like tiny hard spheres; The molecules have no volume -- they may be treated as points; There are no forces of attraction between molecules; Collisions between molecules are elastic (kinetic energy is conserved); The molecules are in ceaseless random motion, frequently colliding with the container and walls with each other. A collection of gas molecules is highly disordered. 6. The average kinetic energy of the molecules is proportional to Kelvin temperature. We now pursue the consequences of these postulates. Postulate 6, the origin of which is not obvious, will be considered later. Our aim is to develop an expression for pressure. At the molecular level, pressure must result from collisions of gas molecules with the container walls. A molecule hitting the wall exerts a force on it. The collection of all such forces on a unit area of wall during a given time interval constitutes the pressure. We should be able to calculate P from the force exerted by each collision, multiplied by the number of collisions per unit area of wall: (5-3-1): P = Force/Area = Force/collision x Collisions/area From Newton's second law, force is the rate of change of momentum. Thus F = momentum change/time-collision x collisions/area We move the time factor over to the second term to obtain equation 5-3-2. (5-3-2): F = momentum change/collision x collisions/time-area To obtain the momentum change per collision, envision a molecule moving directly toward a wall with velocity v, as shown in Figure 5-7. Since momentum is conserved in any collision, the change in momentum of the molecule is momentum change = momentum after collision - momentum before = m(-v) - m(v) = -2mv The change of momentum of the wall must therefore be 2mv to give a total change of zero. We now have the first term in equation 5-3-2: (5-3-3): momentum change/collision = 2mv We take an intuitive approach to the second term. The number of collisions per unit time per unit area of wall should depend on 1) the number of molecules per unit volume in the container, N/V (the more there are, the more collisions there should be with the walls); and 2) the speed v at which a molecule moves (the faster the movement, the more collisions per unit time that should occur). Equation 5-3-4 is based on these ideas: (5-3-4): collisions/time-area = (N/V)(v) If we assume for simplicity that the container is cubical, there are 6 walls over which these collisions must be spread. The collisions with a particular wall are then (5-3-5): collisions/time-area = Nv/6V (Despite our restrictive assumptions about the shape of the container, equation 5-3-5 turns out to be valid for a container of any shape!) We now multiply the expressions in equations 5-3-3 and 5-3-5 to obtain the pressure: (5-3-6): P = (2mv)(Nv/6V) = Nmv2/3V Rearrangment gives equation 5-3-7. (5-3-7) PV = Nmv2/3 This is as far as our molecular model takes us. We are now at the interface between theory and experiment. Experiment (the ideal gas law) relates pressure and volume to temperature: (5-2-1): PV = nRT Theory, equation 5-3-7, relates pressure and volume to the mass and speed of a gas molecule. To bring experiment and theory into correspondence, we must equate the right sides of the preceding two equations. In order for KMT to successfully explain the ideal gas law, it is necessary that (5-3-8): nRT = Nmv2/3 Equation 5-3-8 can be simplified by replacing n with N/No (No is Avogadro's Number); mv2/2 by KE (the kinetic energy of an average gas molecule); and solving for KE. The result is (5-3-9): KEmolecule = 3RT/2No Hence postulate 6! We refine this equation by making two further realizations. First, since R and No are both constants of nature, so must their ratio be. It is symbolized k, and is called Boltzmann's constant. Its value is 1.381 x 10-23 J/K-particle. Second, we have recognized that the kinetic energy of a gas molecule changes frequently as it undergoes collisions. However, its average speed over time is constant. The refined equation, 5-3-10, is among the most important equations in science: (5-3-10): Average KEmolecule = 3kT/2 The average kinetic energy of a gas molecule, no matter what its chemical identity, depends only on the Kelvin temperature. This gives a deep insight into the concept of temperature: it is a measure of the average kinetic energies of the molecules of a substance, hence a measure of their average speed. As T is increased, molecules move faster; as it is lowered, they move slower. The temperature at which molecular motion ceases is absolute zero. Temperatures lower than absolute zero are impossible because a molecule may not have negative kinetic energy. Equation 5-3-10 is simple and profound. Is there an experiment that can be done to test its validity? There is in fact a simple experiment that verifies equation 5-3-10 in a rearranged form. If we replace the average kinetic energy of a molecule with the expression m(v2)avg/2, where (v2)avg is the average of the square of the speed, and solve the resulting expression for (v2)avg, we obtain equation 