The kinetic theory of gases


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The kinetic theory of gases

  1. 1. GASES• Gases are one of the most pervasiveaspects of our environment on theEarth. We continually exist withconstant exposure to gases of allforms.• The steam formed in the air during ahot shower is a gas.• The Helium used to fill a birthdayballoon is a gas.• The oxygen in the air is an essentialgas for life.
  2. 2. GASESA windy day or a still day is a result of the difference in pressure of gasesin two different locations. A fresh breeze on a mountain peak is a study inbasic gas laws.
  3. 3. Important Characteristics of Gases1) Gases are highly compressibleAn external force compresses the gas sample and decreases itsvolume, removing the external force allows the gas volume toincrease.2) Gases are thermally expandableWhen a gas sample is heated, its volume increases, and when it iscooled its volume decreases.3) Gases have high viscosityGases flow much easier than liquids or solids.4) Most Gases have low densitiesGas densities are on the order of grams per liter whereas liquidsand solids are grams per cubic cm, 1000 times greater.5) Gases are infinitely miscibleGases mix in any proportion such as in air, a mixture of many gases.
  4. 4. • Helium He 4.0• Neon Ne 20.2• Argon Ar 39.9• Hydrogen H2 2.0• Nitrogen N2 28.0• Nitrogen Monoxide NO 30.0• Oxygen O2 32.0• Hydrogen Chloride HCl 36.5• Ozone O3 48.0• Ammonia NH3 17.0Substances That Are Gases underNormal ConditionsSubstance Formula MM(g/mol)
  5. 5. Kinetic Molecular Theory• To fully understand the world around usrequires that we have a good understandingof the behavior of gases. The description ofgases and their behavior can be approachedfrom several perspectives.• The Gas Laws are a mathematicalinterpretation of the behavior of gases.• However, before understanding themathematics of gases, a chemist must havean understanding of the conceptualdescription of gases. That is the purpose ofthe Kinetic Molecular Theory.
  6. 6. Kinetic Molecular Theory• The Kinetic Molecular Theory is a single set ofdescriptive characteristics of a substance known asthe Ideal Gas.• All real gases require their own unique sets ofdescriptive characteristics. Considering the largenumber of known gases in the World, the task oftrying to describe each one of them individuallywould be an awesome task.• In order to simplify this task, the scientificcommunity has decided to create an imaginary gasthat approximates the behavior of all real gases. Inother words, the Ideal Gas is a substance that doesnot exist.• The Kinetic Molecular Theory describes that gas.While the use of the Ideal Gas in describing all realgases means that the descriptions of all real gaseswill be wrong, the reality is that the descriptions ofreal gases will be close enough to correct that anyerrors can be overlooked.
  7. 7. The Nature of GasesThree basic assumptions of the kinetictheory as it applies to gases:1. Gas is composed of particles- usuallymolecules or atoms–Small, hard spheres–Insignificant volume; relatively farapart from each other–No attraction or repulsion betweenparticles
  8. 8. The Nature of Gases2. Particles in a gas move rapidly inconstant random motion–Move in straight paths, changingdirection only when colliding with oneanother or other objects–Average speed of O2 in air at 20 oC isan amazing 1660 km/h!(1.6km=1mile)
  9. 9. The Nature of Gases3. Collisions are perfectly elastic-meaning kinetic energy is transferredwithout loss from one particle toanother- the total kinetic energy remainsconstantNewtonian Cradle-Where the collisions between the balls elastic?Yes, because kinetic energy was transferredwith each collision
  10. 10. • Why did the balls eventually stopswinging? The collisions were notperfectly elastic, some kinetic energywas lost as heat during each collision.• At constant temperatures and low tomoderate pressures, collisions betweengas particles are perfectly elastic
  11. 11. THE KINETIC THEORY OF GASES• Gas consists of large number of particles(atoms or molecules)• Particles make elastic collisions with eachother and with walls of container• There exist no external forces (densityconstant)• Particles, on average, separated by distanceslarge compared to their diameters• No forces between particles except whenthey collideRemember the assumptions
  12. 12. What happens to a ball when itdrops?The potential energyof the ballWhich is converted tokinetic energy in theballWhich is convertedto potential energyin the ballIs converted to kineticenergy in the ballWhich is converted intothe potential energy ofthe ball…………..…..but in reality the ballloses height andeventually stops bouncingWhy does thishappen?
