2 Chapter 5: Gases 5.1 Early Experiments The gas laws of Boyle, Charles, and AvogadroThe Ideal Gas LawGas StiochiometryDalton’s Law of Partial PressuresThe Kinetic molecular Theory of GasesEffusion and DiffusionCollisions of Gas Particles with the Container WallsIntermolecular CollisionsReal GasesChemistry in the Atmosphere
3 Hurricanes, such as this one off the coast of Florida, are evidence of the powerful forces present in the earth's atmosphere.
4 Important Characteristics of Gases 1) 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 low 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.
5 Substances that are Gases under Normal Conditions Substance Formula MM(g/mol)Helium HeNeon NeArgon ArHydrogen HNitrogen NNitrogen Monoxide NOOxygen OHydrogen Chloride HCLOzone OAmmonia NHMethane CH
6 Some Important Industrial Gases Name - Formula Origin and useMethane (CH4) Natural deposits; domestic fuelAmmonia (NH3) From N2 + H2 ; fertilizers, explosivesChlorine (Cl2) Electrolysis of seawater; bleachingand disinfectingOxygen (O2) Liquified air; steelmakingEthylene (C2H4) High-temperature decomposition ofnatural gas; plastics
8 Pressure of the Atmosphere Called “Atmospheric pressure,” or the force exerted upon us by the atmosphere above us.A measure of the weight of the atmosphere pressing down upon us.Measured using a Barometer! - A device that can weigh the atmosphere above us!ForceAreaPressure =
13 Construct a Barometer using Water! Density of water = 1.00 g/cm3Density of Mercury = 13.6 g/cm3Height of water column = HwHw = Height of Hg x Density of MercuryHw = 760 mm Hg x 13.6/1.00 = 1.03 x 104 mmHw = 10.3 m = __________ ftHeightWaterHeightMercuryDensityMercuryDensityWater=Density of Water
15 Common Units of Pressure Unit Atmospheric Pressure Scientific Field UsedPascal (Pa); x 105 Pa SI unit; physics,kilopascal (kPa) kPa chemistryAtmosphere (atm) atm ChemistryMillimeters of mercury mmHg Chemistry, medicine(mmHg) biologyTorr torr ChemistryPounds per square inch lb/in Engineering(psl or lb/in2)Bar bar Meteorology, chemistry
16 Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from thedecomposition of Lime stone (CaCO3) in a closed end manometer, theheight of the mercury is 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:converting from mmHg to torr:1 torr1 mm HgPCO2 (torr) = mm Hg x = torrconverting from torr to atm:1 atm760 torrPCO2( atm) = torr x = atmconverting from atm to kPa:kPa1 atmPCO2(kPa) = atm x = _____________ kPa
17 Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container.
18 Figure 5.3: A J-tube similar to the one used by Boyle.
19 Boyle’s Law : P - V relationship Pressure is inversely proportional to VolumeP = or V = or PV=kChange of Conditions Problemsif n and T are constant !P1V1 = k P2V2 = k’k = k’Then :P1V1 = P2V2kVkP