5-3-11. (5-3-11): (v2)avg = 3kT/m The square root of (v2)avg is called the root mean square speed, and is for our purposes approximately equal to the average molecular speed. Making this equality gives equation 5-3-12. (5-3-12): vavg = (3kT/m)1/2 Multiplying both numerator and denominator of the argument on the right side of this equation by Avogadro's number No gives the useful variant in equation 5-3-13. (5-3-13): vavg = (3RT/MM)1/2 The implication of equation 5-3-12 and 5-3-13 is that a heavy gas molecule moves more slowly than a light one, in a quantifiable way. The ratio of the speeds of the light (L) and heavy (H) molecules should be the square root of the inverse ratio of their molar masses: (5-3-14): (vH/vL)avg = (MML/MMH)1/2 In 1846, Thomas Graham measured the rates of diffusion of various gases. Diffusion is the process by which gases disperse in space via random molecular motion. The results of Graham's experiments are summarized in equation 5-3-15: (5-3-15): rate diffusion of gas A/rate of diffusion of gas B = (density of gas B/density of gas A)1/2 But we have previously seen (Example 5-7) that gas density is proportional to MM. If we make the logical assumption that a gas diffuses at a rate that is directly proportional to the average speed of its molecules, then Graham's Law of Diffusion is in exact agreement with equation 5-3-14. Example 5-11. Calculate the ratio of the speeds of H2 and CO2 molecules at 25oC. Solution. As long as temperature is the same for both molecules, its actual value is unimportant. Using equation 5-3-14, vH2/vCO2 = (MMCO2/MMH2)1/2 = (44.0/2.02)1/2 = 4.67 Example 5-12. Calculate the average speed of a molecule of nitrogen gas at room temperature, 25oC. Solution. Use equation 5-3-13. vN2 = (3RT/MM)1/2 = (3(8.314 J/K-mole)(298 K)/(0.02801 kg/mole))1/2 = 5.15 * 102 m/s This is 1150 miles per hour, or somewhere between the speeds of a commercial jet liner and a fighter jet. Please note that in applying equation 5-3-13 in Example 5-12, values of R and MM with appropriate units have been used, so that speed comes out with appropriate units. Use of R in units of, for example, atm-L, produces a speed with absurd units. The Maxwell-Boltzmann Distribution Law. We have said that in a gas the molecules move randomly, and that the speed of a molecule changes frequently. In the mid 1800's, James Clerk Maxwell in England, and Ludwig Boltzmann in Austria, were concerned with using statistical methods to describe, precisely and mathematically, the distribution of molecular speeds. The result of their efforts is called the MaxwellBoltzmann Distribution Law. They showed that at a given temperature, the molecular speed distribution follows a curve like that in Figure 5-8. The curve is a plot of the fraction of molecules having speed s, f(s) (we use s for speed, v for velocity), versus the possible values of speed between zero and infinity. The curve has several noticeable features: The curve has the expected shape -- low on both ends, indicating that few molecules have extremes of speed; and peaking in the middle, indicating that most molecules have speeds somewhere between the extremes. The curve is unsymmetrical because there is a definite lower limit (zero) but no definite upper limit to speed. The total area under the curve is the total number of gas molecules. The height of the curve at a particular speed (say s1) is proportional to the fraction of molecules having speed s1. The area under the curve between two speeds s1 and s2 (shaded in the figure) is the fraction of molecules haveing speeds in the range s1 to s2. The mathematical expression for f(s) obtained by Maxwell and Boltzmann is equation 5-3-16: (5-3-16): f(s) = (constant)(s2) exp(-ms2/2kT) The s2 term increases with increasing s and the exponential term decreases with increasing s. Thus f(s) first increases, but then peaks and decreases with increasing s as the exponential term takes over. The speed at the maximum of the curve, smp, is the most probable speed, because the greatest fraction of molecules have it. As T increases, the curve flattens and the maximum moves to higher speed, because the speeds of all molecules tend to increase with increasing T. The area under the curve stays constant, however, as long as the number of gas molecules is unchanged. We now do two calculations using equation 5-3-16. First, we calculate the most probable speed, smp. This is a good approximation to the average speed, but the average will be somewhat larger because the curve is biased to higher speeds. smp is the speed at which f(s) is maximum. At this maximum, the slope of the plot -the derivative of f(s) -- is zero: df(s)/ds = d/ds [(s2)(exp(-ms2/2kT))] = 0 = ds2/ds exp(-ms2/2kT) + s2 d[exp(-ms2/2kT)]/ds = 0 = 2s exp(-ms2/2kT) - s2(ms/kT)exp(-ms2/2kT) = 0 Dividing by the exponential, we obtain 2s - ms3/kT = 0 This is readily solved for smp: smp = (2kT/m)1/2 As expected, this is a bit smaller than the root-mean-square speed in equation 5-3-12. Next we calculate the average KE of a gas molecule by a method of averaging; we