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Gases Gases Document Transcript

  • AP Chemistry Rapid Learning Series - 16 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … p g Teach Yourself AP Chemistry Visually in 24 Hours 1/78 http://www.RapidLearningCenter.com The Gas Laws AP Ch i t R id Learning Series Chemistry Rapid L i S i Wayne Huang, PhD Kelly Deters, PhD Russell Dahl, PhD Elizabeth James, PhD Debbie Bilyen, M.A. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com Rapid Learning Center www.RapidLearningCenter.com/ © Rapid Learning Inc. All rights reserved. 1
  • AP Chemistry Rapid Learning Series - 16 Learning Objectives By completing this tutorial you will learn… How gases cause pressure The Kinetic Molecular Th Th Ki ti M l l Theory How properties of a gas are related How to use several gas laws The difference between ideal and real gases Diffusion and Effusion 3/78 Concept Map Previous content Chemistry New content Studies Matter Volume One state is Pressure Temperature Rates of Effusion and Diffusion Gas Have properties Moles Density D it Molar Mass Related to each other with Gas Laws 4/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 2
  • AP Chemistry Rapid Learning Series - 16 Kinetic Molecular Theory 5/78 Kinetic Molecular Theory Theory – An attempt to explain why or how behavior or properties are as they are. It’s based on empirical evidence. Kinetic Molecular Theory (KMT) – An attempt to explain gas p p g behavior based upon the motion of molecules. 6/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 3
  • AP Chemistry Rapid Learning Series - 16 Assumptions of the KMT 1 All gases are made of atoms or molecules. 2 Gas particles are in constant, rapid, random motion. motion 3 The temperature of a gas is proportional to the average kinetic energy of the particles. 4 Gas particles are not attracted nor repelled from one another. 5 All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms). 6 The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant. 7/78 Calculating Average Kinetic Energy Temperature is proportional to average kinetic energy…how do you calculate it? Avg. KE = 3 RT 2 Avg. KE = Average Kinetic Energy (in J, Joules) g g gy ( , ) R = Gas constant (use 8.31 J/K mol) T = Temperature (in Kelvin) Example: Find the average kinetic energy of a sample of O2 at 28°C. Avg. KE = ? J R = 8.31 J/K mol T = 28°C + 273 = 301 K Avg. KE = ( ) 3 8.31 J × 301K K × mole 2 Avg. KE = 3752 J 8/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 4
  • AP Chemistry Rapid Learning Series - 16 Gas Behavior 9/78 KMT and Gas Behavior The Kinetic Molecular Theory and its Th d it assumptions can be used to explain gas behavior. 10/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 5
  • AP Chemistry Rapid Learning Series - 16 Definition: Pressure Pressure – Force of gas particles running into a surface. 11/78 Pressure and Number of Molecules If pressure is molecular collisions with the container… As the A th number of b f molecules increases, there are more molecules to collide with the wall Collisions C lli i between molecules and the wall increase Pressure P increases As # of molecules increases, pressure increases. Pressure (P) and # of molecules (n) are directly proportional (∝). P∝n 12/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 6
  • AP Chemistry Rapid Learning Series - 16 Pressure and Volume If pressure is molecular collisions with the container… As l A volume increases, molecules can travel farther before hitting the wall Collisions C lli i between molecules and the wall decrease Pressure P decreases As volume increases, pressure decreases. Pressure and volume are inversely proportional. P∝ 13/78 1 V Definition: Temperature Temperature – Proportional to the average kinetic energy of the molecules. Energy due to motion (Related to how fast the molecules are moving) As temperature increases Molecular motion increases 14/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 7
