1 / 51

Packet 5

Packet 5. Gases. Gas Laws Crossword Handout. Why study gases?. What is our atmosphere composed of? Gas behavior can be described by fairly simple math formulas. Some common elements (oxygen and nitrogen) are gases. Many solvents – like gasoline – easily evaporate. Those vapors are gases!.

irina
Download Presentation

Packet 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Packet 5 Gases

  2. Gas Laws Crossword Handout Why study gases? • What is our atmosphere composed of? • Gas behavior can be described by fairly simple math formulas. • Some common elements (oxygen and nitrogen) are gases. • Many solvents – like gasoline – easily evaporate. Those vapors are gases! CHE 170 Packet 5 - 2 Tro page 179

  3. Concept Area I: Terminology • pressure • atmospheric pressure • millimeters of mercury, mmHg • torr • atmosphere, atm • Boyle’s law • Charles’ law • Avogadro’s law • combined gas law • ideal gas law • universal gas constant, R • standard temperature and pressure, STP • molar volume • partial pressure, Pn • Dalton’s law of partial pressure • vapor pressure • kinetic molecular theory • elastic • Maxwell-Boltzmann distributions • root-mean-square speed • dependent variable • independent variable • direct relationship • inverse relationship Don’t forget to start demos… CHE 170 Packet 5 - 3

  4. Concept Area II: Kinetic Molecular Theory of Gases (KMT) • You should understand and be able to explain the “basics” of KMT. • You should be able to sketch Maxwell-Boltzmann distributions and be able to interpret them. CHE 170 Packet 5 - 4

  5. Kinetic Molecular Theory • It so happens that thermal energy or heat is the kinetic energy of molecules. • Of course, molecules, atoms and ions are always in motion unless we are at . • These particles are too small to see. So, we use Kinetic Molecular Theory to explain what is happening at the submicroscopic level using the macroscopic properties that we can see. • First, what determines if our microscopic particles will be a gas, liquid or solid? absolute zero, 0 K The speed at which the particles are moving and their intermolecular forces! CHE 170 Packet 5 - 5

  6. Reviewing The Three States of Matter • The gas particles are around like crazy. • The liquid particles are around a bit. • The solid particles are a little. • The gas particles are very far apart. • The liquid particles are spaced out a little. • The solid particles are tightly packed. zipping moving jiggling Above, a flask of nitrogen liquid is allowed to evaporate into gas.

  7. Kinetic Molecular Theory • A gas contains a large number of individual particles that are so small as compared with the container size that even though the particles have mass, they have essentially no volume. • Average kinetic energy of a particle is proportional to its temperature in Kelvin. • The gas particles constantly move in random straight line until they collide with each other or container walls. These collisions are completely elastic – no energy is lost. CHE 170 Packet 5 - 7

  8. Implications of Kinetic Molecular Theory • How are speed, kinetic energy and temperature related? • What causes pressure according to this theory? • How can we increase the pressure of a gas? Let’s look at each of these one at a time! CHE 170 Packet 5 - 8

  9. 1. How are speed, kinetic energy and temperature related? The the speed of particles, the kinetic energy they possess, and the the temperature of the sample! greater Which eventually leads to the root-mean-square speed: Where u is speed, R is the gas constant in SI units, T is temperature in Kelvin and M is molar mass. more higher CHE 170 Packet 5 - 9

  10. Let’s calculate the rms speed of oxygen molecules at 25°C! • We’ll have to use , right? • Let’s go! CHE 170 Packet 5 - 10

  11. 2. What causes pressure according to this theory? The particles colliding with the walls of the container. The more collisions, the higher the pressure. CHE 170 Packet 5 - 11 bottom image on Tro page 164

  12. 3. How can we increase the pressure of a gas? Increase number of collisions by • Increasing temperature • Decreasing volume • Increasing number of particles CHE 170 Packet 5 - 12

  13. To summarize, why does heating a sealed container of gas increase the pressure? • We are increasing the temperature of the gas particles, so what is happening to the speed of the particles? . • This will also the number of collisions with the container walls which in turn will the pressure! They are increasing. increase increase

  14. Why is there less pressure at higher altitudes? • Because there are gas particles above items at higher altitudes. • Yes, that’s right. At any given time we have an enormous amount of pressure from all the gases in the atmosphere pushing down on us! • Why aren’t we crushed? fewer

  15. OK, but why does a hot air balloon rise? • Like with the sealed container, we are still increasing the temperature of the gas particles, however, this time the container size is not fixed. • Thus, the gas particles are allowed to the volume they take up. Once they do, the . in the balloon becomes less, and it rises! expand density