multiply each possible kinetic energy by the number of molecules with that kinetic energy, and divide by the total number of molecules. The required mathematical procedure is integration, as shown in equation 5-3-17. (5-3-17): KEmolecule = Integral from 0 to Infinity of f(s)(ms2/2)ds = 3kT/2 Even if you are not yet able to carry out the integration, you should note that the result is the same as equation 5-3-10, which came from KMT. This is gratifying. It reinforces the validities of both KMT and the M-B Distribution Law. Finally, we calculate the KE per mole of gas from the average KEmolecule: (5-3-18): KEmole = No x KEmolecule = 3RT/2 The kinetic energy of a mole of gas depends only on temperature. For this reason, kinetic energy of molecules is called thermal energy. Summary of the Interpretation of the Ideal Gas Law in Terms of Molecular Behavior. The relationship between the macroscopic and microscopic views of gas behavior can be summarized in several statements. Gases are compressible because they consist of small molecules that are far apart. Molecules of a gas are in constant motion, allowing gases to diffuse and to flow. Pressure results from collisions of gas molecules with the container walls. Pressure increases as volume decreases (Boyle's Law). A decrease in V increases the number of molecules per unit volume, leading to more molecular collisions with a unit area of the wall per unit time. An increase of temperature causes an increase in pressure (the Law of Gay-Lussac). KE of the molecules increases with T. Faster molecules cause an increase in pressure for two reasons: There are more collisions with the wall per unit time; Each collision exerts a greater force, since the molecule is moving faster. An increase in the amount (moles) of gas causes an increase in pressure at constant T (Avogadro's hypothesis). More gas means larger N/V. Higher N/V means more collisions per unit time and higher pressure. The ratio of diffusion rates of two gases is proportional to the square root of the inverse ratio of their molar masses (Graham's Law). This is a consequence of the average kinetic energy of a gas molecule being dependent only on T. At a particular T, a heavy molecule moves slower than a light one, but its larger mass offsets its smaller speed, giving the same kinetic energy. An important consequence of this is that a sample of gas with heavy molecules will exert the same pressure as a sample of gas with the same number of light molecules, as long as their temperatures and volumes are the same. Both samples satisfy the ideal gas law, independent of what their molecules are like. Although heavy molecules move slowly and strike the wall less often, they strike it with bigger mass, and cause the same pressure. 5-4 The Historical Role of Gas Phase Chemical Reactions. The first decade of the 19th century must have been both exhilarating and frustrating for chemists. On the one hand, Dalton's atomic theory (1803) rationalized many known facts about chemical elements, compounds, and reactions. In particular, the theory provided an explanation for the laws of conservation of mass, definite proportions, and multiple proportions, and gave a rational basis for the idea of reproducible combining masses of the elements in compound formation (e.g., the combining masses of oxygen and hydrogen in forming water are in the ratio 7.94 to 1). On the other hand, attempts to develop a scale of atomic masses were thwarted by the ignorance of chemical formulas. Chemists, led by Dalton, assumed that the formula for water was HO, and that water was formed from hydrogen and oxygen by equation 5-4-1: (5-4-1): H + O ---> HO This was the simplest assumption to make in the absence of definite knowledge of the formulas of elemental hydrogen, elemental oxygen, and water. In 1809, Joseph Gay-Lussac conducted a series of experiments with gases that ultimately enabled the development of the atomic mass scale. Specifically, for reactions in which gases react to form other gases, he studied the relationships among the volumes of gaseous reactants consumed and the volumes of gaseous products formed. His result is of tremendous importance. Gay-Lussac discovered that, as long as the experiments were conducted at a constant temperature and pressure, the volumes of reactants used and products produced always gave whole-number ratios. One of many reactions that he studied was that between hydrogen and oxygen to form water vapor. He found consistently and reproducibly that for every one volume of oxygen used, two volumes of hydrogen were required, and two volumes of water vapor were produced: (5-4-2): 2 vol hydrogen + 1 vol oxygen ---> 2 vol water vapor (5-4-2) Gay-Lussac was not certain what to make of these results; however, Amadeo Avogadro was. He made the (to him) reasonable proposal that Gay-Lussac's results implied that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. In modern terms, we say that the volume occupied by a gas is proportional to the number of moles of gas present. Operating from this assumption, he concluded and stated that Dalton's view of the hydrogen/oxygen reaction, expressed in 5-4-1, was incorrect, because it was not consistent with the two-to-one-to-two hydrogen/oxygen/water volume ratios that GayLussac observed. He then went on to propose that hydrogen and oxygen occur as diatomic molecules, and that water in fact contains two atoms of hydrogen per atom of oxygen, so that the water formation reaction becomes: (5-4-3): 2H2(g) + O2(g) ---> 2H2O(g) This is the simplest proposal consistent with the combining volumes observed; as we know today, it is entirely correct. 