  • AP Chemistry Rapid Learning Series - 16 Pressure and Temperature If temperature is related to molecular motion… and pressure is molecular collisions with the container… As temperature increases, molecular motion increases Collisions between molecules and the wall increase Pressure increases As temperature increases, pressure increases. Pressure and temperature are directly proportional. P ∝T 15/78 Pressure Inside and Outside a Container 16/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 8
  • AP Chemistry Rapid Learning Series - 16 Definition: Atmospheric Pressure Atmospheric Pressure – Pressure due to the layers of air in the atmosphere. Climb in altitude Less layers of air Lower atmospheric pressure As altitude increases, atmospheric pressure decreases. 17/78 Pressure In Versus Out A container will expand or contract until the pressure inside = atmospheric pressure outside. Expansion will lower the internal pressure. Contraction will raise the internal pressure. (Volume and pressure are inversely related) Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain. Lower pressure Higher pressure Lower pressure The internal pressure is from low altitude (high pressure) ( g p ) The external pressure is high altitude (low pressure). The internal pressure is higher than the external pressure. 18/78 The bag will expand in order to reduce the internal pressure. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 9
  • AP Chemistry Rapid Learning Series - 16 When Expansion isn’t Possible Rigid containers cannot expand. Example: An aerosol can is left in a car trunk in the summer. What happens? The temperature inside the can begins to rise. Lower pressure Can Higher Explodes! pressure As temperature increases, pressure increases. The internal pressure is higher than the external pressure. The can is rigid—it cannot expand, it explodes! Soft containers or “movable pistons” can expand and contract. Rigid containers cannot. 19/78 Attacking Strategy for Gas Law Problems 20/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 10
  • AP Chemistry Rapid Learning Series - 16 General Strategy for Gas Law Problems The following steps are a general way to approach these problems. 1 Identify Id tif quantities by their units. titi b th i it 2 Make a list of known and unknown quantities in symbolic form. 3 Look at the list and choose the gas law that relates all the quantities together. 4 Plug quantities in and solve. Pl titi i d l 21/78 Pressure Units Several units are used when describing pressure Unit Symbol atmospheres atm Pascals, kiloPascals Pa, kPa millimeters of mercury mm Hg pounds per square inch psi 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi 22/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 11
  • AP Chemistry Rapid Learning Series - 16 Definition: Kelvin Scale Kelvin (K) – Temperature scale with an absolute zero. b l t Temperatures cannot fall below an absolute zero. A temperature scale with absolute zero is needed in Gas Law calculations because you can’t have g p negative pressures or volumes. C + 273 = K 23/78 Standard Temperature & Pressure Standard Temperature and Pressure (STP) – 1 atm (or the equivalent in another unit) and 0°C (273 K). Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list! 24/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 12
  • AP Chemistry Rapid Learning Series - 16 Gas Laws 25/78 KMT and Gas Laws The Gas Laws are the experimental observations of the gas behavior that the Kinetic Molecular Theory explains. 26/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 13
  • AP Chemistry Rapid Learning Series - 16 “Before” and “After” in Gas Laws This section has 4 gas laws which have “before” and “after” conditions. For example: P P2 1 = n1 n2 Where P1 and n1 are pressure and # of moles “before” and P2 and n2 are pressure and # of moles “after”. Both sides of the equation are talking about the same sample of gas—with the “1” variables before a change, and the “2” variables after the change. 27/78 Avogadro’s Law Avogadro’s Law relates # of particles (moles) and Volume. Where Temperature & Pressure are held constant. p V1 V2 = n1 n2 Example: The two volume units must match! A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? n1 = 0.15 moles V1 = 2.5 L n2 = 0.55 moles V2 = ? L V = Volume n = # of moles of gas 2 .5 L V2 = 0.15mole 0.55mole 0.55mole × 2.5L = V2 0.15mole V2 = 9.2 L 28/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 14
