Thermodynamics following Callen 1985 book 1The problem and the - - PowerPoint PPT Presentation

Thermodynamics

following Callen 1985 book

1—The problem and the postulates

2

Temporal nature of macroscopic measurements

• Complete (microscopic) description of (macroscopic) matter: positions and

momenta of all molecules … ~1023 values (~NA)

• vs. macroscopic coordinates aka thermodynamic coordinates.
• Macroscopic measurements are extremely slow on the atomic scale of

time (atomic vibrations ~10−15 s), and are extremely coarse on the atomic scale of distance (~10−9 m).

• By definition (← macroscopic observations), thermodynamics describes
• nly static states of macroscopic systems.
• Regarding coarseness of time scale: ~NA molecules and their coordinates

(positions, momenta) → only few are time-independent. Obvious candidates for thermodynamic variables are total E (energy), total p (momentum), total L (angular momentum).

3

Spatial nature of macroscopic measurements

• Macroscopic measurements are extremely slow on the atomic scale of time (atomic

vibrations ~10−15 s), and are extremely coarse on the atomic scale of distance (~10−9 m).

• Macroscopic observations, using “blunt” instruments, only sense coarse spatial averages
• f atomic coordinates.
• For example, of all the vibrational modes at various wavelengths (and frequencies), only

the longest wavelength mode “survives” the spatial and the time averaging → volume in macroscopic description.

• Thermodynamics is concerned with the macroscopic consequences of all those

coordinates, that do not appear explicitly in a macroscopic description of a system.

• These “hidden” modes may act as repositories of energy. Energy transfer via hidden

modes → heat. (Compare to energy transfer associated with a macroscopic coordinate, e.g., changing volume → mechanical work −P dV.)

4

Composition of thermodynamic systems

• We (temporarily) restrict our attention to simple systems, defined as systems that are:
• macroscopically homogeneous,
• isotropic,
• uncharged,
• large enough to neglect surface effects,
• not acted on by electric, magnetic, gravity fields.
• Relevant quantities (parameters):
• V …volume (m3) … mechanical parameter
• Nk (k = 1,…r) … number of molecules (in units of moles) in each of the chemically

pure components of which the system is a mixture … chemical composition

• Both V and Nk are extensive parameters = increase in system’s size proportionally

increases V and Nk

5

V
 Nk V
 Nk 2V 2Nk Avogadro constant

Numerical value 6.022 140 76 x 1023 mol-1 Standard uncertainty (exact) Relative standard uncertainty (exact) Concise form 6.022 140 76 x 1023 mol-1

https://physics.nist.gov/cgi-bin/cuu/Value?na

Internal energy

• Conservation of energy … a principle developed over ~2.5 centuries (1693 Leibnitz

through 1930 Pauli)

• 1798 Count Rumford … thermal effects as he bore cannons → Humphry Davy, Sadi

Carnot, Robert Mayer, James Joule … heat as a form of energy transfer

• Macroscopic systems have definite and precise energies, subject to a definite

conservation principle.

6

Postulate (0) For a macroscopic system there exists an energy function U, the internal

• energy. U is an extensive parameter and is measured relative to the energy in a

fiducial state, the energy of which is arbitrarily taken as zero.

Thermodynamic equilibrium

• Macroscopic systems exhibit some memory. This memory eventually fades (fast or very

slowly) toward a simple state, independent of memory.

• In all systems there is a tendency to evolve toward states in which the properties are

determined by intrinsic factors and not by previously applied external influences. Such simple states are, by definition, time independent. They are called equilibrium states. Thermodynamics describes these equilibrium states.

• Experimental observations as well as formal simplicity suggest (and ultimately verified

by the derived theory’s success):

7

Postulate 1 There exist particular states (equilibrium states) of simple systems that, macro- scopically, are characterized completely by the internal energy U, the volume V, and the mole numbers N1,…,Nr of the chemical components.

• N.B.:
• There will be more parameters for more involved systems.
• Is system in equilibrium? Quiescence not sufficient condition. Inconsistencies with

thermodynamics formalism are a sign of non-equilibrium.

• equilibrium—ergodicity…
• Few real systems (…none?) are in equilibrium.

Walls and constraints

• Manipulations of a wall (separation from surrounding, boundary condition)
• f a thermodynamics system → redistribution of a quantity.
• Walls can be restrictive or non-restrictive to a change in a particular

extensive parameter (i.e., a wall may constrain the parameter or allow it to change). For example, a rigidly fixed piston in a cylinder constitutes a wall restrictive with respect to volume, whereas a movable piston is non- restrictive w.r.t volume.