5-5 Gas Production via Double Displacement Reactions. In Chapter 1, we introduced three broad classifications of chemical reactions: electron transfer processes, proton transfer processes, and double displacement processes. We defined a double displacement reaction as one in which the positive and negative portions of the reactant molecules interchange to form products. It is appropriate in this chapter on gases to focus briefly on double displacement reactions that produce a gas as one of the products. An important series of examples of such reactions involves metal carbonate compounds (called carbonate salts) and acids as reactants. The reaction of a metal carbonate with an acid is shown generically in equation 5-5-1. (5-5-1) MCO3(s or aq) + HX(aq) ---> MX2(aq) + H2CO3(aq) This reaction does not have a gas as one of its products. However, in a subsequent step, carbonic acid, H2CO3, decomposes to water and carbon dioxide, a gas that escapes in part from the aqueous solution in the form of effervescence: (5-5-2): H2CO3(aq) ---> H2O + CO2(g) A specific example of the acid-carbonate reaction is shown in 5-5-3. Here the net reaction, obtained as the sum of 5-5-1 and 5-5-2, is shown. (5-5-3): Na2CO3 + 2 HCl ---> 2NaCl + H2O + CO2 Acid-carbonate reactions are very important in at least two contexts. First, metal carbonates that have low solubility in water make up a substantial fraction of the earth's crust (limestone is an example). Deposits of carbonate rocks in contact with slightly acidic ground waters undergo reactions 5-5-1 and 5-5-2. In so doing, they serve to maintain the acidity level of the water at a fairly constant level, supporting aquatic plant and animal life. The details of this process will be understandable when we reach the end of Chapter 13. Second, reactions 5-5-1 and 5-5-2 are important in the context of acid rain, which results when nitric acid, HNO3, and sulfuric acid, H2SO4, are produced in the atmosphere from SO2, produced in coal combustion, and NO2, a byproduct of the internal combustion engine. Limestone (primarily calcium carbonate) is widely used in construction, and has been used for centuries as a medium for sculptors. Repeated and prolonged contact of building edifices and priceless works of art with acid rain results in their slow destruction as calcium carbonate reacts with nitric and sulfuric acids according to equations 5-5-1 and 5-5-2. Building edifices can be replaced; works of art cannot. Finally, dietary calcium tablets are made primarily of calcium carbonate. Contact of a stomach tablet with stomach acid (hydrochloric acid) causes the tablet to dissolve via reaction 55-4, after which the calcium can be utilized by the body. (5-5-4): CaCO3(s) + HCl ---> CaCl2(aq) + H2O + CO2(g) Metal sulfites (for example, Na2SO3) react with acids in analogous fashion to form aqueous sulfurous acid, H2SO3. This subsequently decomposes to water and sulfur dioxide: (5-5-5): Na2SO3(aq) + HNO3(aq) ---> NaNO3(aq) + H2SO3(aq) (5-5-6): H2SO3(aq) ---> H2O + SO2(g) Example 5-13. A 500-mg calcium tablet dissolves in an excess of stomach acid. What volume of CO2 is produced? Solution. We can use the simple ideas of stoichiometry introduced in Example 1-15. We begin with a balanced chemical equation for the reaction of stomach acid and the calcium tablet, which we assume to be pure calcium carbonate: CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O + CO2(g) To use the numerical information in the equation coefficients, we must work in moles: moles CaCO3 = 0.500 g/(100.086 g/mole) = 5.00 * 10-3 moles According to the equation, 1 mole of CO2 is produced for each mole of calcium carbonate that reacts. It follows that 5.00 x 10-3 moles of CO2 are formed. The ideal gas law enables us to calculate the volume of CO2. We assume the temperature in the stomach to be 98.6 oF, or 37 oC, and that the carbon dioxide is produced at 1 atm pressure. Then V = nRT/P = (5.00 * 10-3 moles)(0.08206 atm-L/mole-K)(310 K)/(1.00 atm) =0.127 L This volume of CO2 is probably not sufficient to cause "indigestion." Applications 5-1. A sample of SO2 has volume 1.45 L at 2.75 atm. Assuming constant temperature, what volume will the gas occupy at 800 torr? 5-2. At 25oC and 1 atm a gas occupies 1.50 L. What volume will it occupy at 100oC and 1 atm? 5-3. A tire is inflated to a gauge pressure of 29 lb/in2 at 65oF (the actual pressure is 29 lb/in2 above atmospheric pressure). After a trip the temperature of the tire is 130oF. What is the gauge pressure of the air in the tire assuming that the volume of the tire has not changed? Atmospheric pressure is 14.7 lb/in2. 