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A sample of nitrogen occupies 10.0 liters at 25ºC and 98.7 kPa. What would be the volume at 20ºC and kPa? A 7.87 L B 9.45 L C 10.2 L D 10.6 L Bell Ringer A sample of nitrogen occupies 10.0 liters at 25ºC and 98.7 kPa. What would be the volume at 20ºC and kPa? A 7.87 L B 9.45 L C 10.2 L D 10.6 L = V1V1 T1T1 P1P1 V2V2 T2T2 P2P2 V2V2 = P 1 V 1 T 2 T 1 P 2 V2V2 = (98.7 kPa) (10.0 L) (293 K) (298 K) (102.7 kPa) 298 K 293 K T2T2 P2P2 T2T2 P2P2

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Gas Laws Quiz

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For Next Class: Homework: –Practice Problems worksheet Quiz next class on Ideal Gas Law, Partial Pressures, and Density –5 questions; 22 points total 2 short answer (2 points each) 3 math problems (6 points each)

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The Ideal Gas Law & Co. The Ups and Downs of Gas Laws

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A Reminder… We that we live in an world where: Gas particles have no mass Gas particles have no volume Gas particles have elastic collisions These assumptions are used when trying to calculate the AMOUNT of a gas we have!

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Why are these assumptions important? PV = nRT Image source: thefreedictionary.com

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Image source: popartuk.com PV = nRT P V n R T RESSURE OLUME MOLES OF GAS GAS CONSTANT EMPERATURE

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The Myste R ious R Image source: toysrus .com R is a constant (doesnt change). Number value of R depends on other units. Units of R are a combination of many units atm · L mol · K 8.31 kPa · L mol · K 62.4 mmHg · L mol · K

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Ummm… What? PV = nRT Solve for R: R = P V n T R = Plug in units: (mm Hg) (L) (mol) (K) (kPa) (atm)

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Gas Laws, Gas Laws Everywhere! V1V1 T1T1 = V2V2 T2T2 P 1 x V 1 = P 2 x V 2 P 1 V 1 P 2 V 2 = T1T1 T2T2 Used with CHANGING CONDITIONS P V = n R T Used with only ONE SET OF CONDITIONS

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When to Use PV = nRT Calculating amount of gas in moles Calculating P, V, or T if moles of gas are known. –IMPORTANT! We must have 3 out of 4 pieces of information: P V n T

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Practice with the Ideal Gas Law 1.A gas sample occupies 2.62 L at 285ºC and 3.42 atm. How many moles are present in this sample? PV = nRT n= P V R T n= (3.42 atm)(2.62 L) L · atm mol · K (558 K) = mol

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But Lets Be Practical… We dont usually measure in moles! We usually measure quantities in GRAMS! PV = nRTPVM = gRT

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RESSURE OLUME OLAR MASS OF GAS (g/mol) GAS CONSTANT EMPERATURE P V M R T g RAMS OF GAS Image source: popartuk.com

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Practice with the Ideal Gas Law A balloon is filled with g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is L, what must the temperature be in ºC? PVM = gRT T= P V M g R T= (1.26 atm)(1.250 L) 4.00 g mol L · atm mol · K ( g) = 308 K 35 ºC - 273

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PV=nRT vs. PVM=gRT Use PV=nRT when: –You are given moles in the problem. –You are searching for moles as an answer. Use PVM=gRT when: –You are given grams in the problem. –You are searching for grams as an answer.

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What Else Happens Under Unchanging Conditions? At constant V and T, pressure is easy to calculate! Total Pressure = Pressure of gas 1 + Pressure of gas 2 + Pressure of gas 3 + Pressure of gas 4 … P total = P 1 + P 2 + P 3 + … The sum of the individual pressures is equal to the total pressure.

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Partial Pressures Practice A sample of hydrogen gas is collected over water at 25ºC. The vapor pressure of water at 25ºC is 23.8 mmHg. If the total pressure is mmHg, what is the partial pressure of the hydrogen? Source: 2003 EOC Chemistry Exam P total = P H 2 + P H 2 O 23.8 mm Hg523.8 mm Hg = PH2PH2 + PH2PH2 = mm Hg

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What do Changing Conditions Affect? If MASS remains constant… …But VOLUME changes… Then DENSITY CHANGES! D = M V We have learned that we can change 3 variables: Temperature, Volume, and Pressure.

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Two Types of Density Problems: At STP:Not at STP: molar volume of any gas at STP = Density at STP = molar volume Determine new volume (V 2 ) using Combined Gas Law molar mass P 1 V 1 P 2 V 2 = T1T1 T2T2 Density at non-STP = molar mass V2V Liters

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Practice with Density Problems: Determine the density of ethane (C 2 H 6 ) at STP: molar mass D (at STP) = molar mass = g molar volume = 22.4 L D = g 22.4 L =1.34 g/L Determine the density of C 2 H 6 at 3.0 atm and 41ºC. molar volume P 1 V 1 P 2 V 2 = T1T1 T2T2 V 2 = P 1 V 1 T 2 T 1 P 2 V 2 = 8.6 L P 1 = 1.0 atm V 1 = 22.4 L T 1 = 273 K P 2 = 3.0 atm V 2 = ? T 2 = 314 K V 2 = (1.0 atm) (22.4 L) (314 K) (273 K) (3.0 atm) V 2 = 8.6 L D = molar mass V2V2 D = g 8.6 L 3.5 g/L =

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For Next Class: Homework: –Gas Laws Packet #2, problems 1-10 Quiz next class on Ideal Gas Law, Partial Pressures, and Density –5 questions; 22 points total 2 short answer/FITB (2 points each) 3 math problems (6 points each)

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What do Changing Conditions Affect? Density Stoichiometry problems So Far:Now…

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Mass-Volume at Non-STP Two parts to solving these problems: Use Stoichiometry Use Gas Law

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