  • AP Chemistry Rapid Learning Series - 16 Boyles’ Law - 1 Boyles’ Law relates pressure and volume. Where temperature and # of molecules are held constant. P = pressure V = volume PV1 = P2V2 1 The two pressure units must match and the two volume units must match! Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg? Pressure units need to match—convert one: P it dt t h t 745 mm Hg 1 atm = ______ atm 0.980 P1 = 1.05 atm 760 V1 = 2.5 L mm Hg P2 = 745 mm Hg =0.980 atm V2 = ? L 29/78 Boyles’ Law - 2 Boyles’ Law relates pressure and volume. Where temperature and # of molecules are held constant. P = pressure V = volume PV1 = P2V2 1 The two pressure units must match and the two volume units must match! Example: P1 = 1.05 atm A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg? 1.05atm × 2.5L = 0.980atm × V2 V1 = 2.5 L P2 = 745 mm Hg =0.980 atm V2 = ? L 1.05atm × 2.5L = V2 0.980atm V2 = 2.7 L 30/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 15
  • AP Chemistry Rapid Learning Series - 16 Charles’ Law - 1 Charles’ Law relates volume and temperature. V1 V2 = T1 T2 Example: Where pressure and # of molecules are held constant. V = Volume T = Temperature The two volume units must match and temperature must be in Kelvin! What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C? Temperature needs to be in K l i ! T t d t b i Kelvin! V1 = 10.5 L T1 = 25°C = 298 K 25°C + 273 = 298 K V2 = ? L T2 = 50°C = 323 K 50°C + 273 = 323 K 31/78 Charles’ Law - 2 Charles’ Law relates temperature and pressure. V1 V2 = T1 T2 Example: Where pressure and # of molecules are held constant. V = Volume T = Temperature The two volume units must match and temperature must be in Kelvin! What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C? V1 = 10.5 L T1 = 25°C = 298 K V2 = ? L T2 = 50°C = 323 K 10.5L V = 2 298K 323K 323K × 10.5 L = V2 298 K V2 = 11.4 L 32/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 16
  • AP Chemistry Rapid Learning Series - 16 The Combined Gas Law The combined gas law assumes that nothing is held constant. PV1 P2V2 1 = n1T1 n2T2 P = Pressure V = Volume n = # of moles T = Temperature Each “pair” of units pair must match and temperature must be in Kelvin! Example: What is the final volume if a 0.125 mole sample of gas at 1.7 atm, 1.5 L and 298 K is changed to STP and particles P1 = 1.7 atm are added to 0.225 mole? V1 = 1.5 L STP is standard temperature (273 K) and pressure (1 atm) n1 = 0.125 mole T1 = 298 K P2 = 1.0 atm V2 = ? L n2 = 0.225 mole 33/78 T2 = 273 K 1.7atm × 1.5 L 1.0atm × V2 = 0.125mole × 298 K 0.225mole × 273K 0.225mole × 273K ×1.7 atm × 1.5 L = V2 1.0atm × 0.125mole × 298 K V2 = 4.2 L Only Really Need One Law The combined gas law can be used for all “before” and “after” gas law problems! PV1 P2V2 1 = n1T1 n2T2 For example, if volume is held constant, then and the combined gas law becomes: V1 = V2 PV1 P2V1 1 = n1T1 n2T2 When two variables on opposites sides are the same, they cancel out and the rest of the equation can be used. P P 1 = 2 n1T1 n2T2 34/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 17
  • AP Chemistry Rapid Learning Series - 16 “Transforming” the Combined Gas Law Watch as variables are held constant and the combined gas law “becomes” the other 3 laws. Hold pressure and temperature constant PV1 P2V2 1 = n1T1 n2T2 Avogadro’s Law Hold moles and temperature constant PV1 P2V2 1 = n1T1 n2T2 Boyles’ Law Hold pressure and moles constant PV1 P2V2 1 = n1T1 n2T2 Charles’ Law 35/78 How to Memorize What’s Held Constant How do you know what to hold constant for each law? Avogadro’s Law Hold Pressure and Temperature constant Avogadro was a Professor at Turin University (Italy) Boyles’ Law Hold moles and Temperature constant The last letter of his first name Robert is T name, Robert, Charles’ Law Hold Pressure and moles constant Charles was from Paris 36/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 18