21 Figure 5.4: Plotting Boyle’s data from Table 5.1.
22 Figure 5.5: Plot of PV versus P for several gases.
23 Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of15.8 cm3, what will be the volume if the pressure is increased to3.16 atm?Plan: We begin by converting the volume that is in cm3 to ml and thento liters, then we do the pressure change to obtain the final volume!Solution:V1 (cm3)P1 = 1.23 atm P2 = 3.16 atmV1 = 15.8 cm V2 = unknownT and n remain constant1cm3 = 1 mLV1 (ml)1 mL1 cm31 L1000mL1000mL = 1LV1 = 15.8 cm3 x x = LV1 (L)x P1/P2P1P21.23 atm3.16 atmV2 = V1 x = L x = ________ LV2 (L)
24 Boyle’s Law - A gas bubble in the ocean! A bubble of gas is released by the submarine “Alvin” at a depth of6000 ft in the ocean, as part of a research expedition to study underwater volcanism. Assume that the ocean is isothermal( the sametemperature through out ),a gas bubble is released that had an initialvolume of 1.00 cm3, what size will it be at the surface at a pressure of1.00 atm?(We will assume that the density of sea water is g/cm3,and use the mass of Hg in a barometer for comparison!)Initial ConditionsFinal ConditionsV 1 = 1.00 cm3V 2 = ?P 1 = ?P 2 = 1.00 atm
25 Calculation Continued m1 ft100 cm1 m1.026 g SH2O1 cm3Pressure at depth = 6 x 103 ft x x xPressure at depth = 187, g pressure from SH2OFor a Mercury Barometer: 760 mm Hg = 1.00 atm, assume that thecross-section of the barometer column is 1 cm2.The mass of Mercury in a barometer is:1.00 atm760 mm Hg10 mm1 cmArea1 cm21.00 cm3 Hg13.6 g HgPressure = x x x x187,635 g =Pressure = _____ atm Due to the added atmospheric pressure = ___ atm!V1 x P1P21.00 cm3 x ____ atm1.00 atmV2 = = = ____ cm3 = liters
26 Boyle’s Law : Balloon A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant(which obviously is not true), what is the final volume of the balloon?P1 = 1.0 atm P2 = 0.40 atmV1 = 0.55 L V2 = ?V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm)V2 = __________ L
27 Figure 5.6: PV plot verses P for 1 mole of ammonia
31 Charles Law - V - T- relationship Temperature is directly related to volumeT proportional to Volume : T = kVChange of conditions problem:Since T/V = k or T1 / V1 = T2 / V or:T1 = V1 xT2V2T1V1T2=V2Temperatures must be expressed in Kelvin to avoid negative values!
34 Figure 5.7: Plots of V versus T (c) for several gases.
35 Figure 5. 8: Plots of V versus T as in Fig. 5 Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.
36 Charles Law ProblemA sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 oC. Calculate the temperature (oC) at which the gas will occupy 1.54 L if the pressure remains constant.V1 = 3.20 L T1 = 125oC = 398 KV2 = 1.54 L T2 = ?T2 = T1 x ( V2 / V1) T2 = 398 K x =______KT2 = ___ K oC = K = ______ - 273oC = ______oC1.54 L3.20 L
37 Charles Law Problem - IA balloon in Antarctica in a building is at room temperature ( 75o F ) and has a volume of 20.0 L . What will be its volume outside where the temperature is -70oF ?V1 = 20.0 L V2 = ?T1 = 75o F T2 = -70o Fo C = ( o F - 32 ) 5/9T1 = ( )5/9 = 23.9o CK = 23.9o C = __________ KT2 = ( ) 5/9 = o CK = o C = ___________ K
38 Antarctic Balloon Problem - II V1 / T1 = V2 / T V2 = V1 x ( T2 / T1 )V2 = 20.0 L xV2 = __________ LThe Balloon shrinks from 20 L to 15 L !!!!!!!Just by going outside !!!!!216.4 K297.0 K
40 Applying the Temperature - Pressure Relationship (Amonton’s Law) Problem: A copper tank is compressed to a pressure of 4.28 atm at atemperature of oF. What will be the pressure if the temperature israised to 95.6 oC?Plan: The volume of the tank is not changed, and we only have to dealwith the temperature change, and the pressure, so convert to SI units,and calculate the change in pressure from the Temp.and Pressure change.Solution:P P2T T2=T1 = (0.185 oF oF)x 5/9 = oCT1 = oC K = KT2 = oC K = KP2 = P1 x = ?T2T1368.8 KKP2 = 4.28 atm x = _______ atm
