= atm. 760 mm Hg. = atm. d. 767 torr = 767 mm Hg. = 1.01 atm


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1 Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by changes in those conditions. Although the particles of matter in solids are essentially fixed in position (the solid is rigid), the particles in liquids and gases are free to move. 2. The pressure of the atmosphere represents the weight of the severalmilethick layer of gases pressing down on every surface of the earth. Pressure, in general, represents a force exerted over a particular area, and the pressure of the atmosphere corresponds to a pressure of nearly 15 pounds per square inch on the surface of the earth. 3. A simple mercury barometer is a tube filled with mercury inverted over a reservoir containing mercury that is open to the atmosphere. When the tube is inverted, the mercury falls to a level at which the pressure of the atmosphere is sufficient to support the column of mercury. One standard atmosphere of pressure is taken to be the pressure capable of supporting a column of mercury to a height of mm above the reservoir level. 1 atm 4. a kpa atm kpa b cm g 10 mm g 1 cm g 1 atm atm 760 mm g 1 atm c. 752 mm g 760 mm g d. 767 torr 767 mm g atm 767 torr 1 atm 760 torr 1.01 atm 760 mm g 5. a atm mm g 1 atm 760 mm g b. 225,400 Pa 1691 mm g 101,325 Pa 760 mm g c kpa 748 mm g kpa 760 mm g d atm mm g 1 atm 101,325 Pa 6. a. 774 torr Pa 760 torr 101,325 Pa b atm Pa 1 atm 130
2 Gases 131 c kpa Pa d. 801 mm g 101,325 Pa 760 mm g Pa 7. a. P mm g P mm g V ml V 2? ml V 2 P 1 V 1 P 2 (53.2 ml)(785 mm g) 700 mm g 59.7 ml b. P atm P 2? atm V L V L P 2 P 1 V 1 V 2 (1.67 atm)(2.25 L) 2.00 L 1.88 atm c. P mm g P atm 1148 mm g V L V 2? L V 2 P 1 V 1 P 2 (695 mm g)(5.62 L) 1148 mm g 3.40 L 8. a. P atm P atm V ml V 2? ml V 2 PV 1 1 (1.07 atm)(291 ml) P atm 146 ml b. P mm g P atm 2668 mm g V L V 2? L V 2 PV (755 mm g)(1.25 L) 1 1 P mm g L c. P kpa mm g P 2? mm g V L V L P 2 PV (760.6 mm g)(2.71 L) 1 1 V 2 (3.00 L) 687 mm g 9. If the pressure exerted on the gas in the balloon is decreased, the volume of the gas in the balloon will increase in inverse proportion to the factor by which the pressure was changed. The factor in this example is (1.01 atm/0.562 atm) P mm 1.00 atm P 2? atm V L V ml L P 2 PV (1.00 atm)(1.00 L) 1 1 V 2 ( L) 20.0 atm
3 132 Chapter Absolute zero is the lowest temperature that can exist. Absolute zero is the temperature at which the volume of an ideal gas sample would be predicted to become zero. Absolute zero is the zeropoint on the Kelvin temperature scale (and corresponds to 273 C). 12. Charles s law states that the volume of an ideal gas sample varies linearly with the absolute temperature of the gas sample. In an experiment performed to determine absolute zero, the volume of a sample of gas is measured at several convenient temperatures (e.g., between 0 C and 100 C) and the data is plotted. The straight line obtained is then extrapolated to the point where the volume of the gas would become zero. The temperature at which the volume of the gas would be predicted to become zero is then absolute zero. 13. V ml V 2? ml 26.5 C 300 K T C 328 K V 2 P 1 T 2 (45.0 ml)(328 K) (300 K) 49.2 ml 14. a. V L V L 0 C 273 K T 2? C T 2 V 2 V 1 (50.0 L)(273 K) (25.0 L) 546 K 273 C b. V ml V ml 25 C 298 K T 2? C T 2 V 2 V 1 (255 ml)(298 K) (247 ml) 308 K 35 C c. V ml V 2? ml 272 C 1 K T 2 25 C 298 K V 2 V 1 T 2 (1.00 ml)(298 K) (1 K) 298 ml 15. a. V L V L 1150 C 1423 K T 2? C T 2 V 2 V 1 (5.00 L)(1423 K) (2.01 x 10 2 L) 35.4 K 238 C b. V ml V 2? ml 298 K T 2 0 K V 2 V 1 T 2 (44.2 ml)(0 K) (298 K) 0 ml (0 K is absolute zero)
