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# Chem 101 week 12 ch 5

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### Chem 101 week 12 ch 5

1. 1. GasesChapter 5Gases
2. 2. GasesCharacteristics of Gases• Unlike liquids and solids, they Expand to fill their containers. Are highly compressible. Have extremely low densities.
3. 3. GasesMeasurements• We have learned in our early chaptersand experienced in the lab, the meansof measurement for solids and liquids• We will begin our study of gases bylooking at the means we employ tomake measurements on gases
4. 4. GasesMeasurements involvingGases• To describe a gas we must specify thefollowing:1.Volume2. Amount3.Temperature4. Pressure
5. 5. GasesVolume of Gases• Gases expand uniformly to fill anycontainer they are placed in• Therefore, the volume of a gas is thevolume of its container• Volumes are expressed in L, cm3or m3
6. 6. GasesAmount• The amount of matter in a sample ofgas is expressed in terms of thenumber of moles(n)• Sometimes the mas in grams is given• n = g/mol mass(M)• n(M)= g
7. 7. GasesTemperature• Temperature of a gas is measured inCelsius• Calculations will require conversion tothe Kelvin scale• TK = Tcelsius + 273.15• Temperature is expressed to thenearest degree (simply add 273 tocelsius temperature)
8. 8. Gases• Pressure is theamount of forceapplied to an area.Pressure• Atmosphericpressure is theweight of air perunit of area.P =FA
9. 9. GasesUnits of Pressure• English System pounds per square inch(psi)• Pascals (SI unit) 1 Pa = 1 N/m2• Bar 1 bar = 105Pa = 100 kPa
10. 10. GasesUnits of Pressure• mm Hg or torrThese units are literallythe difference in theheights measured in mm(h) of two connectedcolumns of mercury.• Atmosphere1.00 atm = 760 torr
11. 11. GasesManometerUsed to measure thedifference in pressurebetween atmosphericpressure and that of agas in a vessel.
12. 12. GasesStandard Pressure• Normal atmospheric pressure at sealevel.• It is equal to1.00 atm760 torr (760 mm Hg)101.325 kPa
13. 13. GasesBoyle’s LawThe volume of a fixed quantity of gas atconstant temperature is inversely proportionalto the pressure.
14. 14. GasesBoyle’s Law
15. 15. GasesAs P and V areinversely proportionalA plot of V versus Presults in a curve.SinceV = k (1/P)This means a plot ofV versus 1/P will bea straight line.PV = k
16. 16. GasesCharles’s Law• The volume of a fixedamount of gas atconstant pressure isdirectly proportional to itsabsolute temperature.A plot of V versus T will be a straight line.• i.e.,VT= k
17. 17. GasesAvogadro’s Law• The volume of a gas at constant temperatureand pressure is directly proportional to thenumber of moles of the gas.• Mathematically, this means V = kn
18. 18. GasesIdeal-Gas EquationV ∝ 1/P (Boyle’s law)V ∝ T (Charles’s law)V ∝ n (Avogadro’s law)• So far we’ve seen that• Combining these, we getV ∝nTP
19. 19. GasesIdeal-Gas EquationThe constant ofproportionality isknown as R, thegas constant.
20. 20. GasesIdeal-Gas EquationThe relationshipthen becomesnTPV ∝nTPV = RorPV = nRT
21. 21. GasesIdeal-Gas Equation
22. 22. GasesDensities of GasesIf we divide both sides of the ideal-gasequation by V and by RT, we getnVPRT=
23. 23. Gases• We know thatmoles × molecular mass = massDensities of Gases• So multiplying both sides by themolecular mass (Μ ) givesn × Μ = mPΜRTmV=
24. 24. GasesDensities of Gases• Mass ÷ volume = density• So,• Note: One only needs to know themolecular mass, the pressure, and thetemperature to calculate the density ofa gas.PΜRTmV=d =
25. 25. GasesMolecular MassWe can manipulate the densityequation to enable us to find themolecular mass of a gas:BecomesPΜRTd =dRTPΜ =
26. 26. GasesThe conditions 0 0C and 1 atm are called standardtemperature and pressure (STP).PV = nRTR =PVnT=(1 atm)(22.414L)(1 mol)(273.15 K)R = 0.082057 L • atm / (mol • K)5.4Experiments show that at STP, 1 mole of an ideal gasoccupies 22.414 L.