  16. According to the Maxwell-Boltzmann distribution, . Notice at 273 K, the most probably speed for N2 is about 500 m/s. At 1000 K, it increases to about 900 m/s. It increases to about 1100 m/s at 2000 K. At all these temperatures, however, there are particles moving at many other speeds – faster & slower. Great! Then, are all particles in a container moving at the same speed? no CHE 170 Packet 5 - 16 Tro page 190

  17. Another question: What if the particles aren’t all the same? Would helium gas and oxygen gas have the same most probable speed? • No, they don’t. Oxygen would move slower than the other gases shown below. Why is this so? Oxygen has more mass than rest. So, the same energy given to each (since same T) would push lighter ones faster. 32 g/mol 28 g/mol 18 g/mol 4 g/mol 2 g/mol Metaphor: say pushing a 3-year old in a swing vs. a 20-year old. If the person isn’t “helping” who can be pushed higher? The 3-year old (as long as they don’t get scared!) CHE 170 Packet 5 - 17 Tro page 190

  18. Another question: in separate containers of equal volume at the same temperature, which container would have the higher pressure, if either? • They would have the same pressure! Why? • He is moving faster and thus has more collisions than O2, but the fewer collisions of the more massive O2 have more force. • If we have the same amount of oxygen and helium in representing He representing O2 CHE 170 Packet 5 - 18 Tro page 190

  19. One last question: • Which of the following samples of ideal gases, all at the same temperature, will have the greatest pressure? • Box c! • Why? • Because it has the greatest number of particles! CHE 170 Packet 5 - 19 Tro page 190

  20. Concept Area III: The Ideal Gas Law • You should know what STP conditions are. • You need to know how to convert between millimeters of mercury, torr and atmospheres. • You should be able to explain and use the following laws: Boyle’s Law, Charles’ Law, Combined Gas Law, and the Ideal Gas Law. • You should be able to use the Ideal Gas Law to calculate the molar mass or density of a gas. • You should be able to combine gas calculations with stoichiometry and solution stoichiometry problems. • You should understand and be able to use Dalton’s law of partial pressures. • Understand that gases may not be ideal, but we will not be learning the real gas law. CHE 170 Packet 5 - 20

  21. STP: Standard Temperature & Pressure • We have defined some standard conditions. They are: • temperature: 0º C or 273 K • pressure: 1 atm • Be sure to memorize these values! CHE 170 Packet 5 - 21

  22. Amount – measured in Temperature – measured in Volume – measured in Pressure – measured in moles, mol degrees, K liters, L atmospheres, atm What four variables will define a gas’ state? CHE 170 Packet 5 - 22

  23. Pressure, P • Where F is in newtons (N) and A is in square meters (m2). So, the SI unit for pressure is N/m2 = pascal (Pa). • For gasses, we typically compare their pressure against that of the atmosphere. CHE 170 Packet 5 - 23 Tro page 165

  24. Pressure • At 0° C and where the force of gravity is 9.80665 m/s2, the atmosphere exerts a pressure that will cause a column of mercury in a barometer to be 760 mm tall. This is defined as one atmosphere, atm. • So, we now know that760 mm Hg = 1 atm. CHE 170 Packet 5 - 24 similar image on Tro page 165

  25. Pressure Conversions And now, time for those demos! • Okay, so we just learned that normal atmospheric pressure creates a column of mercury 760 mmHg tall, so • 760 mmHg = 1 atm • What about other units for pressure? • Well, • 760 mmHg = 760 Torr • Also, just for your information, • 760 mmHg = 101,325 Pa = 29.921 in Hg = 1.01325 bar • How do we convert these equalities to conversion factors? or or or or etc. CHE 170 Packet 5 - 25

  26. Gas Laws Demos Handout Gas Laws Demos CHE 170 Packet 5 - 26

  27. Volume versus PressureBoyle’s Law • At constant T, P1V1 = P2V2. • Boyle noticed that as the volume decreased on a closed container of gas, that the pressure increased. • Upon further investigation, he found their relationship: the volume of gas varies inversely with its pressure. • What demonstration corresponds with this? • Gas canisters use this principle so that we can store the gas in a smaller container. Many times, the gas is under such high pressure that it is condensed! • We also use this principle in bicycle pumps! CHE 170 Packet 5 - 27 top image: Tro page 167

  28. Why is Boyle’s Law important to scuba divers? • This very question was asked of the main character in Men of Honor. • Well, what would happen if someone held his or her breath while going back up to the surface? CHE 170 Packet 5 - 28 top image Tro page 168

  29. Temperature versus VolumeCharles’ Law • At constant P, . • Charles decided to study the effect of temperature on volume. He found their relationship to be linear: V a T. • What demonstration corresponds with this? Remember, we can’t have negative volumes, so can’t use a temperature scale with negative values. Always use Kelvin! CHE 170 Packet 5 - 29 similar images on Tro page 170