4 Gases 133 c. V ml V 2? ml 298 K T 2 0 C 273 K V 2 V 1 T 2 (44.2 ml)(273 K) (298 K) 40.5 ml C 297 K 272 C 1 K 5.00 L 1 K 297 K L 0.02 L 17. You should be able to answer these without having to set up a formal calculation. Charles s law says that the volume of a gas sample is directly proportional to its absolute temperature. So if a sample of neon has a volume of 266 ml at 25.2 C (298 K), then the volume will become half as big at half the absolute temperature (149 K, 124 C). The volume of the gas sample will become twice as big at twice the absolute temperature (596 K, 323 C) C K 54 C K 500. ml 327 K 298 K 549 ml 19. V ml V 2? L n mol n mol 652 ml mol mol 1143 ml 1.14 L 20. V L V 2? L n g/32.00 g mol 1 n g/32.00 g mol 1 V 2 V 1 n 2 n 1 (100. L)(5.00 g/32.00 g mol 1 ) (46.2 g/32.00 g mol 1 ) 10.8 L Note that the molar mass of the O 2 gas cancels out in this calculation. Since the number of moles of Ne (or any gas) present in a sample is directly proportional to the mass of the gas sample, the problem could also have been set up directly in terms of the masses. 21. Real gases most closely approach ideal gas behavior under conditions of relatively high temperatures (0 C or higher) and relatively low pressures (1 atm or lower). 22. For an ideal gas, PV nrt is true under any conditions. Consider a particular sample of gas (so that n remains constant) at a particular fixed temperature (so that T remains constant also). Suppose that at pressure P 1 the volume of the gas sample is V 1. Then for this set of conditions, the ideal gas equation would be given by P 1 V 1 nrt
5 134 Chapter 13 If we then change the pressure of the gas sample to a new pressure P 2, the volume of the gas sample changes to a new volume V 2. For this new set of conditions, the ideal gas equation would be given by P 2 V 2 nrt Since the righthand sides of these equations are equal to the same quantity (since we defined n and T to be constant), then the lefthand sides of the equations must also be equal, and we obtain the usual form of Boyle s law. P 1 V 1 P 2 V a. P 782 mm g 1.03 atm T 27 C 300 K V nrt P (0.210 mol)( L atm mol 1 K 1 )(300. K) (1.03 atm) 5.02 L b. V 644 ml L P nrt V ( mol)( L atm mol 1 K 1 )(303 K) (0.644 L) 3.56 atm P 3.56 atm mm g c. P 745 mm atm T PV nr (0.980 atm)(11.2 L) (0.401 mol)( L atm mol 1 K 1 ) 334 K 24. a. T 25 C 298 K V ( mol)( L atm mol 1 K 1 )(298 K) (1.01 atm) L b. V 602 ml L P (8.01 x 103 mol)( L atm mol 1 K 1 )(310 K) (0.602 L) atm c. V 629 ml L T 35 C 308 K n (0.998 atm)(0.629 L) ( L atm mol 1 K 1 )(308 K) mol
6 Gases molar mass of N g 58.2 C 331 K n 4.24 g N 2 1 mol N g N mol N 2 V nrt/p (0.151 mol)( L atm mol 1 K 1 )(331 K) (2.04 atm) The number of moles of any ideal gas that can be contained in the tank under the given conditions can first be calculated. T 24 C 297 K n PV (135 atm)(200 L) RT ( L atm mol 1 K 1 )(297 K) mol gas Molar masses: e, g; 2, g for e: mol e g g e 4.44 kg e 1 mol for 2 : mol g g kg 2 1 mol 27. Molar mass of N g 16.3 g N 2 1 mol mol N g T PV nr (1.25 atm)(25.0 L) (0.582 mol)( L atm mol 1 K 1 ) 654 K 381 C 28. Molar mass of O g 56.2 kg g g 1 mol g mol T 21 C 294 K P nrt V (1.76 x 10 3 mol)( L atm mol 1 K 1 )(294 K) (125 L) 340 atm 29. P atm P atm V ml V 2? ml 100 C 373 K T 2 25 C 298 K V 2 P 1 V 1 T 2 P 2 (0.981 atm)(125 ml)(298 K) (1.15 atm)(373 K) 85.2 ml 30. P atm P atm V ml V 2? ml
7 136 Chapter C 300. K V 2 P 1 V 1 T 2 P 2 (1.05 atm)(459 ml)(288 K) (0.997 atm)(300 K) T 2 15 C 288 K 464 ml 31. In deriving the ideal gas law, we assume that the molecules of gas themselves occupy no volume, and that the molecules do not interact with each other. Under these conditions, there is no difference between gas molecules of different substances (other than their masses) as far as the bulk behavior of the gas is concerned. Each gas behaves independently of other gases present, and the overall properties of the sample are determined by the overall quantity of gas present. P total P 1 + P P n where n is the number of individual gases present in the mixture 32. As a gas is bubbled through water, the bubbles of gas become saturated with water vapor, thus forming a gaseous mixture. The total pressure in a sample of gas that has been collected by bubbling through water is made up of two components: the pressure of the gas of interest and the pressure of water vapor. The partial pressure of the gas of interest is then the total pressure of the sample minus the vapor pressure of water. 