spi06chap2

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2. Thermodynamic potentials

2.1 Fundamental equation

Thermodynamic potentials are extensive state variables of dimensions of

energy. Their purpose is to allow for simple treatment of equilibrium for

systems interacting with the environment.In thermodynamics all variables are either extensive or intensive.

Mathematically this may expressed in homogeneity relations with re-

spect to the system size. Thus, extensive variables (e.g. N, V, U, S, . . .)are first-order homogeneous functions, whereas intensive variables (like

p, T, , . . .)are independent of the size of the system.

Natural variables. are those whose differentials appear in the differ-

ential form of the first law: dU = T dS p dV + dN so that S, V ja Nare natural variables of internal energy. With all intensive variables fixed,

extensivity of all these variables means

U(S,V,N) =U(S,V,N). (2.1)

Differentiating both sides with respect to the auxiliary parameter andputting = 1thereafter we arive at the identity (Euler equationfor homo-geneous functions):

U=S

U

S

V,N

+ V

U

V

S,N

+ N

U

N

S,V

.

From the first law it follows that

U

S

V,N

=T ,

U

V

S,N

= p ,

U

N

S,V

= .

Thus, we arrive at the fundamental equation

U=T SpV + N . (2.2)

15

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16 2. THERMODYNAMIC POTENTIALS

2.2 Internal energy Uand Maxwell relations

The first lawdU=T dSp dV + dNyields

T =

U

S

V,N

, (2.3a)

p =

U

V

S,N

, (2.3b)

=

U

N

S,V

. (2.3c)

From the definition of heat capacity it follows that

CV = dQdTV,N

= UTV,N . (2.4)Since U may be assumed to be single-valued smooth state variable,result of iterative differentiation does not depend on order T/N =(U/S)/N = (U/N)/S = /S. This procedure gives rise toMaxwell relations:

T

V

S,N

=

p

S

V,N

, (2.5a)T

N

S,V

=

S

V,N

, (2.5b)

pNS,V = VS,N . (2.5c)These and similar relations for other thermodynamic potentials are often

useful in expressing differential relations in terms of response functions

and state variables.

In an irreversible processT S > Q= U+ W N, thereforeU < T Sp V + N. (2.6)

In an irreversible process with fixed S, V and N the internal energy de-creases. Thus, in equilibriumUassumes the mimimal value with S,V andNfixed (implying, of course, that something else may change).

If some other work may be done in a reversible process, then

U=R = Wfree ,where thefree workWfree =R is the work the system may carry out ingiven circumstances.

If the process is irreversible, then

Wfree U (2.7)even if (S,V,N) are kept fixed. Thus, the minimal work needed to bringabout the change of internal energy U isR = U.

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2.3. ENTHALPYH 17

2.3 EnthalpyH

Other thermodynamic potentials are Legendre transformsof the internal

energyU(S,V,N)with respect to natural variablesS,V orN. Enthalpy(orthe heat function) is obtained by usingp instead ofV:

H U+ pV . (2.8)

The differential follows from the definition and the first law:

dH=T dS+ V dp + dN . (2.9)

Natural variables are (S,p,N). From the definition of heat capacity it fol-

lows thatCp=

dQdT

p,N

=H

T

p,N

. (2.10)

From the expression fordHthree more Maxwell relations follow:T

p

S,N

=

V

S

p,N

, (2.11a)T

N

S,p

=

S

p,N

, (2.11b)

V

NS,p=

p S,N. (2.11c)

Ia an irreversible change Q = U +W dN < T S. Substitution ofU=(HpV)yieldsH (pV) + W dN < T S, i.e

H < T S+ V p + N . (2.12)

If in the process S, p and Nremain constant, spontaneous changes drivethe system to the state with mimimum enthalpy.

Many practically important processes (phase transitions, chemical re-

actions etc) take place at constant (ambient) pressure. If the conditions

include also thermal isolation, the enthalpy is the natural energy quantity

to use.

In hydrodynamics adiabatic flow is a popular approximation. Then the

specific (per unit mass) internal energy u appears in the energy equationonly in the combinationu +p/= h, which is the specific enthalpy and thusthe natural energy variable.

When (S,p,N) are fixed, the portion of the energy of the system freelyexchangeable for work obeys the condition

Wfree H . (2.13)

The mimimum work required to bring aboutHis thusR = H.

