T2 V2 T1 78 81 Avogadros Law of Combining Volumes Standard Temperature Pressure

T2 v2 t1 78 81 avogadros law of combining volumes

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T2 = V2 • T1
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78 81. Avogadro's Law of Combining Volumes Standard Temperature & Pressure: 22.4 L at 0 o C and 1 atmosphere pressure Standard Ambient Temperature & Pressure: 24.8 L at 20 o C and 1 atmosphere pressure 1. A sample of carbon dioxide gas has a volume of 55.0 mL at 45 o C and 85.0 kPa. Determine the volume at STP and SATP. 2. What pressure will 37.18 grams of CO 2 gas exert on a container at standard temperature? 3. Find the mass of 543 mL of acetylene gas, C 2 H 2 , collected at a pressure of 85.0 kPa and standard temperature. 4. What is the density of CO 2 gas measured at 5 o C and 200 kPa? 5. A sample of cooking gas, taken from a cylinder, was collected and its density measured at 27 o C and 100 kPa. The density at those conditions was 1.768 g/L. What was the molar mass of the cooking gas? 6. At STP, how many molecules of hydrogen are in 22.4 L? 7. Given 4.80 g of O2gas and 2.80 g of N2gas. Calculate for each of these samples: 82. Answers - Avogadro's Law of Combining Volumes 1. @ STP V 1 = 55.0 mL V 2 = ? P 1 = 85.0 kPa P 2 = 101.325 kPa T 1 = 45 o C = 318.15 K T 2 = 0 o C = 273.15 K V 2 = V 1 • P 1 • T 2 = 55.0 mL • 85.0 kPa • 273.15 K = 39.61 mL at STP P 2 • T 1 101.325 kPa • 318.15 K @ SATP V 1 = 55.0 mL V 2 = ? P 1 = 85.0 kPa P 2 = 100 kPa T 1 = 45 o C = 318.15 K T 2 = 25 o C = 298.15 K V 2 = V 1 • P 1 • T 2 = 55.0 mL • 85.0 kPa • 298.15 K = 42.51 mL at SATP P 2 • T 1 100 kPa • 318.15 K 2. Pressure of 37.18 of CO 2 at standard temperatures Solution steps Step #1 Find the number of moles of carbon dioxide present Step #2 Using the STP definition find the value of pressure Step #3 Convert to all other standard pressures values Step #1 Moles of CO 2 present n = m = 37.18 g = 0.845 moles of CO 2 M 44.01 g/mol Step #2 Pressure at STP @ STP 1 mole = 1 atm therefore 0.845 mol x x = 0.845 atm Step #3 Convert to all other pressure units 1 atm = 101.325 kPa = 760 Torr 0.845 atm x y x = 85.60 kPa y = 642.05 Torr 3. @ STP V 1 = 543 mL V 2 = ? P 1 = 85.0 kPa P 2 = 101.325 kPa T 1 = 0 o C = 273.15 K T 2 = 0 o C = 273.15 K V 2 = V 1 • P 1 = 543 mL • 85.0 kPa = 0.456 L at STP P 2 101.325 kPa @ STP 1 mol = 22.4 L therefore x 0.456 L x = 0.02 mol of gas present m = n • M = 0.02 mol • 26.04 g/mol = 0.52 grams C 2 H 2 = 2 C = 2 • 12.01 = 24.02 g/mol
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79 2 H = 2 • 1.01 = 2.02 g/mol 26.04 g/mol 4. 1 mole of any gas at STP = 22.4 L @ STP V 1 = 22.4 L V 2 = ? P 1 = 101.325 kPa P 2 = 200 kPa T 1 = 0 o C = 273.15 K T 2 = 5 o C = 278.15 K V 2 = V 1 • P 1 = 22 4 L • 101.325 kPa • 278.15 K = 11.56 L at STP P 2 101.325 kPa • 273.15 K D = m = 44.01 g = 3.81 g/L V 11.56 L 5. The gas has a density of 1.768 g/L Therefore two pieces of information are given: m = 1.768 g and V 1 = 1 L @ STP V 1 = 1 L V 2 = ? P 1 = 100 kPa P 2 = 101.325 kPa T 1 = 27 o C = 300.15 K T 2 = 0 o C = 273.15 K V 2 = V 1 • P 1 • T 2 = 1 L • 100 kPa • 273.15 K = 0.90 L P 2 • T 1 101.325 kPa • 300.15 K Therefore @ STP 1 mole = 22.4 L x 0.90 L x = 0.04 mol Therefore M = g/mol = 1.76 g/0.04 mol = 44.01 g/mol 6. At STP 22.4 L = 1 mole and 1 moles has 6.02 X 10 23 molecules
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