By John T. Moore  Richard Langley — McGrawHill Professional
Updated on Feb 9, 2011
Answers and Explanations
 C—This question relates to the combined gas law: P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}. Since the pressure remains constant, the pressures may be removed from the combined gas law to produce Charles's law: V_{1}/T_{1} = V_{2}/T_{2}. This equation may be rearranged to: T_{2} = V_{2}T_{1}/V_{1}. The doubling of the volume means V_{2} = 2 V_{1}. On substituting: T_{2} = 2V_{1}T_{1}/V_{1}; giving T_{2} = 2T_{1}. The identity of the gas is irrelevant in this problem.
 E—This problem depends on the ideal gas equation: PV = nRT. R, V, and T are known, and by using the partial pressure for a gas, the number of moles of that gas may be determined. To convert from moles to mass, the molar mass of the gas is needed.
 B—Since T and P are known, and since the moles (n) can be determined from the masses given, this question could use the ideal gas equation. The number of moles of each gas is 0.50. Equal moles of gases, at the same T and P, have equal volumes. Equal volume eliminates answer choice D. Equal volume also means that the greater mass has the greater density, eliminating choice A. Equal moles means that the numbers of molecules and atoms are equal, eliminating choice E. The average kinetic energy of a gas depends on the temperature. If the temperatures are the same, then the average kinetic energy is the same, eliminating C. Finally, at the same temperature, heavier gases travel slower than lighter gases. Nitrogen is lighter than argon, so it travels at a faster average speed, making B the correct answer. You may find this reasoning process beneficial on any question in which you do not immediately know the answer.
 A—This experiment requires the ideal gas equation. The mass of the solid is needed (to convert to moles); this eliminates answer choice D. The volume, temperature, and pressure must also be measured during the experiment, eliminating choices B, C, and E. The measured pressure is the total pressure. Eventually the total pressure must be converted to the partial pressure of the gas using Dalton's law. The total pressure is the sum of the pressure of the gas plus the vapor pressure of water. The vapor pressure of water can be looked up in a table when the calculations are performed (only the temperature is needed to find the vapor pressure in a table). Answer A is correct.
 C—Real gases are different from ideal gases because of two basic factors (see the van der Waals equation): molecules have a volume, and molecules attract each other. The molecules' volume is subtracted from the observed volume for a real gas (giving a smaller volume), and the pressure has a term added to compensate for the attraction of the molecules (correcting for a smaller pressure). Since these are the only two directly related factors, answers B, D, and E are eliminated. The question is asking about volume; thus, the answer is C. You should be careful of "NOT" questions such as this one.
 C—A balanced chemical equation is needed:
 D—The partial pressure of any gas is equal to its mole fraction times the total pressure. The mole fraction of carbon monoxide is [0.30/(0.60 + 0.30 + 0.10)] = 0.30, and the partial pressure of CO is 0.30 × 0.80 atm = 0.24 atm.
 C—Using Dalton's law (P_{Total} = P_{A} + P_{B} + …), the partial pressure may be found by: 756 mm Hg –41 mm Hg = 715 mm Hg.
 A—The answer may be found using the combined gas law. Removing the constant pressure leaves Charles's law: V_{1}/T_{1} = V_{2}/T_{2}. This is rearranged to: T_{2} = V_{2}T_{1}/V_{1} = (20.00 L × 400. K)/(8.00 L) = 1000 K (= 727°C). The other answers result from common errors in this problem.
 A—The average kinetic energy of the molecules depends on the temperature. The correct answer involves a temperature difference (333 K – 303 K). Do not forget that ALL gas law calculations require Kelvin temperatures.
 A—You can begin by removing the volume (constant) from the combined gas law to produce GayLussac's law = P_{1}/T_{1} = P_{2}/T_{2}. This equation may be rearranged to: P_{2} = P_{1}T_{2}/T_{1} = (0.300 atm × 290. K)/(300. K) = 0.290 atm. The moles are not important since they do not change. Some of the other answers result from common errors.
 B—The molar mass may be obtained by dividing the grams by the number of moles (calculated from the ideal gas equation). Do not forget to convert the temperature to kelvin.
 C—Choice I requires an increase in volume. Choice II requires an increase in temperature. Choice III requires a change in the composition of the gas.
 B—Lighter gases effuse faster. The only gas among the choices that is lighter than methane is helium. To calculate the molar mass, you would begin with the molar mass of methane and divide by the rate difference squared:
 D—A steel tank will have a constant volume, and the problem states that the temperature is constant. Adding gas to the tank will increase the number of moles (molecules) of the gas and the pressure (forcing the molecules closer together). A constant temperature means there will be a constant average speed.
 B—The pressure is calculated using the ideal gas equation. A common mistake is forgetting that 5 mol of gas are produced for every 2 mol of solid reactant. The ideal gas equation is rearranged to P = nRT/V = ()(0.2 mol) (0.082 L atm mol^{–1} K^{–1}). (300. K)/(2 L) = 6 atm.
 E—The lighter the gas, the faster it effuses (escapes). Equal moles of gases in the same container would give equal initial partial pressures. The partial pressures would be reduced relative to the masses of the molecules, with the lightest gas being reduced the most.
 C—Deviations from ideal behavior depend on the size and the intermolecular forces between the molecules. The greatest deviation would be for a large polar molecule. Sulfur tetrafluoride is the largest molecule, and it is the only polar molecule listed.
 D—The molar mass of gas must be determined. The simplest method to find the molar mass is: (5.47 g/L) × (22.4 L/mol) = 123 g/mol (simple factor label). The molar mass may also be determined by dividing the mass of the gas by the moles (using 22.4 L/mol for a gas at STP and using 1 L). If you did not recognize the conditions as STP, you could find the moles from the ideal gas equation. The correct answer is the gaswith the molar mass closest to 123 g/mol.
 D—The hotair balloon rises because it has a lower density than air. Less dense objects will float on more dense objects. In other words "lighter" objects will float on "heavy" objects.
2 Al + 6 HCl → 2 AlCl_{3} + 3 H_{2}
The reaction produced 33.6L/22.4 L or 1.50 mol at STP. To produce this quantity of hydrogen, (2 mol Al/3 mol H_{2}) × 1.50 moles H_{2} = 1.00 mol of Al is needed. The atomic weight of Al is 27.0; thus, 27.0 g of Al are required.
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From 5 Steps to a 5 AP Chemistry. Copyright © 2010 by The McGrawHill Companies. All Rights Reserved.
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