  13. 13. How does the bouncing ball loseenergy?• Through friction with the air (airresistance)• Through sound when it hits the floor• Through deformation of the ball• Through heat energy in the bounce
  14. 14. IDEAL GAS MODEL• The gas consists of objects with a defined mass and ze• The gas particles travel randomly in straight-line motion• All collisions involving gas particles are elastic; the kin• The gas particles do not interact with each other or the• The gas phase system will have an average kinetic ener
  15. 15. Boltzman Distribution. The behaviour ofthe gas molecules under the action ofgravity.
  16. 16. Maxwell Distribution. Experiment withGalton board demonstrates thestatistical sense of Maxwell distribution.
  17. 17. Ideal Gas ModelKinetic Molecular Theory (KMT) for an idealgas states that all gas particles:• are in random, constant, straight-line motion.• are separated by great distances relative totheir size; the volume of the gas particles isconsidered negligible.• have no attractive forces between them.• have collisions that may result in the transferof energy between gas particles, but the totalenergy of the system remains constant.
  18. 18. Brownian motion. Chaotic motion ofminute particle suspended in a gas orliquid
  19. 19. This animation illustrates the concept offree path length of molecules in a gas.
  20. 20. Ideal vs. Non-Ideal Gases• Kinetic Theory Assumptions– Point Mass– No Forces Between Molecules– Molecules Exert Pressure Via ElasticCollisions With Wallsxx(courtesy F. Remer)
  21. 21. Ideal vs. Non-Ideal Gases• Non-Ideal Gas– Violates Assumptions• Volume of molecules• Attractive forces of molecules(courtesy F. Remer)
  22. 22. Deviations from ideal behaviour• A real gas is most like an ideal gas when thereal gas is at low pressure and hightemperature.• At high pressures gas particles are closetherefore the volume of the gas particles isconsidered.• At low temperatures gas particles have lowkinetic energy therefore particles have someattractive force• Example• Dry ice, liquid oxygen and nitrogen
  23. 23. Ideal GasesBehave as described by the ideal gasequation; no real gas is actually idealWithin a few %, ideal gas equation describesmost real gases at room temperature andpressures of 1 atm or lessIn real gases, particles attract each otherreducing the pressureReal gases behave more like ideal gases aspressure approaches zero.
  24. 24. Atmospheric Pressure• Weight of column of air above your head.• We can measure the density of theatmosphere by measuring the pressure itexerts.
  25. 25. Effect of Atmospheric Pressure onObjects at the Earth’s Surface
  26. 26. Atmospheric PressurePressure = Force per Unit AreaAtmospheric Pressure is the weight ofthe column of air above a unit area. Forexample, the atmospheric pressure feltby a man is the weight of the column ofair above his body divided by the areathe air is resting onP = (Weight of column)/(Area of base)Standard Atmospheric Pressure:1 atmosphere (atm)14.7 lbs/in2(psi)760 Torr (mm Hg)1013.25 KiloPascals or Millibars (kPa =N/m2)
  27. 27. Pressure MeasurementTorricellis Barometer• Torricelli determined from thisexperiment that the pressure of theatmosphere is approximately 30inches or 76 centimeters (onecentimeter of mercury is equal to 13.3millibars. He also noticed that heightof the mercury varied with changes inoutside weather conditions.For climatological and meteorological purposes, standard sea-level pressureis said to be 76.0 cm or 29.92 inches or 1013 millibars
  28. 28. The Nature of GasesAtmospheric pressure results fromthe collisions of air molecules withobjects–Decreases as you climb a mountainbecause the air layer thins out aselevation increasesBarometer is the measuringinstrument for atmosphericpressure; dependent upon weather
  29. 29. Common Units of PressureUnit Atmospheric Pressure Scientific Fieldpascal (Pa); 1.01325 x 105Pa SI unit; physics,kilopascal(kPa) 101.325 kPa chemistryatmosphere (atm) 1 atm* Chemistrymillimeters of mercury 760 mmHg* Chemistry, medicine,( mm Hg ) biologytorr 760 torr* Chemistrypounds per square inch 14.7 lb/in2Engineering( psi or lb/in2)bar 1.01325 bar Meteorology,chemistry, physics
  30. 30. Converting Units of PressureProblem: A chemist collects a sample of carbon dioxide from thedecomposition of limestone (CaCO3) in a closed end manometer, theheight of the mercury is 341.6 mm Hg. Calculate the CO2 pressure intorr, atmospheres, and kilopascals.Plan: The pressure is in mmHg, so we use the conversion factors fromTable 5.2(p.178) to find the pressure in the other units.Solution:PCO2 (torr) = 341.6 mm Hg x = 341.6 torr1 torr1 mm Hgconverting from mmHg to torr:converting from torr to atm:PCO2( atm) = 341.6 torr x = 0.4495 atm1 atm760 torrconverting from atm to kPa:PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 kPa1 atm
  31. 31. Change in PressureChange in averageatmospheric pressure withaltitude.
