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Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
Gas lawschem
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Gas lawschem

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  1. High School Chemistry Rapid Learning Series - 18 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … p g Teach Yourself High School Chemistry in 24 Hours 1/66 http://www.RapidLearningCenter.com The Gas Laws HS 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 Rapid Learning Center www.RapidLearningCenter.com/ © Rapid Learning Inc. All rights reserved. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 1
  2. High School Chemistry Rapid Learning Series - 18 Learning Objectives By completing this tutorial you will learn… How gases cause pressure. The Kinetic Molecular Theory (KMT). How properties of a gas are related. How to use several gas laws. The difference between ideal and real gases. 3/66 Concept Map Previous content Chemistry New content Studies Volume Matter One state is Pressure Temperature Gas Have properties # of Moles Density D it Molar Mass Related to each other with Gas Laws 4/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 2
  3. High School Chemistry Rapid Learning Series - 18 Kinetic Molecular Theory 5/66 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. 6/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 3
  4. High School Chemistry Rapid Learning Series - 18 Assumptions of the KMT 6 1 All gases are made of particles. 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 (3RT/2). 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. 7/66 Calculating Average Kinetic Energy Temperature is proportional to average kinetic energy…how do you calculate it? Avg. KE = (3/2) RT Avg. KE = Average Kinetic Energy (in J) g g gy ( ) R = Gas constant (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 Avg. KE = (3/2) x (8.31J/K mol) x (301K) R = 8.31 J/K mol T = 28°C + 273 = 301 K Avg. KE = 3752 J 8/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 4
  5. High School Chemistry Rapid Learning Series - 18 Gas Behavior 9/66 KMT and Gas Behavior The Kinetic Molecular Theory and its assumptions can be used to explain gas behavior. 10/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 5
  6. High School Chemistry Rapid Learning Series - 18 Definition: Pressure Pressure – Force of gas g particles running into a surface. 11/66 Pressure and Number of Molecules If pressure is molecular collisions with the container… As A number of b f molecules increases, there are more molecules to collide with the wall. Collisions between molecules and the wall increase. Pressure increases. As # of molecules increases, pressure increases. Pressure (P) and # of molecules (n) are directly proportional (∝). P∝n 12/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 6
  7. High School Chemistry Rapid Learning Series - 18 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 between molecules and the wall decrease. Pressure decreases. As volume increases, pressure decreases. Pressure (P) and volume (V) are inversely proportional. P ∝ 1/v 13/66 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. 14/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 7
  8. High School Chemistry Rapid Learning Series - 18 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 (P) and temperature (T) are directly proportional (∝). P∝T 15/66 Pressure Inside and Outside a Container 16/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 8
  9. High School Chemistry Rapid Learning Series - 18 Definition: Atmospheric Pressure Atmospheric Pressure – Pressure due to the layers of air in the atmosphere. Climb in altitude… altitude Less layers of air… air Lower atmospheric pressure. As altitude increases (less air), atmospheric pressure decreases. 17/66 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 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 Hi h 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. 18/66 The bag will expand in order to reduce the internal pressure. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 9
  10. High School Chemistry Rapid Learning Series - 18 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. 19/66 Attacking Strategy for Gas Law Problems 20/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 10
  11. High School Chemistry Rapid Learning Series - 18 General Strategy for Gas Law Problems The following four 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 a 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 21/66 Pressure Units Several units are used when describing pressure. Pressure Unit atmospheres Symbol atm Pascals, kiloPascals Pa, kPa millimeters of mercury mm Hg pounds per square inch psi Pressure Unit Conversions: 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi 22/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 11
  12. High School Chemistry Rapid Learning Series - 18 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. K = °C + 273 23/66 Definition: Standard Temp & 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 hidden making your list! 24/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 12
  13. High School Chemistry Rapid Learning Series - 18 Gas Laws 25/66 KMT and Gas Laws The Gas Laws are the experimental observations of the gas behavior that b ti f th b h i th t the Kinetic Molecular Theory explains. 26/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 13
