In the preceding section we found the two quantitiesE−T S, cf. (2.13) andH−T S, cf. (2.98) to be rather useful. We therefore define the two new functions called the free energy11
F =E−T S (2.104)
and the free enthalpy12
G=H−T S. (2.105)
Computing their total differentials we find
d F =d E−d(T S)=d E−Sd T−T d S
(1= −.51) Sd T−Pd V +μdn+. . . (2.106) and
d G=d H−d(T S)=d H−Sd T −T d S
(1= −.51) Sd T +V d P+μdn+. . . , (2.107) where
∂F
∂T
V,n,...= −S, (2.108)
∂F
∂V
T,n,...= −P, (2.109)
∂F
∂n
T,V,...=μ, (2.110)
and analogously
∂G
∂T
P,n,... = −S, (2.111)
11or also Helmholtz free energy; Hermann Ludwig Ferdinand von Helmholtz, german physiologist and physicist, * 31.8.1821 Potsdam, Germany; †8.9.1894 Charlottenburg, Germany.
12or also Gibbs free energy; Josiah Willard Gibbs, american scientist, * 11. 2.1839 New Haven, Connecticut, †28.4.1903 New Haven, Connecticut; the founder of modern thermodynamics.
∂G
∂P
T,n,...=V, (2.112)
∂G
∂n
T,P,...=μ. (2.113)
ObviouslyF =F(T,V,n, . . . )whereasG=G(T,P,n, . . . ). The two are related via
G=F+P V, (2.114)
i.e. they are Legendre transformsof each other (G = F − V∂F/∂V|T and F =G−P∂G/∂P|T).
Remark 1: F is called a thermodynamic potential with respect to the variables T,V,n,. . .. The same is true forGwith respect toT,P,n,. . .. In general we call a thermodynamic quantity a thermodynamic potential if all other thermodynamic quantities can be derived from partial derivatives with respect to its variables.
Remark 2:By straightforward differentiation and knowing that Sand E are state functions it is easy to show thatFandGare state functions also.
2.3.1 Relation to the Second Law
According to the first law we have
d E =δq−δw. (2.115)
Combination of this with Clausius’ statement of the second law, cf. (1.48) yields
d E−T d S≤ −δw (2.116)
or
d F|T +δw≤0. (2.117)
Ifδwstands for volume work only, then we may deduce
d F |T,V≤0. (2.118)
From this follows (cf. p. 286) the attendant relation for the free enthalpy, i.e.
d G|T,P≤0. (2.119)
Fig. 2.13 An illustration of the relation of G to the second law
T G
P
dG| <0 T,P
dG =-SdT + VdP
An illustration in the case of the free enthalpy is shown in Fig.2.13. A system initially may be prepared in a state corresponding to the solid circle. It lowers its free enthalpy as much as possible, which brings it down to a point on the surface shown in the sketch, whered Gis given by Eq. (2.107). This surface represents the states of lowest free enthalpy above theT-P-plane.
Of course there may be other types of work, not just volume work, involving vari- ables, X, other thanV, which are controlled from outside the system (the examples below explain what is meant here). In this case the Eqs. (2.118) and (2.13) still apply if. . .|T,V and. . .|T,P are replaced by. . .|T,V,Xand. . .|T,P,X.
This book contains a number of applications of both Eq. (2.118), e.g. phase sepa- ration in the context of van der Waals theory, and Eq. (2.119), e.g. chemical reactions.
Nevertheless, already at this point we want to look at three instructive examples.
Example: Capillary Rise. Figure2.14shows a tube of radiusRwith one end submerged in a liquid. In the figure the liquid has risen to a heighthagainst the pull of gravity. We want to calculate the equilibrium height of the liquid in the tube.
This system comprises the liquid, the tube, and the earth. Ignoring all other effects we useδF |T=(δE−TδS)|T, and find thatδF |T may be expressed by the following types of work involved whenhchanges by the small amount δh:
δF |T=γT LδA−γT AδA+cghδV. (2.120) HereδA=2πRδhandδV =πR2δh, i.e.