5-4. A 2.00 L sample of gas originally at 25oC and 700 torr expands to 5.00 L at a final pressure of 585 torr. What is the final temperature of the gas in oC? 5-5. The density of CO2 is 1.96 g/L at 0oC and 1 atm. Determine its density at 650 torr and 25oC. 5-6. A 1.00-L flask is filled by placing in it the contents of a 2.00-L flask of N2 at 300 torr and a 2.00-L flask of H2 at 80 torr. What is the pressure of the mixture in the 1.00-L flask? 5-7. A gas collected over water occupies 100-mL at a total pressure of 700 torr and 25oC. What volume would the dry gas occupy at STP? The vapor pressure of water at 25oC is 23.8 torr. 5-8. A mixture of N2 and O2 in a 200-mL container exerts 720 torr of pressure at 35oC. If there are 0.0020 moles N2 present, 1. 2. 3. 4. What is the mole fraction of N2? What is the partial pressure of N2? What is the partial pressure of O2? How many moles of O2 are there? 5-9. What volume is occupied by 12.4 g Cl2 at STP? 5-10. What is the MM of a gas that has density 1.81 g/L at 30oC and 1.026 atm? 5-11. Nitric acid is produced by dissolving NO2 in water according to the equation 3NO2(g) + H2O(l) ---> 2HNO3(l) + NO(g) What volume of NO2 at 25oC and 770 torr will produce 10.0 g HNO3? 5-12. The oxidation of ammonia is important in the production of nitrogen fertilizer: 4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g) What volume of O2 at 25oC and 0.895 atm must be used to produce 100. L of NO at 500oC and 750 torr? 5-13. 8.0 g CO2, 6.0 g O2(g), and an unknown amount of N2(g) were added to a 10.0-L flask to give a 800 torr total pressure at 30oC. Calculate 1. 2. 3. 4. The total moles of gas in the container. The mole fraction of each gas. The partial pressure of each gas. The mass of N2. 5-14. The average speed of CH4 molecules is 1000 miles/hr at some temperature. What is the average speed of CO2 molecules at the same temperature? 5-15. What is the molar mass of an unknown gas that diffuses 2.92 times faster than NH3? 5-16. 0.02 moles of He(g) at 20oC are placed into a 3.00-L tank containing Ne(g) at a pressure of 7.60 torr at 20oC. Calculate the partial pressure of He; the partial pressure of Ne; the total pressure; and the molecular fraction of He(g). 5-17. The molar masses of He and Ne are 4.00 and 20.18 g/mole, respectively. A 1.00-L flask contains equal numbers of He and Ne atoms at a total pressure of 7.00 * 10-3 atm at 20oC. Calculate 1. 2. 3. 4. PHe, the partial pressure of He(g). KE(He)avg, the average kinetic energy of a He molecule. KE(Ne)avg sHe/sNe (ratio of average speeds) 5-18. The atmosphere of Mars is largely CO2. How fast are CO2(g) molecules moving during the day on Mars when the temperature is about 244 K? How much translational kinetic energy does a mole of CO2(g) lose when it cools off at night to 188 K? 5-19. In the earliest light bulbs developed by Edison a high vacuum of about 1.00 x 10-3 torr was used to prevent oxidation of the filament. In modern bulbs an inert gas is used instead of a vacuum. Suppose Edison had a glass sphere of radius 5.0 cm containing 1.00 * 10-3 torr pressure of N2(g) (MW = 28.01 g/mole) at 298 K. 1. How many N2 molecules are in the sphere? 2. How fast would an N2 molecule be moving immediately after striking the filament, at 1600oC? 3. How long would the hot molecule in b take to travel from the filament (at the bulb center) to the glass wall assuming no collisions with other molecules? 5-20. An unknown mass of aluminum is confined with an excess of oxygen at a pressure of 1.05 atm in a cylinder/piston apparatus. The aluminum is burned in the oxygen, and the system is allowed to cool to its original temperature of 298 K. The final volume of the cylinder is 0.25 L less than the initial volume. Calculate the amount of aluminum that was in the cylinder. 5-21. Suppose that you have 1 mole of hydrogen gas in one flask and 1 mole of nitrogen gas in another. The samples are at the same temperature. On one set of axes, draw the speed distribution curves for the two gas samples. On another set of axes, draw the kinetic energy distribution curves for the two gas samples. Explain any differences. 5-22. Automobile air bags contain solid sodium azide, NaN3. Impact of the automobile during collision detonates the sodium azide, which rapidly produces N2(g) according to 2NaN3(s) ---> 2Na(s) + 3N2(g) The N2(g) rapidly inflates the air bag. What mass of NaN3 (in grams) is needed to produce 50.0 L of N2, enough to fill the air bag at a pressure of 0.950 atm and a temperature of 30oC? 5-23. Initially, a quantity of gas is confined in the left box of an apparatus at a pressure of 3.40 atm and a temperature of 286 K. 4 sequential processes, a, b, c, and d, are carried out. Work out the value of each indicated quantity after the completion of each process. 