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Example 5.1 A Canadian weather report gives the atmospheric pressure as kPa. What is the pressure expressed in the unit torr? Strategy To convert a pressure from one unit to another, we need a relationship between the two units that we can use as a conversion factor. Generally we can find such a relationship in Table 5.2. Solution To convert from kPa to Torr, we need this relationship from Table 5.2: 1 atm = 760 Torr = kPa. That is, With this conversion factor, we find that

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Example 5.1 continued Assessment Note that the units kPa cancel and that the pressure expressed in the unit torr is a larger number (by a factor of about 7.5) than the pressure in kilopascals, as expected from the form of the conversion factor. Carry out the following conversions of pressure units. (a) atm to mmHg(c) in. Hg to Torr (b) 98.2 kPa to Torr(d) 768 Torr to atm Exercise 5.1A What is the pressure, in kilopascals and in atmospheres, if a force of 1.00 x 10 2 N is exerted on an area of 5.00 cm 2 ? Exercise 5.1B

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Example 5.2 Calculate the height of a column of water (d = 1.00 g/cm 3 ) that exerts the same pressure as a column of mercury (d = 13.6 g/cm 3 ) 760 mm high. Strategy With the above equation, we can establish the pressures exerted by the columns of mercury and water: When we equate these two pressures, we will find that there is only one unknown, for which we can solve. Solution We start by equating the two liquid pressures. Next we cancel the factor g and substitute the known numerical values into the resulting expression. Then we solve the expression for

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Example 5.2 continued Assessment Because water has a much lower density than mercury (about 1/14), it should take a much taller water column to produce the same pressure as a mercury column, and that is what we found. To measure normal atmospheric pressure with a water-filled barometer, we would have to use one that is more than 10 m tall—as tall as a three- story building! Calculate the height of a column of carbon tetrachloride, CCl 4 (d = 1.59 g/cm 3 ), that exerts the same pressure as a column of mercury (d = 13.6 g/cm 3 ) 760 mm high. Exercise 5.2A A diver reaches a depth of 30.0 m. What is the pressure in atmospheres that is exerted by this depth of water? What is the total pressure the diver experiences at this depth? Explain. Exercise 5.2B

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Example 5.3 A Conceptual Example Without doing calculations, arrange the drawings in Figure 5.5 so that the pressures denoted in red are in increasing order. Analysis and Conclusions The pressure in (a) is expressed as the depth of the liquid mercury, that is, 745 mmHg. The pressure of helium in the open-end manometer (b) is slightly greater than P bar, which is 762 Torr = 762 mmHg and, therefore, greater than the pressure in (a). Although there is much less mercury in (c) than in (a), the pressure of the mercury column in (c) is greater than in (a) because of its greater height: 75.0 cm = 750 mm. Finally, in the closed-end manometer in (d), the difference in the two mercury levels, 735 mm, is the actual gas pressure. (d)<(a)<(c)<(b) P Ne < P Hg (l) < P atm < P He 735 mmHg<745 mmHg<750 mmHg

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Example 5.3 continued Place a barometric pressure of (e) 101 kPa into the order of increasing pressures established in Example 5.3. If possible, also place a barometric pressure of (f) 103 kPa in the order. If not possible, explain why. Exercise 5.3A Use principles from this section to explain (a) how a beverage is transferred from glass to mouth when one uses a soda straw and (b) why an old-fashioned hand-operated pump cannot raise water from a well if the water level is more than about 30 feet below the pump. Exercise 5.3B

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Example 5.4 A helium-filled party balloon has a volume of 4.50 L at sea level, where the atmospheric pressure is 748 Torr. Assuming that the temperature remains constant, what will be the volume of the balloon when it is taken to a mountain resort at an altitude of 2500 m, where the atmospheric pressure is 557 Torr? Strategy We can solve the mathematical expression of Boyle’s law for V final and calculate its value from the information given. Solution Using Equation (5.2), we derive the equation for V final : Then we can substitute known values for the three variables on the right side of the equation: Assessment Because the final pressure is less than the initial pressure, we expect the final volume to be greater than the initial volume, and it is.