  14. High School Chemistry Rapid Learning Series - 18 “Before” and “After” in Gas Laws This section has 4 gas laws which have “before” and “after” conditions. For F example: l 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. 27/66 Avogadro’s Law Avogadro’s Law relates # of moles (n) and Volume (V). Where T and P are held constant.. V1 V2 = n1 n2 V = Volume n = # of moles of gas The two volume units must match! Example: 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 2 .5 L V2 = 0.15mole 0.55mole 0.55mole × 2.5L = V2 0.15mole V2 = 9.2 L 28/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 14
  15. High School Chemistry Rapid Learning Series - 18 Boyles’ Law - 1 Boyles’ Law relates pressure and volume. PV1 = P2V2 1 Where temperature and # of molecules are held constant… P = pressure V = volume The two pressure units must match and the two volume units must match! Example: A gas sample is 1.05 atm when 2.50 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 P1 = 1.05 atm 1 atm = ______ atm 0.980 760 V1 = 2.50 L mm Hg P2 = 745 mm Hg = 0.980 atm V2 = ? L 29/66 Boyles’ Law - 2 Boyles’ Law relates pressure and volume. PV1 = P2V2 1 Where temperature and # of molecules are held constant… P = pressure V = volume 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 30/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 15
  16. High School Chemistry Rapid Learning Series - 18 Charles’ Law - 1 Charles’ Law relates volume and temperature. Where pressure and # of molecules are held constant… V1 V2 = T1 T2 V = Volume T = Temperature The two volume units must match and temperature must be in Kelvin! Example: 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 Kelvin! T t d t b i K l i ! 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 31/66 Charles’ Law - 2 Charles’ Law relates temperature and pressure. Where pressure and # of molecules are held constant… V1 V2 = T1 T2 V = Volume T = Temperature The two volume units must match and temperature must be in Kelvin! Example: 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 32/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 16
  17. High School Chemistry Rapid Learning Series - 18 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 33/66 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 V2 = 4.2 L 1.0atm × 0.125mole × 298 K Why You Only Really Need 1 out of the 4 Laws! 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 34/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 17
  18. High School Chemistry Rapid Learning Series - 18 “Transforming” the Combined Gas Law Watch as variables are held constant and the combined gas law “becomes” the other 3 laws. Hold pressure P and temperature T constant PV1 P2V2 1 = n1T1 n2T2 Avogadro’s Law Hold moles n and temperature T constant PV1 P2V2 1 = n1T1 n2T2 Boyles’ Law Hold pressure P and moles n constant PV1 P2V2 1 = n1T1 n2T2 Charles’ Law 35/66 How to Memorize What’s Held Constant How do you know what to hold constant for each law? Use the mnemonics. 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. Charles’ Law Hold Pressure and moles constant. Charles was from Paris. 36/66 Gas Laws Rhyme: Avogadro hold P, T; Bolyes Charles hold P = “ABC – PreTend To Pee!” © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com hold T; 18
  19. High School Chemistry Rapid Learning Series - 18 Example of 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 37/66 V2 = 14.1 L Mixtures of Gases 38/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 19
  20. High School Chemistry Rapid Learning Series - 18 Definition: Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure – The sum of the pressures of each type of gas p yp g equals the pressure of the total sample. Ptotal = ∑ Ppartial of each gas 39/66 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. G ft ll t d b b bbli th h t And bubbles up to the top (less dense). Reaction producing gas 40/66 Gas travels through tube. Through water This results in a mixture of gases—the one being collected and water vapor. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 20
  21. High School Chemistry Rapid Learning Series - 18 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 Pwater = 0.016 atm Phydrogen = ? Ptotal = Pwater + Phydrogen 0.970 atm = 0.016 atm + Phydrogen 0.970 atm - 0.016 atm = Phydrogen Phydrogen = 0.954 atm 41/66 Definition: Mole Fraction Mole Fraction (χ) – Ratio (fraction) of moles (n) of one type of gas to the total moles of gas. l f χA = nA ntotal Mole fraction has no units as it is “moles/moles”. 42/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 21
  22. High School Chemistry Rapid Learning Series - 18 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 43/66 Dalton’s Law - 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 = 28.7 kPa χhydrogen = ? Phydrogen = χhydrogen x Ptotal 28.7 kPa = χhydrogen x 115.5 kPa (28.7 kPa)/(115.5 kPa) = χhydrogen χhydrogen = 0.248 44/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 22
  23. High School Chemistry Rapid Learning Series - 18 Dalton’s Law - 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 = 0.87 kPa ntotal = 0.556 moles noxygen = ? Poxygen = χoxygen x Ptotal Poxygen = (noxygen/ntotal) x Ptotal Poxygen = (noxygen/0.556mol) x Ptotal (0.87atm x 0.556mol)/1.23atm = noxygen noxygen = 0.39 moles 45/66 Molar Volume of a Gas 46/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 23