  • AP Chemistry Rapid Learning Series - 16 Using only the Combined Law Example: What is the final volume if a 15.5 L sample of gas at 755 mm Hg and 298 K is changed to STP? STP is standard temperature (273 K) and pressure (1 atm). P1 = 755 mm H Hg V1 = 15.5 L T1 = 298 K “moles” is not mentioned in the problem—therefore problem therefore it is being held constant. It is not needed in the combined law formula. P2 = 1.0 atm = 760 mm Hg V2 = ? L T2 = 273 K Pressure units must match! 1 atm = 760 mm Hg PV1 P2V2 1 = n1T1 n2T2 755mm Hg ×15.5 L 760mm Hg × V2 g g = 298 K 273K 273K × 755mm Hg ×15.5L = V2 760mm Hg × 298 K V2 = 14.1 L 37/78 Mixtures of Gases 38/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 19
  • AP Chemistry Rapid Learning Series - 16 Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure – The sum of the pressures of each type of gas equals the pressure of the total sample. Ptotal = ∑ Ppartial of each gas 39/78 Dalton’s Law in Lab Dalton’s Law of Partial Pressure is often used in labs where gases are collected. Gases are often collected by bubbling through water. And bubbles up to the top (less dense) Reaction producing gas Gas travels through tube Through water This results in a mixture of gases - the one being collected and water vapor. 40/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 20
  • AP Chemistry Rapid Learning Series - 16 Dalton’s Law in Lab - Example Example: Hydrogen gas is collected by bubbling through water. If the total pressure of the gas is 0.970 atm, and the partial pressure of water at that temperature is 0.016 atm, find the pressure of the hydrogen gas. Ptotal = 0.970 atm Ptotal = Pwater + Phydrogen Pwater = 0.016 atm 0.970atm = 0.016atm + Phydrogen Phydrogen = ? 0.970atm − 0.016atm = Phydrogen Phydrogen = 0.954 atm 41/78 Definition: Mole Fraction Mole Fraction (χ) – Ratio of moles (n) of one type of gas to the total moles of gas. χA = nA ntotal Mole fraction has no units as it is “moles/moles”. 42/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 21
  • AP Chemistry Rapid Learning Series - 16 Dalton’s Law and Mole Fractions Dalton’s Law of Partial Pressure calculations can be done with mole fractions. Pressure P of gas “A” PA = χ A × Ptotal Pressure of the whole sample l Mole fraction of gas “A” PA = nA × Ptotal ntotal 43/78 Example - 1 Dalton’s Law of Partial Pressure calculations can be done with mole fractions. Example: If the total pressure of the sample is 115.5 kPa, and the a pe pressure of hydrogen gas is 28.7 kPa, what is the mole fraction of hydrogen gas? Ptotal = 115.5 kPa Phydrogen = χ hydrogen × Ptotal Phydrogen = 28.7 kPa 28.7 kPa = χ hydrogen × 115.5kPa χhydrogen = ? 28.7 kPa = χ hydrogen 115.5kPa χhydrogen = 0.248 44/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 22
  • AP Chemistry Rapid Learning Series - 16 Example - 2 Another type of problem: Example: How many moles of oxygen are present in a sample with a total of 0.556 moles, 1.23 atm and a partial pressure for f oxygen of 0 87 atm? f 0.87 t ? Ptotal = 1.23 atm Poxygen = χ oxygen × Ptotal Poxygen = 0.87 kPa ntotal = 0.556 moles noxygen = ? Poxygen = 0.87 atm = noxygen ntotal × Ptotal noxygen 0.556moles ×1.23atm 0.87 atm × 0.556moles = noxygen 1.23atm 45/78 noxygen = 0.39 moles Gas Stoichimetry: Molar Volume of a Gas 46/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 23
  • AP Chemistry Rapid Learning Series - 16 Definition: Molar Volume of a Gas Standard Temperature and Pressure (S ) – 1 atm essu e (STP) at (760 mm Hg) and 273 K (0°C) Molar Volume of a Gas – at STP, 1 mole of any gas STP = 22.4 liters 47/78 Mass-Volume Problems (Gases) Example: If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) From balanced equation: q 1 mole Zn 1 mole H2 Molar volume of a gas: 1 mole H2 = 22.4 L Molar Mass of Zn: 1 mole Zn = 65.39 g K 1.5 g Zn D 1 mole Zn 1 mole H2 65.39 g Zn 1 mole Zn 22.4 1 L H2 mole H2 U 0.51 = ________ L H2 0.51 is a reasonable answer for L (514 mL) “L H2” is the correct unit 2 sf given 2 sf in answer S O 48/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 24