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Example 5.4 continued A sample of helium occupies 535 mL at 988 Torr and 25 °C. If the sample is transferred to a 1.05-L flask at the same temperature, what will be the gas pressure in the flask? Exercise 5.4A A sample of air occupies 73.3 mL at 98.7 kPa and 0 °C. What volume will the air occupy at 4.02 atm and the same temperature? Exercise 5.4B

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Example 5.5 An Estimation Example A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is 2.00 L at 398 Torr. The piston is moved to increase the gas pressure to 5.15 atm. Which of the following is a reasonable value for the volume of the gas at the greater pressure? 0.20 L 0.40 L 1.00 L 16.0 L Analysis and Conclusions The initial pressure (398 Torr) is about 0.5 atm. An increase in pressure to 5.15 atm is about a tenfold increase. As a result, the volume should drop to about one-tenth of its initial value of 2.00 L. A final volume of 0.20 L is the most reasonable estimate. (The calculated value is L.) A gas is enclosed in a 10.2-L tank at 1208 Torr. Which of the following is a reasonable value for the pressure when the gas is transferred to a 30.0-L tank? 0.40 atm 25 lb/in mmHg 3600 Torr Exercise 5.5A Which of the following is a reasonable estimate of the pressure-volume product of gas in the 30.0-L tank of Exercise 5.5A? (a) 3.6 x 10 4 Torr L, (b) 4 x 10 3 mmHg L, (c) 1.6 x 10 4 kg/m 2, (d) 1.6 kPa m 3 Exercise 5.5B

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Example 5.6 A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What will its volume be outdoors, where the temperature is –23 °C? (Assume no change in the gas pressure.) Strategy We can use Charles’s law to determine how the change in temperature affects the volume, but we must consider the temperature change on the absolute (Kelvin) temperature scale. Solution First, we convert both temperatures to the Kelvin scale, using Equation (5.3): Now we apply Charles’s law in the form of Equation (5.5) and solve for V final : Assessment Because the final temperature is lower than the initial temperature, we expect the final volume to be smaller than the initial volume, and it is.

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Example 5.6 continued A sample of hydrogen gas occupies 692 L at 602 °C. If the pressure is held constant, what volume will the gas occupy after cooling to 23 °C? Exercise 5.6A The balloon described in Example 5.6 must be at a particular temperature in order to have a volume of 2.25 L. Assuming that the pressure remains constant, find this temperature in kelvins and in degrees Celsius. Exercise 5.6B

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Example 5.7 An Estimation Example A sample of nitrogen gas occupies a volume of 2.50 L at –120 °C and 1.00 atm pressure. To which of the following approximate temperatures should the gas be heated in order to double its volume while maintaining a constant pressure? –240 °C –60 °C –12 °C 30 °C Analysis and Conclusions The initial temperature is about 150 K (– ≈ 150 K). The final temperature must be twice the initial temperature—about 300 K—if the volume is to double. The Celsius temperature corresponding to 300 K is about 30 °C ( ≈ 300 K). Notice that the four temperatures from which to choose are all multiples or fractions of –12 °C, but this has nothing to do with our answer because the relationship between volume and temperature must be based on the Kelvin scale and not the Celsius scale. If the gas in Example 5.7—initially 2.50 L at –120 °C and 1.00 atm—is brought to a temperature of 180 °C while a constant pressure is maintained, which of the following is the approximate final volume? (a) 3.75 L (b) 5.0 L (c) 7.5 L (d) 10.0 L Exercise 5.7A A sample of gas is initially 1.50 L at 50 °C and 0.60 atm pressure. If this gas is brought to a final temperature of 150 °C while the volume is held constant, which of the following is a close estimate of the final pressure? (a) 1.8 atm (b) 0.8 atm (c) 0.5 atm (d) 0.2 atm Exercise 5.7B

