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Physics I Microscopic Model of Gas Prof. WAN, Xin [email protected] xinwan

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Physics I Physics I Microscopic Model of Microscopic Model of Gas Gas Prof. WAN, Xin [email protected] http://zimp.zju.edu.cn/~xin wan/
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Physics IPhysics I

Microscopic Model of GasMicroscopic Model of Gas

Prof. WAN, Xin

[email protected]://zimp.zju.edu.cn/~xinwan/

The Naïve Approach, AgainThe Naïve Approach, Again

N particles ri(t), vi(t); interaction V(ri-rj)

Elementary Probability TheoryElementary Probability Theory

Assume the speeds of 10 particles are 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0 m/s

m/s5.410

0.90.10.0

N

vv i

m/s5.710

0.90.10.0 2222

N

vv i

rms

When we have many particles, we may denote pa the probability of finding their velocities in the interval [va, va+1].

Elementary Probability TheoryElementary Probability Theory

Now, the averages become

1where, a

aaa ppvv

arms pvva

2

In the continuous version, we may denote p(v)dv the probability of finding particles’ velocities in the interval [v, v+dv].

1)(where,)(

dvvpdvvvpv

dvvpvvrms )(2

Assumptions of the Ideal Gas ModelAssumptions of the Ideal Gas Model

Large number of molecules and large average separation (molecular volume is negligible).

The molecules obey Newton’s laws, but as a whole they move randomly with a time-independent distribution of speeds.

The molecules undergo elastic collisions with each other and with the walls of the container.

The forces between molecules are short-range, hence negligible except during a collision.

That is, all of the gas molecules are identical.

The Microscopic ModelThe Microscopic Model

xF

LWDALV

AWD

xL

A

tvm

A

F

A

Fp

moleculesonxpistononx

,,

Pressure, the Microscopic ViewPressure, the Microscopic View

Pressure that a gas exerts on the walls of its container is a consequence of the collisions of the gas molecules with the walls.

xvmvmtAp 2

2

1 tvA x

22

3mvmvp x

= N / Vhalf of molecules

moving right

Applying the Ideal Gas LawApplying the Ideal Gas Law

22

33mv

nNmv

NVp A

TknN

Vpmv B

nRTpV

A 2

3

2

3

2

1 2

KJmole

KmoleJ

N

Rk

AB /1038.1

)(/1002.6

)/(31.8 2323

Boltzmann’s

constant

TemperatureTemperature

Temperature is a measure of internal energy (kB is the conversion factor). It measures the average energy per degree of freedom per molecule/atom.

Equipartition theorem: can be generalized to rotational and vibrational degrees of freedom.

Tkmvmvmv Bzyx

2

1

2

1

2

1

2

1 222

Heat Capacity at Constant VHeat Capacity at Constant V

We can detect the microscopic degrees of freedom by measuring heat capacity at constant volume.

Internal Energy U = NfkBT/2

Heat capacity

Molar specific heat cV = (f/2)R

degrees of freedomBV Nk

f

T

UC

2Vfixed

Specific Heat at Constant VSpecific Heat at Constant V

• Monoatomic gases has a ratio 3/2. Remember?

• Why do diatomic gases have the ratio 5/2?

• What about polyatomic gases?

Specific Heat at Constant VSpecific Heat at Constant V

A Simple Harmonic Oscillator

2

2)(

dt

xdmkx

dx

xdUF

x

2

2

1)( kxxU

xdt

xd 22

2

FF

m

kω 2

)cos0 t(xx

Two Harmonic Oscillators

)(' 21121

2

xxkxkdt

xdm

1x

)(' 12222

2

xxkxkdt

xdm

)(')(

21221

2

xxm

k

dt

xxd

)(2')(

21221

2

xxm

kk

dt

xxd

2x

Two Harmonic Oscillators

)(' 21121

2

xxkxkdt

xdm

1x

)(' 12222

2

xxkxkdt

xdm 201010

2 )'( kxxkkxm

2010202 )'( xkkkxxm

)cos0 t(xx ii

2x

Assume

Two Harmonic Oscillators

0'

'

20

10

2

2

x

x

mkkk

kmkk

2010102 )'( kxxkkxm 201020

2 )'( xkkkxxm

)cos0 t(xx iiAssume

1x 2x

Vibrational Mode

20100' xxk

2/

2' 0'

m

k

m

kk k

Solution 1:

Vibration with the reduced mass.