  24. High School Chemistry Rapid Learning Series - 18 Definition: Molar Volume of a Gas Standard Temperature and Pressure (STP) – 1 atm (760 mm Hg) and 273 K (0°C). Molar Volume of a Gas – at STP, 1 mole of any gas = 22.4 liters. 47/66 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 The KUDOS Method of Gas Stoichiometry – the calculation of the amounts of reactants and products in gaseous reaction. 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 48/66 0.51 is a reasonable answer for L (514 mL). “L H2” is the correct unit. 2 sf given 2 sf in answer. S © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com O 24
  25. High School Chemistry Rapid Learning Series - 18 Gas Stoichiometry – Example #1 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 x 0.51L = 2.5atm x V2 1.0atm x 0.51L)/(2.5atm) = V2 V2 = 0.20 L 49/66 Gas Stoichiometry – Example #2 Example: What volume of H2 gas is produced at 25°C 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°) 22.4 L H2 1 mole H2 1.88 = ________ L H2 (1.0atm x 1.88L)/273K = (0.97atm x V2)/298K (298K x 1.0atm x 1.88L)/(0.97atm x 273K) = V2 V2 = 2.1 L 50/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 25
  26. High School Chemistry Rapid Learning Series - 18 Ideal Gas Law 51/66 Definition: Ideal Gas Law Ideal Gas – All of the assumptions of the Kinetic Molecular Theory (KMT) are valid. lid Ideal Gas Law – Describes properties of a gas under a set of conditions. PV = nRT 52/66 This law does not have “before” and “after”—there is no change in conditions taking place. © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 26
  27. High School Chemistry Rapid Learning Series - 18 Definition: Gas Constant PV = nRT Gas Constant (R) – constant equal to the ratio of P×V to n×T for a gas. Values for R 8.31 8 31 L•kPa•mol-1K-1 Use this one when the P unit is “mm Hg”. 0.0821 L•atm•mol-1K-1 62.4 L•mmHg•mol-1K-1 Use this one when the P unit is “kPa”. Note: J = L•kPa then R = 8.31 J/mole•K Use this one when the P unit is “atm”. 53/66 Memorizing the Ideal Gas Law PV = nRT Phony Vampires Are(=) not Real Things 54/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 27
  28. High School Chemistry Rapid Learning Series - 18 Ideal Gas Law - Example An example of the Ideal Gas Law: PV = nRT RT 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 0.250 moles and is 325 K? P=? Choose the “0.0821” for “R” since the problem asks for “atm” V = 2.75 L P x 2.75L = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325K n = 0.250 moles P = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325K/2.75L T = 325 K R = 0.0821 (L×atm) / (mol×K) Phydrogen = 2.43 atm 55/66 Definition: Molar Mass Molar Mass (MM) – Mass (m) per moles (n) of a substance substance. MM = Therefore: Th f n= m n m MM 56/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 28
  29. High School Chemistry Rapid Learning Series - 18 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 x 0.936L = (2.889g/MM) x (8.31 L•kPa•mol-1•K-1) x 304K MM = 2.889g x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa x 0.936L ) T = 304 K MM = ? g/mol R = 8.31 (L×kPa) / (mol×K) MM = 79.6 g/mol 57/66 Definition: Density Density – Ratio of mass to volume for a sample. D= m V 58/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 29
  30. High School Chemistry Rapid Learning Series - 18 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. Choose the “8.31” for “R” since the problem uses “kPa” P = 98.0 kPa V = 0.936 L D = 3.09 g/L T = 304 K 59/66 98.0kPa = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/MM MM = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa) MM = ? g/mol R = 8.31 (L×kPa) / (mol×K) MM = 79.6 g/mol Real Gases 60/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 30
  31. High School Chemistry Rapid Learning Series - 18 Definition: Real Gas Real Gas – 2 of the assumptions of the Kinetic Molecular Theory are not valid. #1 - Gas particles are not attracted nor repelled from one another. Gas particles do have attractions and repulsions towards one another. #2 - 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. 61/66 Real Gas Law - Equation 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! 62/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 31
  32. High School Chemistry Rapid Learning Series - 18 Real Gas Law - Example 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 “atm” n = 0.75 mol T=?K a = 3.59 L2atm/mol2 b = 0.0427 L/mol R = 0.0821 (L×atm) / (mol×K) ⎛ n2a ⎞ ⎜ P + 2 ⎟(V − nb ) = nRT ⎜ V ⎟ ⎝ ⎠ [3.45atm + (0.75mol)2 x (3.59 L2•atm•mol-2)/(2.75L)2] x [2.75L – (0.75mol x 0.0427L•mol)] = 0.75mol x (0.0821L•atm•mol-1•K-1) x T T = 164 K 63/66 Learning Summary Real gases do not use 2 g of the assumptions of the KMT. Gas particles cause pressure. Several Gas Laws are used to determine properties under a set of conditions. Ideal gases follow the assumption of the Kinetic Molecular Theory (KMT). 64/66 © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 32
  33. High School Chemistry Rapid Learning Series - 18 Congratulations You have successfully completed the core tutorial The Gas Laws Rapid Learning Center 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! 66/66 http://www.RapidLearningCenter.com © Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 33

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