4/17/2014phy 770 spring 2014 -- lecture 231 phy 770 -- statistical mechanics 12:00 * - 1:45 pm tr...

PHY 770 Spring 2014 -- Lecture 23 14/17/2014

PHY 770 -- Statistical Mechanics12:00* - 1:45 PM TR Olin 107

Instructor: Natalie Holzwarth (Olin 300)Course Webpage: http://www.wfu.edu/~natalie/s14phy770

Lecture 23

Review and perspective Comments about some homework problems Treatment of multicomponent systems

*Partial make-up lecture -- early start time

http://www.wfu.edu/~natalie/s14phy770


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Comments about homework problems:


HW #16 -- continued7.2 A Brownian particle of mass m is attached to a harmonic spring with force constant k, and is driven by an external force F(t). The particle is constrained to move in one dimension. The Langevin equation is

(a) Find the equilibrium correlation function directly from the solution of the Langevin equation assuming that and , , etc.

(b) Find the equilibrium correlation function from the fluctuation-dissipation theorem relation

2202

( ) ( )( ) ( ) ( )

d dxm m x

dt d

x t tt t F

tt

( ) (0)T

t xx

0( )t

20(0) (0) ( )/

Tx T mx k (0) (0) 0

Tx v

( ) (0)T

t xx

2 20

1( ) (0)

where for this case:

( ) cos(

)

1/( )

T

tkTx

m

i

di

m

t x


HW #16 -- continued

2 22 20 0

2202

2 2 2 2

0 0

2

First consider solutions to homogeneous equation:

Imposing boundary conditions:

(0)(0)

( ) ( )( ) 0

( )

:

( )

and

t t t tm m m m

m

x t tt

x t Ae e

x v

x t e

d dxm m x

dt dt

B

dxx

dt

2 20 0

2 20 0 02

20

2cosh sinh2 2

2

tx

mx t tm m

m

v

We can argue that since the random force averages to 0, the correlation function depends only on the homogeneous solution.

Note: we are assuming the over-damped case


HW #16 -- continued

22 20 0

2 220 0 02

20

0 0

20 2

0

The initial values and can be treated as random variabl

2( ) c

es wi

o

th

and

sh sinh2 2

2

tm

tm

T

v

x v

x

x em

x t e x t tm m

m

kTx

m

0 0 0.T

v

2 22 2 220 0 02

20

( ) (0)

2 cosh sinh2 2

2

T

tm

T

x t x

mx e t tm m

m


HW #16 -- continued

2 20

Now consider that dissipation-fluctuation formulation:

1( ) (0)

1 1

( ) cos( )

( ) ( )

1/( )

2 2

with:

T

ti it

x t x di

d e

tkT

kT kT

m

im

d ei i

1

2 20

2

We will need to contour integration methods to evaluate the integrals:

1 1

2 2

Note that the denominator of the i

( ) ( )

( ) :

has po

ntegrals

l

e

ti tiI d e I d ei

im

i

220 whe

s at the

re r

values 0, ,

2 2

P ir ir

m m


HW #16 -- continued

Re(z)

Im(z)

xxx

1

( )1=0

2tiI d e

i

2

1= Res R

(es

2

) i t ir iI d ei

r

2 22 220 02 2

0 20

( ) (0)

2 cosh sinh2 2

2

T

tm

x t x

kT me t tm m m

m


Comments about homework problems:

A dilute gas of density n is contained in a cubic box and is in equilibrium with the walls at temperatureT. Find the number of particles per unit area per unit time which collide with the walls and have magnitude of velocity greater than v0.


HW #18 -- continuedA

2

0

3/2/(2 )

0

3

Maxwell Boltzmann distribution:

( )2

Probability per unit time that particle

will hit top face with velocity magnitude

greater than :

( )( )

p m

z zpv

m

m

p pd

f n e

v

P p fm

p

p


Treatment of multicomponent systems including chemical reactions (Sect. 3.10 of 3rd Ed. Reichl)

Summary of thermodynamic potentials (note XV, Y-P)Potential Variables Total Diff Fund. Eq.