1x 2x

Translational Mode

2010 xx

0' 0' k

m

kSolution 1:

Translation!

1x 2x

Two Harmonic Oscillators

20

102

20

10

'

'

x

xm

x

x

kkk

kkk

1x 2x

In mathematics language, we solved an eigenvalue problem.

The two eigenvectors are orthogonal to each other. Independent!

Mode Counting – 1DMode Counting – 1D

1D: N-atom linear molecule – Translation: 1

– Vibration: N – 1

A straightforward generalization of the two-atom problem.

From 1D to 2D: A Trivial ExampleFrom 1D to 2D: A Trivial Example

rotation

vibration

translation

1x

2x

2y

1y

Mode Counting – 2DMode Counting – 2D

2D: N-atom (planer, nonlinear) molecule– Translation: 2

– Rotation: 1

– Vibration: 2N – 3

Mode Counting – 3DMode Counting – 3D

3D: N-atom (nonlinear) molecule– Translation: 3

– Rotation: 3

– Vibration: 3N – 6

Vibrational Modes of COVibrational Modes of CO22

N = 3, linear– Translation: 3

– Rotation: 2

– Vibration: 3N – 3 – 2 = 4

Vibrational Modes of HVibrational Modes of H22OO

N = 3, planer– Translation: 3

– Rotation: 3

– Vibration: 3N – 3 – 3 = 3

Contribution to Specific HeatContribution to Specific Heat

i

iii i

i qkm

pE 2

2

2

1

2

Equipartition theorem: The mean value of each independent quadratic term in the energy is equal to kBT/2.

Specific Heat of HSpecific Heat of H22

Quantum mechanics is needed to explain this.

Specific Heat of SolidsSpecific Heat of Solids

nRTTNkU B 33

DuLong – Petit law

RdT

dU

nc

VV 3

1

spatial dimension

vibration energy

Molar specific heat

Again, quantum mechanics is needed.

Root Mean Square SpeedRoot Mean Square Speed

m

Tkv B

rms

3root mean square speed

Estimate the root mean square speed of water molecules at room temperature.

m/s6003

m

Tkv B

rms

Distribution of SpeedDistribution of Speed

slow

fast

rotating drum

to pump

oven

Speed SelectionSpeed Selection

Can you design an equipment to select gas molecules with a chosen speed?

to pump

?

Maxwell DistributionMaxwell Distribution

kTmvevkT

mNvN 2/2

2/32

24)(

)(vN

vv dvv

dvvN )(

dvvNv

v 2

1

)(

Maxwell DistributionMaxwell Distribution

)(vN

v1v 2v

kTmvevkT

mNvN 2/2

2/32

24)(

number of molecules v [v1, v2]

Maxwell DistributionMaxwell Distribution

kTmvevkT

mNvN 2/2

2/32

24)(

)(vN

v

Total number of molecules

0

)( dvvNN

Characteristic SpeedCharacteristic Speed

0)(

dv

vdN

02 2/3

2/ 22

kTmvkTmv ekT

mvve

m

kTvp

2

Most probable speed

kTmvevkT

mNvN 2/2

2/32

24)(

Characteristic SpeedCharacteristic Speed

dvevkT

m

N

dvvNv

v kTmv

0

2/42/3

0

2

2 2

24

)(

m

kTvvrms

32

Root mean sqaure speed

kTmvevkT

mNvN 2/2

2/32

24)(

Characteristic SpeedCharacteristic Speed

dvevkT

m

N

dvvvN

v kTmv

0

2/32/3

02

24

)(

m

kTv

8

Average speed

kTmvevkT

mNvN 2/2

2/32

24)(

Varying TemperatureVarying Temperature

kTmvevkT

mNvN 2/2

2/32

24)(

)(vN

v

321 TTT T1

T2

T3

Boltzmann DistributionBoltzmann Distribution

Continuing from fluid statics

The probability of finding the molecules in a particular energy state varies exponentially as the negative of the energy divided by kBT.

