velocity distribution of ideal gas molecules - national …biophysics/pc2267/lecture-05.pdf ·...

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Review of the thermal movement of ideal gas moleculesThe distribution of the ”state” of the gas molecules

The distribution of the ”configuration state” of a system

Velocity distribution of ideal gas molecules

Jie Yan

September 5, 2006

Jie Yan Velocity distribution of ideal gas molecules

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Contents

I Review of the distribution of the velocity of a molecule inideal gas.

I Boltzmann distribution of the state of a system.

I Boltzmann distribution of the configuration of a system.


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Time slots

I Tutorial starts from week 3 (28AUG-1SEP).

I A class tutorial every week, on Friday 9 am - 10 am.

I A small group tutorial every even week (4,6,8,10), on Mon(12.00-12.50pm) or on Wed (5.00-5.50pm). The small grouptutorial classroom is S13-04-02.

I Group A(30%): week 4, 8,12, Mon; group B(70%): week6,10, 12, Wed.


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Tutorial-1, 01SEP06: configuration of freely-joint chainpolymers

I Consider a polymer molecule consisting of N straightsegments. Each segment has a length b and can point to anydirection with equal probability. Please show~rN −~r0 =

∑Ni=1

~bi ;

I Please show (~rN −~r0)2 =

∑Ni ,j=1

~bi · ~bj .

I Please find the solution to < ~rN −~r0 >.

I Please find the solution to (< ~rN −~r0 >)2.


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Maxwell distribution of the velocity of ideal gas molecules

I Velocity distribution of one molecule:

ρ(~v) = ( m2πkBT )3/2e

−mv2

2kBT , where v2 = v2x + v2

y + v2z .

I Speed distribution of one molecule:

ρ(u) = 4π( m2πkBT )3/2u2e

−mu2

2kBT .

I The joint distribution of the system of N molecules:ρ(~v1, ~v2, · · · , ~vN) = ρ(~v1)ρ(~v2) · · · ρ(~vN) =

( m2πkBT )3N/2e

−m(v21 +v2

2 +···+v2N )

2kBT .


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The distribution of the ”state” of a tank of gas

I The velocity distribution of a tank of ideal gas is

ρ(~v1, ~v2, · · · , ~vN) = ( m2πkBT )3N/2e

−m(v21 +v2

2 +···+v2N )

2kBT .

I It can be rewritten as: ρ(~v1, ~v2, · · · , ~vN) =

( m2πkBT )3N/2e

−(εk,1+εk,2+···+εk,N )

kBT = ( m2πkBT )3N/2e

− EtotalkBT .

I Can we generize the argument to ρ(state) ∝ e− Estate

kBT ?


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Specify the ”state” of a freely moving molecule

I The state of a molecule in the tank is specified by its location~r and velocity ~v . Once (~r , ~v) are specified, the movement ofthe molecule is determined.

I Because the molecule is freely moving around, ρ(~r) = 1/V ,where V is the volume.

I Because the position of the molecule and its velocity is

uncorrelated, ρ(~r , ~v) = ρ(~r)× ρ(~v) = 1V ( m

2πkBT )3/2e−mv2

2kBT .


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Specify the ”state” of a tank of ideal gas

I The state of a tank of gas is specified by specifying theposition and the velocity of each individual molecules (~ri , ~vi ).

I The distribution of the state of a tank of ideal gas is thus:ρ(~r1,~r2, · · · ,~rN ;~v1, ~v2, · · · , ~vN) =

1V N ( m

2πkBT )3N/2e−(ε2

k,1+ε2k,2+···+ε2

k,N )

kBT = 1V N ( m

2πkBT )3N/2e− Etotal

kBT


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Boltzmann distribution of the ”state” of a tank of gas

I The generalization of

ρ(~r1,~r2, · · · ,~rN ;~v1, ~v2, · · · , ~vN) ∝ e− Estate (~r1,~r2,··· ,~rN ;~v1,~v2,··· ,~vN )

kBT istrue for any systems!

I It is called the ”Boltzmann distribution”!

I Ideal gas is the simplest example of the General distribution!


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Boltzmann distribution of the ”configuration state” of asystem-a

I The ”state” (~ri , ~vi ) we discussed before denotes the absolutestate of the system. Once it is specified, the system’s state isuniquely specified, and according to Newton’s 2nd law, thefuture movement of the system is determined.

