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April 30, 2013 Positive Attitude Peccadillo: a slight offense or fault; a minor sin

April 30, 2013 Positive Attitude Peccadillo: a slight offense or fault; a minor sin. Do Now: Quad Card Topic: Behavior of Gases. The Ideal Gas Law. The Ideal Gas Law.

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April 30, 2013 Positive Attitude Peccadillo: a slight offense or fault; a minor sin

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  1. April 30, 2013 Positive AttitudePeccadillo: a slight offense or fault; a minor sin • Do Now: Quad Card Topic: Behavior of Gases

  2. The Ideal Gas Law

  3. The Ideal Gas Law • remember, the gas laws are simple, mathematical relationships between the pressure (P), volume (V), temperature (T), and moles (n), of a gas. • so far, we have only dealt with P, V, and T (in the basic gas laws). • today, we will introduce n, the number of moles of a substance. • remember, moles are related to all sorts of good stuff, like grams and molecules! • “n,” however, must always be expressed in moles. This means if you are given grams or molecules, you must convert them into moles before using it in the Ideal Gas Law! • so , what is the Ideal Gas Law? Here it is: • “R” is known as the ideal gas law constant. • its value never changes (duh). • notice the unit for R. It’s complicated! PV = nRT R = .0821 Latm molK

  4. The Ideal Gas Law R = .0821 Latm molK • for the Ideal Gas Law ONLY, all units for P, V, n, and T must match the units in the R value. • this means your P must be in atm, V must be in L, n must be in mol, and of course, T must be in K. • if they don’t match, convert them! • this is ONLY for the Ideal Gas Law! You may still use kPa, mmHg, mL, etc. in the Basic gas laws, just not in the Ideal! • therefore, you should expect to be doing a lot of conversions when doing Ideal Gas Law calculations! Ex1: If 234.68 g of Cl2 was compressed at 3534 mmHg of pressure and -13.8C, what volume would it have? (1 mol Cl2 = 70.90 g)

  5. The Ideal Gas Law Ex1: If 234.68 g of Cl2 was compressed at 3534 mmHg of pressure and -13.8C, what volume would it have? P = V = n = T = 3534 mmHg _____ atm ? 234.68 g  _____ mol -13.8 C  ______ K 4.65 3.31 259.2 3534 mmHg 1 atm = 4.65 atm 760 mmHg 234.68 g Cl2 1 mol Cl2 = 3.31 mol Cl2 70.90 g Cl2 PV = nRT V = nRT P __ ___ P P .0821 Latm V = (3.31 mol) molK (259.2 K) = 4.65 atm = 15.15 L V = 3.31 × .0821 × 259.2 ÷ 4.65

  6. The Ideal Gas Law 528.73 kPa _____ atm 865 mL _____ L ? (in g) 33.7 C  ______ K 5.22 0.865 306.7 528.73 kPa 1 atm = 5.22 atm 101.325 kPa 528.73 kPa 1 atm = 5.22 atm 101.325 kPa 865 mL 1 L = 0.865 L 1000 mL n = (5.22 atm)(0.865 L) = .0821 Latm (306.7 K) molK 0.18 mol C3H8 PV = nRT n = PV RT __ ___ RT RT n = 5.22 × .865 ÷ (.0821 × 306.7) = 0.18 = 7.94 g C3H8 0.18 mol C3H8 44.11 g C3H8 1 mol C3H8 Ex2: How many grams of propane (C3H8) fit into a 865 mL container under 528.73 kPa of pressure at 33.7C? P = V = n = T =

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