  30. Temperature versus PressureAmonton’s Law or Gay-Lussac’s Law • Our last relationship could be derived from the previous to give: • If we think back to KMT, we should have no problem with this relationship! • What demonstration corresponds with this? CHE 170 Packet 5 - 30

  31. Moles versus VolumeAvogadro’s Law • Avogadro hypothesized that different gases if compared at the same Temperature and Pressure would have the same number of molecules in equal Volumes. • So, if we have one mol of He and one mol of CO2, Avogadro believed those two x 2x 2x x • differently sized and massed molecules would take up the same volume (if T and P were constant). • This idea was finally accepted (four years after Avogadro died) in 1860. • So, at STP (Standard Temperature & Pressure) which is 0.00°C and one atm, one mol of gas will have a volume of 22.4 L. Note, only three sig figs. • What demonstration corresponds with this? CHE 170 Packet 5 - 31 top image on Tro page 176

  32. Why only 22.4 L for molar volume? Why not more sig figs? • Check out these molar volumes calculated to four sig figs they are different, but notice almost all would round to same value of 22.4 L. • Why the difference? • Gases aren’t always ideal! CHE 170 Packet 5 - 32 Tro page 193

  33. Graphing! • Let’s graph all these relationships on the second page of the Gas Laws Demos handout! CHE 170 Packet 5 - 33

  34. See the P vs. V graph is not linear. But, if we graph V vs. 1/P, the graph is linear! Why is that? Because P and V are inversely proportional, but 1/P and V (or P and 1/V) are directly proportional! Also, why is the volume on the x-axis on the left but on the y-axis up above? Well, to the left they are saying that they are controlling the volume so pressure is dependent on that volume. Above, they must be controlling the pressure and then volume is dependant on that. CHE 170 Packet 5 - 34

  35. Note with this V vs. T graph that they have extrapolated back to –273.15°C/0 K. Why would that be? Even though this graph is in °C and K, remember in calculations we must use Kelvin! This graph is just to help us visualize where the concept of absolute zero came from. Answer hidden here! CHE 170 Packet 5 - 35 Tro page 169

  36. The Combined Gas Law • What did you find to be the single mathematical expression for our four variables? That’s the third question on our handout. • Hopefully you have noticed that that if we combine Boyle’s Law, Charles’ Law and Avogadro’s Law we get the Combined Gas Law: • We can use this law to determine how one variable changes when the other three are held constant. • Remember, we only care about final and initial states. The initial state is usually called “1” and the final state is usually called “2”. CHE 170 Packet 5 - 36

  37. Practice time! A happy child gets a balloon at the mall. The balloon has a maximum capacity of 5.00 L. To save on helium costs, mall policy dictates that the balloons be filled only to 4.77 L. The mall keeps the temperature at a comfortable 22.00 °C. • Will our child still be happy upon exiting the mall if the outside temperature is 34.20 °C, in other words, will the balloon pop? • If the balloon survives to the car, will it survive the interior temperature of the car, 37.50 °C? The balloon won’t pop. Alas, the balloon will pop and … CHE 170 Packet 5 - 37

  38. Another problem applying the combined gas law: If a 20.0 lb bag of pure NH4NO3 were detonated, what total volume of gases would be produced at 25.0°C and 751 torr?2 NH4NO3(s) → 2 N2(g) + 4 H2O(g) + O2(g) 20.0 lb ? volume of all gas We don’t need moles, so cancel them! Now, plug and chug! CHE 170 Packet 5 - 38 Note: the notes page for this slide show this problem worked out!

  39. The Ideal Gas Law Now in that problem, conditions weren’t changing. We just wanted to know the value of the variable we didn’t know. Is there a formula that can get us to that quicker? • Well let’s combine all of our relationships into one expression: • Then, introduce a proportionality constant, R, to get an equality. R is called the gas constant and is equal to 8.314472 J/mol·K or 0.082057 L·atm/mol·K. or CHE 170 Packet 5 - 39

  40. P is pressure – in atm • V is volume – in L • n is number of moles – in mol • R is the gas constant – 0.082057 L·atm/mol·K • T is temperature – in K Memorize! Or can’t use this formula! CHE 170 Packet 5 - 40

  41. Solving our problem with PV=nRT: If a 20.0 lb bag of pure NH4NO3 were detonated, what total volume of gases would be produced at 25.0ºC and 751 torr?2 NH4NO3(s) → 2 N2(g) + 4 H2O(g) + O2(g) 20.0 lb ? volume of all gas 1st :? mol all gas Note: the notes page for this slide show this problem worked out!

