KineticKinetic TheoryTheory ofof GasesGases
Kinetic Theory: Theory that deals with prediction of transport (µ, κ, D) and thermodynamic properties of gases based on statistical (average) description of the translational motion of its components (atoms & molecules).
velocity statistical distribution description function
Maxwell-Boltzmann speed distribution
3/ 2 2 π m − mv F( v ) = 4 v2 e 2kT 2πkT Probability & Statistics Review
! Suppose we have a series of observation for a value x which can take on a discrete set of values
{x1, x2, x3,…xm} ! Suppose that in a series of N measurements we observe that
N x1 occurs N1 times
x2 occurs N2 times : : :
xm occurs Nm times
x1 x2 x3 x4 …xm
! Average value of x
m N x + N x + N x + ...... + N x N < x >= 1 1 2 2 3 3 m m = i x N ∑ N i i=1 Probability & Statistics Review
! Define the probability that xi will be observed as Pi m m N 1 P = i note P = N =1 i N ∑ i N ∑ i i=1 i=1
m ! The average value of x becomes < >= x ∑ Pi xi i=1 ! Now suppose that x (outcome of an experiment) can take on not discrete but continuous set of values ! Everytime that outcome of the experiment is between x and x+∆x we add 1 to that bin
! N(x) is the number of outcomes out of N total that is between x and x+∆x
xx+∆x Probability & Statistics Review
! Thus, the probability that an experimental outcome will be between x ∆ and x+ x is N( x ) P( x ) = N ! The value P(x) is proportional to the length of segment ∆x. We define probability density function f(x) such that P(x)=f(x)∆x
! The average value of x,
< x >= ∑ xP( x ) = ∑ xf ( x )∆x all segments all segments
! Let ∆x→0 and number of segments →∞ < x >= ∫ xf ( x )dx
! Integral is over the whole domain of x; f(x): distribution function MaxwellMaxwell--BoltzmannBoltzmann velocityvelocity distributiondistribution
vz 1/ 2 2 m − mvx f ( v ) = e 2kT Gaussian probability x π distribution vy 2 kT v 1/ 2 2 x m − mvx f ( v )dv = e 2kT dv x x 2πkT x
Probability that vx is between vx and vx+dvx Example × × -27 SiH4 : mSiH4 = 32 1.66 10 kg × × -27 H2 : mH2 = 2 1.66 10 kg k= 1.38066 ×10-23 J/K T = 300 K MaxwellMaxwell--BoltzmannBoltzmann velocityvelocity distributiondistribution Things to remember/notice " Gaussian distribution " Width of the distribution depends on mass; lighter atoms/molecules have wider distribution " At the same T lighter atoms/molecules are faster " Higher the T faster the molecules
" On average vx=0, (makes sense because on average the gas is stationary.) " Since gas speeds ~1-10 m/s is much smaller than molecular speeds (~500-1000 m/s) we can assume the same distribution even if the gas is flowing. MaxwellMaxwell--BoltzmannBoltzmann speedspeed distributiondistribution
! The three velocity components are independent of each other
3/ 2 2 mv2 2 m − mvx − y − mvz f ( v ) f ( v ) f ( v ) = e 2kT e 2kT e 2kT x y z 2πkT vz dv How many molecules are there in a shell with π 2 v volume 4 v dv ? = f ( vx ) f ( vy ) f ( vz ) dvxdvydvz F( v )dv vy vx Transform into cylindrical coordinates 2 = 2 + 2 + 2 v vx vy vz speed
3 / 2 2 Probability of finding a π m − mv molecule that has speed F( v ) dv = 4 v2 e 2kT dv between v and v+dv 2πkT MaxwellMaxwell--BoltzmannBoltzmann speedspeed distributiondistribution WhyWhy isis MaxwellMaxwell--BoltzmannBoltzmann speedspeed distributiondistribution useful?useful? HowHow dodo youyou useuse it?it?
∞ ψ Calculating an average property <ψ> < > ∫ψF( v ) dv 0 Example:
Example: average kinetic energy and Cv:
ε = 1 < 2 >= 3 K m v kT 2 2 ∞ kT < v2 >= v2 F( v ) dv = 3 ∫ m 3 3 ∂U 3 0 E = N kT = RT = U ⇒ C = = R K Av v ∂ 2 2 T v 2
Example: average flux to a plane vx A
∆ vx t
substrate If a particle has velocity vx it ∆ will strike A if it is vx t away RandomRandom fluxflux toto aa surfacesurface
× ∆ # of collisions = gas density ∞cylinder volume = NAvx t vx A Total # of collisions = NA∆t f ( v )v dv ∫ x x x ∆ vx t ∞ 0 1/ 2 2 Z m − mvx = = ∆w = 2kT Flux F N vxe dvx A t ∫ 2πkT ∞ 0 −ax2 1 π 1/ 2 e dx = m 1 2kT N 8kT ∫ 2a F = N = 0 2 kT 2 m 4 πm
N < v > F = The random flux 4 λλ MeanMean FreeFree Path,Path, mf
Consider a hard sphere traveling at
σ=πd2 d d/2 Collision tube
Atom will sweep out a collision tube whose cross section is σ=πd2 Distance traveled in ∆t is =
σ 2 λ Molecule (cm ) MF (cm) He 2.1×10-15 10 × -15 H2 2.7 10 8 Ar 3.6×10-15 6 Example: Reaction rates of gas phase reactions that proceed at gas phase collision frequencies. Consider for example the following gas phase reaction + →k + SiH 4 O2 SiH 3 OH What is the upper limit for k? Equal if they react everytime = σ < > = Z AB AB vAB N A N B kN A N B they colide GasGas phasephase collisioncollision ratesrates && transporttransport coefficientscoefficients
8kT m m < v >= with = A B AB m + m πµ A B 1 µ σ = ( d + d ) k = < v > AB 2 A B max AB AB σ × −15 2 < σ> AB ~ 3 10 cm vAB ~ 50,000 cm / s ∴ = × −10 3 kmax 1.5 10 cm / s λ 1 3/ 2 = = 1 λ < > T mf D mf v D ~ 2Nσ 3 P κ 1 = λ < v > C N 3 mf V Classic Example: “Monolayer formation time”- Why we use UHV for fundamental surface science studies Example: there are about ~ 1015 cm-2 surface sites on a surface. How much time is required to form a monolayer if all molecules hitting the surface stick to the surface at 300 K and 1 mTorr. (e.g, H2O) 8kT 8×1.38×10−23 J / K ×300 K < v >= π = = 594 m / s m π ×1.66×10−27 kg ×18 5 2 P 0.001Torr 1.013×10 N / m − − N = = = 3.219×1019 m 3 = 3.219×1013 cm 3 kT 1.38×10−23 J / K ×300 K 760 Torr 13 −3 N < v > 3.219×10 cm ×59,400cm / s − − F = = = 4.78×1017 cm 2s 1 4 4 surface site density 1015 cm−2 τ = = = 0.002 s = 2 ms ML Flux 4.78×1017 cm−2s−1 τ Vacuum P (Torr) ML (seconds) High 10-6 2 Very High 10-8 200 UHV 10-10 20,000