1. Valve 1 is opened with valve 2 closed. T is kept constant. V occupied by gas = P of gas = 2. T is raised to 350 K. V occupied by gas = P of gas = 3. Valve 1 is closed, valve 2 is opened. V occupied by gas to the right of valve 1 = P of gas to the right of valve 1 = 4. T is lowered to 300 K. V occupied by gas to the right of valve 1 = P of gas to the right of valve 1 = 5-24. Initially, a quantity of gas is confined in the left box of an apparatus at a pressure of 8.60 atm and a temperature of 274 K. 4 sequential processes, a, b, c, and d, are carried out. Work out the value of each indicated quantity after the completion of each process. 1. T is raised to 350 K. V occupied by gas = P of gas = 2. Valve 1 is opened with valve 2 closed. T is kept constant. V occupied by gas = P of gas = 3. Valve 1 is closed, valve 2 is opened. V occupied by gas to the right of valve 1 = P of gas to the right of valve 1 = 4. Valve 2 is left open while T is lowered to 300 K. V occupied by gas to the right of valve 1 = P of gas to the right of valve 1 = At the end, what is the average kinetic energy of a molecule of the gas? 5-25. Suppose that you have 1 mole of hydrogen gas in one flask and 1 mole of nitrogen gas in another. The samples are at the same temperature. Draw the speed distribution curves for the two gas samples. Draw the kinetic energy distribution curves for the two gas samples. 5-26. What is happening to cause my shower curtain to blow inwards, sticking to my leg, when I turn on the hot water in the shower? 5-27. When I am running relatively fast on a cold day, I can see out of my glasses perfectly well, but if I slow down, my glasses fog badly. What causes this? 5-28. What does it mean to say that the pressure of a gas is inversely proportional to the volume, if the sample size and temperature are kept constant? 5-29. R is called the ideal gas constant. What does the value of this constant represent with respect to a particular gaseous system? 5-30. Which graph best represents the KE distributions of H2 and O2 at the same T? Explain your response? 5-31. Which statements(s) about 1-mole samples of H2 and O2 at room T is true? The average KE per molecule is > in H2. The average speed per molecule is > in O2. The average speed per molecule is the same in both. The average KE per molecule is the same in both. 5-32. Gas A is at a higher temperature than gas B. Which of the following is certainly true? A molecules have greater speed A molecules have greater kinetic energy A molecules have greater speed and kinetic energy A molecules have greater volume A molecules have greater pressure 5-33. 17.24 g of liquid C6H14 (hexane) is enclosed with 3.80 moles of O2(g) in a cylinder fitted with a piston, as shown below. The initial temperature of the mixture is 27 oC, and the external (outside) pressure on the piston is 1.00 atm. The hexane and oxygen are then caused to react according to the following equation: C6H14(l) + O2(g) ---> CO2(g) + H2O(l) All of the hexane is used up. The heat produced by the reaction causes the temperature to rise to 77 oC. The external pressure is maintained at 1.00 atm. 1. Balance the equation 2. Calculate the initial volume of reactants in the cylinder (assume that liquid hexane occupies a negligible volume.) 3. Calculate the final volume of products and left-over reactants in the cylinder (assume that liquid water occupies a negligible volume.) Spreadsheet Applications 5-1. The expression for the Maxwell-Boltzmann distribution is given below. f(s) = 1.556 * 1034 s2 e-ms2/2kT Use a spreadsheet to explore graphically the dependence of f(s) on s. Be sure to express s in m/s. 5-2. The volume occupied by He gas was studied as a function of the mass of gas, the pressure, and the Celsius temperature of the gas. The following data were obtained. Use the data sorting and graphing capabilities of your favorite spreadsheet to determine the relationship between gas volume and each of the other three variables from these data. Make no assumptions about the dependencies in making your analysis. Volume mass, g Pressure, atm Temperature, oC 0.8512 1.0652 1.3033 1.0334 1.62 2.26 0.110 0.110 0.110 0.110 0.110 0.204 0.800 0.650 0.540 0.670 0.420 0.570 25.00 30.00 35.00 30.00 25.00 30.00 5.60 0.578 2.47 1.83 0.615 0.539 0.558 0.714 4.73 8.28 4.00 0.204 0.110 0.110 0.110 0.110 0.110 0.110 0.110 0.316 0.316 0.230 1.20 0.280 0.390 1.20 1.20 1.20 0.970 0.420 0.240 30.00 30.00 30.00 40.00 50.00 10.00 20.00 30.00 30.00 30.00 30.00 0.0762 0.120 Object 1

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George Mason - MATH - 125
Math125-002 Fall 2009Exam 1 ReviewDate: Wednesday, September 23th Book closed. Calculator and any notes may not be used. Sections covered: 2.12.5, 5.15.2 Do not assume that the exam will contain only problems similar to those from quizzes. Average homew
George Mason - MATH - 125
George Mason - MATH - 125
Math125-002 Fall 2009Exam 2 ReviewDate: Wednesday, October 21 Book closed. Calculator and any notes may not be used. New sections covered: 5.3, 6.16.3, 7.17.2 Do not assume that the exam will contain only problems similar to those from quizzes. Average