33. molar masses: O 2, g; e, g 65 C K 4.0 g O 2 1 mol O g O mol O g e 1 mol e g e mol e P oxygen n oxygen RT/V (0.125 mol)( L atm mol 1 K 1 )(338 K) (5.0 L) P oxygen atm 0.69 atm P helium n helium RT/V (0.999 mol)( L atm mol 1 K 1 )(338 K) (5.0 L) P helium 5.54 atm 5.5 atm P total atm atm atm 6.2 atm 34. Total moles of gas 3.0 mol mol mol 6.0 mol 3.0 mol P nitrogen 10.0 atm 5.0 atm 6.0 mol 2.0 mol P oxygen 10.0 atm 3.3 atm 6.0 mol 1.0 mol P carbon dioxide 10.0 atm 1.7 atm 6.0 mol 35. P oxygen P total P water vapor torr
8 Gases P oxygen P total P water vapor mm g atm T 24 C K V 500. ml L n PV/RT ( atm)(0.500 L) ( L atm mol 1 K 1 )(297 K) mol O CaCO 3 (s) CO 2 (g) + CaO(s) molar mass CaCO g 15.2 g CaCO 3 1 mol CaCO g CaCO mol CaCO 3 From the balanced chemical equation, if mol CaCO 3 reacts, mol of CO 2 will result. STP: 1.00 atm, 273 K V nrt/p (0.152 mol)( L atm mol 1 K 1 )(273 K) (1.00 atm) 3.41 L 38. C 3 8 (g) + 5O 2 (g) 3CO 2 (g) O(g) 25 C K molar mass C g 5.53 g C mol C g C mol C mol C mol O 2 1 mol C mol O 2 V nrt/p ( mol O 2 )( L atm mol 1 K 1 )(298 K) (1.04 atm) 14.7 L O C 300 K 26 C 299 K mol N 3 present (1.02 atm)(4.21 L) ( L atm mol 1 K 1 )(300 K) mol N 3 mol Cl present (0.998 atm)(5.35 L) ( L atm mol 1 K 1 )(299 K) mol Cl N 3 and Cl react on a 1:1 basis: N 3 is the limiting reactant. molar mass N 4 Cl g mol N 3 1 mol N Cl g N Cl 4 1 mol N 3 1 mol N 4 Cl 9.31 g N 4 Cl produced
9 138 Chapter Molar mass of Mg 3 N g 10.3 g Mg 3 N 2 1 mol g mol Mg 3N 2 From the balanced chemical equation, the amount of N 3 produced will be mol Mg 3 N 2 2 mol N 3 1 mol Mg 3 N mol N 3 T 24 C 297 K P 752 mm g atm V nrt P (0.204 mol)( L atm mol 1 K 1 )(297 K) (0.989 atm) 5.03 L This assumes that the ammonia was collected dry. 41. Molar masses: e, g; 2, g 14.2 g e 1 mol e g e 3.55 mol e 21.6 g 2 1 mol g mol total moles 3.55 mol mol 14.3 mol 28 C 301 K V nrt P (14.3 mol)( L atm mol 1 K 1 )(301 K) (0.985 atm) 359 L 42. P atm P atm (standard pressure) V ml V 2? ml 44 C 317 K T 2 0 C 273 K (standard temperature) V 2 P 1 V 1 T 2 P 2 (1.47 atm)(145 ml)(273 K) (1.00 atm)(317 K) 184 ml 43. molar masses: O 2, g; N 2, g; CO 2, g; Ne, g 5.00 g O 2 1 mol O g O mol O g N 2 1 mol N g N mol N g CO 2 1 mol CO g CO mol CO 2
10 Gases g Ne 1 mol Ne g Ne mol Ne Total moles of gas mol 22.4 L is the volume occupied by one mole of any ideal gas at STP. This would apply even if the gas sample is a mixture of individual gases mol 22.4 L L 15.6 L 1 mol The partial pressure of each individual gas in the mixture will be related to what fraction on a mole basis each gas represents in the mixture. P oxygen 1.00 atm mol O mol total P nitrogen 1.00 atm mol N mol total atm O atm N 2 P carbon dioxide 1.00 atm mol CO mol total atm CO 2 P neon 1.00 atm mol Ne mol total atm Ne 44. 2Na(s) + Cl 2 (g) 2NaCl(s) molar mass Na g mol Na g Na mol Na mol Na 1 mol Cl 2 2 mol Na mol Cl mol Cl L 1 mol 2.34 L Cl 2 at STP 45. 2K 2 MnO 4 (aq) + Cl 2 (g) 2KMnO 4 (s) + 2KCl(aq) molar mass KMnO g 10.0 g KMnO 4 1 mol KMnO g KMnO mol KMnO mol KMnO 4 1 mol Cl 2 2 mol KMnO mol Cl mol Cl L 1 mol L 709 ml 46. A law is a statement that precisely expresses generally observed behavior. A theory consists of a set of assumptions/hypotheses that is put forth to explain the observed behavior of matter. Theories attempt to explain natural laws.
