chapter 4 thermodynamic potentials. so that also the temperature,pressure,and chemical potential are...

Chapter 4 Thermodynamic Potentials

so that also the temperature,pressure,and chemical potential are known as functions of the natural variables.A similar assertion holds also for the entropy S(U,V,N,...),if we rearrange Equation(4.6):

,...,nvS

UT

,...,nsV

UP

,...,vsN

U

(4.7)

dNPdVTdSdU {4.6}

For example,if we know U(S,V,N,...),it holds that

1. Entropy and energy as thermodynamic potentials

...1

dNT

dVT

PdU

TdS

,...,

1

nvU

S

T

Equations (4.7)and (4.9),respectively, are the equations of state of the system. On the other hand,knowing all equations of state we may calculate the entropy and the internal energy,respectively,as functions of the natural variables by integration.

,...,vUN

S

T

,...,nUV

S

T

P

(4.9)

Example 4.1:

Let us consider the entropy of the ideal gas,as given in Equation (2.40)

p

p

T

TPTNsNkPTNS 0

2/5

00000 ln,,,,

The entropy of the ideal gas

using U = NKT and pV=NKT(U0 = N0KT0 and p0V0=N0KT0 respectively),we obtain

0

2/3

0

2/5

00000 ln,,,,

V

V

U

U

N

NVUNsNkVUNS (4.10)

Knowing Equation (4.10) all equations of state of the ideal gas can be obtained by partial differentiation according to Equation (4.9),

{4.11}

{4.12} {4.13}

NkTUU

NkTU

S

NV 2

31

2

31

,

NkTPVV

NkT

P

V

S

NU

1

,

kV

V

U

U

N

Nsk

TN

S

VU 2

5ln

0

2/3

0

2/5

00

,

Because of these properties the entropy is a so-called thermodynamic potential.

The extensive state variables U,S,V,N,…are very useful for isolated systems,where they assume constant values in equilibrium.

Therefore it is reasonable to look for other thermodynamic potentials which have quite analogous properties to the entropy or the energy,but which depend on the conjugated intensive variables.

2 The free energy

We start from the internal energy U(S,V,N,...)as a function of the natural variables. Euler’s equation is U=TS-PV+μNone defines

which is called the free energy or Helmholtz potential.

F=U-TS =-pV+ μ N {4.36}

du =T dS-p dV+ μ dN+... {4.37}

Correspondingly,the total differential of F isdF=du-S dT-T ds =-S dT-p dV+ μ dN+...{4.38}

Hence,the free energy is a function of T,V,N,...,which contains exactly the same information as the internal energy U,but which now depends on the temperature instead of the entropy.In particular one obtains from Equations(4.38)the equations of state

Here we have employed Euler's equation(2.72).The total differential of U reads

To understand the importance of the free energy we consider a nonisolated system in a heat bath of constant temperature T (Figure 4.3).

TVNTNV N

F

V

FP

T

FS

,,,

,,

dF =-S dT-p dV+ μ dN+...

Figure 4.3 isothermal system.

Let be the heat exchanged with the heat bath (as seen from the system)and the remaining work exchanged with the heat bath.

Then we have,according to the first law,for the change in internal energy of the partial systems:

sysQ

sysW

syssyssys WQdU

we have discussed the following inequalities,which arealso valid for partial systems:

irrrev QQTdS

irrrev WW

This relation we know already from Equation (2.50).As seen from the system we thereforehave

For a given constant temperature we may also write this as follows:

irrsys

revsyssyssys WWTdSdU

irrsys

revsyssyssyssys WWTSUddF

Let the work performed be , then the entropy of the isolated total system has a maximum f and the free energy of the isothermal partial system has a minimum. In particular,processes which diminish the free energy happen spontaneously and irreversibly in an isothermal system.Since

for

0sysirrW

0 TdSdUTSUddF

.0 constTandW irrsys

In general ,an isothermal system which does not exchange work with its surroundings strives for a minimum of the free energy. Irreversible processes happen spontaneously ,until the minimum

min,0 FFdF

Is reached.

Analogously,processes at constant temperature, for which the decrease of the internal energy is larger than the decrease of the entropy,also happen spontaneously.