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JouleThomson process. Consider thermally isolated forced flow of

gas through a throttle valve or a porous wall. Movement of pistons is de-vised to keep pressures p1 > p2 fixed. Although the flow is far from equi-librium and not reversible, a hypothetical reversible process between the

same states is useful, because state variables are process-independent.

For the transfer of an infinitesi-

p1

p2

Figure 21: Flow through porous

wall.

mal quantity of matter the work by

the system is dW =p2 dV2+ p1 dV1.Initially V1 = Viand V2 = 0. FinallyV1 = 0 and V2 = Vf. For constantpressures the work is

W = dW =p2 Vfp1 Vi .

Thermal isolation means Q = 0,therefore U =W. From this it follows Uf +p2 Vf = Ui + p1 Vi . Thus,the quantity U+pV, i.e. the enthalpyHremains constant: the process isisenthalpic,

H=HfHi = 0 . (2.14)Imagine now a reversible isenthalpic process of decreasing the pressure by

infinitesimal steps. The response of the temperature to this is given by the

JouleThomson coefficient T

p

H

. (2.15)

To express this coefficient in terms of already introduced quantities, use the

Jacobi determinant method. It is good policy to introduce variables which

are the natural variables of the thermodynamic potential appearing in this

definition, because then its first derivatives are state variables. Thus,T

p

H

=(T, H)

(p, H) =

(p, S)

(p, H)

(T, H)

(p, S) =

1

T

(T, H)

(p, S)

= 1

T

T

p

S

H

S

p

T

S

p

H

p

S

=

T

p

S

VCp

. (2.16)

Using the Maxwell relation (2.11a) rewrite

Tp S = VSp = TCp VSpSTp = TCp VTp .Thus, we arrive at the expression

T

p

H

= T

Cp

V

T

p

VT

=

V

Cp(T p 1). (2.17)

The latter form follows form the definition of the thermal expansion coeffi-

cient: p= V1(V/T)p.

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2.4. FREE ENERGYF 19

In the process the pressure decreases, so that the gas is cooled, if

T p > 1, or heated, ifT p < 1. For the ideal gas the JouleThomson coef-ficient vanishes, so that the temperature of an ideal gas remains the same.

For real gases the coefficient is positive below a certain pressure-dependent

inversion temperature, so that the gas is cooled. Thus, the Joule-Thomson

process may be and is used for cooling and eventually liquifying gases.

2.4 Free energyF

The Legendre transform of the internal energy with respect to Syields thefree energy(Helmholtz free energy): F =U S(U/S)V,N i.e.

F

U

T S . (2.18)

The corresponding differential is

dF = S dTp dV + dN . (2.19)

The natural variables are T,V andN. The Maxwell relations areS

V

T,N

=

p

T

V,N

, (2.20a)S

N

T,V

=

T

V,N

, (2.20b)

pNT,V = VT,N . (2.20c)As before, for an irreversible process

F < S dTp V + N . (2.21)

Thus, with fixed T, V andN, the system evolves towards the minimum ofthe free energy. For the free work at fixed T , V , N it follows

Wfree F . (2.22)

Free energy is an extremely important tool in statistical mechanics: in

many cases it is the natural macroscopic quantity to calculate for a given

microscopic model.

2.5 Gibbs function G

The Legendre transform of U with respect to both S and V leads to theGibbs function(Gibbs free energy)

G U T S+pV , (2.23)

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with the differential

dG= S dT+ V dp + dN . (2.24)

The natural variables are T,p and Nand the Maxwell relationsS

p

T,N

=

V

T

p,N

, (2.25a)S

N

T,p

=

T

p,N

, (2.25b)V

N

T,p

=

p

T,N

. (2.25c)

With fixedT,p and N, a non-equilibrium system evolves towards the mini-mum of the Gibbs function:

G < S T+ V p + N , (2.26)

and the free work is

Wfree G . (2.27)The Gibbs function is a suitable thermodynamic potential for systems

which change at fixed pressure and temperature (no thermal isolation).

Since these parameters are perhaps most easily of all adjustable, the Gibbs

potential has a wide scope of applications both in physics and chemistry.

