Basic Theoretical Physics: A Concise Overview P41 docx

As an application and generalization we shall presently treat the osmotic pressure and the decrease in boiling and solidification temperatures of a liquid by addition of sugar and de-icin

where without lack of generality it is assumed that only ν k is negative (Of course we choose the smallest possible integral values for the|ν i |).

The concentration ratios c ithen result from (54.3) and (54.4) according to the principle that the entropy should not change in course of the equilibrium reaction In this way from the related free enthalpy condition one obtains the

Law of Mass Action:

c ν1

1 · c ν2

2 · · c νk−1

k −1

c −νk k

where from the ideal gas equation (see the preceding subsection) for the

yield function f (T , p) with a pressure unit p0the following general expression results:

f (T , p) =

k

0

i=1

c ν i

k

0

i=1



p

p0

−νi

· e −νi˜

(0)

i (T )

For further simplification we have written

g(0)i (T , p) = ˜ g(0)(T ) + k B T · ln p

p0

,

which agrees with

s i = k Bln5k B T

2p +

By varying pressure and/or temperature one can thus systematically shift the reaction yield in accordance with (54.6) (e.g., increasing the pressure leads

to an increased fraction of components with negative ν i ).

As an application and generalization we shall presently treat the osmotic pressure and the decrease in boiling and solidification temperatures of a liquid

by addition of sugar and de-icing salt Firstly, however, an unusual topic

54.3 Electron Equilibrium in Neutron Stars

Consider the quantitative answer to the following question: How many elec-trons (or protons!) are there in a neutron star? Firstly we need to know the

fraction:

Nelectrons= Nprotons∼= 10−5 · Nneutrons; i.e., the ratio is extremely small, although different from zero Of course, all absolute numbers would be extremely large3

3 How large is N e? One can estimate this number by inserting typical values for the radius of a neutron star (≈ 10 km) and of a neutron (≈ 10 −13cm).

418 54 Applications II: Phase Equilibria in Chemical Physics

The above result is then obtained Consider the relevant astrophysical

equilibrium reaction, which is the equilibrium inverse β-decay process:

pp + ee ↔ nn + νν e

Here pp stands for the proton4, nn for the neutron, ee for the electron and νν e for the electron-neutrino (In a β-decay process a neutron emitting

an electron plus an electron-antineutrino νν e decays as follows: nn → pp +

ee + νν e ; the inverse β-decay is the fusion of a proton and an electron under

very high pressure, as high as is typical for a neutron star, or even higher, into a neutron plus an outgoing electron-neutrino, i.e., essentially according

to the← part of the previous reaction equation The electron-neutrino and

electron-antineutrino are particles and antiparticles with (almost) vanishing rest mass, i.e., negligible for the present thermodynamics (see below).) In any case, the first reaction can actually be “equilibrated” in both directions; in neutron stars there is thermodynamic equilibrium as above, and additionally the temperature can be neglected, since one is dealing with a degenerate Fermi gas

The equilibrium condition is



i

ν i · μ i (T , p, N1, , N k)= 0 ,!

where the μ i (T , p, N1, ) are the respective chemical potentials, i.e., the free enthalpies per particle for the particle considered, and the ν i the reaction numbers (not to be confused with the neutrinos)

The chemical potentials of neutrino and anti-neutrino, as already men-tioned, can be neglected in the present context However, for the other

parti-cles under consideration, the chemical potential at zero temperature (and in this approximation, our neutron stars – and also “ white dwarfs ”, see above, can always be treated at the temperatures considered, viz as a degenerate Fermi gas) is identical with the nonrelativistic kinetic energy per particle:

μ i ≈ ε kin.

2M n

2

V ,

as was shown for the electron gas in metals (n V is the number density and

M the mass of the considered particles) Thus

− (n V)2/3neutron

Mneutron +

*

(n V)2/3proton

Mproton +

(n V)2/3electron

melectron

+

!

= 0 (54.7)

4 We write pp (instead of p) for the proton, nn (instead of n) for the neutron, ee for the electron, and νν for the neutrino, to avoid confusion with the pressure p, the particle density n or the particle number N as well as with the elementary charge e and the reaction numbers ν mentioned above

However, the electron mass, m e, is 2000 times smaller than the proton

mass M P(≈ neutron mass M N), whereas the number densities of electrons and protons are equal Thus the second term of the previous equation can be neglected As a consequence

(n V)electron

(n V)neutron ≈



m e

M N

3

≈ 10 −5 ,

as stated

54.4 Gibbs’ Phase Rule

In an earlier section we considered the case of a single component (K = 1, e.g., H2O) which could exist in three different phases (P = 3, solid, liquid or

vapor.)