27. 27. GasesWhat is the volume (in liters) occupied by 49.8 g of HCl atSTP?PV = nRTV =nRTPT = 0 0C = 273.15 KP = 1 atmn = 49.8 g x1 mol HCl36.45 g HCl= 1.37 molV =1 atm1.37 mol x 0.0821 x 273.15 KL•atmmol•KV = 30.6 L5.4
28. 28. GasesArgon is an inert gas used in lightbulbs to retard thevaporization of the filament. A certain lightbulb containingargon at 1.20 atm and 18 0C is heated to 85 0C at constantvolume. What is the final pressure of argon in the lightbulb(in atm)?PV = nRT n, V and R are constantnRV=PT= constantP1T1P2T2=P1 = 1.20 atmT1 = 291 KP2 = ?T2 = 358 KP2 = P1 xT2T1= 1.20 atm x 358 K291 K= 1.48 atm5.4
29. 29. GasesDensity (d) Calculationsd = mV=PMRTm is the mass of the gas in gM is the molar mass of the gasMolar Mass (M ) of a Gaseous SubstancedRTPM = d is the density of the gas in g/L5.4
30. 30. GasesA 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and27.0 0C. What is the molar mass of the gas?5.4dRTPM = d = mV4.65 g2.10 L= = 2.21gLM =2.21gL1 atmx 0.0821 x 300.15 KL•atmmol•KM = 54.6 g/mol
31. 31. GasesGas StoichiometryWhat is the volume of CO2 produced at 37 0C and 1.00 atmwhen 5.60 g of glucose are used up in the reaction:C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)g C6H12O6 mol C6H12O6 mol CO2 V CO25.60 g C6H12O61 mol C6H12O6180 g C6H12O6x6 mol CO21 mol C6H12O6x = 0.187 mol CO2V =nRTP0.187 mol x 0.0821 x 310.15 KL•atmmol•K1.00 atm= = 4.76 L5.5
32. 32. GasesDalton’s Law ofPartial Pressures• The total pressure of a mixture of gasesequals the sum of the pressures thateach would exert if it were presentalone.• In other words,Ptotal = P1 + P2 + P3 + …
33. 33. GasesDalton’s Law of Partial PressuresV and TareconstantP1 P2 Ptotal = P1 + P25.6
34. 34. GasesConsider a case in which two gases, A and B, are in a container ofvolume V.PA =nARTVPB =nBRTVnA is the number of moles of AnB is the number of moles of BPT = PA + PB XA =nAnA + nBXB =nBnA + nBPA = XA PT PB = XB PTPi = Xi PT5.6mole fraction (Xi) =ninT
35. 35. GasesA sample of natural gas contains 8.24 moles of CH4, 0.421moles of C2H6, and 0.116 moles of C3H8. If the totalpressure of the gases is 1.37 atm, what is the partial pressureof propane (C3H8)?Pi = Xi PTXpropane =0.1168.24 + 0.421 + 0.116PT = 1.37 atm= 0.0132Ppropane = 0.0132 x 1.37 atm = 0.0181 atm5.6
36. 36. GasesPartial Pressures• When one collects a gas over water, there iswater vapor mixed in with the gas.• To find only the pressure of the desired gas,one must subtract the vapor pressure ofwater from the total pressure.