• Walls can be restrictive / non-restrictive w.r.t.:
• flow of heat … adiabatic vs. diathermal
• flux of matter
• doing work

8

Callen calls this “closed”

Measurability of the energy

• Ok, ∃U (internal energy). But can we control it a measure it?
• Stirring an ice+water system (in a container) causes it to melt faster (transfer of

mechanical energy). Shining sun also causes it to melt faster (inflow of heat); changing the walls (e.g., different material) can decrease the rate of ice melting.

• Wall impermeable (restrictive) to heat flow = adiabatic


Wall permeable to heat flow = diathermal
 Wall allowing the flux of neither work nor heat = restrictive w.r.t. energy
 Wall restrictive w.r.t. U and V and Nk = closed

• Simple system enclosed in an impermeable adiabatic wall: work is the only permissible

type of energy transfer. Since we can quantify mechanical work, we can “measure” the internal energy change of the system.

• In summary: There exists walls, called adiabatic, with the property that the work done in

taking an adiabatically enclosed system between two given states is determined entirely by the states, independent of all external conditions. The work done is the difference in the internal energy of the two states. Thus, changes in internal energy can be measured.

9

1845 Joule: "The Mechanical Equivalent of Heat"

10

An experiment, in which he specified a numerical value for the amount of mechanical work required to produce a unit of heat … to raise the temperature of a pound of water by 1ºF and found a consistent value of 778.24 foot pound force (4.1550 J·cal−1).

Two states, A and B. A → B or B → A possible, but not both. Concept of irreversibility.

Quantitative definition of heat

• ∆U between any two equilibrium states is measurable:
• The heat flux to a system in any process (at constant mole numbers) is simply

the difference in internal energy between the final and initial states, diminished by the work done in that process.

• Quasi-static mechanical work:



 … e.g., compression, displacing a piston in a cylinder
 … work on the system vs. by the system

• Quasi-static heat in an infinitesimal quasi-static process (at constant mole Nj):

11

δWM = − PdV δQ = dU − δWM = dU + PdV δQ, δWM … imperfect differentials, depend on process

Units of energy, heat, work: joule or J or kg⋅m2⋅s−2

The basic problem of thermodynamics

• One central problem that defines the core of thermodynamic theory:

12

The single, all-encompassing problem of thermodynamics is the determination

• f the equilibrium state that eventually results after the removal of internal

constraints in a closed, composite system. Closed composite system Internal constraint Initial state Final equilibrium state

Internal constraint removed Rigid free Adiabatic diathermal permeable

The entropy maximum postulate

• The simplest conceivable formal solution to the basic problem:

extremum principle — U, V, Nk such that they maximize some function.

13

Postulate 2 There exists a function (called the entropy S) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.

• N.B.:
• ∃ of S only postulated for equilibrium states
• If S = S(U,V,Nk) (i.e., the fundamental relation) is known, problem is solved.

14

Postulate 3 The entropy of a composite system is additive over the constituent

• subsystems. The entropy is continuous and differentiable and is

a monotonically increasing function of the energy.

Consequences:

• S = P

α

S(α)

• S(α) = S(α)(U (α), V (α), N(α)

1

, . . . , N (α)

r

)

• S of a simple system is a homogeneous first-order function of the extensive parameters:

S(λU, λV, λNk) = λS(U, V, Nk)

• ∂S

∂U

• V,Nk > 0
• = 1

T

• S can be inverted and solved for U: S(U, V, Nk) → U(S, V, Nk) . . . An alternative form of

the fundamental relation (all info about the system)

• λ = 1

N : S(U, V, Nk) = NS

⇣

U N , V N , Nk N

⌘ = s(u, v, Xk) . . . s, u, v, Xk are per-mole quantities (can be also made per-unit-mass aka specific quantites)

Postulate 4 The entropy of any system vanishes in the state for which ∂U/∂S|V,Nk = 0 (that is, at zero temperature). Postulate 4 is an extension, due to Planck, of the so-called Nernst postulate or third law of thermodynamics.

Summary of thermodynamic framework

15

Postulate 3 The entropy of a composite system is additive over the constituent subsystems. The entropy is continuous and differentiable and is a monotonically increasing function of the energy. Postulate 4 The entropy of any system vanishes in the state for which ∂U/∂S|V,Nk = 0 (that is, at zero temperature). Postulate 2 There exists a function (called the entropy S) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold

• f constrained equilibrium states.

Postulate 1 There exist particular states (equilibrium states) of simple systems that, macroscopically, are characterized completely by the internal energy U, the volume V, and the mole numbers N1,…,Nr of the chemical components. Postulate (0) For a macroscopic system there exists an energy function U, the internal energy. U is an extensive parameter and is measured relative to the energy in a fiducial state, the energy of which is arbitrarily taken as zero.

