Chapter 0 Gases. A sample of gas (4.g) initially at 4.00 atm was compressed from 8.00 L to.00 L at constant temperature. After the compression, the gas pressure was atm. (a). 4.00 (b)..00 (c)..00 (d). 6.0 Explanation: This is an example of using Boyle s law. The temperature and the number of moles of the gas are constant. According to Boyle s law where stands for the initial conditions and for final. Need to find 4.00 atm! 8.00 L.00 L 6.0 atm. A sample of a gas (5.0 mol) at.0 atm is expanded at constant temperature from 0.0 L to 5.0 L. The final pressure is atm. (a)..5 (b). 7.5 (c). 0.67 (d). 3.3 Explanation: This is an example of using Boyle s law. The temperature and the number of moles of the gas are constant. According to Boyle s law where stands for the initial conditions and for final. Need to find by rearranging the above equation as follows:.0 atm! 0.0 L, 5.0 L 0.67 atm 3. A balloon occupies 4.39 L at 44.0 ºC and a pressure of 79.0 torr. What temperature must the balloon be cooled to, to reduce its volume to 3.78 L (at constant pressure)? (a). 38.7 ºC (b). 0.00 ºC (c)..6 ºC (d). 30.9 ºC Explanation: This is an example of using Charles s law. The pressure and the number of moles are held constant while the temperature is lowered. The initial volume and temperature conditions can be labeled as and T while the final Copyright 006 Dr. Harshavardhan D. Bapat
conditions can be labeled as and T. The temperature T (in Kelvin) can be found by rearranging Charles s law as follows:! T 3.78 L! 37.5 K or T 73. K T T 4.39 L This is approximately equal to 0.00 ºC. 4. A gas originally at 7.0 ºC and.00 atm pressure in a 3.90 L flask is cooled at constant pressure until the temperature is.0 ºC. The new volume of the gas is L. (a)..7 (b). 3.69 (c). 3.96 (d). 4. Explanation: This is an example of using Charles s law. The initial volume and temperature conditions can be labeled as and T while the final conditions can be labeled as and T. The final volume (in Liters) can be found by rearranging Charles s law as follows: T! T 3.90 L! 84.5 K or 3.69 L T T 300.5 K 5. If 50.75 g of a gas occupies 0.0 L at ST, 9.3 g of the gas will occupy L at ST. (a). 3.9 (b). 50.8 (c)..9 (d). 5.5 Explanation: Since the pressure and temperature are being held constant the volume occupied by the gas should be directly proportional to the amount of the gas. The initial conditions (of mass and volume) can be labeled as and the final conditions as. The final volume can be now found by: m m m! 9.3 g! 0.0 L or 5.5 L m 50.75 g Copyright 006 Dr. Harshavardhan D. Bapat
6. A sample of He gas (.35 mol) occupies 57.9 L at 300.0 K and.00 atm. What is the volume of this sample at 43.0 K and.00 atm? (a). 0.709 (b). 4. (c). 8.6 (d)..4 Explanation: This is an example of using Charles s law. The initial volume and temperature conditions can be labeled as and T while the final conditions can be labeled as and T. The final volume (in Liters) can be found by rearranging Charles s law:! T 57.9 L! 43.0 K or 8.6 L T T T 300.0 K 7. A sample of H gas (.8g) occupies 00.0 L at 400.0 K and.00 atm. A sample weighing 9.49 g occupies L at 353.0 K and.00 atm. (a). 09 (b). 68. (c). 54.7 (d). 47 Explanation: Convert both the grams of H to moles (divide g by the molar mass of H ) and then rearrange the ideal gas law to calculate the volume. n T 00.0 L! (9.49 g/.06 g)! 353.0 K or 68. L n T n T n T (.8 g/.06 g)! 400.0 K 8. The amount of gas that occupies 60.8 L at 3.0 ºC and 367.0 mmhg is mol. (a)..8 (b). 0.850 (c). 894 (d)..6 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin and the pressure to atm before using these numbers. (367.0 mm Hg/760 mm Hg)! 60.8 L nrt or n.8 moles RT 0.08 Latm/molK! 304.5 K Copyright 006 Dr. Harshavardhan D. Bapat 3