Equation (4.59) defines the enthalpy,which is also a thermodynamic potential,in the variables S,p and N.The total differential of the enthalpy reads

dNVdPTdS

H=U +pV=TS+ μ N {4.59}

dH=dU +p dV+V dp

４．３　 The enthalpy（焓） After extensively discussing the principle of the free energy and the pair of variables T and S, 　　　　　　　　Ｆ＝Ｕ－ＴＳit is not difficult to transfer this method also to other pairs of variables.

If the enthalpy H(S,p,N,...)is known,all other state quantities may be obtained by partial differentiation as for U and F,

VPNSNP N

H

P

HV

S

HT

,,,

,,

As with all other thermodynamic potentials the enthalpy c ａｎ　ｉｎ　 principle be calculated for any system.

dNVdPTdS dH

Especially for isobaric systems (p=const,dp=0) we find with the help of the first law,

PdVWQWQdU other

PPPPP

dVPdUVdPPdVdUPVUddH

p

revotherpP

WQdH

for reversible changes of state at constant pressure, that

Let us consider especially an isobaric and adiabatic system with p=const. And . Then we have according to Equation(4.62)

0Q

adp

revotheradP

WdH,,

irrother

revother WW

adp

irrotheradp

revotheradP

WWdH,,,

If, especially for an irreversible process in the isobaric,adiabatic system ,I.e.,if no utilizable work is performed,we have  

0irrotherW

0dH

Again,for irreversible processes is valid and thus

(1)If we add an amount of heat to a system at constant volume,we have,with ,

Q

0W

VQdU

VVV T

U

dT

QC

so that the amount of heat directly increases the internal energy.For the specific heat at constant volume it holds that

(2)However,if the heat is added under constant pressure,generally the volume of the system will change,and a certain volume work will be performed:

PdVQdUP

Q

The internal energy is not very appropriate for describing this process,since not only the temperature,but also the volume of the system changes.

At constant pressure,however, Equation above can be simply put in the form

PP

QpdVpdVQPVUddH

For an ideal gas

NkTNPTH2

5,,

Which also follows directly from NkTNkTPVUH 2

3

Thus we obtain for an ideal gas

NkCP 2

5

while NkCV 2

3

PPP T

H

dT

QC

PPP T

H

dT

QC

For systems with given temperature and pressure :

The free enthalpy（自由焓，吉布斯函数）

G=U - TS+pV {4.81}

The corresponding thermodynamic potential is the free enthalpy introduced by J.W.Gibbs (1875),for which reason it is also called the Gibbs’ potential.

J.W.Gibbs1839-1903

dG=dU -T dS -S dT+p dV+V dp =-S dT+V dp+ μ dN+... {4.82}

PTNTNP N

G

P

GV

T

GS

,,,

,,

Consequently,G indeed depends only on T,p,and N.If the function G(T,p,N) is known, we can obtain all further quantities by partial differentiation,

(4.83)

The total differential of the free enthalpy reads:

Euler's equation for a system of one particle species which does not exchange any further kinds of work,reads U=TS-pV+ μ N {4.84} From this follows immediately by comparison with Equation (4.81)

Equations above yield again the equation of state of the system.Using Euler's equation, which must be fulfilled in any case,we can identify the Gibbs' free enthalpy somewhat more explicitly.

G=U-TS+pV= μ N

The third of Equations (4.83) is thus trivially fulfilled for the free enthalpy;i.e.,

NGNGPT

//,

Hence,G is directly proportional to the particle number,and the free enthalpy per particle is identical with the chemical potential. The free enthalpy is especially convenient for systems at a given temperature and given pressure.The quantity represents a combination of free energy and enthalpy ,a fact,which is expressed also in its name. G=U-TS+pV

Figure 4.6 isothermal,Isobaric System.

To understand the meaning of the free enthalpy,we form an isolated system consisting of the isothermal,isobaric system shown in Figure 4.6 and its surroundings (the heat bath).