δF |T=
2πR(γT L−γT A)+πR2cgh
δh. (2.121)
The quantities γT L andγT A are the surface tensions of the tube-liquid and tube-air interface, respectively, cf. (1.13). The last term in Eq. (2.120) is the negative of the work done by the system when raising the liquid volumeδV to the heighth. Notice thatcis the mass density of the liquid, andgis the magnitude of the gravitational acceleration.
In the above equations h is a parameter not controlled from outside the system. We are not doing work to the overall system nor do we extract work.
The equilibrium height may be affected from the outside only by altering the temperature, changing the surface tension or the liquid density. But hereT is held constant together with the overall volume of the system. Thus we may apply Eq. (2.118) to the above free energy, i.e. we find the equilibrium value ofhby minimization of the above free energy with respect toh. This means we simply set the term in square brackets equal to zero:
0=(γT L−γT A)2πR+cghπR2. (2.122) Solving forhyields the equilibrium height
h= 2(γT A−γT L)
cgR . (2.123)
Becausehdepends on the sign ofγT A−γT L, it may be positive or negative. A second solution to this problem follows via the force balance 2πRγcosθ =
γT A− γT L =mg, where m =cπR2h (cf. Fig.2.14). This implies that the above equation also can be written in terms of the liquid-air surface tension, γ, and the contact angle,θ,
h= 2γcosθ
cgR (2.124)
(e.g.γ =0.0728 N/m for water-air at 20◦C). A nice discussion of capillarity and wetting phenomena can be found in de Gennes et al (2004) (Pierre-Gilles de Gennes, Nobel Prize in chemistry for his contributions to the theory of polymers and liquid crystals, 1991.).
Example: Dielectric Liquid in a Plate Capacitor. The next, partially re- lated problem is illustrated in Fig.2.15. A plate capacitor is in contact with a liquid possessing again the mass densitycand the dielectric constantεr. If a constant voltage,φ, is applied to the capacitor, the liquid rises to a certain equi- librium height between the capacitor plates. This is what we want to calculate.
Fig. 2.14 Capillary rise
h TL
TA
Fig. 2.15 Dielectric liquid in a plate capacitor
h
However, first we must discuss the relation of the free energy,F, to the electric work expressed in Eq. (1.25)…to be continued…
We start from
δF =δ(E−T S)=δE−SδT −TδS
=δE−SδT −
δE−
d V EãδD 4π +. . .
= −SδT +
d VEãδD 4π , i.e.
δF = −SδT +
d VEãδD
4π . (2.125)
Momentarily we use only the electric field contribution from Eq. (1.21). The dots in the second line indicate that there are other types of work in general, which here either do not occur (e.g., chemical work) or can be neglected (e.g., volume change) or will be added later (work against the gravitational field). Also notice that
E 4π = ∂f
∂D
T,V,..., (2.126)
where f is a free energy density.
Thus we find that in the present case we haveF=F(T,D). However, this is not appropriate here—why? Notice that the scalar potential (voltage drop),φis related to the (mean) electric field in the capacitor via
E= − ∇φ . (2.127) The equation applies to both the filled and the empty part of the capacitor. This means that if we hold the voltage on the capacitor plates constant, we also hold theE-field between the plates constant. If we want to minimize F = F(T,D;h)with respect toh, in order to compute the equilibrium height, we must do this keepingT andD fixed. But D is not the same asE, which is constant in the present setup. Thus we need a new function F˜ = ˜F(T,E; h)instead.
We try the Legendre transformation F˜ =F−
d V Eã D
4π (2.128)
and obtain
δF˜ =δF−
d VδEã D 4π −
d V EãδD 4π
= −SδT −
d VδEã D 4π , i.e.
δF˜ = −SδT −
d VδEã D
4π . (2.129)
Indeed we haveF˜ = ˜F(T,E;h). We now continue our example.
