1Chem 101Fall 2004Gases and Kinetic-Molecular Theory:Chapter 12Chem 101Fall 2004Chapter Outline• Comparison of Solids, Liquids, and Gases• Composition of the Atmosphere and Some CommonProperties of Gases•Pressure• Boyle’s Law: The Volume-Pressure Relationship• Charles’ Law: The Volume-Temperature Relationship;The Absolute Temperature Scale• Standard Temperature and Pressure• The Combined Gas Law EquationChem 101Fall 2004Chapter Outline• Avogadro’s Law and the Standard Molar Volume• Summary of Gas Laws: The Ideal Gas Equation• Determination of Molecular Weights and MolecularFormulas of Gaseous Substances• Dalton’s Law of Partial Pressures• Mass-Volume Relationships in Reactions Involving Gases• The Kinetic-Molecular Theory• Diffusion and Effusion of Gases• Real Gases: Deviations from Ideality2Chem 101Fall 2004Summary of the Ideal Gas Laws n) fixed(for TVP TVP Law Gas Combined nV nV Law sAvogadro' TV TV Law sCharles' VP VP Law sBoyle'222111221122112211====Chem 101Fall 2004Dalton’s Law of Partial Pressures• The pressure exerted by a single component in agas mixture• “What the pressure would be if only that gas werepresent in the container”Chem 101Fall 2004Dalton’s Law of Partial Pressures.... P P P P321tot+++=tot11tot11nnX wherePX P ==total numberof molesnumber ofmoles of 1mole fraction3Chem 101Fall 2004Dalton’s Law of Partial Pressures• If the gases above are allowed to mix, what are thepartial pressures of the gases?He1.0 L0.50 atmNe1.0 L0.40 atmAr2.0 L0.20 atmChem 101Fall 2004Dalton’s Law of Partial Pressures• Each gas has fixed n, R, T• So PV = constant, or PiVi = PfVf• For helium:Pi=0.5 atm,Vi = 1 L, Vf=4L• So: Pf = PiVi /Vf = (0.5)(1)/(4)=0.125 atm• For others: PNe = 0.1 atm = PAr• Total P = sum = 0.325Chem 101Fall 2004Gas Stoichiometry• Chemical equations relate number of moles, asalways!• Gas law relates number of moles to P, V, T• Allows us to find amount of gas produced orconsumed in a reaction in terms of P, V, T.4Chem 101Fall 2004Gas Stoichiometry• If we burn 10 L of ammonia with 15 L of oxygenat 600 oC what volume of nitric oxide can weform?4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)• (Assume the gases are at the same temperature,600 oC, and 1 atm pressure)Chem 101Fall 2004Gas Stoichiometry: Air Bags• Automobile air bag systems involve chemicalgeneration of gas to fill bag.• Some requirements:• Rapid generation of gas(bag fills in 30 ms)• “Safe” gas: cool, non-toxic, etc.• Easily triggered in response to impactChem 101Fall 2004Air Bags: Chemistry• Decomposition of sodium azide (NaN3) togive sodium and nitrogen• Reaction triggered on impact by a spark•Problems?Sodium product.Also, NaN3 is very toxic2 NaN3(s) → 2 Na(s) + 3 N2(g)5Chem 101Fall 2004Gas Stoichiometry: Air Bags• Estimate the mass of solid sodium azide needed toproduce enough N2 to fill a typical air bag• Typical air bag is about 67 liters, and inflates to amaximum P of 1.3 atm (then rapid deflation)• Assume ambient T= 25 °CChem 101Fall 2004Gas Stoichiometry: Air Bags• P = 1.3 atm, V = 40 L, T = 25 oC• Solve for n:333232NaN g 92.3=NaN mol 1NaN g 65xN mol 3NaN mol 2xN mol 2.13gas of mol 13.2=)298)(0821.0()40)(3.1(=RTPV=nChem 101Fall 2004Gas Stoichiometry: Air Bags• Final P = 1.3 atm, V = 67 L, T = 25 oC. Supposewe wanted to store as gas. If V were 1 L, P wouldneed to be (67)x(1.3) = 87 atm, or 1270 psi• 1L would be big tank in steering column. Alsohigh pressure & valve problems.6Chem 101Fall 2004Space Shuttle: Booster Engine•2 NH4ClO4 (s) → N2 (g) + Cl2(g) + 2 O2 (g) + 4 H2O (g)• Volume of gas produced from75,000 kg of NH4ClO4?(assume final gas at 100°Cand 1 atm)Chem 101Fall 2004Space Shuttle: Booster Engine•2 NH4ClO4 (s) → N2 (g) + Cl2(g) + 2 O2 (g) + 4 H2O (g)•HN4ClO4 = 117.5 g/mol• 75,000 kg = 75x106 g = 6.38x105 mol• 2 mol NH4ClO4 → 8 mol gas• Get 2.55 x 106 mol gas• Now use the gas lawChem 101Fall 2004Space Shuttle: Booster Engine• 2.55 x 106 mol gas• P = 1 atm, T = 100 C =398 KL 34x10.81)398)(0821.0)(55x10.2(=PnRT=V767Chem 101Fall 2004Kinetic Theory of Gases• The basic assumptions of kinetic-molecular theory are:• Postulate 1• Gases consist of discrete molecules that are relatively far apart.• Gases have few intermolecular attractions.• The volume of individual molecules is very small compared tothe gas’s volume.• Postulate 2• Gas molecules are in constant, random, straight line motionwith varying velocities.Chem 101Fall 2004Kinetic Theory of Gases• Postulate 3• Gas molecules have elastic collisions with themselvesand the container.• Total energy is conserved during a collision.• Postulate 4• The kinetic energy of the molecules is proportional tothe absolute temperature.• The average kinetic energies of molecules of differentgases are equal at a given temperature.Chem 101Fall 2004Kinetic Energy• SI Units: kg m2s-2 or Joules, J2mv 21 KE =mass velocity8Chem 101Fall 2004Kinetic Energy0 20 40 60 80Energy(kJ/mol)0.00.20.40.60.81.0Number of MoleculesChem 101Fall 2004Average Kinetic Energy• R=8.314 J mol-1 K-1•NA=6.02x1023 mol-1AavgN 2RT 3 KE =Avogadro’sConstantGasConstantChem 101Fall 2004Kinetic Energy and Temperature• At a given temperature, T, different gases haveIDENTICAL average kinetic energy• What is the average kinetic energy of an N2molecule in the room?AavgN 2RT 3 KE =9Chem 101Fall 2004Average Molecular Speed2/1avg2avgavgAavgMMRT 3 vmv 2 1 KE and N 2RT 3 KE=∴==molar massChem 101Fall 2004Molecular Speeds0 500 1000 1500 2000 2500 3000Speed (m/s)0.00.20.40.60.81.0Number of M ol eculesvavg (He)vavg (N2)Maxwell-BoltzmannDistribution•Average speed α (T/m)1/2•Light faster than Heavy•Hot faster than ColdChem 101Fall 2004Molecular Speedsvavg (800K)vavg (300K)0 500 1000 1500 2000 2500 3000Speed (m/s)0.00.20.40.60.81.0Number of MoleculesN2 at two differenttemperatures•Average speed α (T/m)1/2•Light faster than Heavy•Hot faster than Cold10Chem 101Fall 2004Molecular Speeds0 500 1000 1500 2000 2500 3000Speed (m/s)0.00.20.40.60.81.0Number of MoleculesThe ‘tail’ of thedistribution isextremely importantin chemical reactionsChem 101Fall 2004Next Class:Gases and Kinetic-MolecularTheory: Chapter
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