Aaccording to Equation (4.86),

0 revothersyssyssystot WPdVdUTdSTdS

irrother

revother WW

VdPPdVSdTTdSdUPVTSUddG

0 bathsystot dSdSdS

{4.88}which can also be written as follows (the subscript sys is omitted)

0 0

0totdSprocessreversiblefor

Emphasis is put on the reversibility of the process:then the equality sign holds in Equations (4.86)and (4.88),and thus it follows that ;For an irreversible process accordingly less work is released or more work is necessary.This is represented by the inequality in Equation (4.89).

revotherWdG

irrother

revother WWPVTSUddG

Thus,in an isothermal,isobaric system which is left to its own,irreversible processes happen until a minimum of the free enthalpy is achieved, dG=0, G=Gmin

{4.90}

Exercise 4.11: Gibbs-Helmholtz equationShow that the free enthalpy and its derivative with respect to temperature can be related to the enthalpy of the system via the Gibbs-Helmholtz equation.

PP T

TGT

T

GTGH

/2

Solution:

PVUH

TSPVUG

TSGH

From the equations above ,it follows immediately that

dG=dU -T dS -S dT+p dV+V dp

dG=-S dT+V dp+ μdN+... {4.82}

NPT

GS

,

If we insert this into equation TSGH It follows that

NPT

GTGH

,

It also holds that

NP

NPNP

T

GTG

T

TG

T

G

TT

G

T

,2

2,,

1

11

Thus the Gibbs-Helmhotz Equation can be obtained.

PP T

TGT

T

GTGH

/2

4.5 The Maxwell relations

James Clerk Maxwel 1831 ～ 1879

A variety of relations between the thermodynamic state variables can be derived from the fact that the thermodynamic potentials U,F,H,and G are state functions; i.e.,they have exact differentials.

The Maxwell relations

The total differential of the internal energy reads (we now consider only systems which are completely characterized by three state variables),

dU =T dS-p dV+ μ dN

it immediately follows,e.g.,that

dNN

UdV

V

UdS

S

U

SVNSNV ,,,

since

NVNSNSNV V

U

SS

U

V,,,,

)1(,, NVNS S

P

V

T

dU =T dS-p dV+ μ dN

(4.124)

PV

UT

S

U

SV

,

In this manner many relations emerge that may allow for the calculation of unknown quantities from known quantities and which we now want to present in a systematic way

)2(,, NVVS SN

T

)3(,, SVNS N

P

V

At first it follows from Equation(4.124) that

dU =T dS-p dV+ μ dN

dNN

UdV

V

UdS

S

U

SVNSNV ,,,

NVVSVSNV N

U

SS

U

N,,,,

SVNSNSSV V

U

NN

U

V,,,,

Here the coefficients T,p,and μ in Equation(4.124)are of course to be considered asfunctions of S,V,and N.Corresponding relations exist for the free energy F(T,V,N):

dNPdVSdTdF

)4(,, NVNT T

P

V

S

)5(,, NVVT TN

S

)3(,, TVNT N

P

V

Here the coefficients S,p,and μ are functions of the variables T,V,and N.Analogouslyit holds for the enthalpy H(S,p,N) that

dNVdPTdSdH

)6(,, NPNs S

V

P

T

)2(

,, NPPS SN

T

)7(,, NSPs PN

V

The coefficients T,V,and μare now functions of S,p,N.For the free enthalpy G(T,p,N) one has

)8(,, NPNT T

V

P

S

)5(,, NPPT TN

S

)7(,, NTPT PN

V

dG=-S dT+V dp+ μ dN {4.133} 

However,there exists a simple device which yields the Maxwell relations. This device is the thermodynamic rectangle,which is shown in Figure 4.8.

Relations above are called Maxwell relations.In the literature one often considers systems with constant particle number (dN=O);thus the number of relations is considerably reduced.

Figure 4.8 Thermodynamic rectangle for N=const.

)1(,, NVNS S

P

V

T

)6(,, NPNs S

V

P

T

)4(,, NVNT T

P

V

S

)8(,, NPNT T

V

P

S

The thermodynamic rectangle was conceived especially for system with constant particle number and without further state variables. The variables V,T,p,and S,which are the only possible quantities at constant particle number,form the corners of this quadrangle. Along the edges we denote the potentials, which depend on the variables at the corresponding corners , e.g., F(V,T).With this way of presentation partial derivatives are easily read off .

Example 4.12: Heat capacities

A general relationship shall be established between the heat capacities Cv and CP,which should,if possible,contain only quantities that are easy to measure.