42 Avogadro’s Law - Amount and Volume The Amount of Gas (Moles) is directly proportional to the volumeof the Gas.n V orn = kVFor a change of conditions problem we have the initial conditions,and the final conditions, and we must have the units the same.n1 = initial moles of gas V1 = initial volume of gasn2 = final moles of gas V2 = final volume of gasn1V1n2V2V1V2=or: n1 = n2 x
43 Avogadro’s Law: Volume and Amount of Gas Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes inthe atmosphere, if the volume of 2.67 g of SF6 at atm and 28.5 oCis 2.93 m3, what will be the mass of SF6 in a container whose volume is543.9 m3 at atm and 28.5 oC?Plan: Since the temperature and pressure are the same it is a V - nproblem, so we can use Avogadro’s Law to calculate the moles of thegas, then use the molecular mass to calculate the mass of the gas.Solution: Molar mass SF6 = g/mol2.67g SF6146.07g SF6/mol= mol SF6V2V1n2 = n1 x = mol SF6 x = 3.40 mol SF6543.9 m32.93 m3mass SF6 = 3.40 mol SF6 x g SF6 / mol = __________ g SF6
44 Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H2) of Hydrogen andhas a volume of L. What mass of Hydrogen must be added to theballoon to increase it’s volume to Liters. Assume T and P areconstant.Plan: Volume and amount of gas are changing with T and P constant, sowe will use Avogadro’s law, and the change of conditions form.Solution:V1 = LV2 = LT = constantP = constantn1 = 1.14 moles H2n2 = 1.14 moles + ? molesn n2V V2V2V1L28.75 L=n2 = n1 x = 1.14 moles H2 xmass = moles x molecular massmass = 4.46 moles x g / molmass = ___________ g H2 gasn2 = moles = 4.46 molesadded mass = 8.99g g = 6.69g
45 Change of Conditions, with no change in the amount of gas ! = constant Therefore for a changeof conditions :T T2P x VTP1 x V1P2 x V2=
46 Change of Conditions :Problem -I A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure?P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atmP2 = ?V1 = 45.9 L V2 = 3.10 ml = LT1 = 25 oC = 298 KT2 = 155 oC = 428 K
47 Change of Condition Problem I : continued P1 x V1P2 x V2=T1T2P2 ( L)( atm) ( 45.9 L)=( 298 K)( 428 K)( 428 K) ( atm) ( 45.9 L)P2 == ________ atm( 298 K) ( L)
48 Change of Conditions : Problem II A weather balloon is released at the surface of the earth. If the volume was 100 m3 at the surface ( T = 25 oC, P = 1 atm ) what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 oC and the pressure is 15 mm Hg ?Initial Conditions Final ConditionsV1 = 100 m V2 = ?T1 = 25 oC T2 = -90 oC= 298 K = 183 KP1 = 1.0 atm P2 = 15 mm Hg760 mm Hg/ atmP2= ___________ atm
49 Change of Conditions Problem II: continued P1 x V P2 x V2V2 =V2 = =V2 = m3 = __________ m3 or 3.1 times thevolume !!!P1V1T2=T1T2T1P2( 1.0 atm) ( 100 m3) (183 K)(298 K) (0.198 atm)
50 Change of Conditions :Problem III How many liters of CO2 are formed at 1.00 atm and 900 oC if 5.00 L of Propane at 10.0 atm, and 25 oC is burned in excess air?C3H8 (g) + 5 O2 (g) = 3 CO2 (g) + 4 H2O(g)25 oC = 298 K900 oC = 1173 K
51 Change of Conditions Problem III:continued V1 = 5.00 L V2 = ?P1 = 10.0 atm P2 = 1.00 atmT1 = 298K T2 = 1173 KP1V1/T1 = P2V2/T V2 = V1P1T2/ P2T1V2 = = 197 LVCO2 = (197 L C3H8) x (3 L CO2 / 1 L C3H8) =VCO2 = _________ L CO2( 5.00 L) (10.00 atm) (1173 K)( 1.00 atm) ( 298 K)
52 Standard Temperature and Pressure (STP) A set of Standard conditions have been chosen to make it easier tounderstand the gas laws, and gas behavior.Standard Temperature = 00 C = KStandard Pressure = 1 atmosphere = 760 mm MercuryAt these “standard” conditions, if you have 1.0 mole of a gas it willoccupy a “standard molar volume”.Standard Molar Volume = Liters = 22.4 L