From the fundamental equation it follows that

G= N , (2.28)

i.e. the chemical potential is the Gibbs function per particle in a single-

species system. Since from (2.35) it follows that dG = dN +N d and,taking into account the alternative form (2.24), we arrive at the Gibbs

Duhem equation

d= SN

dT+V

Ndp , (2.29)

showing that the natural variables of the chemical potential are T , p.

2.6 Grand potential

Thegrand potentialis also an important quantity for calculations in statis-

tical mechanics when the number of particles cannot be fixed. The defini-

tion is

U T S N (2.30)leading to the differential

d = S dTp dV N d , (2.31)

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2.7. THERMODYNAMIC RESPONSES 21

showing that the natural variables are T,V and. The Maxwell relations

are S

V

T,

=

p

T

V,

, (2.32a)S

T,V

=

N

T

V,

, (2.32b)p

T,V

=

N

V

T,

. (2.32c)

In an irreversible change the inequality

CV.

Construction of potentials. An equation of state like p = p(T, V)anda thermal response, say CV, are required to this end. Consider, for instance,van der Waals matter:

p + aN2

V2

(V N b) =N T .

The heat capacityCVis directly a partial derivative of the internal energy:

CV =

U

TV.

To calculate the other one, change variablesU

V

T

= (U, T)

(V, T) =

(U, T)

(V, S)

(V, S)

(V, T)

=

S

T

V

U

V

S

T

S

V

U

S

V

T

V

S

. (2.47)

Here, derivatives ofU arep and T, this is the point of introducing thenatural variables ofU. Thus

U

V

T

= p + T

S

V

T

=T

p

T

V

p ,

where the free-energy Maxwell (2.20a) has been used once more. This rela-tion shows, in particular, that for the van der Waals equation of state

CVV

T

= 2U

V T = 0 ,

i.e. CVis a function of temperature only! Then the integration is simple:

U(T, V) =

CVdT+

T

p

T

V

p

dV =

CV(T)dT a N

2

V .

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2.8 Thermodynamic stability conditions

pTV

Figure 22: System near equi-

librium

Let the near-equilibrium system be

divided to (semi)macroscopic subsystems

(labeled by index ) each in a local equi-librium, but with different pressure, tem-

perature etc. in neighbouring subsys-

tems. Extensive quantities remain addi-

tive:S=

S,V =

V,U =

U.LetNj be the particle number of speciesj in the subsystem. ThenNj =

Nj

for thejth species.

Due to local equilibriumS= S(U, V, {Nj}) .

For a small change ofS

S= 1

TU+

pT

V j

jT

Nj .

Assume a system isolated as a whole, then U, V and Nj remain constant. Tosimplify notation, consider two subsystems: = A, B. Conservation lawsyieldUB = UA,VB = VA and NjB = NjA . Thus,S =

S

= 1

TA 1

TB

UA+

pATA

pBTB

VA

j

jATA

jBTB

NjA .

At equilibriumS = 0 identically. Since the fluctuations UA, VA andNjA are arbitrary, theequilibrium conditions:

TA = TBpA = pB

jA = jB

(2.48)follow. Thus, in equilibrium the temperature is the same everywhere, as

well as the pressure (provided no external fields impose inhomogeneity) and

the chemical potential for each particle species. The conditions hold also in

the case system consists of different phases (constant pressure requires flat

interfaces, however).

2.9 Stability conditions of matter

In stable equilibrium the entropy must be at maximum. To analyze this,

the second variation of the entropy with respect to{U, V andNj}may be used.

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2.10. THERMODYNAMIC POTENTIALS IN ELECTROMAGNETISM25

Let T, p and

{j}

be the common equilibrium values. For simplicity,

assume one species. In Taylor expansion at the equilibrium pointS=S0+dS+ 1

2d(dS) the linear term vanishes. Sinced2X = 0 for any independent

variableX, we obtain

S= S0+1

2d(dS) =

1

2

d

1

T

(dU+pdV dN) +1

2

1

T(dpdV ddN) .

Here,dU+ pdV dN= TdS, so that (denotedX X)Stot S S0 = 1

2T

(TS pV+ N). (2.49)

The condition of a stable equilibrium is that this expression is negativedefinite.

Since any subsystem is at local equilibrium, only three of fluctuationsof the quantities T, S, p, V, , N are independent, the rest must beexpressed as functions of the chosen three.