37. 37. Gases2KClO3 (s) 2KCl (s) + 3O2 (g)Bottle full of oxygen gasand water vaporPT = PO + PH O2 2 5.6
38. 38. Gases5.6
39. 39. GasesChemistry in Action:Scuba Diving and the Gas LawsP VDepth (ft) Pressure(atm)0 133 266 35.6
40. 40. GasesKinetic-Molecular TheoryThis is a model thataids in ourunderstanding of whathappens to gasparticles asenvironmentalconditions change.
41. 41. GasesKinetic Molecular Theory of Gases1. A gas is composed of molecules that are separated from eachother by distances far greater than their own dimensions. Themolecules can be considered to be points; that is, they possessmass but have negligible volume.2. Gas molecules are in constant motion in random directions, andthey frequently collide with one another. Collisions amongmolecules are perfectly elastic.3. Gas molecules exert neither attractive nor repulsive forces on oneanother.4. The average kinetic energy of the molecules is proportional to thetemperature of the gas in kelvins. Any two gases at the sametemperature will have the same average kinetic energy5.7KE = ½ mu2
42. 42. GasesKinetic theory of gases and …• Compressibility of Gases• Boyle’s LawP α collision rate with wallCollision rate α number densityNumber density α 1/VP α 1/V• Charles’ LawP α collision rate with wallCollision rate α average kinetic energy of gas moleculesAverage kinetic energy α TP α T5.7
43. 43. GasesKinetic theory of gases and …• Avogadro’s LawP α collision rate with wallCollision rate α number densityNumber density α nP α n• Dalton’s Law of Partial PressuresMolecules do not attract or repel one anotherP exerted by one type of molecule is unaffected by thepresence of another gasPtotal = ΣPi5.7
44. 44. GasesMain Tenets of Kinetic-Molecular TheoryThe average kineticenergy of themolecules isproportional to theabsolutetemperature.
45. 45. GasesApparatus for studying molecular speed distribution5.7
46. 46. GasesThe distribution of speedsfor nitrogen gas moleculesat three different temperaturesThe distribution of speedsof three different gasesat the same temperature5.7urms = 3RTM√
47. 47. GasesDiffusionThe spread of onesubstancethroughout a spaceor throughout asecond substance.
48. 48. GasesGas diffusion is the gradual mixing of molecules of one gas withmolecules of another by virtue of their kinetic properties.5.7NH317 g/molHCl36 g/molNH4Clr1r2M2M1√=
49. 49. GasesEffusionThe escape ofgas moleculesthrough a tinyhole into anevacuatedspace.
50. 50. GasesGas effusion is the is the process by which gas under pressureescapes from one compartment of a container to another by passingthrough a small opening.5.7r1r2t2t1M2M1√= =Nickel forms a gaseous compound of the formula Ni(CO)xWhat is the value of x given that under the same conditionsmethane (CH4) effuses 3.3 times faster than the compound?r1 = 3.3 x r2M1 = 16 g/molM2 =r1r2( )2x M1 = (3.3)2x 16 = 174.258.7 + x • 28 = 174.2 x = 4.1 ~ 4
51. 51. Gases
52. 52. Gases
53. 53. GasesReal GasesIn the real world, thebehavior of gasesonly conforms to theideal-gas equationat relatively hightemperature and lowpressure.
54. 54. GasesDeviations from Ideal BehaviorThe assumptions made in the kinetic-molecularmodel break down at high pressure and/or lowtemperature.
55. 55. GasesDeviations from Ideal Behavior1 mole of ideal gasPV = nRTn =PVRT= 1.05.8Repulsive ForcesAttractive Forces
56. 56. GasesEffect of intermolecular forces on the pressure exerted by a gas.5.8
57. 57. GasesCorrections for NonidealBehavior• The ideal-gas equation can be adjustedto take these deviations from idealbehavior into account.• The corrected ideal-gas equation isknown as the van der Waals equation.
58. 58. Gases5.8Van der Waals equationnonideal gasP + (V – nb) = nRTan2V2( )}correctedpressure}correctedvolume
59. 59. Gases
60. 60. Gases