VVVV T

U

T

ST

dT

QC

PPPP T

H

T

ST

dT

QC

In the first equation T and V are supposed to be independent variables,while in the second equation T and p are independent.One can write both equations with S(T,V)or S(T,p), respectively,as follows

In Equation above the pressure can as well be regarded as a function of the variables T and V,

)1(dVV

STdTCdV

V

STdT

T

STTdS

Tv

TV

)2(dPP

STdTCdP

P

STdT

T

STTdS

TP

TP

)2(dPP

STdTCdP

P

STdT

T

STTdS

TP

TP

dVV

PdT

T

P

P

STdTC

TVTP

dVV

STdT

T

P

P

STdTC

TVTP

)1(dVV

STdTCdV

V

STdT

T

STTdS

Tv

TV

By comparison with （ 1 ） , we can immediately obtain

VVT

P CT

P

P

STC

Equation above for the relationship between CV and Cp is,however,still not very useful in practice,since for instance is difficult to measure. The quantity is conveniently measurable for gases,but in fluids or solids processes at constant volume are connected with extreme pressure.Thus,if possible,we want to express bothquantities in terms of easily measurable quantities. The Maxwell relation(4.133)reads

TPS /

VTP /

)8(PT T

V

P

S

The righthand side is the isobaric expansion coefficient α

Also,the factor can be rewritten with an often used trick in terms of other quantities.To this end,we consider the volume as a function of p and T, we have

VT

V

p

VTP /

dPP

VdT

T

VdV

TP

For processes at constant volume (dV=O)it thus holds that

T

P

V

P

V

T

V

T

P

VP

V

T

V

V

T

P

V

Thus:

And because:

In this relation only quantities which are easy to determine appear.

2

TVCC PV

VVT

P CT

P

P

STC

VT

P

VP

S

T

4.6 Jacobi transformation

A frequently occurring problem in thermodynamics is the transformation of variables in state functions. Such transformations must not be confused with the Legendre transformation. For the latter we have not simply replaced one variable in the internal energy by another, but have defined a new physical quantity, which is especially convenient for a certain system. In the following we want to investigate pure transformations of variables in the same physical quantity.

（雅可比）

（勒让德）

, ( , )

.( , )

,

u u

x yu v u v u v

v vx y x y y x

x y

( , ), ( , )u u x y v v x y

Then,the Jacobi determinant can be denoted as

J（ x,y)=

( , )(1)

( , )y

u u y

x x y

Especially useful is this denotation of a derivative,

yx

u

x

u

y

y

x

y

y

u

x

u

yx

yu

,

,

0 1

Certification:

( , ) ( , )(2)

( , ) ( , )

u v v u

x y x y

x

v

y

u

y

v

x

u

y

v

x

v

y

u

x

u

yx

vu

,

,

x

u

y

v

y

u

x

v

y

u

x

u

y

v

x

v

yx

uv

,

,

( , ) ( , ) ( , )(3)

( , ) ( , ) ( , )

u v u v x s

x y x s x y

x

v

y

u

y

v

x

u

y

s

x

v

s

u

s

v

x

u

y

s

x

s

y

x

x

x

s

v

x

vs

u

x

u

yx

sx

sx

vu

,

,

,

,

1 0

( , ) ( , )(4) 1/

( , ) ( , )

u v x y

x y u v

vu

yx

yx

vu

,

,

,

,

vu

yx

yx

vu

,

,

,

,

vu

vx

vx

vu

,

,

,

,

1

vx

vx

vx

vx

,

,

,

,

vv u

x

x

u

Example 4.15 Joule-Thomson coefficient

While discussing the Joule-Thomson experiment we calculated the Joule-Thomson coefficient

HP

T

If is to be expressed by the known enthalpy H(T,P), one obtain this with the help of Jacobi transformations.

HP

HT

HP

HT

P

T

H ,

,

,

,

TP

HP

TP

HT

,

,

,

,

TP

HP

TP

TH

,

,

,

,

PT T

H

P

H

TP

TP

,

,

Exercise1:

Please prove the equality is right. Here is the isothermal compressibility, and is the adiabatic compressibility.

P

V

T

S

C

C

T S

TT P

V

V

1

SS P

V

V

1

Exercise 2:

3

1

3

4

VCSU For one system ,try to formulate it’s pressure P, free energy F and the free enthalpy G, respectively.

Exercise3：

Assume a gas, which pressure is proportional to its temperature at constant volume , then ,it must have that the entropy is also proportional to the volume at constant temperature.