  42. A crazy example with PV=nRT! An experiment was done where the oxygen consumption was measured while male cockroaches were running on treadmills at different speeds. They found that in one hour an average cockroach running at 0.08 km/hr consumed 0.8 mL of oxygen (per gram of their weight) at 1 atm and 24 °C. So, how many moles of oxygen would be consumed in one hour by a 5.2 g cockroach running at that speed? First, we must figure out how much oxygen (in mL) our cockroach used. Now, use PV=nRT to determine moles! CHE 170 Packet 5 - 42 Note: the notes page for this slide show this problem worked out!

  43. Let’s try a more normal problem: Molar mass and gas… • The empirical formula of a gaseous compound is determined to be CHF2. Now, a new experiment show that a 0.100 g sample of the gas exerts a pressure of 70.5 mmHg in a 256 mL container at 22.3 ˚C. What is the molar mass and molecular formula? PV = nRT n = 9.795×10–4 mol C2H2F4 102 g/mol CHE 170 Packet 5 - 43

  44. Your turn! • An unknown gas has a density of 1.104 g/L at 30.0°C and 720 torr. Calculate its molar mass. • Calculate the density of a gas at 52°C and 600 torr if its molar mass is 62.2 g/mol. • Now use preferred equation to calculate the volume • Combined gives 26.2 L • Ideal gives 26.3 L Now use preferred equation to calculate the volume. This time, both give same answer! CHE 170 Packet 5 - 44 Note: the notes page for this slide shows these problems worked out!

  45. 3.83 L 2.19 L (don’t use Celsius or get wrong answer of 2.65 L) 1.43 g 26.2 L 0.470 g C 9.17 L 1.96 g/Lmore 8.73 L of Freon gas has a pressure of 735 mmHg. What will be the volume if the pressure increases to 1675 mmHg? If a gas occupies 2.15 L at 24.5°C, how much volume will it take up at 30.2°C? What’s the weight of 1.00L of O2 at STP? What volume does one mole of gas fill at 30.0°C at 720 torr? Given the following reaction: 2C + O2→ 2 CO, how many grams of carbon are needed to react completely with 500 mL O2 gas at 30.0°C and 740 torr? 22.0 g of CO2 at 286°C and 2.50 atm. What is the volume of the CO2? Calculate the density of CO2 at STP. Is CO2 more or less dens than air (1.2 g/L)? Some problems (with answers) for home practice: CHE 170 Packet 5 - 45 Note: the notes page for this slide shows these problems worked out!

  46. Dalton’s Law of Partial Pressures • Say we have a container that holds the following: 2 H2O2(ℓ)  2 H2O(g) + O2(g) 0.32 atm 0.16 atm • What is the total pressure in the flask? • Well, Ptotal in gas mixture must equation the sum of the pressure of each gas present: Ptotal = PA + PB + … • Therefore, for our situation: Ptotal = P(H2O) + P(O2) Ptotal = 0.32 atm + 0.16 atm = 0.48 atm • Dalton’s Law: total pressure is the sum of all the partial pressures from each gas present. John Dalton (1766-1844) CHE 170 Packet 5 - 46

  47. Dalton’s Law of Partial Pressures • Our book also introduces the concept of mole fractions here. We won’t cover mole fractions until CHE 180. So, don’t worry about homework questions that require the use of mole fractions. You only have to know Dalton’s law, and all of the required problems assigned should be able to be solved without the use of mole fractions. CHE 170 Packet 5 - 47

  48. Why do we care about partial pressures? • Well, say we are collecting a gaseous product from a reaction. Typically we do in this fashion: • We are only going towant to know about our product, but we’ll have to subtract off the partial pressure of the water vapor! CHE 170 Packet 5 - 48 Tro page 182

  49. Let’s look at example 5.11 on page 182 In order to determine the rate of photosynthesis, the oxygen gas emitted by an aquatic plant was collected over water at a temperature of 293 K and a total pressure of 755.2 mmHg. Over a specific time period, a total of 1.02 L of gas was collected. What mass of oxygen gas (in grams) was formed? Step 1: Determine PO2 from Ptotal given. Step 3: Convert moles to grams! PO2 = Ptotal – PH2O • PO2 = 755.2 mmHg – 17.55 mmHg • PO2 = 737.65 mmHg Step 2: Use PV=nRT to determine nO2. CHE 170 Packet 5 - 49 Tro page 182

  50. false true true false true false One Last Problem on Gases: • We have 1 L of N2 and 1 L of CO2 at 50 ºC and 2 atm. Are the following statements true or false about these gases? Why? • Both have the same average velocity. • Both have the same average kinetic energy. • Both contain the same number of molecules. • Both contain the same number of atoms. • Both have the same pressure. • An average CO2 molecule moves faster than an average N2 molecule. CHE 170 Packet 5 - 50

More Related