review of thermodynamics - by j g saven

U. Penn. Chemistry 521. Statistical Mechanics I. Fall 1998 J. G. Saven

9/16/98 1

Thermodynamics

References: Chandler, chapters 1 and 2 [1]; McQuarrie, chapters 1-4 [2]; mostly from Callen,chapters 1-8 [3].

Heat, work, and some other definitionsMacroscopic measurements are slow and coarse on the atomic scale. Typically, we end upaveraging over a huge number ~1023 of molecular degrees of freedom in determining macroscopicquantities. We can treat matter as being essentially continuous and for the most part ignore thefact that molecules exist.

We will consider systems that are homogeneous, isotropic, uncharged, and we neglect surfaceeffects.

Thermodynamics describes only static states of macroscopic systems. That is, those states thatare time-independent.

We will specify the amount of stuff (matter) by chemical composition or by mole numbers, N.Extensive parameters are those parameters that depend on the amount of stuff we have. Morespecifically, if a system can be broken up into subsystems, the value of an extensive quantity isequal to the sum of the values it has in each of the subsystems.

N1, V1 N2, V2 Nm, Vm

Examples include the volume V and the mole number N.

N = Nii = 1

m

V = Vii = 1

m

Macroscopic systems have well-defined energies that are conserved and extensive.

Equilibrium: Tendency to evolve to states that depend only on intrinsic factors and not on thepast history of the system. An equilibrium state may also be called a time-independent state (atleast with regard to macroscopic variables).Equilibrium states are characterized by relatively few variables, e.g., U, V, Ni

We will use the somewhat circular definition that a system is at equilibrium if it obeys the laws andequations of thermodynamics. We note that thermodynamics is also useful for describingmetastable (long lived, but unstable) statesPossible constraints on system

adiabatic: no heat flowdiathermal or diabatic: heat flow permitted


9/16/98 2

closed: no matter flowisolated system: no heat flow, no matter flow

Lets consider a piston in a system with adiabatic walls.

N, V, U N', V', U'

If the process happens really slowly and is reversible, the work done is the difference in internal energy U between the two states. The internal energy depends only on the initial and final statesand not on the path between them. The internal energy is said to be a state variable.

For a quasistatic (reversible) process: dW = - PdV = dUNote that if dV < 0, work is done on system and its energy increases (hence dW = PdV).Let the number of moles of a substance be fixed (constant N). The heat flux dQ is the difference inU between two states diminished by the work done:

dQ = dU dW = dU + PdVWe must emphasize that neither Q nor W are state functions. In general, their values depend on thepath taken between to points that specify the state of a system. Only when a process is quasi-staticor reversible, can we equate these quantities with state functions.

Central problem of thermodynamicsWhat is equilibrium state after the removal of internalconstraints on a closed or isolated system?

For example, what happens to U1, U2, etc. if thepiston is allowed to move? made diathermal? madepermeable to N1?

Postulate I(empirically verified ) S, the entropy, is a state function of U and other extensive variablesS(U,V,N...). The equilibrium values of the unconstrained intrinsic variables are those thatmaximize S in the absence of internal constraints S(U, V, N) = max S(U, V, N,X1,...,Xn), wherewe maximize with respect to the variables X.

Implies the second law of thermodynamics since S without the constraint is at a maximum.Therefore any other state of the system, including one where a constraint is imposed, must be ofequal or lower entropy. After the removal of an internal constraint,

Sfinal Sinit 0

N1, V1, U1 N2, V2, U2

piston


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Postulate IIEntropy is extensive and is a monotonically increasingfunction of the energy.

Homogeneous functionsConsider a system that is broken up into m subsystems.

N1, V1 N2, V2 Nm, Vm

Let all the subsystems be equivalent and identical: Si=So for each i.

S = S(U,V,N) =

i

m

Si (Ui, Ni, Vi)

m different subsystems

S (U,V,N) = m So (Uo No, Vo) S is an extensive functionS (mUo mNo, mVo) = m S(Uo No, Vo)(turns out this is true for any value of m)

Or we could just scale system:

l S (U,V, N) = S ( l U, l V, l N)same entropy function in each case l S (U,V, N) = S ( l U, l V, l N)

Let l = N-1 N-1 S (U,V, N) = S (N-1 U, N-1 V, 1)

S (U,V, N) = NS UN , VN, 1

= NS (u, v, 1) = N s(u,v)Thus S scales like size of system. s(u, u) is entropy per mole.S = Ns, V = Nv, U = Nu

S = (mu, mv, m) = m s (u, v)S (and the other extensive variables) are said to be a homogeneous function of order 1.