55 Table 5.2 (P 149) Molar Volumes for Various Gases at 00 and 1 atm Gas Molar Volume (L)Oxygen (O2)Nitrogen (N2)Hydrogen (H2)Helium (He)Argon (Ar)Carbon dioxide (CO2)Ammonia (NH3)
56 IDEAL GASESAn ideal gas is defined as one for which both the volume of molecules and forces between the molecules are so small that they have no effect on the behavior of the gas.The ideal gas equation is:PV=nRTR = Ideal gas constantR = J / mol K = J mol-1 K-1R = l atm mol-1 K-1
57 Variations on the Gas Equation During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed.Thus, PV=nRT can be rearranged for the fixed variables:for a fixed amount at constant temperatureP V = nRT = constant Boyle’s Lawfor a fixed amount at constant volumeP / T = nR / V = constant Amonton’s Lawfor a fixed amount at constant pressureV / T = nR / P = constant Charles’ Lawfor a fixed volume and temperatureP / n = R T / V = constant Avogadro’s Law
59 Evaluation of the Ideal Gas Constant, R PVnTIdeal gas Equation PV = nRT R =at Standard Temperature and Pressure, the molar volume = 22.4 LP = 1.00 atm (by Definition!)T = 0 oC = K (by Definition!)n = 1.00 moles (by Definition!)(1.00 atm) ( L)( 1.00 mole) ( K)L atmmol KR = =L atmmol Kor to three significant figures R =
60 Values of R in Different Units Atm x LMol x KR* =R = 62.36R = 8.314Torr x LMol x KkPa x dm3Mol x KJ#Mol x K* Most calculations in this text use values of R to 3 significant figures.# J is the abbreviation for joule, the SI unit of energy. The joule is aderived unit composed of the base units kg x m2/s2
61 Gas Law: Solving for Pressure Problem: Calculate the pressure in a container whose Volume is 87.5 Land it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.Plan: Convert all information into the units required, and substitute intothe Ideal Gas equation ( PV=nRT ).Solution:5038 g Xe131.3 g Xe / molnXe = = mol XeT = 18.8 oC K = KPV = nRT P = nRTV(38.37 mol )( L atm)( K)P = = _______ atm = atm87.5 L(mol K)
63 Ideal Gas Calculation - Nitrogen Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is m3 and the temperature is 36.0 oC.n = 375 g N2/ 28.0 g N2 / mol = 13.4 mol N2V = m3 x 1000 L / m3 = 150 LT = 36.0 oC = KPV=nRT P= nRT/VP =P = atm( 13.4 mol) ( L atm/mol K) ( K)150 L
64 Mass of Air in a Hot Air Balloon - Part I Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 oF and the pressure is 748 mm Hg?P = 748 mm Hg x 1atm / 760 mm Hg= atmV = 1.41 x 104 ft3x (12 in/1 ft)3x(2.54 cm/1 in)3 xx (1ml/1 cm3) x ( 1L / 1000 cm3) =3.99 x 105 LT = oC =(86-32)5/9 = 30 oC30 oC = 303 K
65 Mass of Air in a Hot Air Balloon - Part II PV = nRT n = PV / RTn = = 1.58 x 104 molmass = 1.58 x 104 mol air x 29 g air/mol air= 4.58 x 105 g Air= 458 Kg Air( atm) ( 3.99 x 105 L)( L atm/mol K) ( 303 K )
67 Sodium Azide Decomposition-I Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 .2 NaN3 (s) Na (s) + 3 N2 (g)mol NaN3 = 60.0 g NaN3 / g NaN3 / mol == mol NaN3mol N2= mol NaN3 x 3 mol N2/2 mol NaN3= ___________mol N2
68 Sodium Azide Calc: - II PV = nRT V = nRT/P V = V = __________ liters( 1.38 mol) ( L atm / mol K) (294 K)( 823 mm Hg / 760 mmHg / atm )
69 Ammonia Density Problem Calculate the Density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC.Density = mass per unit volume = g / LP = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atmT = 55 oC = 328 Kn = mass / Molar mass = g / Md = =d = __________ g / LP x MR x T( atm) ( g/mol)( L atm/mol K) ( 328 K)