…The explicit E-dependent contribution to F˜ |T is obtained by integration of the differential relation∂F˜/∂E|T = −εrE V/(4π)with respect toEfrom zero to the actual field strength. Notice that we assume a constant field strength between the capacitor plates as well asD =εrE(This equation is used in sev- eral places throughout this book, which means that the attendant calculations do depend on its validity!). Notice also that the field strength is the same with and without the dielectric. Analogous to Eq. (2.120) we collect all relevant work terms expressingδF˜ |T via
δF˜ |T= − r
E
0
dEã E
4π δV +
E
0
dEã E
4π δV +cghδV (2.130) The first term is due to a volume increase,δV, of liquid between the capacitor plates. The second term is due to the corresponding reduction of vacuum (or air). Following the same reasoning as in the previous example, here the system includes the capacitor, the liquid, and the earth, we find the equilibrium height by setting the right side of the above equation equal to zero:
0= − 1
8π(εr−1)E2+cgh. (2.131) Solving forhyields
h= (εr −1)E2
8πcg . (2.132)
Remark:Somebody may comment that the introduction ifF˜ is not necessary, be- cause DandEhere are related via a constantεr. This is not true. UsingF instead of F˜ changes the sign of the first term in brackets in Eq. (2.131) and there would be no sensible solution.
Before leaving this subject we study the same problem from another angle. We detach the capacitor from its voltage supply. This means that we switch fromφ= constto a situation whereQ=const. Here±Qis the charge on the capacitor plates.
What happens to Eq. (2.132)?
Figure2.16illustrates the situation. At (a) we have the boundary conditions (EI− EI I)ì n=0 and (DI− DI I)ã n=0, (2.133) wherenis a unit vector perpendicular to the interface between liquid and vacuum (or air) at (a). At (b) we have instead
EIì n =0 and DIã n=4πσI. (2.134) Nownis perpendicular to the capacitor plates. If we move (b) up into region II the boundary conditions become
EI Iì n =0 and DI Iã n=4πσI I. (2.135)
Fig. 2.16 Different bound- aries in the case of the liquid inside the plate capacitor
h
Q -Q
H
a I b
II
x z
σI and σI I are the surface charge densities in these respective regions. From Eq. (2.133) we can see that Ex,I = Ex,I I. Because DI = εrEI and DI I = EI I
we can conclude from Eqs. (2.134) and (2.135) that
σI =εrσI I, (2.136)
i.e. the surface charge densities in the two regions are not equal. We can use this result to obtain
Q=σILyh+σI
εr
Ly(H−h) or σI = εrQ/Ly
(εr −1)h+H, (2.137) whereLyis the width of the capacitor iny-direction.
We now work out
Vcapaci t or
d VEã D 8π
f illed−
Vcapaci t or
d V Eã D 8π
empt y (2.138)
= A 8π
hεrEx2,I+(H−h)E2x,I I
− A
8πH E2x,I I
= A
8π(εr −1)h E2.
In the last step we have use Ex,I = Ex,I I = E. This shows that we can treat this case analogous to the fixed potential case above. The result of course is the same as before. However, this time we must express E2via the total chargeQ.
Example: Euler Instability. The third example in this series is illustrated in Fig.2.17. The figure shows a thin plate in a vice subject to a compressive force, fc. This systems encompasses the plate only. The pressure exerted by the vice is controlled from the outside. In this case there is an elasticFel|T, i.e.
Fel |T= εh3 24
plat e
d xd y ∂2ζ
∂x2 2
.
The quantityhis the plate’s thickness,εis the elastic modulus of the plate’s material. The functionζ =ζ(x)describes the shape of the plate when looked at along the y-direction. The expression on the right side of the equation is the work done by the plate’s internal forces, when it is bent into the shape described byζ(x). This result of continuum theory of elasticity may be found in Landau et al. (1986).
What is the equilibrium shape of the plate depending on the applied compres- sive force? To find the answer we now employ relation Eq. (2.117), i.e.
δFel |T +δw≤0. (2.139)
Herewis the negative of the work done by the vice, i.e.wc= −w. We compute wc by adding up the total displacement parallel to the direction of the force from infinitesimal increments (cf. the inset in Fig.2.17):
ds−d x=
d x2+(dζ (x))2−d x≈d x
1+1 2
∂ζ (x)
∂x 2
−1
.