  • AP Chemistry Rapid Learning Series - 16 Stoichiometry and the Gas Laws What if you want the volume of a gas not at STP? Example: If you need react 1.5 g of zinc completely, what volume of gas will be produced at 2.5 atm and 273°C? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) P1 = 1.0 atm From balanced equation: 1 mole Zn 1 mole H2 V1 = 0.51 L P2 = 2.5 atm Molar volume of a gas: 1 mole H2 = 22.4 L Molar Mass of Zn: 1 mole Zn = 65.39 g V2 = ? L 1.5 g Zn 1 mole Zn 1 mole H2 65.39 g Zn 1 mole Zn 22.4 1 L H2 mole H2 0.51 = ________ L H2 This is volume at STP (1 atm & 273°) 1.0atm × 0.51L = V2 2.5atm 1.0atm × 0.51L = 2.5atm × V2 V2 = 0.20 L 49/78 Another Example Example: What volume of H2 gas is produced at 25° and 0.97 atm from reacting 5.5 g Zn? 2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g) P1 = 1.0 atm From balanced equation: 1 mole Zn 1 mole H2 V1 = 1.88 L T1 = 273 K Molar volume of a gas: 1 mole H2 = 22.4 L P2 = 0.97 atm V2 = ? L Molar Mass of Zn: 1 mole Zn = 65.39 g T2 = 25°C = 298 K 5.5 g Zn 1 mole Zn 1 mole H2 65.39 g Zn 1 mole Zn This is volume at STP (1 atm & 273°) 1.0atm × 1.88L 0.97atm × V2 = 273K 298K 22.4 1 L H2 mole H2 1.88 = ________ L H2 298 K ×1.0atm × 1.88L = V2 0.97atm × 273K V2 = 2.1 L 50/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 25
  • AP Chemistry Rapid Learning Series - 16 Ideal Gas Law 51/78 Definition: Ideal Gas Law Ideal Gas – all of the assumptions of the Kinetic Molecular Theory (KMT) are valid. Ideal Gas Law – Describes properties of a gas under a set of conditions. PV = nRT This law does not have “before” and “after”—there is no change in conditions taking place. 52/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 26
  • AP Chemistry Rapid Learning Series - 16 Definition: Gas Constant PV = nRT Gas Constant (R) – constant equal to the ratio of P×V to n×T for a gas. Use this one when the P unit is “kPa” Values for R 8.31 8 31 L × kPa mole × K 0.0821 L × atm mole × K 62.4 Use this one when the P unit is “mm Hg” L × mm Hg mole × K Use this one when the P unit is “atm” 53/78 Memorizing the Ideal Gas Law PV = nRT Phony Vampires are not Real Things 54/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 27
  • AP Chemistry Rapid Learning Series - 16 Ideal Gas Law Example An example of the Ideal Gas Law: PV = nRT P = Pressure V = Volume n = # of moles R = Gas constant T = Temperature Choose your “R” based upon your “P” units. P T must be in Kelvin! Example: What is the pressure (in atm) of a gas if it is 2.75 L, has 0.25 moles and is 325 K? Choose the “0.0821” for “R” since the problem asks for “atm”. P=? V = 2.75 L n = 0.25 moles ( P × 2.75 L = 0.25moles × 0.0821 L × atm ( 0.25moles × 0.0821 L × atm P= T = 325 K R = 0.0821 (L×atm) / (mol×K) mole × K mole × K )× 325K )× 325K 2.75 L Phydrogen = 2.43 atm 55/78 Definition: Molar Mass Molar mass (MM) – Mass (m) per moles (n) of a substance. substance MM = Therefore: Th f n= m n m MM 56/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 28
  • AP Chemistry Rapid Learning Series - 16 Ideal Gas Law and Molar Mass The Ideal Gas Law is often used to determine molar mass. PV = nRT and n= m MM PV = m RT MM Example: A gas is collected. The mass is 2.889 g, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass. Choose the “8.31” for “R” since the problem uses “kPa”. P = 98.0 kPa V = 0.936 L m = 2.889 g 98.0kPa × 0.936 L = MM = T = 304 K ( ) 2.889 g kPa 8.31 L × kP × 304 K mole × K MM ( ) 2.889 g 8.31 L × kPa × 304 K mole × K 98.0kPa × 0.936 L MM = ? g/mole R = 8.31 (L×kPa) / (mol×K) MM = 79.6 g/mole 57/78 Definition: Density Density – R i of mass to D i Ratio f volume for a sample. D= m V 58/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 29