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Example 5.8 Calculate the volume occupied by 4.11 kg of methane gas, at CH 4 (g), at STP. Strategy We must first convert the mass of gas to an amount in moles and then use the molar volume as a conversion factor to get the volume of the gas at STP. Solution We can do all this in a single setup, where the conversion factor based on molar volume is shown in red. What is the mass of propane, C 3 H 8, in a 50.0-L container of the gas at STP? Exercise 5.8A Solid carbon dioxide, called dry ice, is useful in maintaining frozen foods because it vaporizes to CO 2 (g) rather than melting to a liquid. How many liters of CO 2 (g), measured at STP, will be produced by the vaporization of a block of dry ice (d = 1.56 g/cm 3 ) that measures 12.0 in. x 12.0 in. x 2.00 in.? Exercise 5.8B

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Example 5.9 The flasks pictured in Figure 5.11 contain O 2 (g), the one on the left at STP and the one on the right at 100 °C. What is the pressure at 100 °C? Strategy We start with Equation (5.8), where the subscript 1 represents the initial condition (STP) and the subscript 2 represents the final condition. Then we solve for the final pressure, P 2. Solution Assessment Because the final temperature is higher than the initial temperature, we expect the final pressure to be greater than the initial pressure; and it is. Note that, although we need to know that the amount and volume of O 2 (g) remain constant, we do not need to know the value of either. Also, we do not need to know that the gas is O 2 ; it could be some other gas.

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Example 5.9 continued Aerosol containers often carry the warning that they should not be heated. Suppose an aerosol container was filled with a gas at 2.5 atm and 22 °C. If there is the possibility that the container may rupture if the pressure exceeds 8.0 atm, at what temperature is rupture likely to occur? Exercise 5.9A In the manner we used to derive Amontons’s law relating gas pressure and temperature, derive an expression relating the pressure and amount of gas when both the temperature and volume remain constant. Rationalize your result in terms of the kinetic-molecular theory and a sketch in the manner of Figure Exercise 5.9B

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Example 5.10 What is the pressure exerted by mol O 2 in a 15.0-L container at 303 K? Strategy Knowing n, V, and T of a gas, we can use the ideal gas law to solve for its pressure. Solution First, we solve the ideal gas law, Equation (5.9), for pressure, P, by dividing both sides of the equation by the volume, V, and then canceling V on the left side of the equation. Because data are given in the units mol, L, and K, we can use R = L atm mol –1 K –1 and substitute the data directly into the equation. Our result will be a pressure in atmospheres.

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How many moles of nitrogen gas, N 2, are there in a sample that occupies 35.0 L at a pressure of 3.15 atm and a temperature of 852 K? Exercise 5.10A If volume and temperature are held constant, how many moles of N 2 (g) should be added to the gas described in Exercise 5.10A to increase the pressure to 5.00 atm? Exercise 5.10B Example 5.10 continued Assessment An important check is for the proper cancellation of units, leading to an answer in atm. Also, compare the O 2 (g) with one mole of gas at STP. The mol O 2 (g), if confined to 22.4 L at 273 K, would exert a pressure of about 0.5 atm. Because the gas is confined to 15.0 L, its pressure should exceed 0.5 atm, but still be less than 1 atm. The pressure increase on raising the temperature to 303 K is small (increasing by 303/273). The answer (0.842 atm) is reasonable.

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Example 5.11 What is the volume occupied by 16.0 g ethane gas (C 2 H 6 ) at 720 Torr and 18 °C? Strategy We will need to solve the ideal gas equation for volume, V, and use an appropriate unit for each variable in the equation. Solution We can use the units that are consistent with the value R = L atm mol –1 K –1, as shown here. Finally, we can substitute the converted data for the variables n, T, and P into the ideal gas equation, which we have solved for V.