  32. 32. The Nature of GasesGas Pressure – defined as theforce exerted by a gas per unitsurface area of an object–Due to: a) force of collisions, and b)number of collisions–No particles present? Then therecannot be any collisions, and thus nopressure – called a vacuum
  33. 33. ManometersRules of thumb: When evaluating, start from the knownpressure end and work towards theunknown end At equal elevations, pressure isconstant in the SAME fluid When moving down a manometer,pressure increases When moving up a manometer,pressure decreases Only include atmospheric pressure onopen endsManometers measure a pressure difference by balancing theweight of a fluid column between the two pressures of interest
  34. 34. Manometers
  35. 35. ManometersFind the pressure atpoint A in this open u-tube manometer with anatmospheric pressure PoPD = γH2O x hE-D + PoPc = PDPB = PC - γHg x hC-BPA = PBExample 2P = γ x h + PO
  36. 36. The Gas Laws• What would PollyParcel look like if shehad no gas moleculesinside?zero molecules = zero pressure insidezero pressure inside = zero force on theinside
  37. 37. Gas Law Variables• In order to describe gases, mathematically, itis essential to be familiar with the variablesthat are used. There are four commonlyaccepted gas law variables• Temperature• Pressure• Volume• Moles
  38. 38. Temperature• The temperature variable is always symbolized as T.• It is critical to remember that all temperature valuesused for describing gases must be in terms ofabsolute kinetic energy content for the system.• Consequently, T values must be converted to theKelvin Scale. To do so when having temperaturesgiven in the Celsius Scale remember the conversionfactor• Kelvin = Celsius + 273• According to the Kinetic Molecular Theory, everyparticle in a gas phase system can have its ownkinetic energy. Therefore, when measuring thetemperature of the system, the average kineticenergy of all the particles in the system is used.• The temperature variable is representing theposition of the average kinetic energy as expressedon the Boltzmann Distribution.
  39. 39. Pressure• The pressure variable is represented by thesymbol P.• The pressure variable refers to the pressurethat the gas phase system produces on thewalls of the container that it occupies.• If the gas is not in a container, then thepressure variable refers to the pressure itcould produce on the walls of a container if itwere in one.• The phenomenon of pressure is really a forceapplied over a surface area. It can best beexpressed by the equation
  40. 40. Pressure• Consider the Pressure equation and the impact ofvariables on it.• The force that is exerted is dependent upon thekinetic energy of the particles in the system. If thekinetic energy of the particles increases, forexample, then the force of the collisions with a givensurface area will increase. This would cause thepressure to increase. Since the kinetic energy of theparticles is increased by raising the temperature,then an increase in temperature will cause anincrease in pressure.• If the walls of the container were reduced in totalsurface area, there would be a change in thepressure of the system. By allowing a given quantityof gas to occupy a container with a smaller surfacearea, the pressure of the system would increase.
  41. 41. Pressure• As this container of gasis heated, thetemperature increases.As a result, the averagekinetic energy of theparticles in the systemincreases.• With the increase inkinetic energy, the forceon the available amountof surface area increases.As a result, the pressureof the system increases.• Eventually,..........................Ka-Boom
  42. 42. Volume• The Volume variable is represented by the symbol V.It seems like this variable should either be very easyto work with or nonexistent.• Remember, according to the Kinetic MolecularTheory, the volume of the gas particles is set at zero.Therefore, the volume term V seems like it should bezero.• In this case, that is not true. The volume beingreferred to here is the volume of the container, notthe volume of the gas particles.• The actual variable used to describe a gas should bethe amount of volume available for the particles tomove around in. In other words
  43. 43. Volume• Since the Kinetic Molecular Theorystates that the volume of the gasparticles is zero, then the equationsimplifies.• As a result, the amount of availablespace for the gas particles to movearound in is approximately equal to thesize of the container.• Thus, as stated before, the variable V isthe volume of the container.
  44. 44. Moles• The final gas law variable is the quantity of gas. This is alwaysexpressed in terms of moles. The symbol that represents themoles of gas is n. Notice that, unlike the other variables, it is inlower case.• Under most circumstances in chemistry, the quantity of asubstance is usually expressed in grams or some other unit ofmass. The mass units will not work in gas law mathematics.Experience has shown that the number of objects in a systemis more descriptive than the mass of the objects.• Since each different gas will have its own unique mass for thegas particles, this would create major difficulties when workingwith gas law mathematics.• The whole concept of the Ideal Gas says that all gases can beapproximated has being the same. Considering the largedifference in mass of the many different gases available, usingmass as a measurement of quantity would cause major errorsin the Kinetic Molecular Theory.• Therefore, the mole will standardize the mathematics for allgases and minimize the chances for errors.
