Ch 5_Part2_102911 - 10/29/2011 Standard T & P (STP) T =...

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10/29/2011 1 Standard T & P (STP) 22.4 L T = 273 K (0 o C) P = 1 atm = 101.325 kPa = 1.01325 bar At STP, 1 mol of any ideal gas occupies 22.4 L Gas Density We can use PV = nRT to determine the density of gases. What are the units of density? mass/volume What does this suggest about gas density? It will depend strongly on T, P, and the mass of the gas molecules. Contrast with liquids and solids, whose densities depend somewhat on T, but far less on P. He balloons Hot-air balloon
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10/29/2011 2 What is the density of O 2 gas in g/L at 25 o C and 0.850 atm? Calculate # moles in 1 L, use MM of O 2 to get g, divide by V. P = 0.850 atm V = 1 L n = ? R = 0.08206 L·atm/mol·K T = 25°C + 273 = 298 K PV n = RT PV = nRT        0.850 atm 1 L n = L atm 0.08206 298 K mol K 2 = 0.0347 mol O 22 ? g/L O = 0.0347 mol O 2 2 32.00 g O 1 mol O    1 1 L 2 o = 1.11 g/L O @ 25 C, 0.850 atm Note : A value for gas density is meaningful only if accompanied by the T and P at which it was measured. Gas Density Example Gas Density and Molar Mass We can develop an expression relating density and MM using PV = nRT. Substituting: Therefore, the density of an ideal gas depends on T, P, V, and Molar Mass We can use the density of a gas to determine its molar mass. m n = M m PV = RT M m PM = RT V Density!! PM = dRT
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10/29/2011 3 PM = dRT Example A 2.30-g sample of an unknown solid was vaporized in a 345-mL vessel. If the vapor has a pressure of 985 mmHg at 148 o C, what is the molecular weight of the solid? Heat applied V = 345 mL mass of solid = 2.30 g P = 985 mmHg T = 148 o C mass of gas = 2.30 g = 1.296 atm = 421 K 2.30 g 0.345 L d = = 6.667 g/L M = ? R = 0.08206 L·atm/mol·K dRT M = P = 177.7 g/mol       L atm 6.667 g/L 0.08206 421 K mol K = 1.296 atm Chemical Equations and Calculations Reactants Products Moles Mass Molar Mass g/mol Atoms (Molecules) Avogadro’s Number 6.022 x 10 23 mol -1 Solutions Molarity moles / L Gases PV = nRT
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10/29/2011 4 Gas Stoichiometry Example What volume of N 2 (g) is produced when 70.0 g NaN 3 is decomposed? P = 735 mmHg, T = 26 o C 2 NaN 3 (s) 2 Na( l ) + 3 N 2 ( g ) 70.0 g ? L 64.99 g/mol 28.02 g/mol 735 mmHg 26 o C 1. The Stoichiometry Part : mass NaN 3 mol NaN 3 mol N 2 2. The Gas Law Part : mol N 2 , P, T V of N 2 (g) Example: The Stoichiometry Part What volume of N 2 (g) is produced when 70.0 g NaN 3 is decomposed? P = 735 mmHg, T = 26 o C 2 NaN 3 (s) 2 Na( l ) + 3 N 2 ( g ) ? mol N 2 = 1.616 mol N 2 1 mol NaN 3 64.99 g NaN 3 = 70.0 g NaN 3 3 mol N 2 2 mol NaN 3
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10/29/2011 5 What volume of N 2 (g) is produced when 70.0 g NaN 3 is decomposed? (P = 735 mmHg, T = 26 o C) P = 735 mmHg = 0.9671 atm V = ? L n = 1.616 mol N 2 R = 0.08206 L.atm/mol.K T = 26 o C + 273 = 299 K nRT V = P PV = nRT       2 L atm 1.616 mol N 0.08206 299 K mol K V = 0.9671 atm V = 40.999 L Yay!!! = 41.0 L
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Ch 5_Part2_102911 - 10/29/2011 Standard T & P (STP) T =...

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