PHYS 342 Fall 2011
Lecture 23: Kinetic Theory of Gases (review) Maxwell-Boltzmann Distribution
Ron ReifenbergerBirck Nanotechnology Center
Purdue Universityy
no particles 1 particle 2 particles ……….. N particles
Lecture 23
Historically, (Boyle, 1662; Charles, 1787; Gay-y, ( y , , yLussac, 1802) it was known that for many common
gasses
PV/T = constant
Thi ti lit b t d t This proportionality can be converted to an equation (Clapeyeron, 1834):
PV=NkBT (N=number of gas atoms)or
PV=nRT (n=number of moles)PV=nRT (n=number of moles)
This is an empirical law; there was no microscopic d t di f h it h ld b t understanding of why it should be true.
kB= 1.38 x 10-23 J/K; R =8.314 J/K = 0.082 liter atm mol-1 K-1
Realize that N is a really large number, like 1023. It is remarkable that science learned to deal
i h h i f b f with the properties of enormous numbers of particles BEFORE it discovered how to deal with
individual atoms! The reason is that the thermodynamic properties of a system like P and T are AVERAGE values over statistically
large assemblies of particles large assemblies of particles.
As a common example, it’s always easier to deal ith th t d t th it i t d l ith with the average student than it is to deal with
any one individual student in this class.
So let’s study one gas molecule. Confine one air molecule into very narrow tube with moveable piston
Moveable i t ith
(fi l) (i iti l)
For elastic collision, magnitude of velocity does not change, so . . . .
ONLY one air l l
piston with area A
pmolecule = py(final) – py(initial)
= -m|vy| - (m|vy|)
molecule, mass m, velocity +vy,
momentum L= -2m|vy|
ppiston = -pmolecule
py=mvy
x
yL
pp p
t tFAt = 2L|v |
roundtrip transit time t
Fav
F y|vy| trans t t me
p tWhat is Fav?
Impulsive ForceF =
pt
How to find average force value of N pulses in a time T?t
tFo
. . . . .1 2 3 4 5 6 N
0 time
time =T
Sample F at k discrete values and calculate an average:
1
k
ii o o o
F p t tN F t F t F t pF 1i o o oa v
p pFk T T N t t t
p Fav =
tp
Evaluating the Average Force and Pressure
Fav = = t
ppiston 2m|vy|2L/|v |
= + m|vy|2 / L
F m|v |2 m|v |2
t 2L/|vy|
P=Pressure (one molecule) = = = Fav
Am|vy|
ALm|vy|
V
PV = 2*KE
Generalization to 3D: ONE gas molecule in a cylinder with a frictionless piston with area A
h l l hi ll l i lli i
y m
• when molecule hits wall, elastic collision
• collisions with top and bottom walls produces no change in magnitude or direction of vx e g
x Adirection of vx, e.g.
lli i i h l f i h ll
b e fo r e x y z
a fte r x y z
v v i v j v k
v v i v j v k
L • collisions with left or right wall reverses direction of vx
befo re x y zv v i v j v k
• no forces acting on molecule between collisions
a fter x y zv v i v j v k
For N molecules in the cylinder
For N non-interacting molecules, with identical mass, each with different velocity vi,x How to
l l t ?2, 2
,1 1 1
( )N N N
i xi i x
i i i
N
mv mP total P P vV V
calculate?
