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Homework7 - S117 Homework 7 Out of 65 Points Chapter 5 (not...

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Unformatted text preview: S117 Homework 7 Out of 65 Points Chapter 5 (not to turn in) 21. Recall that units of torr are equivalent to mm Hg. In other words, since the substance in the manometer is mercury, the pressure in torr is simply the height difference between the two arms of the flask in units of mm. Therefore, the pressure is . Under atmospheric pressure, the height of the mercury would be 76cm or 760mm, so . Therefore, , so 23. unit area is simply pressure, so , where is density and . for a cylinder. The force per . The final conversion factor is . For mercury, , approximately one atmosphere. For water to exert the same pressure, . It would be a good idea for you to check the units in these problems and make sure that all the proper conversion factors are present. 25. a. c. b. d. . Let's first consider the flask containing . Recall that we don't have to 27. Pressure and volume are inversely related; that is, . Therefore, H2. When the stopcock is opened, the flask holding H2 changes from 2.00L to 3.00L, so worry about the effect of N2 on the H2 because we're assuming that these gases are ideal and thus independent. Now consider the flask containing N2. By a similar argument, . 31. Carbon dioxide has a molecular mass of . After the carbon dioxide vaporizes, . Assuming ideal gases, it is irrelevant what the pressure is in the container already because ideal gases are independent. The partial pressure of carbon dioxide will still be 3.08atm, so the total pressure will be . , so 22.0g of it is 37. There will only be half as many moles of N2O4 as there were of NO2 because of the stoichiometry of the problem, so assuming that these are ideal gases, . 40. Assuming ideal gases, each component is independent of each other component, so we can calculate the partial pressures of each component without regard to what else is in the flask. , , and . This makes the total pressure 45. a. The mole fraction of CH4 is . , making the mole fraction of O2 . The total number of . . b. The total pressure is the sum of the partial pressures, or moles, then, is . c. Using the mole fractions, the number of grams of CH4 is The number of grams of O2 is . 51. How about this? Since there are so many "not to turn in" questions, why don't you send me (David) an email if you have questions on how to do any of them? I'll be happy to help you in that case, but I do have a suspicion that very few people look at this part of the answer key, so I'm inclined not to write out solutions to all of these unless I know that people will use them. 59. 67. 70. 71. 73. 74. 75. 83. 85. 86. Chapter 5 (to turn in) 26. (out of 5 points) a. PV b. P c. T V d. P P T V PV/T V 34. (out of 5 points) n, R, and P are constant here, so , and 1/V P , so . Now , so . 38. (out of 5 points) n and R are constant here, so , so . 44. (out of 5 points) Knowing the partial pressure of oxygen or of water vapor in the initial case would certainly help. Let's work on this. When the O2 was dried, Therefore, the partial pressure of O2 in the initial case was , so the partial pressure of water was , which is water's vapor pressure at 25C. . 50. (out of 5 points) By the ideal gas law, compound, the molar mass is . Since there are 0.800g of the . One "unit" of CHCl has a molar mass of , so there are compound. Therefore, the compound is C2H2Cl2. 56. (out of 10 points) We need to know the number of moles of ammonia and of carbon dioxide that flow per minute to determine which is the limiting reagent. , and 100% yield, every two moles of NH3 produces one mole of urea, so , so . . Since two moles "units" of CHCl in this of NH3 are required to react with each mole of CO2, NH3 is the limiting reagent here. Assuming 72. (out of 5 points) a. We could try to use van der Waals equation (page 174) to calculate the pressure of these real gases, but we are not given information about the parameters a and b for these gases, so we are forced to use the ideal gas law. According to the ideal gas law, the pressure is independent of the identity of the gas, and since temperature is constant. Containers ii and vi both have only four particles in the larger volume, so they have the lowest pressure. Containers iv and viii have twice as many molecules as containers i and v, but they also have twice the volume, so the ratio of moles to volume is the same in all four containers. Containers iii and vii have the same number of particles as containers iv and viii, but their volume is smaller, so they have a higher pressure. Combining this information, the pressure goes as ii=vi<iv=viii=i=v<iii=vii. b. According to the kinetic molecular theory, . Since the temperature is the same in all samples, the average kinetic energy is the same in all samples. c. Density is mass divided by volume. Since we're assuming the mass of an argon atom is twice the mass of a neon atom, each argon atom is worth two neon atoms in terms of density. Therefore, the density goes as vii>iii=v=viii>iv=vi=i>ii. d. The root mean square velocity is given by , where M is the mass of one mole of a substance in kilograms. Since the temperature is the same in all cases, the only variable is M, which is essentially a molar mass. Since argon has a greater mass than neon, , so . Therefore, the root mean square velocity goes as i=ii=iii=iv>v=vi=vii=viii. 76. (out of 5 points) a. According to the kinetic molecular theory, average kinetic energy is a function of temperature only, so all of these gases would have the same average kinetic energy in this case. b. According to the kinetic molecular theory, , where M is the molar mass in kilograms. Therefore, the only variable is M when we are at constant temperature. As M increases, uavg decreases. Therefore, average velocity goes as Xe<Cl2<O2<H2. c. Since the samples are separate, the temperature does not have to be the same. We want , so we want . Solving for , . Therefore, when the temperatures are set such that , . 78. (out of 5 points) a. All of those will increase. b. All of those will decrease. c. The frequencies of collision will increase, but the others will stay the same in the case that the pressure increases as the volume is halved. In the case that the temperature decreases as the volume is halved, the frequencies of collision will stay the same, but the others will decrease. d. The frequencies of collision will increase, but the others will stay the same in the case that the pressure increases as the number of moles doubles. In the case that the temperature decreases as the number of moles doubles, the frequencies of collision will stay the same, but the others will decrease. 84. (out of 10 points) a. b. For N2, , and . Solving for . The van der Waals equation is and substituting in appropriate values, . c. The ideal gas law is in good agreement with van der Waals equation, but note that van der Waals equation gives a slightly lower pressure. Recognize based on the form of that equation that the volume correction (the ) decreases the volume which increases the pressure, whereas the intermolecular attraction correction (the ) decreases the pressure. At this high volume, the volume correction is not as important as the attraction correction, so the lower value of pressure with the van der Waals equation tells us that intermolecular attractions occur here. d. By the same equations, the ideal pressure in problem 83 is , and the van der Waals pressure is . In problem 83, the ideal gas law is not as accurate as it was in problem 84. Recall that the ideal gas law works best in the low-pressure, high-temperature limit. Since the volume was higher in problem 84, the pressure was lower, so the ideal gas law worked better. 110. (out of 5 points) a. The reaction must be balanced, so . By the ideal gas law, the number of moles of N2 is equal to the number of moles of H2 because their partial pressures are the same. By the stoichiometry of the problem, H2 is the limiting reagent since more moles of H2 are required than moles of N2. If the reaction goes to completion, then the number of moles of H2 will be 0; the number of moles of N2 will be a 2/3 what it initially was (since three moles of H2 are used for every mole of N2), and the number of moles of NH3 will be the same as the number of moles of N2, so . Since the pressure is constant, , so . b. P, T, and of course R are constant here, so Call the initial number of moles H2 (or N2). The final number of moles is formula, . , where n is the total number of moles. , where x is the initial number of moles of + . Rearranging the earlier ...
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This note was uploaded on 01/17/2012 for the course S 117 taught by Professor Stephenjacobson during the Fall '11 term at Indiana.

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