11 140 Chapter A theory is successful if it explains known experimental observations. Theories that have been successful in the past may not be successful in the future (for example, as technology evolves, more sophisticated experiments may be possible in the future). 48. We assume that the volume of the molecules themselves in a gas sample is negligible compared to the bulk volume of the gas sample: this helps us to explain why gases are so compressible. 49. Chemists believe the pressure exerted by a gas sample on the walls of its container arises from collisions between the gas molecules and the walls of the container. 50. kinetic energy 51. The temperature of a gas reflects, on average, how rapidly the molecules in the gas are moving. At high temperatures, the particles are moving very fast and collide with the walls of the container frequently, whereas at low temperatures, the molecules are moving more slowly and collide with the walls of the container infrequently. The Kelvin temperature is directly proportional to the average kinetic energy of the particles in a gas. 52. If the temperature of a sample of gas is increased, the average kinetic energy of the particles of gas increases. This means that the speeds of the particles increase. If the particles have a higher speed, they will hit the walls of the container more frequently and with greater force, thereby increasing the pressure. 53. Any gas that does not follow the ideal gas law is not behaving ideally. In addition, if there are reasons to believe that the assumptions of the kineticmolecular theory are poor assumptions, the gas sample is not behaving ideally. The fact that gases condense into liquids, for example, shows nonideal behavior. 54. When the volume of a gas is decreased, the gas particles take up a greater percentage of the volume of the container. The assumption in the KMT that gas particles take up a negligible volume is less correct. 55. First determine what volume the helium in the tank would have if it were at a pressure of 755 mm g (corresponding to the pressure the gas will have in the balloons) atm 6384 mm g V 2 (25.2 L) 6384 mm g 755 mm g 213 L Allowing for the fact that 25.2 L of e will have to remain in the tank, this leaves L of e for filling the balloons L e 1 balloon 1.50 L e 125 balloons 56. A decrease in temperature would tend to make the volume of the weather balloon decrease. Since the overall volume of a weather balloon increases when it rises to higher altitudes, the contribution to the new volume of the gas from the decrease in pressure must be more important than the decrease in temperature (the temperature change in kelvins is not as dramatic as it seems in degrees Celsius).
12 Gases S(s) + 3O 2 (g) 2SO 3 (g) 350. C K molar mass S g 5.00 g 1 mol S g S mol S mol 3 mol O 2 2 mol S mol O 2 V nrt/p ( mol)( L atm mol 1 K 1 )(623 K) (5.25 atm) 2.28 L O Assume the pressure at sea level to be 1 atm (760 mm g). Calculate the volume the balloon would have if it rose to the point where the pressure has dropped to 500 mm g. If this calculated volume is greater than the balloon's specified maximum volume (2.5 L) the balloon will burst. 2.0 L 760 mm g 500 mm g 3.0 L > 2.5 L. The balloon will burst C K 100 C K 729 ml 373 K 295 K 922 ml 60. P atm P torr atm V L V 2? 23 C K T 2 31 C 242 K V 2 T 2 P 1 V 1 P 2 (242 K)(1.0 atm)(1.0 L) (295 K)(0.289 atm) 2.8 L 61. 2Cu 2 S(s) + 3O 2 (g) 2Cu 2 O(s) + 2SO 2 (g) molar mass Cu 2 S g 25 g Cu 2 S 1 mol Cu 2 S g Cu 2 S mol Cu 2S mol Cu 2 S 3 mol O 2 2 mol Cu 2 S mol O C K V oxygen ( mol)( L atm mol 1 K 1 )(301 K) (0.998 atm) 5.8 L O mol Cu 2 S 2 mol SO 2 2 mol Cu 2 S mol SO 2