In contrast, in the second-from-last subsection we treated the case of two

or more components (K ≥ 2) reacting with each other, but in a single phase (P = 1), according to the reaction equation k

i=1 ν i A i = 0

We shall now consider the general case, and ask how many degrees of

freedom f , i.e., arbitrary real variables, can be chosen, if K different compo-nents are in thermodynamic equilibrium for P different phases The answer

is found in Gibbs’ phase rule:

As a first application we again consider the (p, T ) phase diagram of H2O

with the three phases: solid, liquid and vapor Within a single phase one has

f ≡ 2 (= 1 − 1 + 2) degrees of freedom, e.g., G = G(T, p); on the boundary

lines between two phases there remains only 1 degree of freedom (= 1−2+2)

(e.g., the saturation pressure is a unique function of the temperature only, and cannot be varied by choosing additional variables), and finally at the

triple point we have f = 0, i.e., at this point all variables, temperature and pressure are completely fixed (f = 1 − 3 + 2 ≡ 0).

As a second application consider a system with K = 2, e.g., two salts.

In one solvent (i.e., for P = 1) in thermal equilibrium one would have f =

2− 1 + 2 = 3 degrees of freedom For example, one could vary T , p and

c1, whereas c2 = 1− c1 would then be fixed If there is thermodynamic

equilibrium with P = 2 phases, e.g., a solid phase plus a fluid phase, the number of degrees of freedom is reduced to f = 2; e.g., only T and p can be varied independently, in contrast to c i If there is thermodynamic equilibrium

of the two components in three phases, then one has only one free variable

(f = 1).

420 54 Applications II: Phase Equilibria in Chemical Physics

We shall now proceed to a proof of Gibbs’ phase rule:

a) Firstly, consider a system with P = 1 Then K + 1 variables can be freely chosen, e.g., T , p, c1, c2, , c k −1 , whereas c k= 1− c1− c2− − c k −1is

dependent; thus the Gibbs’ phase rule is explicitly satisfied: f = K −1+2 b) Now let P ≥ 2 ! Then at first one should consider that for every compo-nent, k = 1, , K, there are P −1 additional degrees of freedom, i.e the

ratios

c(2)k /c(1)k , c(3)k /c(1)k , , c (P ) k /c(1)k Thus one has (P −1)·K additional variables But there are also (P −1)·K

additional constraints, i.e

μ(2)k = μ! (1)k , , μ (P ) k = μ! (1)k

c) A last sequence of constraints must be considered:

p(2)k = p! (1)k , , p (P ) k = p! (1)k This gives P − 1 constraints on the pressure.

Hence

f = K − 1 + 2 + (P − 1) · K − (P − 1) · K − (P − 1) ≡ K − P + 2 , q.e.d.

54.5 Osmotic Pressure

Again assume, as in section 54.2, that we are dealing with a semipermeable membrane (as commonly occurs in biological cells) Let the semipermeable membrane be nonpermeable for the solute, 2, ( ˆ= salt or sugar), but permeable for the solvent, 1

Now consider a U-tube, which is separated into two parts at the center

by a semipermeable membrane and filled with a liquid (water) to different heights in the respective parts

In the left-hand part of the U-tube the water level is h; in the enriched right-hand part (enriched by salt or sugar, Δc2 > 0) the level is enhanced,

h + Δh, with Δh > 0.

This corresponds to a pressure difference5Δp, the so-called osmotic pres-sure posmotic p., which is given by

Δp ≡ posmotic p.= Δc2· N k B T

ΔN2

V k B T (54.9)

5 The solvent concentration is only slightly diminished on the right-hand side, in favor of the enhanced solute concentration on this side, see the text

Fig 54.2 Osmotic pressure (schematically) The

volume V is divided into two parts by a

semiper-meable membrane (the Y -axis between 0 and 1)

The membrane is permeable for the solvent ( −

sym-bols), but non-permeable for the solute molecules

( symbols) As a result, the pressure on the r.h.s.

is enhanced by an amount called the osmotic

pres-sure Δp.

(This expression is analogous to an effective ideal gas of ΔN2 solute molecules suspended in a solvent For a true ideal gas the molecules would

be suspended in a vacuum; but there the pressure does not depend on the

mass of the molecule This is also true for the present situation.)

Figure 54.2 above schematically shows a semi-permeable membrane through which solute molecules cannot pass

In the above derivation, it does not matter that the vacuum mass of the molecule is replaced by an effective mass, for which, however, the value is

unimportant if the solute concentrations c2 (on the l.h.s.) and c2+ Δc2 (on the r.h.s.) are small enough, such that only the interactions with the solvent come into play (These arguments make a lot of ‘microscopic’ calculations unnecessary.)

