equal volumes of gas have the same number of molecules atoms V n where n moles

Equal volumes of gas have the same number of

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equal volumes of gas have the same number of molecules (atoms) V n where n = moles See practice problems in book. Fig 5.10. Volume relationships of gases. Volume relationship is the same as the mole relationship.
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Chapter 5 Chemistry 1303 7 5.4 Ideal Gas Equation Ideal gas - v olume of molecules is negligible; applies in most temperature and pressure ranges. combination of gas laws and Avogadro's Principle Standard Molar Volume ( STP ) Standard Temperature ( 0°C or 273 K ) Standard Pressure ( 1 atm ) 1 mole of any gas occupies 22.4 L or 22.4 L = 1 mole but ONLY at STP PV = nRT where R = "universal gas constant" = 0.0821 L atm / mole K
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Chapter 5 Chemistry 1303 8 Where does R come from? R = PV/nT (Ideal gas law rearranged) R = (1 atm)(22.414 L) / (1 mol)(273.15K) = 0.082057 L atm / mole K { Useful in many different kinds of calculations involving gases!!! } Example: At STP , the density of a certain gas is 4.29 g/L. What is the molecular mass of the gas? 4.29 g x 22.4 L = 96.0 g / mole L mole Example : What is the density (g/L) of UF 6 at 779 mm Hg and 62 ° C? Note: alternate way shown in book means memorizing yet another equation. Just think density - mass /volume
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Chapter 5 Chemistry 1303 9 5.5 Gas Stoichiometry Example: Acetylene (welding gas), C 2 H 2 , is produced by hydrolysis of calcium carbide. CaC 2 (s) + 2 H 2 O → Ca(OH) 2 (s) + C 2 H 2 (g) Starting with 50.0 g of CaC 2 , what is the theoretical yield of acetylene in liters, collected at 24°C and a pressure of 745 Torr? CaC 2 (s) + 2 H 2 O → Ca(OH) 2 (s) + C 2 H 2 (g) 1st find yield in moles : 50.0 g CaC 2 x x = 0.780 mole C 2 H 2 1 mole C 2 H 2 1 mole CaC 2 64.10 g CaC 2 1 mole CaC 2 2 now use ideal gas law to find volume of C 2 H 2 : PV = nRT P V = nRT V = (0.780 mole) x (0.0821 L atm / mole K) x (297 K) (745 torr) x (1 atm / 760 torr) = 19.4 L
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Chapter 5 Chemistry 1303 10 3 A + 2 B → 4 C + D
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Chapter 5 Chemistry 1303 11 5.6 Dalton's Law of Partial Pressures A. For a mixture of gases: P T = P a + P b + P c + . . . . . P T = P a + P b = n a RT / V + n b RT / V P T = (n a + n b ) (RT / V) P a = n a RT / V = n a = X a P t (n a + n b )RT / V n a + n b X a = mole fraction ( dimensionless quantity) X a = moles of one component / total moles present P i = X i . P T
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Chapter 5 Chemistry 1303 12 Example: A sample of natural gas contains 8.24 moles of CH 4 , 0.421 moles of ethane, 0.116 moles of C 3 H 8 . If the total pressure of the gas is 1.37 atm., what is the partial pressure of each gas?
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