U S,X,Ni

H S,Y,Ni

A T,X,Ni

G T,Y,Ni

W T,X,mi

i

iidNYdXTdSdU i

iiNYXTSU

i

iidNXdYTdSdH

i

iidNYdXSdTdA

i

iidNXdYSdTdG

i

iidNYdXSdTd

YXUH

TSUA

YXTSUG

i

iiNTSU


Derivative relationships of thermodynamic potentials

jii NXSiNSNX

i

N

U

X

UY

S

UT

NXSU

,,

i,,

:),,(energy Internal

jii NYSiNSNY

i

N

H

Y

HX

S

HT

NYSH

,,

i,,

:),,( Enthalpy

jii NXTiNTNX

i

N

A

X

AY

T

AS

NXTA

,,

i,,

:),,(energy free Helmholz

jii NYTiNTNY

i

N

G

Y

GX

T

GS

NYTG

,,

i,,

:),,(energy free sGibb'


Properties of the Gibb’s free energy

i ii

U TS PV N i i

i

G U TS PV N

i, , , ,

, ,

Further analysis from Gibb's free energy ( , , ) :

i i j

j j

i i

i

P N T N i T P N

i i Ti i

P N N

G T P N

G G GS V

T P N

S N V NdT dP


Properties of the Gibb’s free energy -- Gibbs phase rule and phase equilibria

Consider a system at uniform temperature T and pressure P with n chemical constituents. How many distinct phases r of this system can exist in equilibrium?

liquid

solid1

solid21

, ,

Enumeration of all of possible components and phases:

where number of component in phase ij

i j

r

i jN i jN N

, ,,

,1 ,2 ,

Generalization of Gibbs free energy:

At equilibrium: = ....

Accounting -- # independent variables: 2 ( 1)

# independent constraints: ( 1)

i j i ji j

i i i r

G N

r n

n r

# degrees of freedom: 2 ( 1) ( 1)

2

f r n n r

f n r


Properties of the Gibb’s free energy -- Gibbs phase rule and phase equilibria -- continued

# degrees of freedom: 2

Examples

1:

for 1, 2 , free to vary

for 2, 1 ( ) on phase boundary curve

for 3, 0 can occur at special ( , ) 'triple point'

f n r

n

r f T P

r f P P T

r f P T

n

2 : such as water and amonium cloride at concentration c

From Schwabl, Statistical Mechanics


Properties of the Gibb’s free energy -- multicomponent ideal gas

i ii

G N

2 2 2

2 2 21

For example, consider a reaction at fixed and :

2H O 2H O

2H O+2H +O 0 General notation: 0i

n

ii

T P

A

1

1

Change in Gibbs free energy with fixed and :

0 at equilibrium

Because of the relationships between the components

it follows that 0

i

i

n

ii

n

ii

T P

dG dN


Multicomponent ideal gas and possible chemical reactions -- continued

1

Change in Gibbs free energy with fixed and :

, 0

Estimation of the chemical potentials:

For each , assume independent ideal gas particles with internal

energies determined by electronic

i

n

ii

T P

T P

i

1/22

1

, internal and kinetic energies

Kinetic energy contributions expressed in terms of thermal wavelength

2

Canonical partition function:

For each species :

n

i

ii

iZ Z

m kT

i

i3

nt ln 1 lneli i i i

i i

VT

NNk Z kT



i3

3

nt

,

1

,

1

int

Helmholz free energy for this system

ln 1 ln

ln

For an ideal gas: where

Let ,

j

eli i i

i i

eli i i

i i iT V N

n

ii

ii

n

i

i

VA kT Z kT

N

VkT

N

PV NkT

T

N

A

N

N N

Nc

NP

int3

0

/, ln

, , ln

eli i i

i i

i i i i

kT Pc kT

c

T P c T kT c P



1

1

1

1

1

0

0

0

0

1

1

Recall: 0

, , ln

ln 0

ln 0

Define: ln ( , ) ln

ln ln ln ( , )

( , )

i i

i

i

i i i i

i i i

i ii i

i

n

ii

n

i

n

i

ni

ii

i

nn

i ii

n

ii

T P c T kT c

T kT c

Tc

kT

TK T P

kT

c c K T P

c K T P

P

P

P

P



1

( , )i

n

ii

c K T P

2 2 21

Example:

2H O+2H +O 0 General notation: 0 n

ii iA

1

2

2 22

2

H O( , )

H Oi

n

ii

c K T P



1

2

2 22

2

H O( , )

H Oi

n

ii

c K T P

2 2 2

2

2

2

2H O 2H O

Suppose H O 1

H

O / 2

x

x

x

3

2( , )

2(1 )

xK T P

x

4/17/2014phy 770 spring 2014 -- lecture 231 phy 770 -- statistical mechanics 12:00 * - 1:45 pm tr...

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