Composite system given, where the fundamental relation of each subsystem is known
 ⇒ S(α) known ∀α ⇒ S(U,V,Nk) known and we characterize the extrema

2—The conditions of equilibrium

16

• where

Intensive parameters

• S = S(U,V,N1,…,Nr) … fundamental relation in entropy representation


U = U(S,V,N1,…,Nr) … fundamental relation in energy representation

• We want to investigate changes of S or U, find extrema ⇢ express the differential (here for U):

17

dU = ∂U ∂S

V,Nk

dS + ∂U ∂V

S,Nk

dV +

r

∑

j=1

∂U ∂Nj

V,Nk≠j

dNj

• Identify (give names) to these intensive parameters:

∂U ∂S

V,Nk

≡ T − ∂U ∂V

S,Nk

≡ P ∂U ∂Nj

V,Nk≠j

≡ μj … temperature … pressure … chemical potential of jth component

∂A ∂B

C

≡ ( ∂A ∂B )C Notation: Partial derivative of A w.r.t. B while keeping C constant

• to then write:

dU = TdS − PdV +

r

∑

j=1

μjdNj

• r:

dU = δQ + δWM + δWC δWM = − PdV δQ = TdS δWC ≡

r

∑

j=1

μjdNj … quasi-static mechanical work … quasi-static heat … quasi-static chemical work

Equations of state

• T, P

, μk also functions of the extensive parameters (generally)

18

T = T(S, V, N1, …, Nr) P = P(S, V, N1, …, Nr) μk = μk(S, V, N1, …, Nr) Equations of state (EOS)

• Knowledge of all EOS ⇒ knowledge of fundamental relation
• Fundamental relation is a homogeneous first-order function 


⇒ EOS are homogeneous zeroth-order functions T(λS, λV, λN1, …, λNr) = T(S, V, N1, …, Nr) (λ > 0)

… homog. 0th order

• Temperature of the whole system ⇔ T is an intensive quantity (and same for P

, μk)

Consider molar (or specific) quantities in a single-component system:

u = u(s, v) … = U N ( S N , V N ,1) du = ∂u ∂s

v

ds + ∂u ∂v

s

dv

∂u ∂s

v

= ∂ (

U N )

∂ (

S N ) v

= ∂U ∂S

V

= T

du = Tds − Pdv

• Fundamental relation
• Intensive parameters
• Therefore
• Fundamental relation
• Differential
• Intensive parameters
• Therefore
• And obviously

General notation

19

U(S, V, N1, …, Nr) ⟶ U(S, X1, X2, …, Xt) ∂U ∂S

Xk

≡ T = T(S, X1, X2, …, Xt) ∂U ∂Xj

S,Xk≠j

≡ Pj = Pj(S, X1, X2, …, Xt) dU = TdS +

t

∑

j=1

PjdXj

In entropic representation:

S(U, V, N1, …, Nr) ⟶ S(X0, X1, X2, …, Xt) (X0 ≡ U) ∂S ∂Xk

Xj≠k

≡ Fk = Fk(X0, …, Xt) dS =

t

∑

k=0

FkdXk dS =

t

∑

k=0

∂S ∂Xk

Xj≠k

dXk F0 = 1 T

In energy representation:

Fk = − Pk T and obviously P1 = − P dU = TdS − PdV +

r

∑

j=1

μjdNj dS = 1 T dU + P T dV −

r

∑

j=1

μj T dNj

Thermal equilibrium

20

1 2

Let us look at the implications of the extremum principle for entropy.

“Closed” composite system. Two simple systems separated by a rigid impermeable diathermal wall.

V(1), V(2) and N(1), N(2) are fixed. U(1), U(2) free to change, subject to U(1) + U(2) = const . Postulate 2: Values of U(1), U(2) such as to maximize the entropy: dS = 0 Postulate 3: Additivity and differentiability of entropy: S = S(1) (U(1), V(1), N(1)

k ) + S(2) (U(2), V(2), N(2) k )

dS = dS(1) + dS(2) = ∂S(1) ∂U(1)

V(1),N(1)

k

dU(1) + ∂S(2) ∂U(2)

V(2),N(2)

k

dU(2) ⟶ dS = 1 T(1) dU(1) + 1 T(2) dU(2) But from constraint: U(1) + U(2) = const . ⇒ dU(2) = − dU(1) ⇒ dS = ( 1 T(1) − 1 T(2)) dU(1) dS must vanish for arbitrary value of dU(1): 1 T(1) = 1 T(2) Condition of equilibrium Assumes the existence of fundamental relations, resp. EOS (either explicit forms or not). Two equations for two unknowns, U(1) and U(2): U(1) + U(2) = const . 1 T(1)(U(1)) = 1 T(2)(U(2))

Thermal equilibrium

21

1 2

Let us look at the implications of the extremum principle for entropy.