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What is the temperature, in degrees Celsius, at which 15.0 g O 2 exerts a pressure of 785 Torr in a volume of 5.00 L? Exercise 5.11A How many grams of N 2 (g), at 25.0 °C and 734 Torr, will occupy the same volume as 25.0 g O 2 (g) at 30.0 °C and 755 Torr? Exercise 5.11B Example 5.11 continued Assessment For 0.5 mol C 2 H 6 at STP, the volume would be 11.2 L. The actual amount of C 2 H 6 is slightly more than 0.5 mol; the temperature is somewhat higher than 273 K; and the pressure (720 Torr) is somewhat less than 1 atm. All of these factors would make the answer somewhat larger than 11.2 L, and we find that it is (13.4 L).

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Example 5.12 If g of a gas occupies L at atm and 289 K, what is the molecular mass of the gas? Strategy Here, the mass (m) of gas is given, and in an approach similar to that used in Example 5.11, we will solve the ideal gas equation for the number of moles (n). The molar mass is M = m/n, and the molecular mass (in atomic mass units) is numerically equal to the molar mass (in g/mol). Solution First, let’s calculate the amount of gas, in moles, from the ideal gas equation. Now we can use the known mass and the calculated number of moles to determine the molar mass. The molar mass of the gas is 67.4 g/mol, and the molecular mass is therefore 67.4 u.

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Example 5.12 continued Assessment Our answers have the correct units for molar mass and molecular mass. Moreover, the values seem to be in the correct range. The molecular mass of a gaseous substance at about STP cannot be less than 1 u (that of H atoms), nor is it likely to be very large (say, greater than 100–200 u). At about STP, substances that have high molecular masses are most likely to be liquids or solids (as discussed in Chapter 11). Calculate the molar mass of a gas if g occupies 179 mL at 741 mmHg and 86 °C. Exercise 5.12A If 8.0 g CH 4 (g) occupies the same volume at 0 °C as 8.0 g O 2 (g) at STP, what is the pressure of the CH 4 (g)? Exercise 5.12B

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Example 5.13 Calculate the molecular mass of a liquid that, when vaporized at 100 °C and 755 Torr, yields 185 mL of vapor that has a mass of g. Strategy We can use Equation (5.10) to calculate molar mass from our knowledge of m, T, P, and V. Then we can establish molecular mass from molar mass. Solution We must convert from mL to L and from °C to K, but we can leave the pressure in Torr and use the corresponding value of R from Table 5.3, that is, L Torr mol –1 K –1. Now we can substitute these data into Equation (5.10). The molar mass is 87.1 g/mol, and the molecular mass is 87.1 u.

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Example 5.13 continued Calculate the molecular mass of a liquid that, when vaporized at 98 °C and 715 mmHg, yields 121 mL of vapor having a mass of g. Exercise 5.13A Diacetyl, a substance that contributes to the characteristic flavor and aroma of butter, consists of 55.80% C, 7.03% H, and 37.17% O by mass. In the gaseous state at 100 °C and 747 mmHg, a g sample of diacetyl occupies a volume of 111 mL. Establish the molecular formula of diacetyl. Exercise 5.13B

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Example 5.14 Calculate the density of methane gas, CH 4, in grams per liter at 25 °C and atm. Strategy We could do this as a two-step problem in which we (1) determine the number of moles of CH 4 in 1.00 L of the gas at the stated temperature and pressure and then (2) determine the mass of this gas. To do so, however, is really the same as solving Equation (5.11). So let’s do it the (simpler) latter way. Solution Calculate the density of ethane gas (C 2 H 6 ) in grams per liter at 15 °C and 748 Torr. Exercise 5.14A What is the molar mass of a gaseous alkane hydrocarbon having a density of 2.42 g/L at 20.0 °C and 762 Torr? What is the molecular formula of the gas? Exercise 5.14B