70 Like Example 5.6 (P 149)Problem: Calculate the volume of carbon dioxide produced by thethermal decomposition of 36.8 g of calcium carbonate at 715mm Hg,and 370 C according to the reaction below:CaCO3(s) CaO(s) + CO2 (g)Solution:1 mol CaCO3100.1g CaCO336.8g x = mol CaCO3Since each mole of CaCO3 produces one 1 mole of CO2 , molesof CO2 at STP, we now have to convert to the conditions stated:nRTP(0.368 mol)( L atm/mol K)(310.15K)(715mmHg/760mm Hg/atm)V = =V = __________ L
71 Calculation of Molar Mass P x VMassR x TMolar MassMass x R x TMolar Mass = = MMP x V
73 Dumas Method of Molar Mass Problem: A volatile liquid is placed in a flask whose volume is mland allowed to boil until all of the liquid is gone, and only vapor fills theflask at a temperature of oC and 736 mm Hg pressure. If the massof the flask before and after the experiment was g and g,what is the molar mass of the liquid?Plan: Use the gas law to calculate the molar mass of the liquid.Solution:1 atm760 mm HgPressure = 736 mm Hg x = atmmass = g g = g(1.082 g)( Latm/mol K)(373.2 K)Molar Mass = = g/mol( atm)(0.590 L)note: the compound is acetone C3H6O = MM = 58g mol !
74 Calculation of Molecular Weight of a Gas Natural Gas - Methane Problem: A sample of natural gas is collected at 25.0 oC in a mlflask. If the sample had a mass of g at a pressure of Torr,what is the molecular weight of the gas?Plan: Use the Ideal gas law to calculate n, then calculate the molar mass.Solution:1mm Hg1 Torr1.00 atm760 mm HgP = Torr x x = atm1.00 L1000 mlV = ml x = LP VR Tn =T = 25.0 oC K = K(0.724 atm)(0.250 L)n = = mol( L atm/mol K) (298.2 K)MM = g / mol = ____________ g/mol
75 Gas MixturesGas behavior depends on the number, not the identity, of gas molecules.Ideal gas equation applies to each gas individually and the mixture as a whole.All molecules in a sample of an ideal gas behave exactly the same way.
76 Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles of that gas.
77 Dalton’s Law of Partial Pressures - I Definiton: In a mixture of gases, each gas contributes to the total pressure the amount it would exert if the gas were present in the container by itself.To obtain a total pressure, add all of the partial pressures: P total = p1+p2+p3+...pi
78 Dalton’s Law of Partial Pressure - II Pressure exerted by an ideal gas mixture is determined by the total number of moles:P=(ntotal RT)/Vn total = sum of the amounts of each gas pressurethe partial pressure is the pressure of gas if it was present by itself.P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ...the total pressure is the sum of the partial pressures.
79 Dalton’s Law of Partial Pressures - Prob #1 A 2.00 L flask contains 3.00 g of CO2 and g of Helium at a temperature of 17.0 oC.What are the Partial Pressures of each gas, and the total Pressure?T = 17 oC = 290 KnCO2 = 3.00 g CO2/ g CO2 / mol CO2= mol CO2PCO2 = nCO2RT/VPCO2 =PCO2 = _____________ atm( mol CO2) ( L atm/mol K) ( 290 K)(2.00 L)
80 Dalton’s Law Problem - #1 cont. nHe = 0.10 g He / g He / mol He= mol HePHe = nHeRT/VPHe =PHe = 0.30 atmPTotal = PCO2 + PHe = atm atmPTotal = 1.11 atm(0.025 mol) ( L atm / mol K) ( 290 K )( 2.00 L )
81 Dalton’s Law Problem #2using mole fractions A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ?Total # moles = = 7.35 molXNe = 4.46 mol Ne / 7.35 mol = 0.607PNe = XNe PTotal = ( 2.00 atm) = 1.21 atm for NeXAr = 0.74 mol Ar / 7.35 mol = 0.10PAr = XAr PTotal = 0.10 (2.00 atm) = 0.20 atm for ArXXe = 2.15 mol Xe / 7.35 mol = 0.293PXe = XXe PTotal = (2.00 atm) = atm for Xe
82 Relative Humidity Pressure of Water in Air Rel Hum = x 100% Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the Relative Humidity?Rel Hum =(6.54 mm Hg/ mm Hg )x100%= ___________ %Pressure of Water in AirMaximum Vapor Pressure of Water
