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: we’ll derive the equation(s) of state for an ideal gas from the principles of statistical mechanics. We will follow the path 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 with translational degrees of freedom, we need to learn how to calculate the multiplicity.

Multiplicity for a Single particle- is more complicated than that for an Einstein solid, because it depends on three rather than two macro parameters (e.g., N, U, V).

Example: particle in a one-dimensional “box”

-L L

-L L x

px

px

-px x

The total number of microstates: x

L

px

pL

x

x

x

x

x pL

px

pL

The number of microstates:

Q.M.

xpx

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 “momentum” cells can be filled

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 6D space - its position (x,y,z) and its momentum (px,py,pz). The number of “space” microstates is:

There is some momentum distribution of molecules in an ideal gas (Maxwell), with a long “tail” that goes all the way up to p = (2mU)1/2 (U is the total energy of the gas). However, the momentum vector of an “average” molecule is confined within a sphere of radius p ~ (2mU/N)1/2 (U/N is the average energy per molecule). Thus, for a single “average” molecule:

3x

V

zyx

Vspace

zyxp ppp

p

3

34

The total number of microstates for N molecules:

NN

x

pspace

pV

px

pV

3

3

33

3

For N molecules: N

space x

V

3

p

n

p

However, we have over-counted the multiplicity, because we have assumed that the atoms are distinguishable. For indistinguishable quantum particles, the result should be divided by N! (the number of ways of arranging N identical atoms in a given set of “boxes”):

N

ishableindistingu

pV

N

3

3

!

1

More Accurate Calculation of p

Momentum constraints:

mUppp zyx 2222

mUpppppp zyxzyx 222

22

22

21

21

21

1 particle -

2 particles -

The accessible momentum volume for N particles = the “area” of a 3N-dimensional hyper-sphere p

132/3

!12

32

area""

NN

rN

pmU

N

V

N

NN

N

2

2

!2/3!

12/3

2

Monatomic ideal gas:

(3N degrees of freedom)

N =1

222/3

42/!2/1!2/1

2rr

f N- the total # of “quadratic” degrees of freedom 2/NfUU

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

p

N

U

N

V

h

meNVU

N

ishableindistingu

2

3,,

2/3

3

2/5

Entropy of an Ideal Gas

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

Monatomic ideal gas:

2

5

3

4ln),,(

2/3

2 N

U

h

m

N

VkNUVNS B

),(ln2

3ln

3

4ln

3

5

2

3ln

2

2/3

mNN

UkN

N

VkN

h

mkN

N

U

N

VkN BBBB

the Sackur-Tetrodeequation

In general, for a gas of polyatomic

molecules:),(ln

2ln),,( mNTkN

f

N

VkNTVNS BB

pmU

N

V

N

NN

N

2

2

!2/3!

12/3

2

pkNN

U

h

m

N

VkNUVNS BB

2ln

2

5

3

4ln),,(

2/3

2

an average volume per molecule

an average energy per molecule

Problem Two cylinders (V = 1 liter each) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105

Pa, T = 200C , in another one – Helium (He) at P = 3·105 Pa, T=1000C. Find the entropy change after mixing and equilibrating.

i

fB

i

fB T

TkN

f

V

VkNTVNS ln

2ln),,(

H2 :

ftotal TUTUTU 2211

2

2

1

1

21

21

2211

35

35

2

3

2

52

3

2

5

T

P

T

PPP

kNkN

TkNTkNT

BB

BB

f

1

ln2

52ln

2 T

TkNkNS fBBH He :

222 ln

2

32ln

T

TkNkNS fBBHe

22

1121 ln3ln5

22ln

2 T

TN

T

TN

kkNNSSS ffBBHeHtotal J/K 67.0total S

The temperatureafter mixing: fBBBB TkNkNTkNTkN

212211 2

3

2

5

2

3

2

5

For each gas:

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:

2

5

3

4ln),,(

2/3

2 N

U

h

m

N

VkNUVNS B

The total entropy of the system is greater after mixing – thus, mixing is irreversible.

22

11 lnln

V

VN

V

VN

k

SS

k

S

B

BA

B

total

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

ln22/

ln2

NV

VN

V

VN

k

S

B

total

Gibbs Paradox

If two mixing gases are of the same kind (indistinguishable molecules):

V

N

V

N

V

N

V

N

N

VN

V

N

N

VN

N

VN

N

VN

NN

VVNNSSkS BAB

2

2

1

1

2

22

1

11

2

22

1

11

21

2121

0lnln

lnlnln/

if

total

NN

N

N

N mUNh

V

N

32/3

32

!2

32

!

1

BB kNN

U

h

m

N

VkNUVNS

2

5

3

4ln),,(

2/3

2

Quantum-mechanical indistinguishability is important! (even though this equation applies only in the low density limit, which is “classical” in the sense that the distinction between fermions and bosons disappear.