  45. 45. ConclusionsThere are four variables used mathematically for describing agas phase system. While the units used for the variables maydiffer from problem to problem, the conceptual aspects of thevariables remain unchanged.1. T, or Temperature, is a measure of the average kinetic energy ofthe particles in the system and MUST be expressed in theKelvin Scale.2. P, or Pressure, is the measure of the amount of force per unitof surface area. If the gas is not in a container, then Prepresents the pressure it could exert if it were in a container.3. V, or Volume, is a measure of the volume of the container thatthe gas could occupy. It represents the amount of spaceavailable for the gas particles to move around in.4. n, or Moles, is the measure of the quantity of gas. Thisexpresses the number of objects in the system and does notdirectly indicate their masses.
  46. 46. Gas Laws• (1) When temperature is held constant, the density of agas is proportional to pressure, and volume is inverselyproportional to pressure. Accordingly, an increase inpressure will cause an increase in density of the gas anda decrease in its volume. – Boyles’s Law• (2) If volume is kept constant, the pressure of a unitmass of gas is proportional to temperature. Iftemperature increase so will pressure, assuming nochange in the volume of the gas.• (3) Holding pressure constant, causes the temperature ofa gas to be proportional to volume, and inverselyproportional to density. Thus, increasing temperature ofa unit mass of gas causes its volume to expand and itsdensity to decrease as long as there is no change inpressure. - Charles’s Law
  47. 47. Boyle’s Law• Hyperbolic Relation Between Pressure andVolumepVp – V Diagramp – V DiagramisothermsT1 T2 T3 T3 >T2>T1(courtesy F. Remer)
  48. 48. Charles’ Law• Linear Relation Between Temperature andPressurePT (K)0 100 200 300P – T DiagramP – T DiagramisochorsisochorsV1V2V3V1 <V2 <V3(courtesy F. Remer)
  49. 49. Charles’ LawReal data must beobtained aboveliquefactiontemperature.Experimental curves fordifferent gasses,different masses,different pressures allextrapolate to acommon zero.
  50. 50. Another version of Charles Law
  51. 51. Compression and expansion ofadiabatically isolated gas isaccompanied by its heating and cooling.
  52. 52. The Gas Laws• What would PollyParcel look like if shehad a temperature ofabsolute zero inside?absolute zero = no molecular motionno molecular motion = zero force onthe inside
  53. 53. Ideal Gas LawThe equality for the four variables involvedin Boyle’s Law, Charles’ Law, Gay-Lussac’sLaw and Avogadro’s law can be writtenPV = nRTR = ideal gas constant
  54. 54. PV = nRTR is known as the universal gas constantUsing STP conditionsP VR = PV = (1.00 atm)(22.4 L)nT (1mol) (273K)n T= 0.0821 L-atmmol-K
  55. 55. Learning CheckWhat is the value of R when the STP valuefor P is 760 mmHg?
  56. 56. SolutionWhat is the value of R when the STP valuefor P is 760 mmHg?R = PV = (760 mm Hg) (22.4 L)nT (1mol) (273K)= 62.4 L-mm Hgmol-K
  57. 57. Learning CheckDinitrogen monoxide (N2O), laughing gas, isused by dentists as an anesthetic. If 2.86mol of gas occupies a 20.0 L tank at 23°C,what is the pressure (mmHg) in the tank inthe dentist office?
  58. 58. SolutionSet up data for 3 of the 4 gas variablesAdjust to match the units of RV = 20.0 L 20.0 LT = 23°C + 273 296 Kn = 2.86 mol2.86 molP = ? ?
  59. 59. Rearrange ideal gas law for unknown PP = nRTVSubstitute values of n, R, T and V andsolve for PP = (2.86 mol)(62.4L-mmHg)(296 K)(20.0 L) (K-mol)= 2.64 x 103mm Hg
  60. 60. Learning CheckA 5.0 L cylinder contains oxygen gasat 20.0°C and 735 mm Hg. How manygrams of oxygen are in the cylinder?