9. The pressure of a sample of 6.0 g of CH 4 gas in a 30.0 L vessel at 40.5 K is atm. (a)..4 (b). 6.6 (c). 0.44 (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the pressure () of the gas. To use the ideal gas law convert the grams of methane to moles (n), before using this number. nrt (6.0 g/6.043 g)! 0.08 Latm/molK! 40.5 K nrt and 0.44 atm 30.0 L 0. A 0.35 L flask filled with gas at 0.94 atm and 9.0 ºC contains mol of gas. (a).4 x 0 - (b)..48 x 0 - (c). 9.4 (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin before using the number. nrtand n RT 0.94 atm " 0.35 L 0.08 Latm/molK" 9.5 K.4 " 0! moles. A gas in a 350.0 ml container has a pressure of 695.0 torr at 9.0 ºC. Calculate the number of moles of the gas in the flask. (a)..48 x 0 - (b)..9 x 0 - (c). 9.4 (d)..4 Explanation: This problem involves using the ideal gas law and rearranging it to find the number of moles (n) of the gas. To use the ideal gas law convert the temperature to Kelvin, the pressure to atm and volume to liters before using these numbers. nrt and n RT atm 695 torr " " 0.350 L 760 torr 0.08 Latm/molK" 30.5 K.9" 0! moles Copyright 006 Dr. Harshavardhan D. Bapat 4
. Calculate the volume occupied by a sample of gas (.30 mol) at.0 ºC and.50 atm. (a). 0.079 (b). 0.94 (c)..8 (d). 3 Explanation: This problem involves using the ideal gas law and rearranging it to find the volume () of the gas. To use the ideal gas law convert the temperature to Kelvin, the before using this number. nrt.30 mol! 0.08 Latm/molK! 99.5 K nrt and.8 L.50 atm 3. What is the volume of 0.65 mol of an ideal gas at 365.0 torr and 97.0 ºC? (a). 0.054 (b). 9.5 (c). (d). 4. Explanation: This problem involves using the ideal gas law and rearranging it to find the volume () of the gas. To use the ideal gas law convert the temperature to Kelvin and the pressure to atm before using these numbers. nrt 0.65 mol! 0.08 Latm/molK! 370.5 K nrt and 4.L atm 365.0 torr! 760 torr 4. The density of ammonia gas in a 4.3 L container at 837.5 torr and 45.0 ºC is g/l. (a). 3.86 (b). 0.78 (c). 0.43 (d). 0.94 Explanation: The density of a gas can be calculated by using a formula that relates the density (g/l), pressure ( in atm), molar mass (M in g/mol) and temperature (T) of the gas. The molar mass of NH 3 is 7.03 g/mol. The pressure must be converted to atm and the temperature to Kelvin before using these numbers. atm 837.5 torr!! 7.03g/mol M d 760 torr 0.78 g/l RT 0.08 Latm/molK! 38.5 K Copyright 006 Dr. Harshavardhan D. Bapat 5
5. The density of N O at.53 atm and 45. ºC is g/l. (a). 8. (b)..76 (c). 0.88 (d)..58 Explanation: The density of a gas can be calculated by using a formula that relates the density (g/l), pressure ( in atm), molar mass (M in g/mol) and temperature (T) of the gas. The molar mass of N O is 44.03 g/mol. The temperature must be converted to Kelvin before using this number. M.53 atm! 44.03 g/mol d.58 g/l RT 0.08 Latm/molK! 38.35 K 6. The molar mass of a gas that has a density of 6.70 g/l at ST is g/mol. (a). 496 (b). 50 (c). 73.0 (d). 3.35 Explanation: The molar mass of a gas is related to its density (in g/l), pressure (in atm) and its temperature (in K). drt 6.70 g/l! 0.08L atm/molk! 73.5 K M 50. g/mol.00 atm 7. What is the volume of hydrogen gas that can be produced by the reaction of 4.33 g of zinc with excess sulfuric acid at 38.0 ºC and 760.0 torr? (a)..69 (b)..7 x 0-4 (c). 3.69 x 0 4 (d)..84 Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Need to write a balanced chemical equation for the reaction and then calculate the number of moles of hydrogen that can be produced from the moles of the limiting reagent (Zinc in this case). Zn + H SO 4 ZnSO 4 + H 4.33 g Zn " mol Zn 65.39 g mol H " mol Zn! 6.6" 0 moles H Copyright 006 Dr. Harshavardhan D. Bapat 6