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Example 5.15 Under what pressure must O 2 (g) be maintained at 25 °C to have a density of 1.50 g/L? Strategy We can use Equation (5.11), d = MP/RT, to solve for any one of the five quantities (here, P) if the other four are known. Solution First, we solve Equation (5.11) for the unknown pressure: Then, we substitute the known quantities into the right side: Assessment To ensure that we have used a correct variation of the ideal gas equation, we can use unit cancellation as a check. Here, all units cancel except for the desired atm. To what temperature must propane gas (C 3 H 8 ) at 785 Torr be heated to have a density of 1.51 g/L? Exercise 5.15A At what temperature will the density of a sample of O 2 (g) at 725 Torr be the same as the density of NH 3 (g) at 22.5 °C and 1.45 atm? Exercise 5.15B

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Example 5.16 How many liters of O 2 (g) are consumed for every 10.0 L of CO 2 (g) produced in the combustion of liquid pentane, C 5 H 12, if all volumes are measured at STP? Strategy The fact that pentane and water are liquids does not affect our calculation because the substances of interest, CO 2 and O 2, are both gases. The actual temperatures and pressures before, during, and after the combustion reaction are not relevant, as long as each gas volume is measured at the same temperature and pressure. The condition of STP is relevant only in establishing the particular temperature and pressure at which the two gas volumes are compared. Solution We begin by writing a balanced equation for the combustion reaction: The fundamental equivalence in the reaction is Based on the equal volume-equal numbers hypothesis, we can substitute a volume unit for the unit mol: This equivalence provides the stoichiometric factor we need. Thus,

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Example 5.16 continued What volume of oxygen gas is required to burn L of propane gas, C 3 H 8, if both gas volumes are measured at the same temperature and the same pressure? Exercise 5.16A What volume of O 2 (g) is consumed in the combustion of 125 g of gaseous dimethyl ether, (CH 3 ) 2 O, if both gas volumes are measured at the temperature and pressure at which the density of dimethyl ether is 1.81 g/L? Exercise 5.16B

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Strategy For this reaction, we must relate the number of moles of N 2 (g) to the number of moles of NaN 3 (s), using their molar ratio. Then, using the ideal gas equation, we can determine the volume of nitrogen from the number of moles of nitrogen. Example 5.17 In the chemical reaction used in automotive air-bag safety systems, N 2 (g) is produced by the decomposition of sodium azide, NaN 3, at a somewhat elevated temperature: What volume of N 2 (g), measured at 25 °C and atm, is produced by the decomposition of 62.5 g NaN 3 ? Solution First, we can determine the amount of N 2 (g) produced when 62.5 g NaN 3 decomposes. Here we use the molar mass (1 mol NaN 3 /65.01 g NaN 3 ) and a stoichiometric factor from the balanced equation (3 mol N 2 /2 mol NaN 3 ) as conversion factors: Then, we can use the ideal gas equation to determine the volume of 1.44 mol N 2 (g) under the stated conditions: The solution is also outlined in the stoichiometry diagram of Figure 5.13.

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Example 5.17 continued Assessment We can establish an approximate answer rather easily. Because the quantity of NaN 3 is about 1 mol, the number of moles of N 2 should be about equal to the stoichiometric factor, that is, 3/2 = 1.5. Because the gas conditions are not far removed from STP, the 1.5 mol of gas should occupy a volume of about 1.5 x 22.4 L = 33.6 L. This is close to our calculated value of 35.9 L. How many liters of cyclopentane, C 5 H 10 (d = g/mL), must be burned in excess O 2 (g) to produce 1.00 x 10 6 L of CO 2 (g) measured at 25.0 °C and 736 Torr? Exercise 5.17B Quicklime (CaO), used in the construction industry, is manufactured by heating limestone (CaCO 3 ) to decompose it: Exercise 5.17A How many liters of CO 2 (g) at 825 °C and 754 Torr are produced in the decomposition of 45.8 kg CaCO 3 (s)?