  • AP Chemistry Rapid Learning Series - 16 Ideal Gas Law and Density Using the density equation with the Ideal Gas Law: m m RT m and D = P= PV = RT V MM MM V P=D RT MM Example: A gas is collected. The density is 3.09 g/L, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass. P = 98.0 kPa Choose the “8.31” for “R” since the problem uses “kPa”. (8.31 L × kPa mole × K ) × 304K V = 0.936 L D = 3.09 g/L 98.0kPa = 3.09 g L T = 304 K MM = ? g/mole R = 8.31 (L×kPa) / (mol×K) MM = 3.09 g ( MM 8.31 L × kPa L 59/78 mole × K 98.0kPa ) × 304 K MM = 79.6 g/mole Real Gases 60/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 30
  • AP Chemistry Rapid Learning Series - 16 Definition: Real Gas Real Gas – 2 of the assumptions of the Kinetic Molecular Theory are not valid. Gas particles are not attracted nor repelled from one another. Gas particles do have attractions and repulsions towards one another. The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant. Gas particles do take up space - thereby reducing the space available for other particles to be. 61/78 Real Gas Law The Real Gas Law takes into account the deviations from the Kinetic Molecular Theory. PV = nRT Ideal Gas Law ⎛ n2a ⎞ ⎜ P + 2 ⎟(V − nb ) = nRT ⎜ V ⎟ ⎠ ⎝ Real Gas Law Also called “van der Waals equation” Take into account the change in pressure due to particle attractions and repulsions Takes into account the space the particles take up “a” and “b” are constants that you look up for each gas! 62/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 31
  • AP Chemistry Rapid Learning Series - 16 Real Gas Law Example ⎛ n2a ⎞ ⎜ P + 2 ⎟(V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ Example: At what temperature would a 0.75 mole sample of CO2 be 2.75 L at 3.45 atm? (van der Waals constants for CO2: a = 3.59 L2atm/mol2 P = 3.45 atm b = 0.0427 L/mol V = 2.75 L Choose the “0.0821” for “R” since the problem uses n = 0.75 mole “atm”. T=?K a = 3.59 L2atm/mol2 b = 0.0427 L/mol R = 0.0821 (L×atm) / (mol×K) ( 2 ⎛ (0.75mol ) 2 × 3.59 L atm ⎜ mol 2 ⎜ 3.45atm + (2.75 L) 2 ⎜ ⎝ )⎞⎟(2.75L − 0.75mol × 0.0427 L ) = 0.75mol × (0.0821 L × atm )× T ⎟ mol mol × K ⎟ ⎠ T = 164 K 63/78 Diffusion and Effusion 64/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 32
  • AP Chemistry Rapid Learning Series - 16 Definition: Diffusion & Effusion Diffusion – A gas spreads throughout a space. Perfume is sprayed in one corner of the room and a person on the other side smells it after a moment. Effusion – A gas escapes through a tiny hole. Air leaks out of a balloon overnight and is flat the next day. 65/78 Diffusion, Effusion and Mass The mass of a particle affects the rate of diffusion and effusion. Temperature is proportional to average kinetic energy A heavier object with the same kinetic energy as a lighter object moves slower than the lighter object Heavy molecules move slower than smaller molecules Diffusion: If molecules move slower, it will take them longer to reach the other side of the room. Effusion: If molecules move slower, it will take them longer to find the hole to escape through. 66/78 Both rates of diffusion and effusion are inversely proportional to molecular mass. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 33
  • AP Chemistry Rapid Learning Series - 16 Effusion and Graham’s Law - 1 Effusion rates are related by Graham’s Law. r1 = r2 MM 2 MM 1 r1 = Rate of Effusion for molecule 1 r2 = Rate of Effusion for molecule 2 MM1 = M l Molecular mass for molecule 1 l f l l MM2 = Molecular mass for molecule 2 Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule? If the unknown molecule is 0.355 Molecule 1 = O2 times as fast as O2, Molecule 2 = unknown molecule then make the rate of O2 = 1 and r1 = 1 the rate of unknown = 0.355 r2 = 0.355 Molecular mass of O2: MM1 = 32.00 g/mole O 2 × 16.00 = 32.00 g/mole MM2 = ? g/mole 67/78 Effusion and Graham’s Law - 2 Effusion rates are related by Graham’s Law. r1 = r2 MM 2 MM 1 r1 = Rate of Effusion for molecule 1 r2 = Rate of Effusion for molecule 2 MM1 = M l Molecular mass for molecule 1 l f l l MM2 = Molecular mass for molecule 2 Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule? Molecule 1 = O2 1 MM 2 = 0.355 32.00 g / mole Molecule 2 = unknown molecule r1 = 1 r2 = 0.355 MM1 = 32.00 g/mole MM2 = ? g/mole 2 MM 2 ⎛ 1 ⎞ ⎜ ⎟ = ⎝ 0.355 ⎠ 32.00 g / mole (32.00 g / mole)× ⎛ ⎜ 2 1 ⎞ ⎟ = MM 2 ⎝ 0.355 ⎠ MM = 254 g/mole 68/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 34