The volume occupied by these moles of H can now be calculated using the ideal gas law: nrt 6.6! 0 "! 0.08 Latm/molK! 3.5 K atm 760.0 torr! 760 torr.69 L 8. What is the volume of HCl gas required to react with excess magnesium metal to produce 6.8 L of hydrogen gas at.9 atm and 5.0 ºC? (a). 6.8 (b)..9 (c). 3.6 (d). 4.38 Explanation: This problem uses a combination of stoichiometry and the ideal gas law. The number of moles of the hydrogen can be calculated from the ideal gas law and then used with the stoichiometry of the reaction..9 atm! 6.8 L n 0.60 moles H RT 0.08 Latm/molK! 98.5 K The balanced equation for this reaction is: Mg + HCl MgCl + H The equation shows that moles of HCl are needed for every mole of H produced. Thus (0.60 x. moles) of HCl would be required for this reaction. The volume occupied by these moles of HCl can then be calculated by using the ideal gas law: nrt.! 0.08Latm/molK! 98.5 K 3.6 L.9 atm 9. The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO) 4 (l) Ni (s) + 4CO(g) What volume (L) of CO is formed from the complete decomposition of 444.0 g of Ni(CO) 4 at 75.0 torr and.0 ºC? (a). 0.356 (b). 63.7 (c). 55. (d). 0. Copyright 006 Dr. Harshavardhan D. Bapat 7
Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Using the reaction stoichiometry from the above balanced equation, need to calculate the number of moles of CO produced, and then calculate the volume occupied using the ideal gas law. 444.0 g Ni(CO) nrt 4! mole 70.733 g! 4 moles CO mole Ni(CO) 0.40! 0.08 Latm/molK! 95.5 K atm 75.0 torr! 760 torr 4 0.40 moles of CO 55. L 0. Ammonium nitrite undergoes thermal decomposition to produce only gases: NH 4 NO (s) N (g) + H O (g) What volume (L) of gas is produced by the decomposition of 35.0 g of NH 4 NO (s) at 55. ºC and.50 atm? (a). 47 (b). 60 (c). 5 (d). 7. Explanation: This problem uses a combination of stoichiometry and the ideal gas law. Using the reaction stoichiometry from the above balanced equation, need to calculate the number of moles of gases formed and then calculate the volume occupied by these gases. It is not necessary to calculate the moles of the products individually. mole 3 moles (N + HO) 35.0 g NH4NO!!.64 moles 64.044 g mole nrt.64! 0.08 Latm/molK! 798.65 K.50 atm 7. L. The pressure in a. L vessel that contains.34 g of carbon dioxide,.73 g of sulfur dioxide and 3.33 g of argon, all at 4.0 ºC is mmhg. (a). 63. (b). 34 (c). 395 (d). 6 Explanation: Since this is a mixture of gases need to find the total number of moles of gases (n t ) and then calculate the pressure of this gas mixture. Copyright 006 Dr. Harshavardhan D. Bapat 8
.34 g CO.73 g SO + + 44.009 g/mol 64.063 g/mol nrt 3.33 g Ar 39.948 g/mol 0.63! 0.08 Latm/molK! 35.5 K!. L 0.63 moles 760 mm atm 63. mm. A sample of He gas (3.0 L) at 5.6 atm and 5.0 ºC was combined with 4.5 L of Ne gas at 3.6 atm and 5.0 ºC at constant temperature in a 9.0 L flask. The total pressure in the flask was atm. Assume the initial pressure in the flask was 0.00 atm. (a)..6 (b). 9. (c)..0 (d). 