“Closed” composite system. Two simple systems separated by a rigid impermeable diathermal wall.

We have used: dS = 0 … extremum Not yet: d2S < 0 … maximum

Agreement with intuitive concept of temperature

Consider initially T(1) > T(2) Entropy difference between initial and final states (final minus initial): ΔS > 0 But approximately where T(1) and T(2) are the initial temperatures But if the expression in parentheses is negative, then ΔU must also be negative. That is, heat flows from system with a high T to a system with a low T. ΔS ≈ ( 1 T(1) − 1 T(2)) ΔU(1)

Temperature scale and units

Absolute zero (0 K) and triple point of H2O (273.16 K) define the thermodynamic temperature scale.

Note the conventional method (Kelvin and Caratheodory) of introduction of temperature−1 as the integrating factor of Pfaffian form δQ.

Mechanical equilibrium

22

1 2

“Closed” composite system. Two simple systems separated by an impermeable movable diathermal wall.

N(1), N(2) are fixed. U(1), U(2) and V(1), V(2) free to change, subject to U(1) + U(2) = const . & V(1) + V(2) = const . Values of U(1), U(2), V(1), V(2) such as to maximize the entropy (Postulate 2): dS = 0 Additivity and differentiability of entropy (Postulate 3): S = S(1) (U(1), V(1), N(1)

k ) + S(2) (U(2), V(2), N(2) k )

dS = dS(1) + dS(2) = ∂S(1) ∂U(1)

V(1),N(1)

k

dU(1) + ∂S(1) ∂V(1)

U(1),N(1)

k

dV(1) + ∂S(2) ∂U(2)

V(2),N(2)

k

dU(2) + ∂S(2) ∂V(2)

U(2),N(2)

k

dV(2) & dU(2) = − dU(1) & dV(2) = − dV(1) ⇒ dS = ( 1 T(1) − 1 T(2) ) dU(1) + ( P(1) T(1) − P(2) T(2) ) dV(1) = 0 1 T(1) = 1 T(2) & P(1) T(1) = P(2) T(2) for arbitrary values of dU(1) and dV(1): Formally, 4 equations for 4 variables (T(1), T(2), P(1), P(2)). T(1) = T(2) P(1) = P(2)

• r &

Example

23

Callen 1985 book, pages 51–52

Equilibrium w.r.t. matter flow

24

1 2

“Closed” composite system. Two simple systems separated by a rigid diathermal wall, permeable to material 1 (N1) but impermeable to others (N2,N3,…).

V(1), V(2) and Nk(1), Nk(2), k>1 are fixed. U(1), U(2) and N1(1), N1(2) free to change, subject to U(1) + U(2) = const . & N(1)

1

+ N(2)

1

= const . dS = 0 dS = dS(1) + dS(2) = ∂S(1) ∂U(1)

V(1),N(1)

k

dU(1) + ∂S(1) ∂N(1)

1 U(1),V(1),N(1)

k≠1

dN(1)

1

+ ∂S(2) ∂U(2)

V(2),N(2)

k

dU(2) + ∂S(2) ∂N(2)

1 U(2),V(2),N(2)

k≠1

dN(2)

1

& dU(2) = − dU(1) & dN(2)

1

= − dN(1)

1

⇒ dS = ( 1 T(1) − 1 T(2) ) dU(1) − ( μ(1)

1

T(1) − μ(2)

1

T(2)) dN(1)

1

= 0 1 T(1) = 1 T(2) & μ(1)

1

T(1) = μ(2)

1

T(2) for arbitrary values of dU(1) and dN1(1): T(1) = T(2) μ(1)

1

= μ(2)

1

• r &

T … “potential” for heat flux P … “potential” for volume changes μ … “potential” for matter flux (also phase changes and chemical reactions) Matter tends to flow from regions of high chemical potential to regions of low chemical potential.

Chemical reactions and equilibrium

25

Mg2SiO4 ⇌ MgSiO3 + MgO 2H2 + O2 ⇌ 2H2O Stochiometric coefficients νj reflect the changes in mole number. … changes in mole numbers in ratios −1 : 1 : 1 … changes in mole numbers in ratios −2 : −1 : 2 0 ↔

r

∑

j=1

νjAj Chemical reaction for a system with r componenets, generally: ⇌ Fundamental equation: S = S(U, V, N1, …, Nr) Entropy change in a virtual chemical process: dS = −

r

∑

j=1

μj T dNj For now, assume an adiabatic rigid reaction vesser (i.e., U and V fixed) … even if not typical dS = − d ˜ Nj T

r

∑

j=1

μjνj Aj … symbol for chemical component but dNj proportional to νj dNj = νjd ˜ Nj writing dS = 0 ⇒

r

∑

j=1

μjνj = 0 Extremum principle: Example

2 independent reactions (equations) 3 additional constraints: amount of H, C, O Total 2 + 3 = 5 constraints. 5 unknown parameters: mole numbers of H2, O2, H2O, CO, CO2

