1. gibbs paradox 1

8/13/2019 1. Gibbs Paradox 1

1/19

Lecture 6. Entropy of an Ideal Gas (Ch. 3)

Find (U,V,N,...)the most challenging step

S(U,V,N,...)= kB ln (U,V,N,...)

Solve for U= f(T,V,N,...)

1

,...),,(

U

NVUST

Today we will achieve an important goal: well derive the equation(s) of state

for an ideal gas from the principles of statistical mechanics. We will follow thepath outlined in the previous lecture:

So far we have treated quantum systems whose states in the configuration

(phase) space may be enumerated. When dealing with classical systems withtranslational degrees of freedom, we need to learn how to calculate the

multiplicity.


2/19

Multiplicity for a Single particle

- depends on threerather than twomacro parameters (e.g., N, U, V).

Example: particle in a one-dimensional box

-L L

-L L x

px

px-px

x

2x x

x

L p L p

x p h

The number of microstates:

Q.M.

hpx x

Quantum mechanics (the

uncertainty principle) helps

us to numerate all different

states in the configuration

(phase) space:

pspaceThe total number of ways of filling up the cells in phase space is the

product of the number of ways the space cells can be filled times

the number of ways the momentumcells can be filled


3/19

Multiplicity of a Monatomic Ideal Gas (simplified)

For a molecule in a three-dimensional box: the state of the

molecule is a point in the 6Dspace - its position (x,y,z) and its

momentum (px,py,pz). The number of spacemicrostates is:

There is some momentum distribution of molecules in an ideal

gas (Maxwell), with a long tailthat goes all the way up to p =

(2mU)1/2 (U is the total energy of the gas). However, themomentum vector of an averagemolecule is confined within

a sphere of radius p ~ (2mU/N)1/2(U/N is the averageenergy

per molecule). Thus, for a single averagemolecule:

3x

V

zyx

Vspace

zyx

pppp

p

3

34

The total number of microstates

for Nmolecules:

N

p

N

x

pspace

h

VV

px

pV

333

3

For Nmolecules:

N

spacex

V

3

p

n

p

However, we have over-counted the multiplicity,

because we have assumed that the atoms are

distinguishable. For ind is t ingu ishab le quantum

particles, the result should be divided by N! (the

number of ways of arranging N identical atoms in a

given set of boxes):

3

1

!

N

p

indistinguishable

V V

N h


4/19

mUppp zyx 2222

mUpppppp zyxzyx 2222222212121

1 particle -

2 particles -

The accessible momentum volume for Nparticles = the area of a 3N-dimensional hyper-

sphere p13

2/3

!12

3

2area""

NN

rN

p

h

mU

N

V

N

NN

N

2

2

!2/3!

1 2/3

2

Monatomic

ideal gas:

(3Ndegrees

of freedom)

N =1

22

2/3

42/!2/1!2/1

2rr

f N- the total # of quadratic degrees of freedom 2/Nf

UU

The reason why m matters: for a

given U, a molecule with a larger

mass has a larger momentum, thus a

larger volume accessible in the

momentum space.

px

py

pz

pN

Um

N

V

h

eNVU

N

ishableindistingu

2

3

4,,

2/3

3

2/5

Momentum constraints:

More Accurate Calculation of N(I)


5/19

More Accurate Calculation of N(II)

For a particle in a box (L)3: (Appendix A)

If p>>p, the total degeneracy (multiplicity)

of 1 particle with energy U is:

Plug in the area of the hyper-sphere:

px

py

pz

3/ 25/ 2

3

4, , 2

3

N

indistinguishable

e V mU U V N p

h N N

2 2 2 22 2 2

2

, ,2 8

x y z

x y z x y z

p p p hE n n n n n n

m mL

2x

hp

L

3

31 3 333 3 3 3

21 1

2 2

pp pLS p VU S p S p

h hp

If p>>p, the total degeneracy (multiplicity) of N

indistinguishable particle with energy U is:

331

!