I In biology, we are more interested in the configuration state ~ri ,because the interactions among the molecules depend on theconfiguration of a system.

I can we have ρs ∝ e− Es

kBT , where Es is the total interactionenergy among the molecules in the configuration state ”s”?


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Boltzmann distribution of the ”configuration state” of asystem - b

I We know that: the total kinetic energy is Ek =∑N

i=112m~v2

i ,the total interaction is Es =

∑i<j φ(~rj −~ri ), and the total

energy of the state is Estate = Ek + Es .

I we know thatρ(~r1,~r2, · · · ,~rN ;~v1, ~v2, · · · , ~vN) ∝

e− Estate (~r1,~r2,··· ,~rN ;~v1,~v2,··· ,~vN )

kBT = e− Ek (~v1,~v2,··· ,~vN )

kBT e− Es (~r1,~r2,··· ,~rN )

kBT .

I We know the configuration distribution: ρ(~r1,~r2, · · · ,~rN) =∫d~v1

∫d~v2 · · ·

∫d~vNρ(~r1,~r2, · · · ,~rN ;~v1, ~v2, · · · , ~vN) ∝

(∫

d~v1

∫d~v2 · · ·

∫d~vNe

− Ek (~v1,~v2,··· ,~vN )

kBT )e− Es (~r1,~r2,··· ,~rN )

kBT .


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The normalization factor of the Boltzmann distribution ofthe ”configuration state” of a system

I So we do have ρs(~r1,~r2, · · · ,~rN) = (Z )−1e− Es (~r1,~r2,··· ,~rN )

kBT .

I The quantity Z is determined by:

Z =∫

d~r1d~r2, · · · , d~rNe− Es ((~r1,~r2,··· ,~rN ))

kBT . Obviously, Z−1 is thenormalization factor.

I Z is called the partition function. It contains importantinformation of the system. We will discuss the importance ofZ in the future classes.


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Example 0: one free particle system in 1-d box of size L

I Its configuration state is specified by its coordinate x .

I It is a free particle, so no interaction, Es(x) = 0.

I The partition function Z = L, and the distribution isρ(x) = 1/L.


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Example 1: one harmonically bound particle system in 1-d

I Now let’s consider that the particle is bound to the center bya spring, so Es(x) = 1

2kx2, where k is the force constant of

the spring, and −L2 < x < L

2 .

I Z =∫ L/2−L/2 e

− 12 kx2

kBT ≈∫∞−∞ e

− 12 kx2

kBT =√

2πkBTk .

I ρ(x) =√

k2πkBT e

− kx2

2kBT .

I σ2 = kBTk . P(−σ < x < σ) =?


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Example 2: one trapped particle system in 1-d

I Now let’s consider that the particle is bound to the center bya short range interaction, so Es(x) = −ε when−d/2 < x < d/2, and Es(x) = 0 when x < −d/2 or x > d/2.

I Z =∫ −d/2−L/2 dx1 +

∫ d/2−d/2 dxe

− −εkBT +

∫ L2

d/2 dx1 = L− d + deε

kBT .

I ρ(x) = 1

L−d(1−eε

kBT )e− Es (x)

kBT .

I Ptrap =∫ d/2−d/2 dxρ(x) = de

εkBT

L−d(1−eε

kBT )= γe

εkBT

1−γ(1−eε

kBT ), where

γ = d/L.


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Example 2: continue


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Example 2: continue

I ρ(x) =? for ε →∞.

I ρ(x) =? for ε → −∞.

I ρ(x) =? for ε → 0.

I ρ(x) =? for T →∞.

I ρ(x) =? for T → 0.

I ρ(x) =? for γ → 0.

I ρ(x) =? for γ → 1.


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Example 2: one trapped particle system in 1-d continue

I Re-understanding: Ptrap = deε

kBT

L−d(1−eε

kBT )= ne

εkBT

(N−n)+neε

kBT ),

where N = L/l , n = d/l , while l is the occupation length (thesize) of the particle.

I Re-understanding of two region binding problem:

Pi = nieεi

kBT

n2eε2

kBT +n1eε1

kBT, where −εi is the interaction energy of

the particle in region i , and ni is the number of rooms inregion i . The previous example corresponds to ε1 = ε, and

ε2 = 0. Show that P1 + P2 = 1, and P1/P2 = n1n2

eε1−ε2kBT


velocity distribution of ideal gas molecules - national …biophysics/pc2267/lecture-05.pdf ·...

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