Chapter 03 - Kinetic Theory Of Gases - Grandinetti

Chapter 03Kinetic Theory of GasesP. J. GrandinettiChem. 4300P. J. GrandinettiChapter 03: Kinetic Theory of Gases

History of ideal gas law1662: Robert Boyle discovered with changing pressure at constant temperature thatproduct of pressure and volume of a gas at equilibrium is constant,pV constant at constant T1780s: Jacques Charles found that ratio of volume to temperature was also invariantwhen temperature was changed with pressure kept constant,V T constant at constant p1811: Amedeo Avogadro found ratio of volume to amount remained constant withchanging amount at fixed pressure and temperature,V n constant at constant p and T1834: Emile Clapeyron combined gas laws of Boyle, Charles, and Avogadro into idealgas equation of state,pV nRTwhere R is gas constant.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Bernoulli’s derivation of Boyle’s law, pV constant.As early as 1738 Daniel Bernoulli proposed a microscopic kinetic explanation of Boyle’s law,but only after Clapeyron’s work did Bernoulli’s kinetic theory gain widespread acceptance.Bernoulli’s derivationzyxmRemember pressure is defined as force per unit area.What is the force of one gas molecule hitting a wall?P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Force on wall is momentum change when a molecule hits it.Force along y is given by ratio of change in momentum to time betweencollisions.ΔpyFy Δtvy-vyLinear momentum is conserved in collision with wallΔpy py,final py,initial ( mvy ) (mvy ) 2mvyTime to travel length of box, hit wall, and travel back is Δt 2𝓁 vyAverage force of 1 molecule hitting 1 wall of box is Fy P. J. GrandinettiΔpyΔt 2mvy2𝓁 vyChapter 03: Kinetic Theory of Gases mvy2𝓁

Force of N molecules hitting all walls of the box.Sum over N molecules hitting one wall ism 2v𝓁 𝛼 1 y𝛼NFyN Add magnitude (i.e., ignore signs) of all forces on all 6 walls (top, bottom, left, right, front, back)Ftotal 2NNNN (N)m 2m 2m 2m 2m 2vx v2y v2z 2vx 2vy 2vz 2v𝛼𝛼𝛼𝓁 𝛼 1 𝛼𝓁 𝛼 1 𝛼𝓁 𝛼 1 𝛼𝓁 𝛼 1𝓁 𝛼 1 𝛼 v2𝛼Define mean square velocity as1 2vN 𝛼 1 𝛼Nv2 orNv2 N 𝛼 1v2𝛼and write total force on all 6 walls of box asmFtotal 2 Nv2𝓁P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Pressure from N molecules inside the box.Pressure is force per unit area. Total area of box walls is 6 times area of 1 wall: Atotal 6Awall .p Ftotal Atotal Ftotal (6Awall )Substituting previous result:mFtotal 2 Nv2𝓁givesNmv2Nmv2m p 2 Nv2 (6Awall ) 3Awall 𝓁3V𝓁where V Awall 𝓁 is volume of boxRearranging gives Boyle’s law (pV constant)pV ()Nmv2212 N mv2 N𝜖 k ,3323where 𝜖 k is mean kinetic energy of moleculeRewriting in terms of moles, i.e., N nNA where NA is Avogradro constant,pV P. J. Grandinetti22nNA 𝜖 k nEk33where Ek is kinetic energy of 1 mole of gasChapter 03: Kinetic Theory of Gases