2 2,
1
1 N
i x x avi
N NN
m mv vV V
PV = Nm[v2x]av = 2N(½ m [v2
x]av )
KEx
[ x]av (½ [ x]av )x-direction only
Molecular Interpretation of Temperature
• Comparing to Ideal Gas Law
22 12B x av
PV Nk T N m v
• In general, 22 2 2x y zav av avav v v vv
2
21 12 x x avav
m v KE kT
2 22
23
x y zav av av
x
but v v v
v
(there’s nothing special about x-direction)
x av
2 2
3 2 31 12 22
aB B
v avv vPV Nk T N m k T m
21 12 23 B
avm k Tv
From this result we infer that 21 ( )
232av av av BKE K m k T for one mv olecule
equipartition theorem
Estimating the average speed in a gas of N particlesin a gas of N particles
2 31 ( )2 2tot av BK N m v Nk T
2 2
2 33 3( ) A Bav
A
N k TkT RTvm N m M
Am N m M
2 3 3( ) Brms av
k T RTv vm M
m M
2 2rms[Note that v ( ) ]avv
Two ChoicesHow would you
statistically characterize a gas of N atoms in
A. List of numbersAt some t= t
B. Probability Distribution
equilibrium at temperature T?
Particle 1: r1 v1 E1
Particle 2: r2 v2 E2
At some t= t1
r1
v1
Particle 3: r3 v3 E3
r1
..
P ti l N E
.
Particle N: rN vN EN
Need a different list for each time!
<Kav> = 3/2 kBT
from equipartition theorem we know for each time! from equipartition theorem, we know that each molecule has (on average) 3/2 kBT of KE
Maxwell-Boltzmann Distribution
The Probability Distribution function for an ideal (classical) gas is called the Maxwellideal (classical) gas is called the Maxwell-
Boltzmann distribution function. It specifies the probability that a single gas particle from a large distribution of particles in equilibrium at
some temperature T has a speed v or a momentum p or a kinetic energy E.momentum p or a kinetic energy E.
There are many ways to derive the Maxwell-Boltzmann (M-B) Distribution Law:
Calculate the number of particles with velocities
Maxwell Boltzmann (M B) Distribution Law:
pbetween v and v+dv as a function of energy E. Maxwellvelocity distribution results.
Enumerate all possible ways that N particles can bedistributed in energy. Use variational calculus to find thedi t ib ti l ( b f ti l ) th tdistribution law (number of particles vs. energy) thatmaximizes the system weight W. The M-B Distributionresults (see Appendix GG).
Maximize the Entropy S=kBlnW. The M-B Law results.
Use microscopic reversibility arguments.
The Distribution of Molecular Speeds in a Gasfirst derived by J.C. Maxwell in 1852
23 2
2 24 ( )( )2
Bmv k Tm n vP v dv v e dv dvk T N
2 Bk T N • Normalized
2 2 2•
• Probability that molecule has speed v e-mv2/2kB
T
N b f l l s ith s d 2
2 2 2x y zv v v v
• Number of molecules with speed v v2
• Note that P(v) is a continuous probability distribution function
• Note that only v2 appears not the components of velocity• Note that only v appears, not the components of velocity
• Connects macroscopic Thermodynamic properties with microscopic models 14
Maxwell Speed Distribution Function
Note asymmetry – there are more ways to get a large
d th ll
( )P v
speed than a small one.
The inherent asymmetry The inherent asymmetry gives rise to different
values for vmax, vav, and vrms.vv
Check out: http://www.chm.davidson.edu/ChemistryApplets/KineticMolecularTheory/Maxwell htmlets/KineticMolecularTheory/Maxwell.html
Shaded area represents probability that molecule will have a speed v ± dv/2
15
The Average Speed
23
2234( )
2B
mvk T
avgmv v P v dv v e dvk T
2
0 0
2 1
( )2
! 1
avgB
n ax
k T
n md
2 11
0
! ; 1;2 2
n axn
B
n mx e dx n aa k T
8 Bavg
k Tvm
(mean speed)
On earth, T=300K, atmosphere is mostly N2. Mass of N2 molecule is 4.7x10-26 kg. What is vavg?
