#Chem 161-2007 Homework 6th week

#Chem 161-2007 Homework 6th week - Chem 161-2007 Homework 6...

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Unformatted text preview: Chem 161-2007 Homework 6 th Week Hill & Petrucci Homework Chapter 5 Chem 161-2007 Hill & Petrucci Chapter 5 Homework Problems: 53,57,59,71,75,79,81,85,89,91,93,94,100,102,110,117 53. Calculate (a) the volume, in liters, of 1.88 mol of an ideal gas at 55 o C and 1.08 atm; (b) the pressure, in atmospheres, of 137 g CO(g) in a 4.49-L tank at 28 o C; (c) the mass, in milligrams, of 97.4 mL H 2 (g) at 744 Torr and -2 o C; and (d) the pressure, in kilopascals, of 19.6 g N 2 (g) in a 6.41-L flask at 0 o C. (a) PV = nRT V = nRT/P V = (1.88mol x 0.08205Latmdeg-1 mol-1 x (273.15+55))/1.08atm V = 46.9 L (b) PV = nRT PV = (g/MW)RT P = (gRT)/VMW P = (137g x 0.08205Latmdeg-1 mol-1 x (273.15+28))/(4.49L x 28.01gmol-1 ) P = 26.92 atm (c) PV = nRT PV = (g/MW)RT g = PVMW/RT g = ((740Torr x (1atm/760Torr)) x 0.0974L x 2.02gmol-1 )/(0.08205Latmdeg-1 mol-1 x (273.15 2)) g = 8.61 x 10-3 = 8.61mg (d) PV = nRT PV = (g/MW)RT P = (gRT)/VMW P = (19.6g x 0.08205Latmdeg-1 mol-1 x (273.15K + 0K))/(6.41L x 28.02gmol-1 ) P = 2.45 atm 2.45 atm x (101kPa/atm) = 247 kPa 57. A hyperbaric chamber is an enclosure containing oxygen at higher-than-normal pressures used in the treatment of certain heart and circulatory conditions. What volume of O 2 (g) from a cylinder at 25 o C and 151 atm is required to fill a 4.20 x 10 3 L hyperbaric chamber to a pressure of 2.50 atm at 17 o C? 1 Cylinder Hyperbaric chamber ?L 4200 L 25 o C 17 o C 151 atm 2.50 atm We cant determine the volume of gas released from the cylinder without knowing the number of moles of gas released from the cylinder. However, the number of moles of gas released from the cylinder is equal to the number of moles of gas required to fill the hyperbaric chamber. Solution I: Use the ideal gas law twice. Plan: V 2 ,T 2 ,P 2 n 2 n 1 V 1 P 2 V 2 = n 2 R 2 T 2 n 2 = (P 2 V 2 )/(RT 2 ) = (2.50 atm x 4200 L)/(0.08205 Latmdeg-1 mol-1 x 290K) = 441.28 mol P 1 V 1 = n 1 RT 1 V 1 = n 1 RT 1 /P 1 = (441 mol x 0.08205 Latmdeg-1 mol-1 x 298K)/151 atm = 71.45 L Solution II: Use the combination gas law. P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 n 1 = n 2 P 1 V 1 /T 1 = P 2 V 2 /T 2 ((151 atm x V 1 )/(273.15+25)) = ((2.50 atm x 4200L)/(273.15+17)) V 1 = 71.45L 59. Calculate the molecular mass of a liquid that, when vaporized at 98 o C and 756 Torr, gives 139 mL of vapor with a mass of 0.808 g. PV = nRT PV = (g/MW)RT MW = gRT/PV MW = (0.808g x 0.08205Latmdeg-1 mol-1 x (273.15K + 98))/((756Torr x (1atm/760Torr)) x 0.139L) MW = 177.96 gmol-1 71. What volume of nitrogen gas can be produced from the decomposition of 37.6 L of ammonia, with both gases measured at 725 o C and 5.05 atm pressure? 2 NH 3 (g) N 2 (g) + 3 H 2 (g) 37.6L ?L 725 o C 725 o C 5.05 atm 5.05 atm 2 2X moles X moles P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 Temperatures and pressures are constant. Therefore, using the combination gas law, the temperature and pressure are irrelevant....
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This homework help was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

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#Chem 161-2007 Homework 6th week - Chem 161-2007 Homework 6...

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