Let N = 0. Then only two independent variables remain. ChooseTand Vand expressSas pfunctions thereof. Maxwell relationsallow for simplification and the result is

Stot= 12T

CV,

T (T)

2 + 1

TV(V)

2

. (2.50)

Another possibilityV = 0 with T andN as independent variablesleads to

Stot= 12T

CV,T

(T)2 +

N

T,V

(N)2

. (2.51)

From these expressions it is readily seen that the total entropy is at maxi-

mum, when the followingstability conditionshold: :

CV > 0

T > 0

N

T,V

> 0

. (2.52)Otherwise the equilibrium is unstable and small spontaneous disturbances

give rise to growing changes which lead to another state.

2.10 Thermodynamic potentials in electro-

magnetism

The starting point here is the basic differential form of work (1.17)

dW =

d3r (E dD +H dB),

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whose addition to previously introduced differentials gives rise to differen-

tials of thermodynamic potentials in electromagnetism, say (relevant para-meters only explicit)

dU=T dS+

d3r (E dD +H dB) .

Material parameters contained in vectors E and B like the permittivity and permeability should be expressed here as functions of the entropyS. This is inconvenient, therefore a preferable choice is the free energy, forwhich

dF = SdT+

d3r (E dD +H dB) . (2.53)

Here, and are functions of the temperature.

Thermodynamics potentials assume minimum values at equilibrium,when their natural variables are fixed. Since free charges are sources of

the electric induction D and the vector potential A the source of the mag-

netic induction B, the free energy (2.53) is the choice for problems with

fixed charges of conductors and fixed vector potentials (the latter might be

difficult to control in real world, though).

For other cases, new potentials should be formed by suitable Legendre

transforms. For instance, the potentialFE = F d3rE D gives rise tothe differential

d FE= SdT d3rD dE, (2.54)which reveals that the natural variables are T and E. Thus, this potential

minimizes at equilibrium when the field E (or the electric potential) is keptconstant.

Similarly, the potentialFH=F d3rH B with the differentiald FH= SdT d3rB dH , (2.55)

and natural variables T and H is suitable for cases with fixed currents.Combinations of these transform may appear useful as well. Unfortunately,

there seems to be no standard nomenclature of the different potentials in

the electromagnetic case (cf. the Helmholtz free energy and the Gibbs func-

tion of anS,V,Nsystem).

Example 2.1. Consider a vertical parallel-plate capacitor in contactwith a liquid reservoir. Let us calculate, how high the liquid with the

dielectric constantr rises between the vertical plates, when the capac-

itor is charged and disconnected from any voltage source.

The potential energy of the liquid in the gravitational field is

Wg = 1

2gwdy

2,

whereg is the acceleration of gravity, the density of the liquid, y the

height of the liquid slab between the plates, d the separation of the

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(b) Calculate the heat capacity Cas a function of temperature T. Plot

it and find the position of the peak, known as the Schottky anom-aly. Such a peak is characteristic of a system in which atoms have

a few low-lying closely spaced energy levels, and at low tempera-

tures may dominate all other contributions to the heat capacity of

the solid.

Problem 2.5. Show (N is kept fixed) that the internal energy of the

Clausius gas is

U(T, V) =N u1(T) 2aN2

T(V + N c),

whereu1(T)is a function of temperature only, whose explicit form can-not be determined thermodynamically. However, there is another re-

lation (show this as well) U(T, V) =

CV(T, V) , dT, establishing a

connection between u1 and the isochoric heat capacity of the gas. The

equation of state of Clausiuss gas is

p + aN2

T(V + cN)2

(V bN) = N T .

Problem 2.6. For a unit volume of dielectric at constant density find

the difference cE cD between the heat capacities of a homogeneous

isotropic dielectric at constant electric field strength E and electric in-

ductionD.

Problem 2.7. Show without resorting to the connection between Cp

andCV that in a stable thermodynamic equilibrium Cp > 0 and S >0

Problem 2.8. By minimizing a suitable thermodynamic potential, find

how high dielectric liquid rises between the vertical plates of a parallel-

plate capacitor connected to a voltage source with the constant electro-

motive force E. Express your answer in terms of the electric field in the

capacitor rather than the emf.

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