16
23
26
8 8(1.38 10 )(300) 474 /3.14(4.7 10 )
Bavg
k Tv m sm
Most Probable Speed
( ) 0most probabledP vv
d
2 2
2 22 22 ( ) 0B B
most probable
mv mvk T k T
dvmvve v e
2
2 ( ) 02
2B
B
ve v ek T
k T
2 2
2
Bk Tvm
k T
2 Bmost probable
k Tvm
17
Square Root of the Mean Squared Speed
232
22 2 4
0 0
4( ) ( )2
Bmv
k Tavg
B
mv v P v dv v e dvk T
2
0 0
21
2
1 3 5 (2 1) ; 2;2 2
B
n axn n
k T
n mx e dx n ak T
1
0
2
2 23( )
n nB
B
a a k Tk Tv
2
( )
3( )
avg
B
vm
k Tv v
2
( )
31 1 3
rms avg
B
v vm
k Tmv m k T
(equipartition
182 2 2rms Bmv m k T
m ( q p
theorem)
Maxwell Boltzmann Speed Distribution for Maxwell Boltzmann Speed Distribution for different T
N2 molecule
19
What is the probability that a gas atom with mass m has a momentum between p and p+dp at temperature T?a momentum between p and p+dp at temperature T?
2 23 2 3 2
2 22 24( ) 4B Bmv mvk T k Tm mP v dv v e dv v e dv
22
( ) 42 2
1 12 2
B B
P v dv v e dv v ek T k T
p pdp pdpmv mvdv
dv
dpdv
23 222
22
2 2
42
Bp mk T
m m mv m
v m ek T
m
mm
dpm
2 23 2 3 22
23 2
2 2
2
4 142 2
B B
B
p pmk T mk T
k T
p m me dp p e dpm k T m k T
m m
2 2B Bm k T m k T
22
24 ( )( ) Bp mk Tp n pP d d d
20 3 2
2 ( )( )2
B
B
p mk Tp pP p dp e dp dpNmk T
What is the probability that a gas atom with mass m has an energy between E and E+dE at temperature T?
Change Variables:
an energy between E and E dE at temperature T?
2
2
2pEm
g
2
2
222
2p d
p mEp dp mdE
p pdp mE m Ep d
2
3 2
22( ) 4
2B
B
p mk Tn p dp eN mk T
p dp
3 2 1 21 22
2p d
m E d
p pdp mE m Ep d
E
2
3
3 22
2 1 2
242
4 1
B
B
p mk T
E k T
emk T
dp p
3
32 1 2
2
1 2
3
42
( ) 2( )
1 22
B
B
B
E k T
E k T
emk T
n E EP E dE dE e d
m E dE
E
32
( )B
d d e dN k T
21
What is the Average Energy of Gas Atom at Temperature T?
1 2
320 0
2( ) Bavg
B
E k TEE P E dE e dEk T
E E
3 2
03
2
2 1 B
B
E k TE e dEk T
3 2
0 0
1 ; 3 / 2B n
B
E k T aEa E e dE E e dE nk T
5
52
21
( 1) 3 14 1
34n B
na
k T
25
1
6 332 1Bk T
E k T Tk kT
22
32 4 24avg
B
B B BE k Tk T
Tk kT
Probability of finding gas atom with energy between E and E+dEgy
0.6 200K
0.4
0.5
)
300K
500K
0 2
0.3
0.4
P(E)
0 0
0.1
0.2
0.0
0.00 0.03 0.06 0.09 0.12 0.15E (in eV)E (in eV)
23
Experimental Verification of Maxwellian Distribution Function
APPLICATION
The Maxwell speed distribution serves as the basic input for computer calculations of molecular
24
dynamic (MD) simulations of gas flow, gas cooling, gas heating, flames, etc.
Summary
23 2
2 2( ) 4( ) Bmv k Tn v mP d d d
Velocity distribution
2 2( )( )2
Bmv
B
k TP v dv dv v e dvN k T
22
3 22( ) 4( ) Bp mk Tn p pP p dp dp e dp
Momentum distribution
3 2( )2 B
p p p pN mk T
Energy distribution
1 2
32
( ) 2( ) BE k Tn E EP E dE dE e dEN k T
Energy distribution
25
2B
N k T