3.7 Explanation: Calculate the number of moles of each of the gases and find the total number of moles of the gas mixture. The pressure of this gas mixture can now be calculated using the ideal gas law. n RT The number of moles of He using this formula 0.686 and of Ne 0.66. The moles of gas mixture.348. Now using the appropriate rearrangement of the ideal gas law the pressure of the gas mixture 3.7 atm. 3. A flask contains a mixture of He and Ne at a total pressure of.6 atm. There are.0 mol of He and 5.0 mol of Ne in the flask. The partial pressure of He is atm. (a). 9. (b). 6.5 (c)..04 (d). 0.74 Explanation: Calculate the mole fraction of He and then the partial pressure of He. The partial pressure of a gas in a mixture is directly proportional to its mole fraction. n He.0 He 0.86. Total moles.0 + 5.0 artial pressure p 0.86!.6atm 0.74atm He Copyright 006 Dr. Harshavardhan D. Bapat 9
4. SO (5.00 g) and CO (6.00 g) were placed in a 750.0 ml container at 50.0 ºC. The total pressure in the container was atm. (a). 0.9 (b). 4.0 (c)..76 (d). 6.78 Explanation: Since each gas behaves as if the other was absent, need to calculate the partial pressure of each gas and then add them together to find the total pressure. 5.00 g SO 0.078 moles 64.063 g/mol nrt 0.0780! 0.08 Latm/molK! 33.5 K SO.76 atm 0.750 L 6.00 g CO 0.36 moles 44.009 g/mol nrt 0.36! 0.08 Latm/molK! 33.5 K CO 4.8 atm 0.750 L Total pressure + 6.78 atm SO CO 5. CO (5.00 g) and CO (5.00g) were placed in a 750.0 ml container at 50.0 ºC. The total pressure in the container was atm. (a). 0.3 (b). 4.0 (c). 6.3 (d). 0.9 Explanation: Since each gas behaves as if the other was absent, need to calculate the partial pressure of each gas and then add them together to find the total pressure. 5.00 g CO 0.78 moles 8.0g/mol nrt 0.78! 0.08 Latm/molK! 33.5 K SO 6.3atm 0.750 L 5.00 g CO 0.4 moles 44.009 g/mol nrt 0.4! 0.08 Latm/molK! 33.5 K CO 4.0 atm 0.750 L Total pressure + 0.3 atm SO CO Copyright 006 Dr. Harshavardhan D. Bapat 0
6. In ideal gas equation calculations, expressing pressure in ascals (a), necessitates the use of the gas constant, R, equal to. (a). 0.0806 atm L mol - K - (b). 8.34 J mol - K - (c). 6.36 L torr mol - K - (d)..987 cal mol - K - Explanation: When the unit a is used for pressure the volume is measured in m 3. The product of these units () then has units of energy which is Joule. 7. Of the following, is a correct statement of Boyle s law. (a). constant (b). constant (c). constant (d). T constant Explanation: According to Boyle s law at constant temperature, the pressure and volume of an ideal gas are inversely related ( a /) to each other. To convert this proportionality to an equality the product must be a constant. 8. Of the following, is a valid statement of Charles law. (a). T constant (b). T constant (c). constant (d). constant x n Explanation: According to Charles s law, at constant pressure, the temperature and volume of an ideal gas are directly related ( a T) to each other. To convert this proportionality to an equality the /T must be a constant. Copyright 006 Dr. Harshavardhan D. Bapat
9. Which one of the following is a valid statement of Avogadro s law? (a). T constant (b). T constant (c). constant (d). constant x n Explanation: According to Avogadro s law, at constant pressure and temperature the volume of an ideal gas is directly related ( a n) to the number of moles. This proportionality can also be expressed as x constant n. 30. The molar volume of a gas at ST is L. (a). 0.0806 (b). 73.5 (c)..00 (d)..4 Explanation: This is a fact. 3. Standard temperature and pressure (ST), in the context of gases, refers to. (a). 98 K and atm (b). 73.5 K and atm (c). 98 K and torr (d). 73 K and pascal Explanation: This is a fact. Copyright 006 Dr. Harshavardhan D. Bapat