  • AP Chemistry Rapid Learning Series - 16 Diffusion and Graham’s Law Distances traveled during diffusion: d1 = d2 MM 2 MM 1 d1 = Distance traveled for molecule 1 d2 = Distance traveled for molecule 2 MM1 = M l Molecular mass for molecule 1 l f l l MM2 = Molecular mass for molecule 2 Example: A gas molecule is 4 times as heavy as O2. How far does it travel in the time that oxygen travels 0.25 m? Molecule 1 = unknown molecule d1 32.00 g / mole = 0.25m 128.00 g / mole Molecule 2 = O2 d1 = ? 1 d1 = 0.25m 4 d2 = 0.25 m MM1 = 4×32.00 g/mole = 128 g/mole MM2 = 32.00 g/mole 1 d1 = × 0.25m 2 Molecular mass of O2: O 2 × 16.00 = 32.00 g/mole d1 = 0.125 m 69/78 Gases & The AP Exam 70/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 35
  • AP Chemistry Rapid Learning Series - 16 Gases in the Exam Common Gases problems: Gas stoichiometry at STP (1 mole = 22.4 L) Using the ideal gas law Using the combined gas law Dalton’s law of partial pressure (and using with mole fractions). 71/78 Multiple Choice Questions Dalton’s Law of Partial Pressure is a common question topic. Example: A flask contains 2.5 moles gas particles and has a total pressure of 1.5 atm. How many moles of O2 are in the flask if the partial pressure of O2 is 0.3 atm? A. B. C. D. E. E 0.20 moles O2 0.50 moles O2 2.5 moles O2 12.5 moles O2 0 mole O2 l Partial pressure = mole fraction × total pressure 0.30 atm = mole fraction × 1.5 atm Mole fraction O2 = 1/5 Total moles = 2.5 moles 1/5 of 2.5 = 0.5 moles O2 Answer: B 72/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 36
  • AP Chemistry Rapid Learning Series - 16 Free Response Questions Gases are often used in conjunction with other topics in Free Response questions. Given the hydrocarbon butane (C4H10), answer the following: A. Write the balanced equation for the combustion of butane to produce carbon dioxide and water. B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane? C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass of the unknown gas? g D. Draw 2 structural isomers of butane. These are the gases sub-questions in this problem. 73/78 Answering Free Response Questions Answer sub-question B: It’s stoichiometry, which requires a balanced equation, which you would have written in sub-question A: ou d a e tte sub quest o 2 C4H10 + 13 O2 8 CO2 + 10 H2O B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane? 3.75 g C4H10 1 mole C4H10 8 mole CO2 22.4 L CO2 58.14 58 14 g C6H10 2 mole C4H10 1 mole CO2 = _____ L CO2 at STP 5.78 L CO2 at STP PV1 P2V2 1 = T1 T2 (1atm)(5.78L) (0.97 atm)V2 = 273K 303K V = 6.61 L CO2 74/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 37
  • AP Chemistry Rapid Learning Series - 16 Answering Free Response Questions Answer sub-question C: C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass of the unknown gas? MM C4H10 = 58.14 g/mole Rate C4H10 = 1 Rate ? = 3 MM ? = ? g/mole r1 = r2 MM 2 MM 1 1 MM 2 = 3 58.14 2 MM 2 ⎛1⎞ ⎜ ⎟ = 58.14 ⎝ 3⎠ MM2 = 6.49 g/mole 75/78 Learning Summary Real gases do not use 2 of the assumptions of the KMT Gas particles cause pressure ca se press re Several Gas Laws are used to determine properties under a set of conditions Rates of Effusion and Diffusion are y inversely proportional to molecular mass Ideal gases follow the assumption of the Kinetic Molecular Theory (KMT) 76/78 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 38
  • AP Chemistry Rapid Learning Series - 16 Congratulations You have successfully completed the core tutorial The Gas Laws Rapid Learning Center 77/78 Rapid Learning Center Chemistry :: Biology :: Physics :: Math What’s N t Wh t’ Next … Step 1: Concepts – Core Tutorial (Just Completed) Step 2: Practice – Interactive Problem Drill Step 3: Recap – Super Review Cheat Sheet Go for it! 78/78 http://www.RapidLearningCenter.com © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 39