  61. 61. SolutionSolve ideal gas equation for n (moles)n = PVRT= (735 mmHg)(5.0 L)(mol K)(62.4 mmHg L)(293 K)= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O21 mol O2
  62. 62. Molar Mass of a gasWhat is the molar mass of a gas if 0.250 g ofthe gas occupy 215 mL at 0.813 atm and30.0°C?n = PV = (0.813 atm) (0.215 L) = 0.00703 molRT (0.0821 L-atm/molK) (303K)Molar mass = g = 0.250 g = 35.6 g/molmol 0.00703 mol
  63. 63. Density of a GasCalculate the density in g/L of O2 gas at STP.From STP, we know the P and T.P = 1.00 atm T = 273 KRearrange the ideal gas equation for moles/LPV = nRT PV = nRT P = nRTV RTV RT V
  64. 64. Substitute(1.00 atm ) mol-K = 0.0446 mol O2/L(0.0821 L-atm) (273 K)Change moles/L to g/L0.0446 mol O2 x 32.0 g O2 = 1.43 g/L1 L 1 mol O2Therefore the density of O2 gas at STP is1.43 grams per liter
  65. 65. Formulas of GasesA gas has a % composition by mass of85.7% carbon and 14.3% hydrogen. AtSTP the density of the gas is 2.50 g/L.What is the molecular formula of thegas?
  66. 66. Formulas of GasesCalculate Empirical formula85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C12.0 g C14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H1.0 g HEmpirical formula = CH2EF mass = 12.0 + 2(1.0) = 14.0 g/EF
  67. 67. Using STP and density ( 1 L = 2.50 g)2.50 g x 22.4 L = 56.0 g/mol1 L 1 moln = EF/ mol = 56.0 g/mol = 414.0 g/EFmolecular formulaCH2 x 4 = C4H8
  68. 68. Gases in Chemical EquationsOn December 1, 1783, Charles used 1.00 x 103lb of iron filings to make the first ascent in aballoon filled with hydrogenFe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)At STP, how many liters of hydrogengas were generated?
  69. 69. Solutionlb Fe → g Fe → mol Fe → mol H2 → LH21.00 x 103lb x 453.6 g x 1 mol Fe x 1 mol H21 lb 55.9 g 1 mol Fex 22.4 L H2 = 1.82 x 105L H21 mol H2Charles generated 182,000 L of hydrogen to fill hisair balloon.
  70. 70. Learning CheckHow many L of O2 are need to react 28.0 gNH3 at24°C and 0.950 atm?4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
  71. 71. SolutionFind mole of O228.0 g NH3 x 1 mol NH3 x 5 mol O217.0 g NH3 4 mol NH3= 2.06 mol O2V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 LP 0.950 atm
  72. 72. Mixture of gases
  73. 73. Reacting mixture of gases
  74. 74. Learning CheckA.If the atmospheric pressure today is 745mm Hg, what is the partial pressure (mmHg) of O2 in the air?1) 35.6 2) 156 3) 760B. At an atmospheric pressure of 714, what isthe partial pressure (mm Hg) N2 in the air?1) 557 2) 9.14 3) 0.109
  75. 75. SolutionA.If the atmospheric pressure today is 745mm Hg, what is the partial pressure (mmHg) of O2 in the air?2) 156B. At an atmospheric pressure of 714, what isthe partial pressure (mm Hg) N2 in the air?1) 557
  76. 76. Partial PressurePartial PressurePressure each gas in a mixture would exertif it were the only gas in the containerDaltons Law of Partial PressuresThe total pressure exerted by a gas mixtureis the sum of the partial pressures of thegases in that mixture.PT = P1 + P2 + P3 + .....
  77. 77. Partial PressuresThe total pressure of a gas mixture dependson the total number of gas particles, not onthe types of particles.STPP = 1.00 atm P = 1.00 atm1.0 mol He0.50 mol O2+ 0.20 mol He+ 0.30 mol N2
  78. 78. Health NoteWhen a scuba diver is several hundred feetunder water, the high pressures cause N2 fromthe tank air to dissolve in the blood. If thediver rises too fast, the dissolved N2 will formbubbles in the blood, a dangerous and painfulcondition called "the bends". Helium, which isinert, less dense, and does not dissolve in theblood, is mixed with O2 in scuba tanks used fordeep descents.
  79. 79. Learning CheckA 5.00 L scuba tank contains 1.05 mole ofO2 and 0.418 mole He at 25°C. What is thepartial pressure of each gas, and what isthe total pressure in the tank?
  80. 80. Solution G20P = nRT PT = PO + PHeV 2PT = 1.47 mol x 0.0821 L-atm x 298 K5.00 L (K mol)= 7.19 atm
  81. 81. Micro Effusion
  82. 82. Macro Effusion