3—Some formal relationships, and sample systems

26

Now use the Euler’s theorem for which is a homog. function of order 1:

The Euler equation

27

Euler’s homogeneous function theorem: Let f(x) be a homogeneous function of order n, that is Then U = U(S, V, N1, …, Nr) f(λx) = λnf(x) x ⋅ ∂f ∂x = nf(x) Proof: Differentiate with respect to lambda:

• Spec. for λ=1 the proof is completed.

f(λx) = λnf(x) ∂f ∂(λx) ⋅ ∂(λx) ∂λ = ∂f ∂(λx) ⋅ x = nλn−1f(x) U = ∂U ∂S

V,Nk

S + ∂U ∂V

S,Nk

V +

r

∑

j=1

∂U ∂Nj

V,Nk≠j

Nj U = TS − PV +

r

∑

j=1

μjNj S = 1 TU + P T V −

r

∑

j=1

μj T Nj Euler equation for U Euler equation for S

The Gibbs–Duhem relation

28

We have: (1) dU = TdS − PdV +

r

∑

j=1

μjdNj Also: (Euler) U = TS − PV +

r

∑

j=1

μjNj Therefore: dU = d(TS) − d(PV) +

r

∑

j=1

d(μjNj) dU = TdS + SdT − PdV − VdP +

r

∑

j=1

μjdNj +

r

∑

j=1

Njdμj (2) Subtracting eqn. (1) from eqn. (2): Gibbs–Duhem relation 0 = SdT − VdP +

r

∑

j=1

Njdμj Relation between the intensive parameters. A simple system of r components has r+2 intensive parameters, but only r+1 can vary independently (i.e., r+1 thermodynamic degrees of freedom). For a single component system: dμ = − S N dT + V N dP = − sdT + vdP

Simple ideal gas

29

PV = NRT S = Ns0 + NR ln ( U U0 )

c

( V V0 ) ( N N0 )

−(c+1)

U = cNRT s0 = (c + 1)R − ( μ T)0 s = s0 + cR ln ( u u0 ) + R ln ( v v0 )

Multicomponent simple ideal gas

30

Ideal van der Waals fluid

Molar heat capacity and other derivatives

31

First derivatives of fundamental equation … T, P, μk Second derivatives … material properties α ≡ 1 v ∂v ∂T

P

= 1 V ∂V ∂T

P

κT ≡ − 1 v ∂v ∂P

T

= − 1 V ∂V ∂P

T

cP ≡ T ∂s ∂T

P

= T N ∂S ∂T

P

= 1 N δQ dT

P

coefficient of thermal expansion isothermal compressibility molar heat capacity at constant pressure For a simple system of constant N, all other second derivatives can be expressed in terms of these 3.

κS ≡ − 1 v ∂v ∂P

S

= − 1 V ∂V ∂P

S

cV ≡ T ∂s ∂T

V

= T N ∂S ∂T

V

= 1 N δQ dT

V

adiabatic compressibility molar heat capacity at constant volume cP = cV + TVα2 NκT κT = κS + TVα2 NcP

4—Reversible processes and the maximum work theorem

32

33

redo slides for this chapter 4

34

Thermodynamic configuration space for a simple system S, U, V, N1,…,Nr directions (axes) Fundamental relation S = S(U, V, Nk) is a (r+2)dim hyper-surface in a dim = (r+3)dim space Such that ∂S/∂U| > 0 (… 1/T > 0) By definition, each point in configuration space is an equilibrium state. Non-equilibrium states have no representation in the thermodynamic configuration space. And not dealing with rates, velocities, time, …

35

Thermodynamic configuration space for a composite system S, U(1), V(1), Nk(1), U, V, Nk S = S(U(1), V(1), Nk(1), U, V, Nk)

36

Quasi-static process = succession of equilibrium states

• vs. real process, succession of equilibrium

AND non-equilibrium states −PdV ≡ mechanical work
 TdS ≡ heat transfer


• nly valid for quasi-static processes

Closed system starts in A. Some restriction is

• removed. There are newly accessible

equilibrium states. System proceeds to B if (and only if) B has maximum S among all newly accessible states. ΔS(A→B) ≥ 0 … non-decreasing ΔS>0 … irreversible
 ΔS=0 … reversible

37

Reversible (quasi-static) process ΔS=0

38

Reversible work source Reversible work sources are defined as systems enclosed by adiabatic impermeable walls and characterized by relaxation times sufficiently short that all processes within them are essentially quasi-static. … a repository system into which work is delivered Reversible heat source Reversible heat sources are defined as systems enclosed by rigid impermeable walls and characterized by relaxation times sufficiently short that all processes within them are essentially quasi-static. … a repository system into which heat is delivered Thermal reservoir A thermal reservoir is defined as a reversible heat source that is so large that any heat transfer of interest does not alter the temperature of the thermal reservoir. …a thermal reservoir characterized by a fixed temperature. Relaxation time τ – processes short compared to τ … not quasi-static – processes long compared to τ … quasi-static

39

The maximum work theorem. For all processes leading from the specified initial state to the specified final state of the primary system, the delivery of work is maximum (and the delivery of heat is minimum) for a reversible process.

