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Mastering the Ideal Gas Law: Gas Behavior Calculation

Understand the Ideal Gas Law and its components (P, V, T, n), perform gas law calculations, and convert units for accurate results. Practice with examples to master gas behavior calculations effortlessly.

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Mastering the Ideal Gas Law: Gas Behavior Calculation

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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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