TkmgyV

Tknppgh BBV ennepp /0

/0

00

potential energy

TkhvEV

Benhvn /),,(0),,(

Boltzmann distribution law

How to cool atoms?How to cool atoms?

Laser CoolingLaser Cooling

Figure: A CCD image of a cold cloud of rubidium atoms which have been laser cooled by the red laser beams to temperatures of a millionth of a Kelvin. The white fluorescent cloud forms at the intersection of the beams.

Bose-Einstein CondensationBose-Einstein Condensation

Velocity-distribution data for a gas of rubidium atoms, confirming the discovery of a new phase of matter, the Bose–Einstein condensate.

Left: just before the appearance of a Bose–Einstein condensate. Center: just after the appearance of the condensate. Right: after further evaporation, leaving a sample of nearly pure condensate.

Earlier BEC ResearchEarlier BEC Research

BEC in ultracold atomic gases was first realized in 1995 with 87Rb, 23Na, and 7Li. This pioneering work was honored with the Nobel prize 2001 in physics, awarded to Eric Cornell, Carl Wieman, and Wolfgang Ketterle.

For an updated list, check http://ucan.physics.utoronto.ca/

BEC of Dysprosium BEC of Dysprosium

Strongly dipolar BEC of dysprosium, Mingwu Lu et al., PRL 107, 190401 (2011)

Brownian MotionBrownian Motion

Mean Free PathMean Free Path

dd

v

d

v

d

Average distance between two collisions

During time interval t, a molecule sweeps a cylinder of diameter 2d and length vt.

Mean Free PathMean Free Path

pd

Tk

dnvtdn

vtl B

VV222

1

Tknp BV

Tkpn BV /

vtdnz V2

Volume of the cylinder

vtdV 2Average number of collisions

Mean free path

During time interval t, a molecule sweeps a cylinder of diameter 2d and length vt.

Mean Free PathMean Free Path

pd

Tk

dntvdn

vtl B

VV222 22

1

)2(

vtdnz V2

Average number of collisions

Mean free path Relative motion vv 2

Q&A: Collision FrequencyQ&A: Collision Frequency

Consider air at room temperature. – How far does a typical molecule (with a

diameter of 2 10-10 m) move before it collides with another molecule?

Q&A: Collision FrequencyQ&A: Collision Frequency

Consider air at room temperature. – How far does a typical molecule (with a

diameter of 2 10-10 m) move before it collides with another molecule?

Q&A: Collision FrequencyQ&A: Collision Frequency

Consider air at room temperature. – Average molecular separation:

Q&A: Collision FrequencyQ&A: Collision Frequency

Consider air at room temperature. – On average, how frequently does one

molecule collide with another?

m

kT

m

kTv ~

8

l

vf

Expect ~ 500 m/s

Expect ~ 2109 /s

Try yourself!

Fluid flows layer by layer with varying v.

F = A dv/dy : coefficient of viscosity

Transport: Viscous FlowTransport: Viscous Flow

A

A

F, vy

Cylindrical Pipe, NonviscousCylindrical Pipe, Nonviscous

v 2Rr

constvrv 0)(

02vRQ (volumetric flow rate)

Cylindrical Pipe, ViscousCylindrical Pipe, Viscous

V(r) 2Rr

22

4)( rR

L

Prv

L

RPrdrrvQ

82)(

4 (Poiseuille Law)

“current”“voltage”

HomeworkHomework

CHAP. 22 Exercises 7, 8, 10, 21, 24 (P513)


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