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Ch 5_extra_111111

# Ch 5_extra_111111 - Gas Law Solving for Pressure Problem...

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Gas Law: Solving for Pressure Problem: Calculate the pressure in a 87.5 L container filled with 5.038 kg of xenon at a temperature of 18.8 o C. Solution: n Xe = = 38.37014471 mol Xe 5038 g Xe 131.3 g Xe / mol T = 18.8 o C + 273.15 K = 291.95 K PV = nRT so P = nRT V P = = 10.5 atm (38.37 mol )(0.0821 L·atm )(291.95 K) 87.5 L (mol·K)

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Ammonia Density Problem Calculate the density of ammonia gas (NH 3 ) in grams per liter at 752 mm Hg and 55 o C . density = d = mass per unit volume = m / L P = 752 mm Hg x (1 atm/ 760 mm Hg) = 0.989 atm T = 55 o C + 273.15 = 328.15 K PV = nRT so… n/V = = n/V = 0.03674 mol/L density = 0.03674 mol/L x (17.03 g/mol NH 3 ) d = 0.626 g / L (more than 1000x less dense than water!) P R x T ( 0.989 atm ) ( 0.08206 L·atm / mol·K) ( 328.15 K)
Ammonia Density Problem – a different approach Calculate the density of ammonia gas (NH 3 ) in grams per liter at 752 mm Hg and 55 o C. density =d = mass per unit volume = m / L P = 752 mm Hg x (1 atm/ 760 mm Hg) = 0.989 atm T = 55 o C + 273.15 = 328.15 K n = mass / Molar Mass = m / MM PV = nRT so n = m = P V MM·V RT therefore, d = m/V = (P·MM)/(RT) d = = d = 0.626 g / L P x MM R x T ( 0.989 atm) ( 17.03 g/mol) ( 0.08206 L · atm/mol · K) ( 328 K)

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Change of Conditions A gas sample in the laboratory has a volume of 45.9 L at 25 o C and a pressure of 743 mm Hg. If the temperature is increased to 155 o C by compressing the gas to a new volume of 3.10 mL, what is the pressure? Convert all units to K, L, atm Constant moles Set up “before” and “after” equations Solve for new pressure Plan:
Change of Condition = = nR = constant T 1 T 2 P 1 x V 1 P 2 x V 2 ( 0.978 atm) ( 45.9 L) P 2 (0.00310 L) ( 298.15 K) (428.15 K) = P 2 = ( 428.15.15 K) ( 0.978 atm) ( 45.9 L) = 2.08 x 10 4 atm ( 298 K) ( 0.00310 L) P 1 = 743 mm Hg x 1 atm/ 760 mm Hg = 0.978 atm V 1 = 45.9 L V 2 = 3.10 mL = 0.00310 L T 1 = 25 o C + 273.15 = 298.15 K T 2 = 155 o C + 273.15 = 428.15 K

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Problem: A sample of natural gas is collected at 25.0 o C in a 250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of 550.0 torr, what is the molecular weight of the gas? Plan: Convert all units to K, L, atm Calculate mass of vapor Use ideal gas law to calculate moles Use mass / mole relationship to calculate molar mass Example: Molar Mass of a gas from its mass and P, V, and T
Example: Molar Mass of a gas from its mass and P, V, and T Problem: A sample of natural gas is collected at 25.0 o C in a 250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of 550.0 torr, what is the molecular weight of the gas? Plan: Use the ideal gas law to calculate n, then calculate the molar mass. Solution: P = 550.0 torr x x = 0.724 atm 1mm Hg 1 torr 1.00 atm 760 mm Hg V = 250.0 mL x = 0.250 L 1.00 L 1000 mL T = 25.0 o C + 273.15 K = 298.2 K n = P V R T n = = 0.007393 mol (0.0821 L atm/mol K) (298.2 K) (0.724 atm)(0.250 L) MM = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol

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Molar Mass of a Gas (like lab #4) Problem:
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Ch 5_extra_111111 - Gas Law Solving for Pressure Problem...

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