13 142 Chapter 13 V sulfur dioxide ( mol)( L atm mol 1 K 1 )(301 K) (0.998 atm) V sulfur dioxide 3.9 L SO molar masses: e, g; Ar, g; Ne, g 5.0 g e 1.0 g Ar 3.5 g Ne 1 mol e g e 1 mol Ar g Ar 1 mol Ne g Ne mol e mol Ar mol Ne Total moles of gas mol 22.4 L is the volume occupied by one mole of any ideal gas at STP. This would apply even if the gas sample is a mixture of individual gases mol 22.4 L 32 L total volume for the mixture 1 mol The partial pressure of each individual gas in the mixture will be related to what fraction on a mole basis each gas represents in the mixture. P e 1.00 atm P Ar 1.00 atm P Ne 1.00 atm mol e mol total mol Ar mol total mol Ne mol total 0.86 atm atm 0.12 atm atm P 1 72 cm g 720 mm g 760 mm g 0.95 atm P atm V ml V 2? n 1 x n 2 2/3x (1/3 of gas removed) 27 C 300 K T K PV nrt or R PV nt Thus, P 1 V 1 n 1 P 2 V 2 n 2 T 2 constant (0.95 atm)(350 ml) (x)(300. K) (1.00 atm)(v 2 ) (2/3x)(600. K) Thus, V ml
14 Gases A balloon is essentially a constant pressure container. Thus, P and n are constant. PV nrt or nr P V T constant Thus, V 1 V 2 T L 293 K 2.00 L T 2 T K 508 C 65. PV nrt or T PV nr P atm Assume V 1.00 L 1.00 L 1000 cm L x 1.4 g 1400 g gas 3 1 cm 1400 g 1 mol 2.00 g 700 mol gas Thus, T (1.3 x 10 9 atm)(1.00 L) (700 mol)( L atm mol 1 K 1 ) K 66. If we can determine the molar mass of X 4 10, we can determine X. PV nrt or n PV RT P 801 mm g 1 atm 760 mm g 1.05 atm V 30.0 cm L L 1000 cm 3 T 20 C K n (1.05 atm)( L) ( L atm mol 1 K 1 )(293 K) mol g molar mass of X g/mol 1.31 x 103 mol Thus, 4(atomic mass of X) + 10(1.008) 54.3 atomic mass of X 11.1 This is closest to boron (B), which has an atomic mass of 10.8.
15 144 Chapter PV nrt or n PV RT 1 atm P 802 mm g 1.06 atm 760 mm g V 618 cm L L 1000 cm 3 T 75 C K n (1.06 atm)(0.618 L) ( L atm mol 1 K 1 )(348 K) mol gas n O2 20% of mol O 2 Number of O 2 molecules mol O x 1023 molecules 1 mol O 2 molecules 68. We are looking for the subscripts for C x y O z For every g of the compound, we have 48.6 g C 1 mol C g C 4.05 mol C 8.18 g 1 mol g 43.2 g O 1 mol O g O 8.12 mol 2.70 mol O 4.05 : 8.12 : : 3 : 1 3 : 6 : 2 The empirical formula is C 3 6 O 2, which has a molar mass of g/mol [3(12.01) + 6(1.008) + 2(16.00)] If we assume 1.00 L, we can solve for n n PV (1.00 atm)(1.00 L) RT ( L atm mol 1 K 1 )(150 o C + 273) 2.13 g molar mass 74.0 g/mol mol Thus, the molecular formula is C 3 6 O 2 Two possible Lewis structures include mol O O C C C O C C O C and
16 Gases ml 1 L 1000 ml PV nrt 2.00 mol Cl 1 L 1 mol 2 2 mol Cl mol 2 gas V nrt/p [(0.100 mol)( L atm/mol K)(298 K)]/(1.00 atm) 2.45 L The balloon will not pop since it can expand to 3.00 L and the hydrogen gas expands to a volume of 2.45 L at 1.00 atm and 25 o C. 70. a. PV nrt V n RT constant V 1 RT P n 1 P V 2 n 2 V 1 V 2 n 1 n 2 b. An increase in number of moles of gas increases the number of collisions with the container (if volume stays constant), which would increase the pressure (since pressure is due to collisions of gas particles with the walls of the container). In order for the pressure to remain constant, the volume must increase so that the distance the particles must travel increases and thus the collision rate overall stays the same (thus the pressure stays the same). Since temperature is a measure of the average kinetic energy of the particles, and changing the temperature would thus change the speed of the particles, the temperature must remain constant for this argument to hold. c g e 2.50 mol e 20.0 g Ar mol Ar V e 2.50 mol V Ar mol V e V Ar The container with e is 4.99 times as large as the container with Ar.
7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.
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