84 Figure 5.10: Production of oxygen by thermal decomposition of KCIO3.
85 Molecular sieve framework of titanium (blue), silicon (green), and oxygen (red) atoms contract on heating–at room temperature (left) d = 4.27 Å; at 250°C (right) d = 3.94 Å.
86 Collection of Hydrogen gas over Water - Vapor pressure - I 2 HCl(aq) + Zn(s) ZnCl2 (aq) + H2 (g)Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg.PTotal = P H2 + PH2O PH2 = PTotal - PH2OPH2 = 769 mm Hg mm Hg= 752 mm HgT = 20oC = 273 = 293 KP = 752 mm Hg /760 mm Hg /1 atm = atmV =___________ L
87 COLLECTION OVER WATER CONT.-II PV = nRT n = PV / RTn =n = molmass = mol x 2.01 g H2 / mol H2mass = ______________ g Hydrogen(0.989 atm) (0.156 L)( L atm/mol K) (293 K)
88 Chemical Equation Calc - III MassAtoms (Molecules)MolecularWeightAvogadro’sNumberg/mol6.02 x 1023MoleculesReactantsProductsMolesMolarityPV = nRTmoles / literSolutionsGases
89 Gas Law Stoichiometry - I - NH3 + HCl Problem: A slide separating two containers is removed, and the gasesare allowed to mix and react. The first container with a volume of 2.79 Lcontains Ammonia gas at a pressure of atm and a temperature of18.7 oC. The second with a volume of 1.16 L contains HCl gas at apressure of atm and a temperature of 18.7 oC. What mass of solidammonium chloride will be formed, and what will be remaining in thecontainer, and what is the pressure?Plan: This is a limiting reactant problem, so we must calculate the molesof each reactant using the gas law to determine the limiting reagent. Thenwe can calculate the mass of product, and determine what is left in thecombined volume of the container, and the conditions.Solution:Equation: NH3 (g) + HCl(g) NH4Cl(s)TNH3 = 18.7 oC = K
90 Gas Law Stoichiometry - II - NH3 + HCl PVn =RT(0.776 atm) (2.79 L)nNH3 = = mol NH3( L atm/mol K) (291.9 K)limiting(0.932 atm) (1.16 L)nHCl = = mol HCl( L atm/mol K) (291.9 K)reactantTherefore the product will be mol NH4Cl or g NH4ClAmmonia remaining = mol mol = mol NH3V = 1.16 L L = 3.95 LnRT( mol) ( L atm/mol K) (291.9 K)PNH3 = = = ____ atmV(3.95 L)
91 Postulates of Kinetic Molecular Theory I The particles are so small compared with the distances betweenthem that the volume of the individual particles can be assumedto be negligible (zero).II The particles are in constant motion. The collisions of the particleswith the walls of the container are the cause of the pressureexerted by the gas.III The particles are assumed to exert no forces on each other;they are assumed to neither attract nor repel each other.IV The average kinetic energy of a collection of gas particles isassumed to be directly proportional to the Kelvin temperature ofthe gas.