Np

N NN

VU S p

N h

, , 0x y zn n n


6/19

Entropy of an Ideal Gas

f 3 (monatomic), 5 (diatomic), 6 (polyatomic)

(Monatom ic id eal g as)

2

5

3

4ln),,(

2/3

2

N

U

h

m

N

VkNUVNS B

3/ 2

2

3 5 4 3ln ln ln ln ( , )

2 3 3 2B B B B

V U m V U Nk Nk Nk Nk N m

N N h N N

The Sackur-Tetrode equation:

In general, for a gas of

polyatomic molecules:),(ln

2ln),,( mNTkN

f

N

VkNTVNS BB

, , ln , ,BS U V N k U V N

3/ 2

2

4 5

( , , ) ln ln 23 2B BV m U

S N V U Nk Nk pN h N

an average volume

per molecule

an average energy

per molecule


7/19

Problem Two cylinders (V = 1 liter each) areconnected by a valve. In one of the

cylinders Hydrogen (H2) at P= 105

Pa, T = 200C , in another one

Helium (He) at P = 3105 Pa,

T=1000C. Find the entropy change

after mixing and equilibrating.

i

f

B

i

f

B

T

TkN

f

V

VkNTVNS ln

2

ln),,(

H2 :

ftotal TUTUTU 2211

2

2

1

1

21

21

2211

35

35

2

3

2

52

3

2

5

T

P

T

P

PP

kNkN

TkNTkNT

BB

BB

f

1

ln2

52ln

2 T

TkNkNS

f

BBH He:2

22 ln2

32ln

T

TkNkNS

f

BBHe

2

2

1

121 ln3ln5

2

2ln2

T

TN

T

TN

kkNNSSS

ffBBHeHtotal J/K67.0totalS

The temperature

after mixing: fBBBB TkNkNTkNTkN

212211

2

3

2

5

2

3

2

5

For each gas:


8/19

Entropy of MixingConsider two different ideal gases (N1,

N2) kept in two separate volumes (V1,V2)

at the same temperature. To calculate the

increase of entropy in the mixing process,

we can treat each gas as a separate

system. In the mixing process, U/N

remains the same (T will be the same

after mixing). The parameter that

changes is V/N:

2534ln),,(

2/3

2 NU

hm

NVkNUVNS B

The total entropy of the system is greater after mixingthus,

mixing is irreversible.

2

2

1

1 lnlnV

VN

V

VN

k

SS

k

S

B

BA

B

total

if N1=N2=1/2N, V1=V2=1/2V 2ln2/ln

22/ln

2N

V

VN

V

VN

k

S

B

total


9/19

Gibbs Paradox

If two mixing gases are of the same kind (indist inguishablemolecules):

1 2 1 2total 1 2 1 21 2 1 2

1 2 1 21 2

1 2 1 2

/ ln ln ln

ln ln 0 if

B total A B B

V V V V S k S S S k N N N N

N N N NV N V N N N N

N NN V N V V V V

NN

N

N

N mUNh

V

N

32/3

3 2

!2

3

2

!

1

BB kN

N

U

h

m

N

VkNUVNS

2

5

3

4ln),,(

2/3

2

Quantum-mechanical ind ist inguishabi l i ty is important! (even though this equation

applies only in the low density limit, which is classicalin the sense that the distinction

between fermions and bosons disappear.

Stotal = 0because U/Nand V/Navailable for each molecule remain the same after mixing.

2

2

1

1 lnlnV

VN

V

VN

k

SS

k

S

B

BA

B

total

- applies only if two gases are different!


10/19

Problem

Two identical perfect gases with the same pressure P and the same number of

particles N, but with different temperatures T1and T2, are confined in two vessels,

of volume V1and V2 , which are then connected. find the change in entropy after

the system has reached equilibrium.

BBBBB kNTkh

m

N

VkNkN

N

U

h

m

N

VkNUVNS

2

5

2

3

3

4ln

2

5

3

4ln),,(

2/3

2

2/3

2

BfBf kNTN

VVkNS 2

2

5

2ln2

2/321

21

2

212121

21

2

21

21

2

21

21

2

21

2

212/3

2

2/3

1

3

21

22

21

4ln

2

5

22

4ln

2

3

4ln

ln23

4lnln

2ln

TT

TTTTNkVVP

TT

TT

VV

VV

TT

T

VVVV

TT

T

VVN

NVV

kNS

B

ff

B

at T1=T2, S=0, as it should be (Gibbs paradox)

BBBBi kNTNV

kNkNTN

VkNSSS 2

5ln2

5ln

2/3

222/3

1121

221 TTTf

- prove it!


11/19

An Ideal Gas: from S(N,V,U) - to U(N,V,T)

2/,, NfNUVNfNVU Ideal gas:(fNdegrees of freedom)

),(lnln2

,, mNN

VNk

N

UNk

fNVUS BB

U

Nkf

U

S

T

B

2

1

TkNfTVNU B

2),,(

The heat capacity for

a monatomic ideal gas:B

NV

V kNf

T

UC

2,

- in agreement with the equipartition theorem, the total energy should be

kBT times the number of degrees of freedom.