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Example 5.18 A 1.00-L sample of dry air at 25 °C contains mol N 2, mol O 2, mol Ar, and mol CO 2. Calculate the partial pressure of N 2 (g) in the mixture. Strategy We are given a rather complete description of the composition of air, but since each component gas expands to fill the 1.00-L volume, the basic question becomes, “What is the pressure exerted by mol N 2 in a 1.00-L container at 25 °C?” Calculate the partial pressure of each of the other components of the air sample in Example What is the total pressure exerted by all the gases in the sample? Exercise 5.18A What is the total pressure exerted by a mixture of 4.05 g N 2, 3.15 g H 2, and 6.05 g He when the mixture is confined to a 6.10-L container at 25 °C? Exercise 5.18B Solution We apply Equation 5.13 for N 2 (g) in the dry air mixture,

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Example 5.19 The main components of dry air, by volume, are N 2, 78.08%; O 2, 20.95%; Ar, 0.93%; and CO 2, 0.04%. What is the partial pressure of each gas in a sample of air at atm? Strategy Volume percent is the same as mole percent, and from the mole percents we can write the mole fractions. Thus, volume percent N 2 is the same as mole percent N 2, which, in turn, is the same as a mole fraction of N 2. We can apply this reasoning to the other gases in the mixture, too: The mole fraction for O 2 is ; for Ar, ; and for CO 2, A sample of expired (exhaled) air is composed, by volume, of the following main gaseous components: N 2, 74.1%; O 2, 15.0%; H 2 O, 6.0%; Ar, 0.9%; and CO 2, 4.0%. What is the partial pressure of each gas in the expired air at 37 °C and atm? Exercise 5.19A A 75.0-L sample of natural gas at 23.5 °C contains methane (CH 4 ) at a partial pressure of 505 Torr, ethane (C 2 H 6 ) at 201 Torr, propane (C 3 H 8 ) at 43 Torr, and 10.5 g of butane (C 4 H 10 ). Calculate the mole fraction of each component. Exercise 5.19B Solution Each partial pressure can be obtained using Equation (5.16):

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Example 5.20 A Conceptual Example Describe what must be done to change the gaseous mixture of hydrogen and helium shown in Figure 5.15a to the conditions illustrated in Figure 5.15b. Analysis and Conclusions First, we assess the situation in (a): Because the chosen volume is 22.4 L, the temperature is 273 K (0 °C), and the total amount of gas is 1.00 mol, the total pressure must be 1.00 atm. This condition is the familiar molar volume of an ideal gas at STP (22.4 L). Next, we appraise the situation in (b): The volume and temperature remain as they were in (a), but the pressure increases to 3.00 atm. This tripling of the pressure while the volume and temperature are fixed requires a tripling of the amount of gas. Therefore, there must be 3.00 mol gas in (b) instead of the 1.00 mol in (a). Because the partial pressure is 2.00 atm, the partial pressure of the other gas or gases must be 1.00 atm. The corresponding mole fractions are 2/3 for H 2 and 1/3 for the other gas or gases. The 3.00-mol mixture of gases in (b) must consist of 2.00 mol H 2 and 1.00 mol of another gas or gases. In order to get from (a), where we have 0.50 mol H 2 and 0.50 mol He, to (b), where we have 2.00 mol H 2 and 1.00 mol of some other gas or gases, we must add 1.50 mol H 2 and 0.50 mol of the other gas or gases to the flask. Note that this 0.50 mol could be 0.50 mol He, but it could also be any single gas or combination of gases other than hydrogen.

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Example 5.20 continued Why is it not possible to achieve the change between (a) and (b) in Figure 5.15 by adding hydrogen gas alone? Why can it not be achieved by adding helium gas alone? Why is it necessary that some hydrogen be added but not necessary that any helium be added? Exercise 5.20A Which of the following actions would you take to change the pressure of 2.2 L of N 2 at STP to 2.0 atm at 400 °C while maintaining a constant volume: (a) add 0.10 mol N 2 ; (b) add mol N 2 ; (c) release 0.10 mol N 2 ; (d) release 0.46 L of N 2, measured at STP? Exercise 5.20B