40

Engine Refrigerator Heat pump Thermodynamic engine efficiency Coefficient of refrigerator performance Coefficient of heat pump performance

41

Carnot cycle

A → B … isothermal expansion … heat flows to aux., work transferred to RWS B → C … adiabatic expansion … work transferred to RWS C → D … isothermal compression … heat flows to RHS, work transferred from RWS D → A … adiabatic compression … work transferred from RWS Primary system … a hot thermal reservoir at Th Reversible heat source (RHS) … a cold thermal reservoir at Tc Reversible work source (RWS) Auxiliary system … the physical engine … a tool to extract/deliver heat and work Net work transferred to RWS: (Th-Tc)ΔS

5—Alternative formulations and Legendre transformations

42

The energy minimum principle

• So far: principle of maximum entropy
• Several equivalent forms … convenient (simpler) in particular types of problems


(For example, in classical mechanics: Newton ⟷ Lagrange ⟷ Hamilton)

• Already had: energy vs. entropy representation

43

✔max,principle ?extremum Entropy Maximum Principle. The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy. Energy Minimum Principle. The equilibrium value of any unconstrained internal parameter is such as to minimize the energy for the given value of the total entropy.

Postulate 2 There exists a function (called the entropy S) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.

Physics proof of the equivalence (max-S vs. min-U)

max S ⇒ min U

• Assume system in equilibrium but U not minimized (given S).
• Then we could withdraw some energy from system in the form of work, maintaining S constant.
• We could return the energy to the system in the form of heat.
• Now the system is restored to its original energy, but entropy has increased. This is a

contradiction with the initial assumption of equilibrium (i.e., max S). Therefore, at equilibrium, U must be minimized for a given S.

44

min U ⇒ max S

• Inversely, assume system in equilibrium but S not maximized (given U).
• Then we could extract heat from a RHS (initially at Tsystem) and deposit it in a RWS. The RHS is

thereby cooled. The entropy of the system therefore increases, while U remains unchanged.

• Now we could restore the RHS to its original temperature by flow of heat from the system.
• The system is restored to its original entropy, but energy has decreased. This is a contradiction

with the initial assumption of equilibrium (i.e., min U). Therefore, at equilibrium, S must be maximized for a given U.

“−1 rule”

45

Lemma (“−1 rule”) Three thermodynamic functions A, B, C. Then −1 = ∂A ∂B

C

∂B ∂C

A

∂C ∂A

B

. ∂A ∂B

C

= − ∂C ∂B

A

∂A ∂C

B

, ∂A ∂B

C

= −

∂C ∂B A ∂C ∂A B

. Or equivalently, Proof C = C(A, B) … dC = ∂C ∂A

B

dA + ∂C ∂B

A

dB C = const . ⇒ 0 = ∂C ∂A

B

dA + ∂C ∂B

A

dB ⇔ − ∂C ∂B

A

dB = ∂C ∂A

B

dA @ C = const . ⇔ − ∂C ∂B

A

= ∂C ∂A

B

∂A ∂B

C

⇔ − 1 = ∂A ∂B

C

∂B ∂C

A

∂C ∂A

B

Formal proof of the equivalence (max-S vs. min-U)

46

Assume entropy maximum principle: ∂S ∂X

U

= 0 & ∂2S ∂X2

U

< 0 (X ≡ X(1)

j )

P ≡ ∂U ∂X

S

= −

∂S ∂X U ∂S ∂U X

= − T ∂S ∂X

U

= 0 ✔ extremum ∂2U ∂X2

S

= ∂P ∂X

S

= ∂P ∂U

X

∂U ∂X

S

+ ∂P ∂X

U

= ∂P ∂U

X

P + ∂P ∂X

U

= ∂P ∂X

U

(at P = 0) = ∂ ∂X

S

−

∂S ∂X U ∂S ∂U X U

= −

∂2S ∂X2 ∂S ∂U X

+ ∂S ∂X

U ∂2S ∂X∂U

(

∂S ∂U X) 2 = − T ∂2S

∂X2 > 0 at ∂S ∂X

U

= 0. ✔ minimum

Note:

• 1. Analogy to isoperimetric problem in geometry:


A circle may be characterized either as the two-dimensional figure of maximum area for a given perimeter, 


• r as the two-dimendional figure of minimum perimeter for a given area.
• 2. Using the energy minimum principle results in the same conditions of equilibrium as we had before using the

entropy maximum principle (equality of T, P, μj).