Temperature is a quantity derived from energyFinally connect to Ideal gas law:2pV nRT nEk3and we discover3Ek RT2kinetic energy of 1 mole of ideal gasEquation reveals true nature of temperature—reflects kinetic energy of atoms andmolecules.Can’t have negative temperatures because can’t have negative kinetic energy.Raising gas temperature increases kinetic energy of gas molecules and vice versa.Dividing by NA we obtain relationship on per molecule basis𝜖k Ek3 R3 T kB TNA2 NA2kB R NA 1.38064852 10 23 J/K is defined as Boltzmann constant.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Average molecular speedGivendefine root mean square speed, crms1 2 3mv kB T22 v2 , and obtain crms 3kB T m 3RTMcrms is related to temperature and molecular mass, m, or molar mass, MMolecular speeds increase with increasing temperature.Molecular speeds decrease with increasing molecular mass.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Average molecular speedExampleCalculate the rms speed for a mole of vanillin molecules at room temperature.SolutionSince Vanillin has chemical formula C8 H8 O3 with a molecular weight of 152.1 g/mol we obtain 3R(300 K)3RT 221 m/s 500 mphcrms M152.1 g/molIf vanillin has such a high speed why does it take so long for the scent to travel across a room?P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Maxwell Distribution LawsWhat is parent probability distribution function formolecular velocities, p(⃗v)molecular speeds, p(c)molecular energies, p(E)?James Clerk Maxwell1831-1879P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Maxwell Distribution LawsIn 1859 James Clerk Maxwell worked out the probability distribution of molecular velocities,f (⃗v), for gas molecules as perfectly elastic spheres.Maxwell assumed that distribution of velocities in each direction were uncorrelated, that is,f (⃗v) can be written as product of 3 independent distributionsf (⃗v) f (vx ) f (vy ) f (vz )He also reasoned that distribution of velocities is independent of direction, implying that f (⃗v)should only depend on magnitude of velocities,f (vx ) f (vy ) f (vz ) 𝜙(v2x v2y v2z )This is an example of a functional equation: an equation in which the unknowns are functions.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

How do we solve this functional equation?f (vx ) f (vy ) f (vz ) 𝜙(v2x v2y v2z )Product of functions on left must give sum of their variables as function argument on right.A function, f (vi ), that satisfies this functional equation is2f (vi ) ae bviPutting this function into functional equation gives222𝜙(v2x v2y v2z ) a3 e b(vx vy vz )Need to determine a and b.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Normalizing Maxwell’s distribution for molecular velocitiesAs the f (vi ) are probability distributions we require them to be normalized, requires that a sof (vi )dvi 1b 𝜋 and f (vi ) 2ae bvi dvi 1b bv2ie𝜋Taken together Maxwell’s probability distribution then becomes( )3 2222be b(vx vy vz )f (⃗v) f (vx ) f (vy ) f (vz ) 𝜋Still need to determine b.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Maxwell’s distribution law for molecular velocitiesFrom Bernoulli’s kinetic theory we learned that v2 3kB T m.This mean square speed should also be obtained from probability distributionv2 (v2x v2y v2z ) f (⃗v) d⃗v 3kB T mSubstituting our normalized solution for f (⃗v) we obtain ( )3 2222be b(vx vy vz ) dvx dvy dvz 3kB T mv2 (v2x v2y v2z ) 𝜋Solving this 3D integral equation requires b m (2kB T), and finally obtainf (⃗v) 1(2𝜋)3(mkB T)3 21222e 2 (vx vy vz ) (kB T m)This is Maxwell’s distribution law for molecularvelocities: A 3D Gaussian probability distribution with a standard deviation of kB T m.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Maxwell’s distribution law for molecular velocitiesMaxwell’s distribution law is a 3D Gaussian distribution centered on v⃗ 0.Distribution for one component of velocity vector for N2 gas at 3 different temperatures.0.0025100 K/ s/m0.0020N2 gas0.00150.0010300 K0.00051000 K0.0000P. J. Grandinetti-1500-1000-5000500velocity/ m/sChapter 03: Kinetic Theory of Gases10001500

Maxwell’s distribution law for molecular speedsP. J. GrandinettiChapter 03: Kinetic Theory of Gases

Maxwell’s distribution law for molecular speedsSpeed is magnitude of velocity vector. To get speed distribution transform Maxwell’s velocitydistribution into spherical coordinates, vyvzc v2x v2y v2z , cos 𝜃 , tan 𝜙 .cvxWith this change of variables we findf (⃗v) f (c, 𝜃, 𝜙) 1((2𝜋)3mkB T)3 21 2 (kB T m)e 2 cThis is independent of 𝜃 and 𝜙 so if we put it into the normalization 0𝜋 0 02𝜋f (c, 𝜃, 𝜙)c2 dc sin 𝜃 d𝜃 d𝜙 1we can integrate away 𝜃 and 𝜙 and obtain ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB TP. J. GrandinettiChapter 03: Kinetic Theory of GasesMaxwell’s distribution lawfor molecular speeds.