George Mason - MATH - 125
George Mason - MATH - 125
Math125-002 Fall 2009Exam 3 ReviewDate: Wednesday, December 2 Book closed. Calculator and any notes may not be used. New sections covered: 7.5-7.71 , 9.1-9.3, 10.1-10.32 , 11.2 Do not assume that the exam will contain only problems similar to those from
George Mason - MATH - 125
George Mason - MATH - 125
Math125-002 Fall 2009Final Exam ReviewDate: Monday, December 14th 4:307:15 Book closed. Calculator and any notes may not be used. Exam will cover the topics covered by Exams 1,2,3, and sections 12.1 12.2. In section 12.1 you may skip the discussion of g
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - MATH - 125
George Mason - ECE - 331
ECE 331 Digital System DesignCourse Introduction and Introduction to Logic Circuit DesignCourse Introduction (see Syllabus)ECE 331 - Digital System Design2Silicon TechnologyECE 331 - Digital System Design3ECE wafer (courtesy Figure 1.1. A silicon
George Mason - ECE - 331
ECE 331 Digital System DesignIntroduction to Logic Circuit Design and Introduction to VHDLIntroduction to Logic Circuit DesignECE 331 - Digital System Design2Switching CircuitsAlso known as Combinational Logic CircuitsCombination of logic gates Tru
George Mason - ECE - 331
ECE 331 Digital System DesignCMOS Circuit Design of Logic GatesSwitching CircuitsAlso known as Combinational Logic CircuitsCombination of logic gates Truth Table Boolean Expression Interconnection of discrete components Synthesis from a Hardware Descr
George Mason - ECE - 331
ECE 331 Digital System DesignCircuit Design Considerations and Programmable Logic DevicesRepresenting Logic Values with Voltage RangesECE 331 - Digital System Design2Voltage Levels in Logic CircuitsVoltage V DD Logic value 1 V 1,min Undefined V 0,ma
George Mason - ECE - 331
ECE 331 Digital System DesignPower Dissipation and Propagation DelayPower DissipationECE 331 - Digital System Design2Power Consumption Each integrated circuit (IC) consumes power P T = PS + PD PT = total power consumed by IC PS = static or quiescen
George Mason - ECE - 331
ECE 331 Digital System DesignBoolean Algebra, Truth Tables, and Logic Function (Circuit) SimplificationMessage from an AuthorThe method which you use to get your answer is very important in this unit. If it takes you two pages of algebra and one hour o
George Mason - ECE - 331
ECE 331 Digital System DesignKarnaugh Maps and Logic Circuit MinimizationKarnaugh MapsGraphical representation of a truth tableCan be used to minimize logic functionsUses logic adjacency Does not produce unique results Not directly transferable to co
George Mason - ECE - 331
ECE 331 Digital System DesignSwitching (Logic) Circuit DesignDesign ConceptsCombinational Logic Circuits Outputs are functions of (present) inputs No memory Can be described using Boolean expressions Used to solve large design problems Break problem
George Mason - ECE - 331
ECE 331 Digital System DesignLogic Circuit Design and AnalysisLogic Circuit Design using NAND and NOR GatesLogic GatesAND and OR Gates 2-input gates realized with 6 CMOS transistors 3-input gates realized with 8 CMOS transistors 2-input gates realiz
George Mason - ECE - 331
ECE 331 Digital System DesignNumbers (Lecture #10)52What does this number represent? What does it mean?ECE 331 - Digital System Design 21011001.101What does this number represent? Consider the base (or radix) of the number.ECE 331 - Digital System
George Mason - ECE - 331
ECE 331 Digital System DesignBinary Arithmetic (Lecture #11)Binary AdditionECE 331 - Digital System Design2Binary Addition+0 0 0 +1 1Sum0+0 11+1 10Carry1Sum1001 (910) + 0101 (510) 1110 (1410)0111 (710) + 0001 (110) + 1000 (810)1100 (1210)
George Mason - ECE - 331
ECE 331 Digital System DesignAdder Circuits (Lecture #12)Half AdderECE 331 - Digital System Design2Half Adderx 0 0 1 1 y 0 1 0 1 Carry c 0 0 0 1 Sum s 0 1 1 0(b) Truth tablex yS = X.Y + X.Y C = X.Ys x y c HA s c(c) Circuit(d) Graphical symbol
George Mason - ECE - 331
ECE 331 Digital System DesignMulti-bit Adder Circuits (Lecture #13)Implementations of Multi-bit Adders: 1. Ripple Carry Adder 2. Carry Lookahead AdderRipple Carry AdderRipple Carry AdderCarry ripples from one column to the next1 +1Carry-out11 01 0
George Mason - ECE - 331
ECE 331 Digital System DesignMultiplexers (Lecture #14)MultiplexerA multiplexer switches (or routes) data from 2N inputs to the output, where N is the number of select inputs. A multiplexer (mux) is a digital switch.A 2-to-1 Multiplexer1 selects w0
George Mason - ECE - 331
ECE 331 Digital System DesignHazards (Lecture #16)HazardsHazardsExtra (and undesirable) state change(s) in the output of a logic function between one steadystate value of the output and the next steady-state value of the output when the input(s) to th
George Mason - ECE - 331
ECE 331 Digital System DesignLatches(Lecture #17)Sequential Logic CircuitsSequential Logic CircuitsCombinational Logic CircuitsOutput is a function of the inputs only. Output is a function of the inputs and the present state. Have history Maintain s
George Mason - ECE - 331