Perhaps simply eliminate X between these two equations, and write ?

Toward Legendre transformations

47

• In both U and S-representations: 


extensive parameters … independent, intensive parameters … derived

• In laboratory: Often some intensive parameters easier to measure, control (esp. S vs. T)
• Is it possible to recast the formalism with a different set of independent parameters?
• That is, given the fundamental relation Y = Y(X0, X1, …, Xt), find a method where Pk = ∂Y/∂Xk can be

considered as independent variables, without any loss of information contained in the original fundamental relation. Now, for simplicity:

Y = Y(P) P ≡ dY dX Y = Y(X)

But we would sacrifice some

• f the mathematical content.

Better way: Exploit the duality between conventional point geometry and Pluecker line geometry. Y(X) envelope of family

• f tangent lines

Point in a plane: [X,Y] Line in a place: slope P, intercept ψ

• Fund. rel. Y=Y(X)
• Fund. rel. ψ=ψ(P)

Legendre transformations

48

Y = Y(X) ψ = ψ(P)

Legendre transformation

• Line with slope P through a point [X,Y]:
• Assume we know
• We find
• Eliminate X and Y using these 3 equations

Y = Y(X) P ≡ dY dX P = Y − ψ X − 0 ⟶ ψ = Y − PX ⇒ ψ = ψ(P)

• Inverse transform:
• Eliminate X and Y using these 3 equations

ψ = ψ(P) known … dψ = d(Y − PX) = dY − PdX − XdP = − XdP

• r

− X = dϕ dP

Legendre transformations

49

• Generalization to more than 1

independent variable

• Fund. rel. is a hypersurface in a

(t+2)-dimensional space

• Represented in point geometry…
• …or equally well as the envelope of

tangent hyperplanes

• Helmholtz free energy
• Enthalpy

Thermodynamic potentials

50

Y = Y(X0, X1, …, Xt) ⟷ U = U(S, V, N1, …, Nr) P0, P1, …, Pt ⟷ T, −P, μ1, …, μr F ≡ U[T] … F = F(T, V, N1, …, Nr) U(S, V, N1, …, Nr) F(T, V, N1, …, Nr) T = ∂U ∂S −S = ∂F ∂T F = U − TS U = F + TS dF = d(U − TS) = dU − d(TS) = TdS − PdV + ∑ μjdNj − TdS − SdT = − SdT − PdV + ∑ μjdNj H ≡ U[P] … H = H(S, P, N1, …, Nr) = U + PV dH = TdS + VdP + ∑ μjdNj

• Gibbs free energy

G ≡ U[T, P] … G = G(T, P, N1, …, Nr) = U − TS + PV dG = − SdT + VdP + ∑ μjdNj

• Grand canonical potential

U[T, μ] … U[T, μ](T, V, μ) = U − TS − μN dU[T, μ] = − SdT − PdV − Ndμ

Note: Generalized Massieu functions

Thermodynamic potentials

51

Helmholtz potential (Helmholtz free energy) F Enthalpy H Gibbs potential (Gibbs free energy) G

6—The extremum principle in the Legendre transformed representations

52

U ⟶ F,H,G … extremum principles for F,H,G ?

The Helmholtz potential

53

The minimum principles for the potentials

54

Helmholtz Potential Minimum Principle. The equilibrium value of any unconstrained internal parameter in a system in diathermal contact with a heat reservoir minimizes the Helmholtz potential over the manifold of states for which T = Tr. Enthalpy Minimum Principle. The equilibrium value of any unconstrained internal parameter in a system in contact with a pressure reservoir minimizes the enthalpy potential over the manifold of states for which P = Pr. Gibbs Potential Minimum Principle. The equilibrium value of any unconstrained internal parameter in a system in contact with a thermal and a pressure reservoir minimizes the Gibbs potential at constant temperature and presure (T = Tr and P = Pr). The General Minimum Principle for Legendre Transforms of the Energy. The equilibrium value of any unconstrained internal parameter in a system in contact with a set of reservoirs (with intensive parameters Pr1, Pr2, …) minimizes the thermodynamic potential U[P1,P2,…] at constant P1,P2,… (equal to Pr1, Pr2, …).