92 Figure 5.11: An ideal gas particle in a cube wholse sides are of length L (in meters).
93 Figure 5.12: (a) The Cartesian coordinate axes.
94 Figure 5.12: (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, ux, uy, and uz.
95 Figure 5.12: (c) In the xy plane, ux2 + uy2 = uyx2 by the Pythagorean theorem.
96 Figure 5.13: (a) Only the x component of the gas particle’s velocity affects the frequency of impacts on the shaded walls. (b) For an elastic collision, there is an exact reversal of the x component of the velocity and of the total velocity.
102 Velocity and Energy Kinetic Energy = 1/2mu2 Average Kinetic Energy (KEavg)add up all the individual molecular energies and divide by the total number of molecules!The result depends on the temp. of the gasKE avg=3RT/2NaT=temp. in Kelvin , Na =Avogadro’s number, R=new quantity (gas constant)kEtotal = (No. of molecules)(KEavg) = (NA)(3RT/2NA) = 3/2RTSo, 1 mol of any gas has a total molecular Kinetic Energy = KE of 3/2RT!!!
103 Figure 5.15: A plot of the relative number of O2 molecules that have a given velocity at STP.
104 Figure 5.16: A plot of the relative number of N2 molecules that have a given velocity at three temperatures.
105 Figure 5.17: A velocity distribution for nitrogen gas at 273 K.
107 Molecular Mass and Molecular Speeds - I Problem: Calculate the molecular speeds of the molecules of Hydrogen,Methane, and carbon dioxide at 300K!Plan: Use the equations for the average kinetic energy of a molecule,and the relationship between kinetic energy and velocity to calculate theaverage speeds of the three molecules.Solution: for Hydrogen, H2 = g/mol3 R2 NA8.314 J/mol KEK = x T = 1.5 x x 300K =6.022 x 1023 molecules/molEK = x J/molecule = 1/2 mu22.016 x kg/mole6.022 x 1023 molecules/molem = = x kg/molecule6.213 x J/molecule = x kg/molecule(u2)u = _______________ m/s = ___________ m/s
108 Molecular Mass and Molecular Speeds - II for methane CH4 = g/mole3 R2 NA8.314 J/mol K6.022 x 1023 molecules/moleEK = x T = 1.5 x x 300K =EK = x J/molecule = 1/2 mu216.04 x kg/mole6.022 x 1023 moleules/molem = = x kg/molecule6.213 x J/molecule = x kg/molecule (u2)u = ___________________ m/s = __________ m/s
109 Molecular Mass and Molecular Speeds - III for Carbon dioxide CO2 = g/moleEK = x T = 1.5 x x 300K =3 R2 NA8.314 J/mol K6.022 x 1023 molecules/moleEK = x J/molecule = 1/2mu244.01 x kg/mole6.022 x 1023 molecules/molem = = x kg/molecule6.213 x J/molecule = x kg/molecule (u2)u = ___________________ m/s = _________ m/s
110 Molecular Mass and Molecular Speeds - IV Molecule H CH CO2MolecularMass (g/mol)Kinetic Energy(J/molecule)6.213 x x xVelocity(m/s)1,
111 Important Point !At a given temperature, all gases have the same molecular Kinetic energy distributions.orThe same average molecular Kinetic Energy!
112 Diffusion vs EffusionDiffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules.Effusion - A gas escaping from a container into a vacuum. There are no other (or few ) for collisions.
113 Figure 5.18: The effusion of a gas into an evacuated chamber.
123 Figure 5.20: Uranium-enrichment converters from the Paducah gaseous diffusion plant in Kentucky.
124 Gaseous Diffusion Separation of Uranium 235 / 238 235UF6 vs 238UF6Separation Factor = S =after Two runs S =after approximately runs235UF6 is > 99% Purity !!!!!Y Plant at Oak Ridge National Lab( (6 x 19))0.5( (6 x 19))0.5
125 Example 5.9 (P167)Calculate the impact rate on a 1.00-cm2 section of a vessel containingoxygen gas at a pressure of 1.00 atm and 270C.Solution:N RTV 2pMCalculate ZA : ZA = AA = 1.00 x 10-4 m21.00 atmN PV RT( )L atmK mol= =(300. K)NVMolLmoleculesmol1000 Lm3= x x x x == 2.44 x 1025 molecules/m31 kg1000gM = 32.0 g/mol x = 3.2 x 10-2 kg/molJK mol8.3145(300. K)ZA = (1.00 x 10-4 m2) (2.44 x 1025 m-3) x2(3.14)( 3.20 x 10-2 kg/mol)ZA = 2.72 x 1023 collisions/s