Maxwell’s distribution law for molecular speeds ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB T0.0035N2 gas100 K0.0030/ s/m0.00250.0020300 K0.00150.00100.00050.0000P. J. Grandinetti1000 K05001000speed/ m/s1500Chapter 03: Kinetic Theory of Gases2000

Maxwell’s distribution law for molecular speeds ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB T0.00250.0020/ s/m300 KAr0.0015Ne0.0010He0.00050.0000P. J. Grandinetti050010001500speed/ m/s2000Chapter 03: Kinetic Theory of Gases2500

Mean speedWith Maxwell’s distribution law for molecular speeds ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB Twe can calculate the mean speed c 0 ()3 2()22m1 2kB T cf (c)dc 𝜋 kB T2mwhich simplifies to c 8kB T 𝜋m 8RT𝜋MNote, mean speed, c, is smaller than root mean square speed, crms P. J. GrandinettiChapter 03: Kinetic Theory of Gases 3kB T.m

Most probable speed ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB TTo calculate most probablespeed need to find c value wheref (c) is maximum.Set df (c) dc 0, solve for c toobtain cmode P. J. Grandinetti2kB T m 2RTM0.60.50.40.30.20.10.0012speed /Chapter 03: Kinetic Theory of Gases345

Fraction of molecules with speeds between c1 and c2 .Fraction of molecules with speeds between c1 and c2 is obtained by integrating Maxwell speeddistributions between these two limits,c2𝛿N f (c)dc c1N0.60.50.40.30.20.10.001234speed /P. J. GrandinettiChapter 03: Kinetic Theory of Gases5

Web Apps by Paul FalstadKinetic Theory of Gases SimulationP. J. GrandinettiChapter 03: Kinetic Theory of Gases

1955 - Miller and Kusch experimentally verify Maxwell’s molecular speed distribution.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Miller and Kusch experimentsIn 1955 Miller and Kusch published firstconvincing measurements of speeddistribution for K and Tl atoms in gasphase.For each fixed rotation speed onlymolecules with a small range of speeds cantravel from the furnace to the detector.With dimensions given in instrumentdiagram the selected speed is v0 𝜔l 𝜙.Measuring intensity as a function ofrotation speed gives the atomic speeddistribution.P. J. GrandinettiChapter 03: Kinetic Theory of Gases

Miller and Kusch experiments20Potassium vapor20Run 31Run 60Run 57Run 99Run 9715IntensityIntensity1510P. J. GrandinettiThallium vapor105500.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0speed /00.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0speed /Chapter 03: Kinetic Theory of Gases

Miller and Kusch experimentsP. J. GrandinettiChapter 03: Kinetic Theory of Gases

Distribution of kinetic energiesHomework0.35derive distribution of kinetic energies,(P. J. Grandinetti1kB T)3 2𝜖k 1 2 e 𝜖k (kB T)0.2510-21 J ()3 21 22mf (c) c2 e 2 c (kB T m)𝜋 kB T2f (𝜖k ) 𝜋100 K0.30Given0.200.15300 K0.100.050.001000 K051015Energy/ 10-21 JChapter 03: Kinetic Theory of Gases202530

Equation reveals true nature of temperature—reflects kinetic energy of atoms and molecules. Can’t have negative temperatures because can’t have negative kinetic energy. Raising gas temperature increases kinetic energy of gas molecules and vice versa. Dividing by NA we obtain relat