Thus
wc= 1 2 fc
d x
∂ζ
∂x 2
=1 2σ
V
d V ∂ζ
∂x 2
.
V is the volume of the plate, andσ is the stress equal to fc/(h Ly), whereLy
is the extend of the plate iny-direction.
We search for the minimum of the left side in Eq. (2.139) by “offering” suitable shape alternatives to the plate, i.e. we carry out a variation (or minimization) in terms ofζ:
0=δζ(Fel −wc)=δζ
d V ε
2z2 ∂2ζ
∂x2 2
−σ 2
∂ζ
∂x 2
.
We writeh1h/2
−h/2d zz2= 121h2≡Iand thus δζ1
2
d xd y
εI ∂2ζ
∂x2 2
−σ ∂ζ
∂x 2
=0. (2.140)
Carrying out the variation we obtain
d xd y εI
∂2ζ
∂x2
∂2
∂x2δζ
−σ ∂ζ
∂x
∂
∂xδζ =0.
Partial integration using δζ (x) = 0 at±L/2 finally yields the differential equation for the “shape function”:
εI∂4ζ
∂x4+σ∂2ζ
∂x2 =0.
Making the Ansatzζ (x)=ζ0sin(q x)we find
εIq4−σq2=0 or σcr i t =εIqmi n2 .
The planar solution isq=0, whereas the bent plate corresponds toq >0. The smallest possible non-vanishingq-value,qmi n, defines the critical stress limit, σcr i t, at which the transition from planar to bent occurs spontaneously. This value ofqmi ndepends on the boundary conditions. Here we haveqn=(π/L)n, wheren=1,2, . . . and therefore
σcr i t = π2εI
L2 . (2.141)
In the literature this phenomenon is called Euler buckling. The problem may be modified by embedding the plate into an elastic medium. This results in larger values forqmi ndepending on the medium’s stiffness (Young’s modulus).
Remark:Notice that the above system has the freedom to decide to which side it buckles. This phenomenon is a spontaneous symmetry breaking.
z y
x L
dx x
ds d fc
Fig. 2.17 Buckling of a thin plate
2.3.2 Maxwell Relations
Equating the right sides of Eqs. (2.16) and (2.17) as well as the right sides of Eqs. (2.101) and (2.102) we have used that bothF andG are state functions. The resulting formulas, Eqs. (2.18) and (2.103), are examples of so called Maxwell rela- tions.
It is easy to construct more Maxwell relations via the following recipe. Take any state functiongand any pair of variables,xandy, it depends upon. Ifg= pd x+qd y then
∂p
∂y
x =∂q
∂x
y (2.142)
yields a Maxwell relation. In general we may use the differential relations in Ap- pendix A.2 to generate even more differential relations between thermodynamic quantities, i.e. more Maxwell relations.
Example: Relating CVand CP. A nice exercise useful for practicing the
“juggling” of partial derivatives, which is so typical for thermodynamics, is the derivation of the general relation betweenCV andCP. We start from
CP−CV = ∂H
∂T
P−∂E
∂T
V. (2.143)
UsingG=H−T Sand
−S=∂G
∂T
P = ∂H
∂T
P−S−T ∂S
∂T
P (2.144)
as well asF=E−T Sand
−S= ∂F
∂T
V =∂E
∂T
V−S−T∂S
∂T
V (2.145)
yields
CV =T∂S
∂T
V (2.146)
and
CP =T∂S
∂T
P. (2.147)
With
∂S
∂T
P
(A=.1) ∂S
∂T
V + ∂S
∂V
T
∂V
∂T
P (2.148)
we have
CP−CV =T ∂S
∂V
T
∂V
∂T
P (2=.18)T∂P
∂T
V
∂V
∂T
P (A= −.3) T∂P
∂V
T
∂V
∂T
P (2.5),(=2.6)T Vα2P
κT
. (2.149)
Using Eq. (2.9) we find for an ideal gas
CP−CV =n R, (2.150)
which we had used in the calculation of the air temperature profile on p. 48.
