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Gases

# Gases - GASES(Chapter 5 I Ideal gases A Ideal gas law...

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1 GASES (Chapter 5) I. Ideal gases. A. Ideal gas law review. 1. PV = nRT Ideal gases obey this equation under all conditions. It is a combination of- a. Boyle's Law: P 1/V at constant n and T b. Charles's Law: V T at constant n and P c. Avogadro's Law: equal numbers of moles of gases at the same T and P occupy the same volume or V n at constant T and P ) 2. Note that according to this equation, 1 mole of an ideal gas at STP (Standard Temperature and Pressure, that is, 273 K and 1.00 atm or 760 Torr ) will occupy 1.00 atm (1.00 mol)(0.08206 mol K L atm )(273 K) = 22.414 L This is called the molar volume at STP. The symbol for the molar volume is V B. Uses. 1. PV = nRT = gRT/M There are five variables in this expression. If any four are known, the fifth can be calculated. 2. This expression can be used a. in stoichiometric calculations— SEE PREVIOUS UNIT b. in molar mass calculations from gas densities. M= PV gRT = V g P RT = d P RT . M d at constant T and P. c. Example: A compound was analyzed and found to contain 53.3% C, 11.1% H, and 35.6% O by mass. In another experiment it was found that a 0.503g sample of the gaseous compound occupied a volume of 697 mL at 127 °C and 200 Torr pressure. Calculate the molecular formula of the compound. Steps: a. Determine the empirical formula from the analysis data. b. Determine the Molar Mass c. Deduce the molecular formula.

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2 a. Empirical formula. mole of C = 53.3g 12.0g/mol = 4.442 divide by 2.225 ! " ! ! ! ! 2.0 mole of H = 11.1g 1.0g/mol = 11.111 divide by 2.225 ! " ! ! ! ! 5.0 mole of O = 35.6g 16.0g/mol = 2.225 divide by 2.225 ! " ! ! ! ! 1.0 Empirical formula C 2 H 5 O 2 b. Molar Mass (MM) From ideal gas law: M = gRT PV = (0.503)(62.4)(400) (200)(0.697) = 90 EFM = mass of C 2 H 5 O 2 = 2(12)+5(1)+1(16) = 45 molecular formula is C 4 H 10 O 2 3. Mixtures of gases- Dalton's Law of Partial Pressures. a. The partial pressure of a component, i , of a mixture of gases is P i . P i = n i V RT P Tot = ! i P i = ! i n i V RT = n Tot V RT b. Partial pressure and mole fraction ( X i ) P Tot P i = n Tot n i = X i = Mole fraction of i c. Example: Suppose that a mixture of 0.80 g of He and an unknown amount of O 2 when placed in a 3.0 L container at 500 K exerted a pressure of 3.59 atm. 1) What are the partial pressures of the two gases? n He = 0.80g 4.0g/mol = 0.20 mol P He = (0.20 mol)(0.08206 atm L mol K )(500K) 3.0 L = 2.74 atm P O 2 = 3.59 atm – 2.74 atm = 0.85 atm 2) What is the mass of O 2 present? n O 2 n He = P O 2 P He ! n O 2 = P O 2 P He n He = 0.85 atm 2.74 atm (0.20mol) = 0.062 mol g O 2 = (0.062mol)(32 g mol ) = 1.99 g
3 4. Gas behavior and the ideal gas law can be understood using the Kinetic Molecular Theory. This is one of the most frequently used theories in Chemistry II. Kinetic Molecular Theory as Applied to Ideal Gases. A. Postulates for ideal gases.

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Gases - GASES(Chapter 5 I Ideal gases A Ideal gas law...

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