126 Figure 5.21: The cylinder swept out by a gas particle of diameter d.
127 Like Example 5.10(P 169)Problem: Calculate the collision frequency for a Nitrogen molecule ina sample of pure nitrogen gas at 280C and 2.0 atm. Assume that thediameter of an N2 molecule is 290 pm.Solution:2.0 atmnVPRT= = = x 10-2 mol/L( )L atmK mol(301. K)NV= (8.097 x 10-2 mol/L)( x 1023 molecules/mol)(1000L/m3)NV= x 1025 molecules/m3d= 290 pm = 2.90 x mAlso, for N2, M = 2.80 x 10-2 kg/molp( J K-1mol-1)(301 K)Z = 4(4.876 x 1025 m-3)(2.90 x m)2 xZ = collisions/sec2.80 x 10-2 kg/mol
128 Like Example 5.11 (P171)Problem: Calculate the mean free path in a sample of nitrogen gasat 280 C and 2.00 atm.Solution: Using data from the previous example, we have:1l =2 (N/V)(pd2)1l = = 5.49 x 10-82 ( x 1025 )(p)(2.90 x )2Note that a nitrogen molecule only travels 0.5 nano meters beforeit collides with another nitrogen atom.
129 Figure 5.22: Plots of PV/nRT versus P for several gases (200 K).
130 Figure 5.23: Plots of PV/nRT versus P for nitrogen gas at three temperatures.
131 Figure 5.24: (a) gas at low concentration - relatively few interactions between particles (b) gas at high concentration - many more interactions between particles.
132 Figure 5.25: Illustration of pairwise interactions among gas particles.
133 ( ) ( ) Table 5.3 (P172) Values of Van der Waals Constants for some Common Gases( )( )atm L2mol2LmolGas a bHeNeArKrXeHNOClCOCHNHH2O
134 Figure 5.26: The volume occupied by the gas particles themselves is less important at (a) large container volumes (low pressure) than at (b) small container volumes (high pressure).
135 Van der Waals Calculation of a Real gas Problem: A tank of 20.0 liters contains Chlorine gas at a temperatureof C at a pressure of atm. if the tank is pressureized to a newvolume of L and a temperature of C. What is the newpressure using the ideal gas equation, and the Van der Waals equation?Plan: Do the calculations!Solution:PV (2.000 atm)(20.0L)RT ( Latm/molK)( K)n = = = molnRT (1.663 mol) ( Latm/molK) ( K)V (1.000 L)P = = = atmnRT n2a (1.663 mol) ( Latm/molK)( K)(V-nb) V (1.00 L) - (1.663 mol)(0.0562)P = =(1.663 mol)2(6.49)(1.00 L)2= = atm
136 Table 5.4(P173) Atmospheric Composition near Sea Level (dry air)# Component Mole FractionNOArCONeHeCHKrHNOXe# The atmosphere contains various amounts of water vapor, depending on conditions.
137 Figure 5.27: The variation of temperature and pressure with altitude.
138 Severe Atmospheric Environmental Problems we must deal with in the next period of time:Urban Pollution – Photochemical SmogAcid RainGreenhouse EffectStratospheric Ozone Destruction
139 Figure 5.28: Concentration (in molecules per million molecules of “air”) of some smog components versus time of day.
140 Figure 5.29: Our various modes of transportation produce large amounts of nitrogen oxides, which facilitate the formation of photochemical smog.
141 This photo was taken in 1990. Recent renovation has since replaced the deteriorating marble.
142 Figure 5.31: Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.
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