Example: Adiabatic Compressibility. Another exercise similar to the previ- ous one is the derivation ofκS, the adiabatic compressibility. This quantity is defined in Eq. (2.89), i.e.
κS= −1 V
∂V
∂P
S. We transform the right side via
∂V
∂P
S
(A=.1) ∂V
∂T P (2=.5)VαP
∂T
∂P
S+ ∂V
∂P T (2= −.6) VκT
. (2.151)
The remaining unknown derivative is
∂T
∂P
S
(A= −.3) ∂T
∂S
P
∂S
∂P
T = ∂T
∂S P
(2.147)=
T CP
∂
∂P
∂G
∂T
P
T
=∂∂VT
P=VαP
. (2.152)
Putting everything together yields
κS=κT −T Vα2P CP
(2.153) or if we combine this equation with Eq. (2.149)
κS
κT =CV
CP. (2.154)
Again we can calculateκSfor an ideal gas. The result is κS= z−1
z 1
P. (2.155)
The quantityz =CV/(n R)+1 was introduced in the context of Eq. (2.77) (cf. its discussion in the footnote on p. 48).
Example: Electric Field Effect on CV. What is the effect of an electric field onCV? We find the answer by working from
∂
∂E
∂2F˜
∂T2
ρ,E
(2.= −146) T−1CV,E
T,ρ= ∂2
∂T2
∂F˜
∂E
T,ρ
(2.129)= −VD/(4π)
ρ,E, (2.156)
i.e.
∂CV,E
∂E
T,ρ = T V 4π
∂2D
∂T2
ρ,E, (2.157)
whereρ =N/V and assuming constant fields throughoutV. WithD =εrE we may easily integrate this equation from zero to the final field strength, which yields
CV(E)=CV(0)+T V E2 8π
∂2εr
∂T2
ρ. (2.158)
Remark:A similar strategy helps to find corresponding expressions forκT orαP. Example: Electrostriction. Here we want to show the validity of
1 ρ
∂ρ
∂E
μ,T = E 4π
∂εr
∂P
E,T (2.159)
(cf. Frank 1955). Imagine a plate capacitor completely submerged in a dielec- tric liquid. The dielectric constant of the liquid isεr. The size of the capacitor is small compared to the extend of the liquid reservoir. This means that far from the capacitor the latter has no effect on the chemical potential,μ, of the liquid, i.e.μis constant. In addition the temperature, T, is held constant as well. The mean electric field,E, between the capacitor plates will affect the density,ρ=N/V, inside the capacitor. This is the expression on the left. The change does depend on electric field strength,E, and the derivative ofεr with respect to pressure,P. Our starting point is relation (A.3), i.e.
1 ρ
∂ρ
∂E
μ,T = −1 ρ
∂ρ
∂μ
E,T
(∗)=ρκT
∂μ
∂E
ρ,T. (2.160)
Note that (*) follows viaρ∂/∂ρ = −V∂/∂V and using ∂G/∂V|T,N,E =
−1/κT derived below, cf. (2.183). We continue with
∂μ
∂E ρ,
T = ∂
∂E
∂F˜
∂N
T,V,E
ρ,
T = ∂
∂N
∂F˜
∂E ρ,
T
T,V,E
(2.= −129) E 4π
∂εr
∂ρ
E,T.(2.161) The final ingredient is
ρ∂εr
∂ρ
E,T
(A=.2)ρ∂P
∂ρ
E,T
∂εr
∂P
E,T = 1 κT
∂εr
∂P
E,T. (2.162) Combination of the last three equations yields the desired result.
In order to estimate the magnitude of this effect we integrate Eq. (2.159) assuming that the derivative on the right side can be replaced by its value at zero field strength, i.e.
ρ ρ ≈ E2
8π
∂εr
∂P
o,T. (2.163)
The transition to SI-units is accomplished by replacing E with√
4πεoE, i.e. in SI-units Eq. (2.163) becomesρ/ρ ≈(εo/2)E2∂εr/∂P|o,T. Becauseεois a small constant (∼10−11 in these units), an appreciable effect requires rather high fields and/or conditions for which the derivative becomes large.13