55

56

57

Joule-Thomson (“throttling”) process

uf + Pfvf = ui + Pivi

• r

hf = hi αTinversion = 1 dT = ∂T ∂P

• H,N1,N2,...

dP = v cP (αT − 1) dP αT>1 ⇒ a small dP<0 cools the gas αT<1 ⇒ a small dP<0 heats the gas

58

Joule-Thomson (“throttling”) process

59

Joule-Thomson: Inversion temperature for van der Waals

P = RT v − b − a v2 α = 1 v ∂v ∂T

• P

αTinv = 1 α = 1 T " 1 1 − b

v

− 2a

• 1 − b

v

• RTv

#−1 ✏1 ≡ b v ∼ 10−3 ✏2 ≡ a RTv ∼ 10−3—10−4 ↵ = 1 T  1 1 − ✏1 − 2(1 − ✏1)✏2 −1 1 = ↵Tinv = 1 − ✏1 + 2✏2 + o(✏) ⇒ ✏1 = 2✏2 ⇒ Tinv = 2a bR ↵T − 1 = = −✏1 + 2✏2(Tinv) + 2✏2(Tinv) ✓Tinv T − 1 ◆ + o(✏) T>Tinv ⇒ αT−1<0 ⇒ heating of gas (for dP<0) T<Tinv ⇒ αT−1>0 ⇒ cooling of gas (for dP<0)

60

U F = U − TS H = U + PV G = U − TS + PV

internal energy Helmholtz free energy enthalpy Gibbs free energy

Valid Facts and Theoretical Understanding Generate Solutions to Hard Problems

Maxwell relations

61

• In practical applications, often partial derivatives have to be

evaluated (e.g., ∂T/∂P|S,N, …)

• Can involve both extensive an intensive parameters,

thermodynamic potentials.

• Of all such derivatives, only three can be independent, and any

given derivative can be expressed in terms of an arbitrarily chosen set of three basic derivatives.

• Conventionally, one chooses: cP, α, κT.
• That is, all first derivatives can be written in terms of second

derivatives of the Gibbs potential, of which cP, α, and κT constitute a complete independent set (at constant mole numbers).

• The procedure: “reduction of derivatives”

Reduction of derivatives in single-component system

62

coefficient of thermal expansion isothermal compressibility molar heat capacity at constant pressure

dg = ∂g ∂T

• P

dT + ∂g ∂P

• T

dP = −sdT + vdP vα = ∂2g ∂T∂P −vκT = ∂2g ∂P 2 −cP T = ∂2g ∂T 2

Choice of α, κT, cP is implicit to Gibbs representation

63

• 1. If the derivative contains any potentials, bring them one by one to the

numerator and eliminate by the thermodynamic square.

• 2. If the derivative contains the chemical potential, bring it to the

numerator and eliminate by means of the Gibbs–Duhem relation.

• 3. If the derivative contains entropy, bring it to the numerator. If one of the

four Maxwell relations of the thermodynamic square now eliminates the entropy, invoke it. If the Maxwell relations do not eliminate the entropy, put a ∂T under ∂S. The numerator will then be expressible as

• ne of the specific heats (cV or cP).
• 4. Bring the volume to the numerator. The remaining derivative will be

expressible in terms of α and κT.

• 5. The originally given derivative has now been expressed in terms os

the four quantities cV, cP ,α and κT. The specific heat at constant volume is eliminated by the equation cV = cP − Tvα2/κT.

Procedure for reduction of derivatives in single-component system

64

∂x ∂y

• z

= 1

∂y ∂x

• z

∂x ∂y

• z

=

∂x ∂w

• z

∂y ∂w

• z

∂x ∂y

• z

∂y ∂z

• x

∂z ∂x

• y

= −1 ↔ ∂x ∂y

• z

= −∂z ∂y

• x

∂x ∂z

• y

↔ ∂x ∂y

• z

= −

∂z ∂y

• x

∂z ∂x

• y

When reducing derivatives, one uses Maxwell relations, Gibbs-Duhem (to eliminate dμ), and these identities:

“−1 rule”

65

1. If the derivative contains any potentials, bring them one by one to the numerator and eliminate by the thermodynamic square. Reduction of derivatives – examples

66

2. If the derivative contains the chemical potential, bring it to the numerator and eliminate by means of the Gibbs–Duhem relation. Reduction of derivatives – examples 3. If the derivative contains entropy, bring it to the numerator. If one

• f the four Maxwell relations of the thermodynamic square now

eliminates the entropy, invoke it. If the Maxwell relations do not eliminate the entropy, put a ∂T under ∂S. The numerator will then be expressible as one of the specific heats (cV or cP).

67

4. Bring the volume to the numerator. The remaining derivative will be expressible in terms of α and κT. Reduction of derivatives – examples 5. The originally given derivative has now been expressed in terms

• f the four quantities cV, cP ,α and κT. The specific heat at

constant volume is eliminated by the equation

cV = cP − Tvα2 κT

68