ECE 331 Digital System DesignFlip-Flops and Registers(Lecture #18)Basic Memory ElementsBasic Memory ElementsBasic LatchA feedback connection of two NOR gates or two NAND gates, which can store one bit of information.Can be set to 1 or reset to 0.G
George Mason - ECE - 331
ECE 331 Digital System DesignCounters(Lecture #19)State Diagram: 3-bit Counter000 111 001110010101 100011Asynchronous CountersDesigning a 3-bit (up) CounterLet each bit in the counter be represented by the output of a flip-flop.Count 0 1 2 3 4
George Mason - ECE - 331
ECE 331 Digital System DesignFinite State Machines(Lecture #20)Logic CircuitsLogic CircuitsCombinational Logic Circuits Output is dependent solely on the current inputs. Does not have memory. Output is a function of the current inputs as well as th
George Mason - ECE - 331
ECE 331 Digital System DesignFinite State Machines(Lecture #21)State Assignment ProblemState Assignment ProblemSome state assignments are better than others. The state assignment influences the complexity of the state machine.The combinational logic
George Mason - ECE - 331
ECE 331 Digital System DesignFinite State Machines(Lecture #22)Designing a Mealy MachineFSM: Design Steps (Mealy)1. Understand specifications 2. Derive state diagramFirst state: start state3. Create state table 4. Perform state minimization 5. Enco
George Mason - ECE - 331
ECE 331 Digital System DesignFinite State Machines(Lecture #23)Finite State Machine AnalysisFSM: Analysis Steps1. Determine output and next state equations 2. Derive the state-assigned table 3. Construct the state table 4. Draw the corresponding stat
George Mason - ECE - 331
ECE 331 Digital System DesignFinite State Machines(Lecture #24)Finite State Machine Designusing D-type Flip-FlopsFSM: Design Steps1. Understand specifications 2. Derive state diagram First state: start state Last state: end state (Moore machine on
George Mason - ECE - 331
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #2due: Thursday, September 16, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Sho
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #2 Solutions1. Roth textbook problem 2.52. Roth textbook problem 2.6, parts (b) and (c)Fall 20101 of 4ECE 331 Digital System Design3. Roth textbook problem 2.13, parts (c) and (d)4. Roth textbook problem 2.28
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #3due: Thursday, September 23, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Sho
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #3 Solutions1. Roth textbook problem 4.72. Roth textbook problem 4.8Fall 20101 of 5ECE 331 Digital System Design3. Roth textbook problem 4.94. Roth textbook problem 4.10Fall 20102 of 5ECE 331 Digital System
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #4 Solutions1. Roth textbook problem 5.17Fall 20101 of 7ECE 331 Digital System Design2. Roth textbook problem 5.19Fall 20102 of 7ECE 331 Digital System Design3. Roth textbook problem 5.20, parts (a) and (c)
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #5due: Thursday, October 14, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #5 Solutions1. Roth textbook problem 1.1(c) 356.8910 356 / 8 = 44 44 / 8 = 5 5/8=0 0.89 * 8 = 7.12 0.12 * 8 = 0.96 0.96 * 8 = 7.68 0.68 * 8 = 5.44 356.8910 = 544.70758 = 101 100 100 . 111 000 111 101 Rem = 4 Rem =
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #6due: Thursday, October 21, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #6 Solutions1. Roth textbook problem 1.112. Roth textbook problem 1.17Fall 20101 of 6ECE 331 Digital System Design3. Roth textbook problem 1.36Fall 20102 of 6ECE 331 Digital System Design4. Roth textbook pr
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #7due: Thursday, October 28, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #8due: Thursday, November 4, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #8 Solutions1. Roth textbook problem 9.1 Solution provided on page 703 of the textbook. 2. Roth textbook problem 9.2 Solution provided on page 703 of the textbook. 3. Roth textbook problem 9.4 Solution provided on p
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #9due: Thursday, November 11, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #9 Solutions1. Roth textbook problem 9.8 (a) and (b) as specified. (c) Using Figure 9.23 as an example, design (and draw) the ROM that can be used to implement X, Y, and Z. Be sure to clearly indicate all connection
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #10due: Tuesday, November 23, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #10 Solutions1. Roth textbook problem 12.3 See Appendix E, page 717, of the textbook for the solution.2. Roth textbook problem 12.7 (a)Fall 20101 of 5ECE 331 Digital System Design3. Roth textbook problem 12.8 (
George Mason - ECE - 331
ECE 331 Digital System DesignHomework #11due: Thursday, December 2, 2010 Write your name at the top of each page of your solutions. Clearly indicate the start of the solution for each problem. Properly order and staple all pages of your solution. Show