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Strategy The “wet” gas is the mixture of hydrogen gas and water vapor. However, because the two gases are found in the same collection vessel, they occupy the same volume, which means we need to calculate the volume of only one of them. Although we can easily get the pressure of the water vapor from Table 5.4, that information alone is not enough to calculate the volume of the water vapor using the ideal gas law because, in order to work the problem that way, we also need to know the number of moles of water vapor, and we do not. We can, however, gather the data necessary to determine the volume of hydrogen, by using Dalton’s law of partial pressures and the ideal gas law. Example 5.21 Hydrogen produced in the following reaction is collected over water at 23 °C when the barometric pressure is 742 Torr: What volume of the “wet” gas will be collected in the reaction of 1.50 g Al(s) with excess HCl(aq)? Solution To begin, we know the temperature, 23 °C, and we can calculate the partial pressure of hydrogen using Dalton’s law of partial pressures.

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Hydrogen gas is collected over water at 18 °C. The total pressure inside the collection jar is set at the barometric pressure of 738 Torr. If the volume of the gas is 246 mL, what mass of hydrogen is collected? What is the mass of the wet gas? (Hint: What is the mass of the water vapor?) Exercise 5.21A A sample of KClO 3 was heated to decompose it to potassium chloride and oxygen. The O 2 was collected over water at 21 °C and a barometric pressure of 746 mmHg. A 155-mL volume of the gaseous mixture was obtained. What mass of KClO 3 was decomposed? Exercise 5.21B Example 5.21 continued Solution continued We can determine the number of moles of hydrogen from the stoichiometry of the reaction. With this information, we can solve the ideal gas equation for V and then calculate the volume of hydrogen, which is equal to the volume of the “wet” gas.

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Example 5.22 Without doing detailed calculations, determine which of the following is a likely value for u rms of O 2 molecules at 0 °C, if u rms of H 2 at 0 °C is 1838 m/s. (a) 115 m/s (b) 460 m/s (c) 1838 m/s (d) 7352 m/s (e) 29,400 m/s Analysis and Conclusions We could calculate u rms for O 2 at 0 °C from Equation (5.23), but let’s see how we can come up with a reasonable answer with a minimum of calculation. We must not be misled by response (c)—that the H 2 and O 2 molecules travel at the same speed. Because they are at the same temperature, they do have equal average translational kinetic energies 1/2(mu). However, because O 2 molecules have a greater mass (m) than H 2 molecules, the O 2 molecules must have a lesser velocity (u). Thus O 2 molecules, on average, should move more slowly. This fact eliminates responses (d) and (e), as well as (c). The molar mass ratio of O 2 to H 2 is 32/2 = 16. Response (a), 115 m/s, is one- sixteenth of 1838 m/s, but this answer is incorrect because the molecular speeds are inversely related not to the molar mass ratio but to the square root of this ratio: Thus the u rms speed of molecules is one-quarter that for hydrogen, or about 460 m/s. The correct response is (b). Without doing detailed calculations, determine which of the following gases has the greatest u rms : (a) CH 4 (g) at 0 °C,(c) SO 2 (g) at 750 °C, (b) O 2 (g) at 250 °C, (d) NH 3 (g) at 35 °C. Exercise 5.22A

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At which temperature(s) listed here will u rms of O 2 be greater than u rms of H 2 at 0 °C, that is, greater than 1838 m/s? (a) 1000 K(c) 3000 K(e) 5000 K (b) 2000 K(d) 4000 K Exercise 5.22B Example 5.22 continued

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Example 5.23 If compared under the same conditions, how much faster than helium does hydrogen effuse through a tiny hole? Strategy To answer this question, we can set up a ratio of the two effusion rates in accordance with Graham’s law (Equation 5.24). Which effuses faster, N 2 or Ar, when the two gases are compared under the same conditions? How much faster? Exercise 5.23A Which effuses faster, nitrogen monoxide or dimethyl ether, (CH 3 ) 2 O, when the two gases are compared under the same conditions? How much faster, expressed as a percentage? Exercise 5.23B, must go into the denominator under the square-root sign on the, in the numerator on the left, its Solution If we place the rate of effusion of hydrogen, molar mass, right. The situation is reversed for helium: The ratio of the two rates is 1.41, which means 1.41 times faster than helium. = 1.41 x Hydrogen effuses