Stotal = 0 because U/N and V/N available for each molecule remain the same after mixing.

22

11 lnln

V

VN

V

VN

k

SS

k

S

B

BA

B

total

- applies only if two gases are different !

ProblemTwo identical perfect gases with the same pressure P and the same number of particles N, but with different temperatures T1 and T2, are confined in two vessels, of volume V1 and 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

22121

2121

221

21

221

21

2

21

221

2/32

2/31

3

21

22

21

4ln

2

5

22

4ln

2

3

4ln

ln2

3

4lnln

2ln

TT

TTTTNkVVP

TT

TT

VV

VV

TT

T

VV

VV

TT

T

VV

N

N

VV

kN

S

B

ff

B

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

BBBBi kNTN

VkNkNT

N

VkNSSS

2

5ln

2

5ln 2/3

222/3

11

21

221 TT

T f

- prove it!

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

2/,, NfNUVNfNVU Ideal gas:(fN degrees of freedom)

),(lnln2

,, mNN

VNk

N

UNk

fNVUS BB

U

Nkf

U

S

TB

2

1

TkNf

TVNU B2),,(

The heat capacity for a monatomic ideal gas: B

NVV kN

f

T

UC

2,

- in agreement with the equipartition theorem, the total energy should be ½kBT times the number of degrees of freedom.

- the “energy” equation of state

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 just one parameter to specify a macrostate, e.g for an ideal gas

Today we will explore what happens if we let the other two parameters (V and N) vary, and analyze the physical meaning of the other two partial derivatives of the entropy:

NUV

S

,

VUN

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,U are allowed to vary:

NdN

SVd

V

SUd

U

SdS

VUUNVN ,,,

Thermodynamic Identity for dU(S,V,N)

NVUSS ,, if monotonic as a function of U

(“quadratic” degrees of freedom!), may be inverted to give

NVSUU ,,

dNN

UdV

V

UdS

S

UdU

VSNSNV ,,,

compare withTU

S

VN

1

,

SVSNVNN

UP

V

UT

S

U

,,,

pressure chemical potential

This holds for quasi-static processes (T, P, are well-define throughout the system).

NddVPSdTUd

shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J.

- the so-called thermodynamic identity for U

The Exact Differential of S(U,V,N)

NdT

dVT

PUd

TdS

1

The coefficients may be identified as:

NdN

SVd

V

SUd

U

SNVUdS

VUUNVN ,,,

,,

TN

S

T

P

V

S

TU

S

UVUNVN

,,,

1

Again, this holds for quasi-static processes (T and P are well defined).

NddVPSdTUd

Type of interaction

Exchanged quantity

Governing variable

Formula

thermal energy temperature

mechanical volume pressure

diffusive particleschemical potential

NVU

S

T ,

1

NUV

S

T

P

,

VUN

S

T ,

connection between

thermodynamics and statistical

mechanics

Mechanical Equilibrium and Pressure

Let’s fix UA,NA and UB,NB , but allow V to 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 mechanical equilibrium (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

V

Nk

T

P

V

S B

NU

,

- the sub-system with a smaller volume-per-molecule (larger P at the same T) will have a larger S/V, it will expand at the expense of the other sub-system.

In mechanical equilibrium:

- the volume-per-molecule should be the same for both sub-systems, or, if T is the same, P must be the same on both sides of the membrane.

S AB

VAVAeq

S A

S B

NVNUNUU

S

V

S

V

STP

,,,

/

The stat. phys. definition of pressure:

The “Pressure” Equation of State for an Ideal Gas

Ideal gas:(fN degrees of freedom)

V

kNT

V

STP B

NU

,

TkNPV B

UkN

f

U

S

T BNV

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!

),(ln2

ln),,( mNTkNf

N

VkNTVNS BB

Quasi-Static Processes

constVTS f 2/0 constVT 1

The quasi-static adiabatic process with an ideal gas :

constVT 1constPV - we’ve derived these equations from the 1st Law and PV=RT

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

Quasistatic adiabatic (Q = 0) processes: 0Sd isentropic processes

(all processes)

dVPSdTUd WQUd

(quasi-static processes with fixed N)

SdTQ Thus, for quasi-static processes :T

QSd

T

QSd

- 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 Functions : P

V

0

0

Q

S

Problem:

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

(b) Calculate the entropy increase of an ideal gas in an isochoric process.

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

fBconstT V

VNkS ln

i

fBconstV T

TNk

fS ln

2

(all the processes are quasi-static)

Let’s verify that we get the same result with approaches a) and b) (e.g., for T=const):

Since U = 0,

i

fB

V

V

B

V

VTkNdV

V

TkNWQ

f

ln T

QS

(Pr. 2.34)

Problem:

A body 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<<T0, the increase S in the entropy of the entire system (body+heat bath) when equilibrium is reached is proportional to (T)2.