- the energy

equation of state


12/19

Partial Derivatives of the Entropy

We have been considering the entropy changes in the processes where two

interacting systems exchanged the thermal energy but the volume and the number of

particles in these systems were fixed. In general, however, we need more than justone parameter to specify a macrostate, e.g. for an ideal gas

Today we will explore what happens if we let the V

vary, and analyze the physical meaning of the

other two partial derivatives of the entropy:NUV

S

,

We are familiar with the physical meaning only

one partial derivative of entropy:

TU

S

NV

1

,

NVUkNVUSS B ,ln, ,,

When all macroscopic quantities S,V,N,Uare allowed to vary:

NdN

SVd

V

SUd

U

SdS

VUUNVN ,,,


13/19

Mechanical Equilibrium

and Pressure

Letsfix UA,NAand UB,NB, but allow Vto vary

(the membrane is insulating, impermeable for

gas molecules, but its position is not fixed).

Following the same logic, spontaneous

exchange of volume between sub-systems

will drive the system towards mechanicalequilibrium (the membrane at rest). The

equilibrium macropartition should have the

largest (by far) multiplicity (U, V) and

entropy S(U, V).

UA , VA , NA

UB, VB, NB

0

B

B

A

A

A

B

A

A

A

AB

V

S

V

S

V

S

V

S

V

S

BA VV B

B

A

A

V

S

V

S

,

B

U N

k NS PV V T

In mechanical equilibrium:

- the volume-per-molecule should be the same

for both sub-systems, or, if Tis the same, P must

be the same on both sides of the membrane.

S AB

VAVAeq

S A

S B

NVNUNU U

S

V

S

V

STP

,,,

/

The stat. phys.

definition of pressure:

e.g. ideal gas


14/19

The Pressure Equation of State for an Ideal Gas

Ideal gas:(fNdegrees of freedom)

V

kNT

V

STP B

NU

,

TkNPV B

UkN

f

U

S

T B

NV

1

2

1

,

The energy equation of state (U T):

TkNf

U B2

The pressure equation of state (P T):

- we have finally derived the equation of state of an ideal gas from first principles!

),(ln

2

ln),,( mNTkNf

N

VkNTVNS BB


15/19

Thermodynamic identity I

, ,N V N U

S SdS dU dV U V

Lets assume N is fixed,

thermal equilibrium:TU

S

NV

1

,

mechanical equilibrium:,U N

S P

V T

1 PdS dU dV

T T dU TdS PdV i.e.


16/19

Quasi-Static Processes

constVTS f

2/

01

1VT const

The quasi-stat ic adiabatic p rocess with an ideal gas :

constVT 1constPV - weve derived these equations from the 1stLaw and PV=RT

On the other hand, from the Sackur-Tetrode equation for an isentropic process :

Quasist atic adiabatic(Q= 0) processes: 0Sd isentropic processes

(al lprocesses)

dVPSdTUd WQUd

(quasi-stat icprocesses with fixed N)

SdTQThus, for quasi-stat ic p rocesses:T

QSd

TQSd - is an exact differential (S is a state function).Thus, the factor 1/T converts Q into an exact

differential for quasi-static processes.

Comment on State Funct ions : P

V

0

0

Q

S


17/19

Problem:

(a) Calculate the entropy increase of an ideal gas in an isothermalprocess.

(b) Calculate the entropy increase of an ideal gas in an i sochor icprocess.

2/,, NfNUVNfNVU

You should be able to do this using (a) Sackur-Tetrode eq. and (b)T

QSd

dV

V

TdT

fNkPdVdUQ B

2

V

dV

T

dTfNk

T

QdS B

2

2/ln fB VTNgNkS

i

f

BconstTV

VNkS ln

i

f

BconstVT

TNk

fS ln

2

(all the processes are quasi-stat ic)

Lets verify that we get the same result with approaches a)and b) (e.g., for T=const):

Since U = 0,

i

f

B

V

V

B

V

VTkNdV

V

TkNWQ

f

ln T

QS

(Pr. 2.34)


18/19

Problem:

A bacterias of mass M with heat capacity (per unit mass) C, initially at temperature

T0+T, is brought into thermal contact with a heat bath at temperature T0..(a) Show that if T


19/19

Problem (cont.)

(b) for the (non-equilibrium) state with Tbacteria= 300.03K is greater than

in the equilibrium state with Tbacteria= 300K by a factor of

6301450

23

20

10/1038.1

/102expexp

0

0

eKJ

KJ

k

S

B

total

TT

T

The number of 1ps trials over the lifetime of the Universe: 3012

18

1010

10

Thus, the probability of the event happening in 1030trials:

30 630# events probability of occurrence of an event 10 10 0

1. gibbs paradox 1

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