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Example 5.24 One percent of a measured amount of Ar(g) escapes through a tiny hole in 77.3 s. One percent of the same amount of an unknown gas escapes under the same conditions in 97.6 s. Calculate the molar mass of the unknown gas. Strategy Here we are given information regarding effusion times. From Equation (5.25), we can relate effusion times to effusion rates and, in turn, establish a relationship between effusion time and molar mass. Thus, the molar mass of the unknown gas can be calculated from the ratio of the effusion times. Solution First, let’s get the appropriate form for an expression relating effusion times and molar masses. To do this, we can combine Equation (5.25), relating effusion time to effusion rate, and Equation (5.24), relating effusion rate to molar mass. This result tells us that effusion time is directly proportional to the square root of molar mass. Now the only variable that is not known is M unk.

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Five percent of a sample of O 2 (g) effuses from an orifice in 123 s. How long should it take five percent of the same amount of butane, C 4 H 10, to effuse under the same conditions? Exercise 5.24B Two percent of a sample of N 2 (g) effuses from a tiny opening in 57 s. Two percent of the same amount of an unknown gas escapes under the same conditions in 83 s. Calculate the molar mass of the unknown gas. Exercise 5.24A Assessment We can see that this is a reasonable answer. Because the unknown gas effuses more slowly than does Ar, the unknown molar mass must be greater than that of Ar. Example 5.24 continued Solution continued We solve for M unk by squaring both sides of the final equation.

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Cumulative Example Two cylinders of gas are used in welding. One cylinder is 1.2 m high and 18 cm in diameter, containing oxygen gas at 2550 psi and 19 °C. The other is 0.76 m high and 28 cm in diameter, containing acetylene gas (C 2 H 2 ) at 320 psi and 19 °C. Assuming complete combustion, which tank will be emptied, leaving unreacted gas in the other? Strategy Our basic task is to determine the limiting reactant in the combustion, that is, C 2 H 2 or O 2. For this, we must compare the numbers of moles of reactants in the two cylinders to the requirements dictated by the stoichiometric relationship between them (that is, the stoichiometric factor). We obtain the stoichiometric factor from the balanced equation for the combustion reaction. To determine the number of moles of O 2 and of C 2 H 2 available for reaction, we must use the ideal gas law; but to obtain the necessary gas volumes we need first to calculate the volumes of the cylinders from the cylinder dimensions. When we have both the stoichiometric factor and the numbers of moles of the two gases, we can determine the limiting reactant. The cylinder containing the limiting reactant will be emptied in the reaction. Solution We begin with the unbalanced equation for the combustion, which forms carbon dioxide and water. Stoichiometric coefficients are added to balance the equation.

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Cumulative Example continued Solution continued Beginning with the oxygen cylinder, the volume in cubic centimeters is calculated from the data given. We now solve for the number of moles of oxygen that occupies this volume. Table 5.2 gives the conversion factor we need between psi (pounds per square inch) and atmospheres. We repeat the calculations of volume and number of moles for acetylene. Because the question merely asks which gas is completely consumed, we do not need to determine the amount of product. We just need to find which reactant is limiting.

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Cumulative Example continued Solution continued A simple calculation of the number of moles of C 2 H 2 needed to react with all the available O 2 will suffice. We have available only 43 mol of C 2 H 2, meaning all the C 2 H 2 will be used up. Assessment We can check several parts of our work. The volume of the oxygen cylinder (30.5 L) is somewhat more than the molar volume at STP (22.4 L), and the pressure is much higher than 150 atm, telling us there should be much more than 150 mol of oxygen gas in the cylinder. Likewise, the acetylene cylinder holds about two molar volumes at roughly 20 atm, or approximately 40 mol of gas. The amount of C 2 H 2 that could be consumed by the available O 2 is more than 60 mol (2/5 x 150), but with only about 40 mol C 2 H 2 available, the acetylene tank will be emptied.

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