A precise proof again uses the entropy of mixing and the partial pres-sure p i :

s i (T , p i ) = s i (T , p) − k B · ln c i Thus we have the molecular free enthalpy

g i (T , p i ) = g i(0)(T , p) + k B T · ln c i ,

and because of the equality of chemical potential and the molecular free enthalpy:

μ i (T , p, c i ) = μ i (T , p) + k B T · ln c i The two equilibrium conditions (not three!) for each side of the

semiper-meable membrane are:

(i) μ1(T , p, c1(= 1− c2))≡ μ1(T , p + Δp, c1− Δc2) , and

(ii) T1≡ T2(= T )

Thus

g1(0)(T , p)+k B T ·ln(1−c2)≡ g(0)

1 (T , p)+Δp ·

∂g(0)1

∂p +k B T ·ln(1−c2−Δc2)

422 54 Applications II: Phase Equilibria in Chemical Physics

With

∂g1(0)

∂p = v1

and the linearizations

ln(1− c2)≈ −c2 and ln(1− c2− Δc2)≈ −c2− Δc2

we obtain for the osmotic pressure:

Δp = k B T

v1

Δc2,

and finally with

v1:= V

N1+ N2

, for N2 N1: Δp ≈ ΔN2

V k B T

This derivation is, in principle, astonishingly simple, and becomes even simpler by using the “effective mass” argument above; a microscopic statistical-mechanical calculation would, in contrast, be unnecessarily complicated

54.6 Decrease of the Melting Temperature Due

to “De-icing” Salt

In the preceding section the addition of sugar on the nonpermeable side

(nonpermeable for the sugar but not for the solvent) of a semipermeable membrane led to a pressure difference, i.e., a higher pressure, higher by the osmotic pressure, on the sugar-enriched side However the temperatures were

identical on both sides of the interface

In contrast, we now consider the changes in the melting temperature of ice (and the boiling temperature of a liquid ) by addition of soluble substances,

e.g., again some kind of salt or sugar, to the liquid phase6, i.e., the interface

is here the surface of the liquid

We thus consider, e.g., the phase equilibria A) solid-liquid and B)

liquid-vapor, i.e., a) without addition and b) with addition of the substance

consid-ered For simplicity we only treat case A)

In case Aa) we have:

μsolid1 (T , p, c1= 1)= μ! liquid1 (T , p, c1= 1) ,

whereas for Ab):

μsolid1 (T − ΔT, p, c1= 1)= μ! liquid1 (T − ΔT, p, c1= 1− Δc2)

6 We assume that the added substance is only soluble in the liquid phase, but this can be changed, if necessary

Forming a suitable difference we obtain

− ∂μsolid1

∂T ΔT = − ∂μ

liquid 1

∂T ΔT + k B T · ln(1 − Δc2) ,

and with ln(1− x) ≈ −x we have:

ΔT = − k B T · Δc2

∂μ l.

1

∂T − ∂μ s.

1

∂T

.

Hence one obtains, with

μ = G

∂G

∂T =−S , as well as S

N ≡ s :

ΔT = − k B T · Δc2

s s.

1 − s l.

1

≡ + k B T2· Δc2

where additionally the molecular heat of melting

l s. →l. := T · s l.1− s s.

1

has been introduced

As a result we can state that the melting temperature (and analogously

the boiling temperature) of the ice (and of the heated liquid) is decreased by

the addition of salt (or sugar) to the liquid phases In the boiling case the relevant equation is

μ l.1(T − ΔT, p, c1− Δc2)= μ! vapor1 (T − ΔT, p, c1))

In this case too it is essentially the enhancement of the entropy by mixing which is responsible for the effect.

54.7 The Vapor Pressure of Spherical Droplets

Now consider an ensemble of spherical droplets of radius R with a sufficiently large value of R such that the number N of particles within a droplet is Outside such droplets the saturation pressure p Ris enhanced w.r.t the value

p ∞ for a planar surface, i.e., for R → ∞.

For the Helmholtz free energy of a droplet we obtain

dF = −p · dV + σ · dO − S · dT , with dV = 4πR2· dR

and dO = 8πR · dR , where dV is an infinitesimally small increment of volume and dO an

incre-ment of surface area of the droplet

424 54 Applications II: Phase Equilibria in Chemical Physics

The above relation for dF defines the surface tension σ, an energy per surface area This is best explained in the framework of the physics of soap

bubbles.7

We thus obtain for the Helmholtz free energy of the droplet:

Fdroplet(T , R) = Ndroplet· f(T, R) + 4πR2σ(T , R)

Here f (T , R) represents the volume part of the Helmholtz free energy per atom, and σ(T , R) is the above-mentioned surface tension, which we now

want to calculate Thermal phase equilibrium gives

μliquid droplet= ∂G

∂N = f + p R · v l. + 4πσ · ∂R2

∂N

!

