Lecture 7. Thermodynamic Identities (Ch. 3)

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

, , ,N V N U U V

S S SdS dU dV dN

U V N

TU

S

NV

1

,

,U N

S P

V T

,

?U V

S

N

Diffusive Equilibrium and Chemical Potential

0,

AA NVA

AB

U

S0

,

AA VUA

AB

N

S

B

B

A

A

N

S

N

S

VUN

ST

,

BA

Sign “-”: out of equilibrium, the system with the larger S/N will get more particles. In other words, particles will flow from from a high /T to a low /T.

Let’s fix VA and VB (the membrane’s position is fixed), but assume that the membrane becomes permeable for gas molecules (exchange of both U and N between the sub-systems, the molecules in A and B are the same ).

UA, VA, NA UB, VB, NB

TN

S

VU

,

For sub-systems in diffusive equilibrium:

In equilibrium, BA TT

- the chemicalpotential

0.2

0.4

0.6

0.8

0.2

0.4

0.6

0.8

-6

-5

-4

0.2

0.4

0.6

0.8

S

AN

AU

Chemical Potential: examples

Einstein solid: consider a small one, with N = 3 and q = 3.

10

!)1(!

!1)3,3(

Nq

NqqN 10ln)3,3( BkqNS

let’s add one more oscillator: 20ln)3,4( BkqNS

NdT

UdT

dS

1

SVN

U

,

SN

U

To keep dS = 0, we need to decrease the energy, by subtracting one energy quantum.

Thus, for this system

S

N

U

2

5ln

3

4ln),,( 2/5

2/3

2NU

h

mVkNUVNS B

N

VTmgTkTk

h

m

N

VTk

N

ST BBB

VU

2/32/3

2,

ln2

ln

Monatomic ideal gas:

At normal T and P, ln(...) > 0, and < 0 (e.g., for He, ~ - 5·10-20 J ~ - 0.3 eV.

Sign “-”: usually, by adding particles to the system, we increase its entropy. To keep dS = 0, we need to subtract some energy, thus U is negative.

The Quantum Concentration

At T=300K, P=105 Pa , n << nQ. When n nQ, the quantum statistics comes into play.

2/3

2

2

Tkh

mn BQ

- the so-called quantum concentration (one particle per cube of side equal to the thermal de Broglie wavelength). When nQ >> n, the gas is in the classical regime.

2/52/3

32/322/3

2 2ln

2ln

2ln

Tk

P

m

hTk

Tmk

hnTkTk

h

m

N

VTk

B

BB

BBB

n=N/V – the concentration of molecules

The chemical potential increases with the density of the gas or with its pressure. Thus, the molecules will flow from regions of high density to regions of lower density or from regions of high pressure to those of low pressure .

0

QB

BB n

nTk

Tmk

hnTk ln

2ln

2/32

when n nQ, 0

2/3

23

1

h

Tmkn

Tmk

h

p

h B

dB

Q

B

dB

when nincreases

Entropy Change for Different Processes

Type of interaction

Exchanged quantity

Governing variable

Formula

thermal energy temperature

mechanical volume pressure

diffusive particles chemical potential

NVU

S

T ,

1

NUV

S

T

P

,

VUN

S

T ,

The last column provides the connection between statistical physics and thermodynamics.

The partial derivatives of S play very important roles because they determine how much the entropy is affected when U, V and N change:

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

Thermodynamic Identities

is an intensive variable, independent of the size of the system (like P, T, density). Extensive variables (U, N, S, V ...) have a magnitude proportional to the size of the system. If two identical systems are combined into one, each extensive variable is doubled in value.

With these abbreviations:

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 coefficients may be identified as: TN

S

T

P

V

S

TU

S

UVUNVN

,,,

1

This identity holds for small changes S provided T and P are well defined.

- the so-called thermodynamic identity

The 1st Law for quasi-static processes (N = const):

The thermodynamic identity holds for the quasi-static processes (T, P, are well-define throughout the system)

dU TdS PdV dN

dU TdS PdV

1 PdS dU dV dN

T T T

, , ,N V N U U V

S S SdS dU dV dN

U V N

The Equation(s) of State for an Ideal Gas

2/,, NfNUVNgNVU Ideal gas:(fN degrees of freedom) / 2, , ln f

BS U V N Nk g N VU

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! In other words, we can calculate the thermodynamic information for an isolated system by counting all the accessible microstates as a function of N, V, and U.

Ideal Gas in a Gravitational Field

Pr. 3.37. Consider a monatomic ideal gas at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy. Assuming that the atmosphere is isothermal (not quite right), find and re-derive the barometric equation.

NmgzUU kin

NddVPSdTUd

3/ 2

2,

2( ) lnB B

S V

U V mz mgz k T k T mgz

N N z h

In equilibrium, the chemical potentials between any two heights must be equal:

2/3

2

2/3

2

2

)0(ln

2

)(ln Tk

h

m

N

VTkmgzTk

h

m

zN

VTk BBBB

)0(ln)(ln NTkmgzzNTk BB Tk

mgz

BeNzN

)0()(

note that the U that appears in the Sackur-Tetrode equation represents only the kinetic energy

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 (A = 0.01 m2, F = 2000N, x = 10-3m). Calculate W, Q, U, and S.

VPSTU

WQU holds for all processes, energy conservation

quasistatic, 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:

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

5 5 5

ln ln2 2

1

2 10 10 10 1J/K

300 300

f f f i f iB B B B

i i i i

i i i

i i i i

i

i

V U V V U Uf fS Nk Nk Nk Nk

V U V U

P U P V P V P VV U

T T T T

P P V

T

J 250m 10Pa 102

5

2

5

233-5 iiiBi VPTkN

fUJWU 2210

V

V

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 adiabata 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):

K/J300

1

K300

1J2J(adiabata) 0

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

ii

ii

f

iB

revrevrevrev V

V

T

VP

V

VNk

T

W

T

QSWQ

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