(b) Find S if the body is a bacteria of mass 10-15kg with C=4 kJ/(kg·K), T0=300K, T=0.03K.

(c) What is the probability of finding the bacteria at its initial T0+T for t =10-12s over the lifetime of the Universe (~1018s).

0ln0

00

0

0

0

TT

TC

T

TCd

T

QΔS

T

TT

T

TT

body

(a)

0000

0

0

T

TC

T

TCd

T

QΔS

T

TTbathheat

02

...32

1lnln2

0

32

0

0

0

T

TC

T

TTC

T

TCΔSΔSΔS bathheatbodytotal

(b)

KJKJ

T

TCΔStotal /102

300

03.0

2

/101104

220

21532

0

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

630145023

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 1030 trials:

01010# 63030 event an of occurrence ofy probabilitevents

Pr. 3.32. A non-quasistatic compression. A cylinder with air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very fast, non-quasistatic) by a piston (S = 0.01 m2, F = 2000N, x = 10-3m). Calculate W, Q, U, and S.

VPSTU

WQU holds for all processes, energy conservationquasistatic, T and P are well-defined for any intermediate state

quasistatic adiabatic isentropic non-quasistatic adiabatic

JmPa 11010101

111

11

11

1

1)(

2335

1

1

11

x

xVP

x

xVP

V

VVP

VV

VP

dVV

VPconstPVdVVPW

ii

ii

f

iii

if

ii

V

V

ii

V

V

f

i

f

i

fi

V

V

VVPdVPWf

i

J2

m10m10Pa102 23225

The non-quasistatic process results in a higher T and a greater entropy of the final state.

S = const along the isentropic

line

P

V Vi Vf

1

2

2*

Q = 0 for both

Caution: for non-quasistatic adiabatic processes, S might be non-zero!!!

An example of a non-quasistatic adiabatic process

NfkUkNf

VkNNVUS BBB lnln2

ln,, Direct approach:

JWU 2210V

V

J/K K

10m 10Pa 10 2-33-5

300

1

300

250

2

2

510

2

ln2

ln

2

i

ii

i

if

i

ifB

i

fB

i

fB

T

VP

U

UUf

V

VVkN

U

UkN

f

V

VkNS

J 250m 10Pa 102

5

2

5

233-5 iiiBi VPTkN

fU

adiabatic quasistatic isentropic 0Q WU VPTkNf

B 2

VV

TkNTkN

f BB

2constTV f 2/

f

f

i

i

f

V

V

T

T/2

0ln2

2ln

i

fB

i

fB V

V

fkN

f

V

VkNS

adiabatic non-quasistatic

K/J300

2

K300

2J

T

Q

T

US

The entropy is created because it is an irreversible, non-quasistatic compression.

P

V Vi Vf

To calculate S, we can consider any quasistatic process that would bring the gas into the final state (S is a state function). For example, along the red line that coincides with the adiabat and then shoots straight up. Let’s neglect small variations of T along this path ( U << U, so it won’t be a big mistake to assume T const):

KJK

1J2J(adiabata) /

300

1

3000

T

QS

U = Q = 1J

For any quasi-static path from 1 to 2, we must have the same S. Let’s take another path – along the isotherm and then straight up:

1

2

P

V Vi Vf

1

2

U = Q = 2J

K/J300

1

K300

10m 10Pa 10

ln )(1

2-33-5

x

x

T

VP

V

V

T

VP

V

dV

T

VPdVVP

TS

ii

i

fii

V

V

ii

V

V

f

i

f

i

isotherm:

“straight up”:

Total gain of entropy: K/J300

1K/J

300

2K/J

300

1S

The inverse process, sudden expansion of an ideal gas (2 – 3) also generates entropy (adiabatic but not quasistatic). Neither heat nor work is transferred: W = Q = 0 (we assume the whole process occurs rapidly enough so that no heat flows in through the walls).

i

fBrev V

VTkNW ln

The work done on the gas is negative, the gas does positive work on the piston in an amount equal to the heat transfer into the system

J/K300

1ln0

x

x

T

VP

V

VkN

T

W

T

QSWQ

i

ii

f

iB

revrevrevrev

P

V Vi Vf

1

2

3 Because U is unchanged, T of the ideal gas is unchanged. The final state is identical with the state that results from a reversible isothermal expansion with the gas in thermal equilibrium with a reservoir. The work done on the gas in the reversible expansion from volume Vf to Vi:

Thus, by going 1 2 3 , we will increase the gas entropy by J/K300

2321 S