= μvapor.

With

N ≡ 4πR3

3v l.

, hence dN = 4πR2dR

v l.

and dR2= 2R · dR ,

it follows that:

μliquid= f + v l. ·



p R+2σ

R



!

= μvapor≡ k B T · ln p R

p0 + B(T ) ,

8

where p0 is an arbitrary unit pressure and B(T ) a temperature-dependent constant of physical dimension energy per atom.

For R < ∞ we thus have:

f + v l. ·



p R+2σ

R



= k B T · ln p R

p0

+ B(T ) ,

and for R = ∞:

f + v l. · p ∞ = k B T · ln p ∞

p0

+ B(T )

By subtracting the second equation from the first, we obtain:

k B T · ln p R

p ∞ = v l. ·



2σ

R + p R − p ∞



, and neglecting (p R − p ∞)9:

p R

p ∞ ≈ e kB T vl. · 2σ

7

For the physics of soap bubbles see almost any textbook on basic experimental physics

8

The last term on the r.h.s is reminiscent of the analogously defined free enthalpy

of mixing: μvapor= k B T · ln c i + , with the partial pressure p i = c i · p.

9

The result reminds us (not coincidentally) of the Clausius-Clapeyron equation

The saturation vapor pressure outside a spherical droplet of radius R

is thus greater than above a planar surface With decreasing radius R the

tendency of the particles to leave the liquid increases

A more detailed calculation shows that a critical droplet radius R c exists, such that smaller droplets, R < R c , shrink, whereas larger droplets, R > R c , increase in size Only at R c is there thermal equilibrium For R c the following rough estimate applies:

σ · 4πR2

c ≈ k B T

In this context the analogy with the Ising model (cf (42.5)) is again

help-ful For a cubic lattice with nearest-neighbor separation a and ferromagnetic nearest-neighbour interaction J (> 0)) in an external magnetic field h0(> 0)

one has:

Ising model: H = −J 

|rl−rm|=a

s l s m − h0



l

s l , (54.12)

with s l and s m =±1, corresponding to ↑ and ↓ spins.

We shall now consider the following situation: embedded in a ferromag-netic “lake” of ↓ spins is a single “island” (approximately spherical) of ↑ spins The volume V of the 3d-“island” is assumed to be

V ≈ 4πR3

3 ,

while the surface area of the island is

O ≈ 4πR2 The energy of formation ΔE of the island in the lake is then given by

ΔE ≈ O

a2 · 2J − V

a3 · 2h0,

as one can easily see from a sketch

The first term on the r.h.s of this equation (∝ J) corresponds to the

surface tension, while the second term (∝ h) corresponds to the difference of the chemical potentials, μ R − μ ∞ From ΔE !

= 0 it follows that

R c = 3aJ/h0.

In this way one can simultaneously illustrate nucleation processes (e.g.,

nucleation of vapor bubbles and of condensation nuclei) in the context of

over-heating or supercooling, e.g., in the context of the van der Waals equation.

The analogy is enhanced by the above lattice-gas interpretation.10

10A reminder: In the lattice-gas interpretation s l =±1 means that the site l is

either occupied, (+), or unoccupied, (-).

55 Conclusion to Part IV

For all readers (not only those who understand German1 or wish to prac-tise their knowledge of the language) for all four parts of our compendium there are (translated) exercises to complement this book These are sepa-rately documented on the internet, see [2], and purposely not integrated into this script

In Thermodynamics and Statistical Physics both types of learning mate-rial, i.e., textbook and corresponding exercises, have been centered around

a) Phenomenological Thermodynamics, with the four quantities

F (T , V, N, ) (Helmholtz f ree energy) ,

U (T , V, N, ) (internal energy) , S(T , V, N, ) (entropy)

and the absolute (or Kelvin) temperature T , and

b) Statistical Physics, for historical reasons usually called Statistical Mechan-ics, although this name is unnecessarily restrictive.

a) and b) are closely interdependent, due essentially to the fundamental relation

F (T , V, N, ) ≡ −k B T · ln Z(T, V, N, ) , (55.1) where

Z(T , V, N, ) =

i

e− Ei(V,N, )

is the partition function.

This relation between a) and b), (55.1), applies for so-called canonical ensembles, i.e., when the particle number N is fixed and the heat bath only

exchanges energy with the system considered – defining the reciprocal Kelvin temperature

β



k B T



as conjugate parameter (conjugate to the energy)

1

Sometimes even a partial understanding may indeed be helpful

422 54 Applications II: Phase Equilibria...

incre-ment of surface area of the droplet

424 54 Applications II: Phase Equilibria in Chemical Physics

The...