Entropy maximization and intensive variables

S/ N

U/ N


9/16/98 4

Lets return to considering the internal energy U. Compute the first differential since we areinterested in small changes.

dU US

dS UV

dV UN

dNV N S N S V

=

+

+

, , ,

Define intensive parameters

U S V,N

= T temperature

= UV

PS N,

pressure

=UN S V,

chemical potential

These definitions agree with our picture of each quantity, and this will be shown when wediscuss equilibria. Note that these quantities are independent of the amount of matter N. Suchquantities are called intensive properties or intensive variables.

The following is also known as the first law of thermodynamics: energy can be converted intodifferent forms but it is conserved.

dU = TdS - PdV + m dN

Note for a quasi-static process with N = constant

dU = dQ + dW dQ = TdS

= TdS - PdV reversible heat flow dW

Heat flow is associated with a change in the entropy S. Note that each intensive variable isassociated with a derivative with respect to an extensive variable e.g.,

1T

SU V N

=

,

U and T are said to be conjugate variables (in the entropy representation). We now apply theentropy extremum principle (Postulate I) and note its implications.


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Equilibria

In maximizing S, we must find a state wheredS = 0.

The system is isolated so each of the naturalinternal variables are conserved for any process

dN = 0, matter is conserveddV = 0, the total volume does not changedU = 0, energy is conserved

Conservation Laws of the Composite SystemU1 + U2 = constant. dU1 = - dU2 V1 + V2 = constant. dV1 = - dV2 N1 + N2 = constant. dN1 = - dN2

dS = ( S U1

)V1N1

( S U2

)V2N2

dU1 + ( S V1

)U2V2

( S V2

)U2N2

dV1

+ ( S N1

)U1V1

( S N2

)U2V2

dN1

S1 U1 V1N1

= 1T1

S1 V1 U1N1

=

S U1 V1N1

U1 V1 S1N1

= -1T1

(-P1) = P1T1

S N1 U1V1

=

S1 U1 V1N1

U1 N1 V1S1

= -1T1

m 1

dS = (( 1T1

) ( 1T2

) dU1 + (P1T1 P2T2

) dV1 + (( m 1T1 + m 2T2

)) dN1

Thermal Equilibrium(dV1 = 0, dN1 = 0)

N1, V1, U1 N2, V2, U2

piston

dS = S1 U1 V1N1

dU1 + S1 V1 U1N1

dV1 + S1 N1 V1N1

dN1

+ S2 U2 V2N2

dU2 + S2 V2 U2N2

dV2 + S2 N2 V2U2

dN2


9/16/98 6

We will assume the wall is adiabatic and switched to diathermal. By so doing, we remove aninternal constraint. What is final state of the system?

We have a system with a diathermal wall that permits heat flow between the two subsystems.

V and N can not change since the wall is fixed and impermeable.

diathermal wall

heat

N1, V1, E1 N2, V2, E2

adiabatic wall

adiabatic wall

adiabatic wall

N2, V2, E2N1, V1, E1

Energy is conserved: U1 + U2 = constant = U

Therefore, dU1 + dU2 = 0

dN1 = dN2 = 0 since there is no matter flow.dV1 = dV2 = 0 No volume change

dS = 0 (entropy at maximum) = d(S1 + S2)

dS = S1 U1 V5N1

dU1 + S2 U2 V2,N2

dU2

But dU1 = -dU2So

0 = 1T1

1T2

dU1

dS must vanish for arbitrary changes dU1. Thus

1T1

= 1T2

At equilibrium, the temperatures of the two subsystems are equal.

Initially let T1 > T2. When remove internal constraint, the entropy must increase D S > 0. Entropywill at a maximum for the new equilibrium conditions. Any other state must be of lower S, e.g,the initial conditions.

D S 1T1

1T2

D U1< 0 < 0


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D U1 < 0 energy flows from hot (high T1) to cold (low T2). Since V is fixed, energy mustflow in the foam of heat Q. Vice versa for T2 > T1 , (D U1 > 0).Mechanical Equilibrium(dN1 = 0)

moveable, diathermal wall

heat

N1, V1, E1 N2, V2, E2

adiabatic wall

fixed, adiabatic wall

adiabatic wall

N1, V1, E1 N2, V2, E2

U1 + U2 = constantV1 + V2 = constant

dS = Si Ui ViWi

i = 1

2dUi +

Si Vi UiNg

dVi

Si Vi UiNi

=

Si Ui NiVi

Ui Vi SiNi

= - ( 1Ti

) (- Pi)

dS = 1T1

1T2

dU1 + +P1T1 + -P2

T2 dV1 = 0

Since dU1 = - dU2, dV1 = - dV2

Since dU1 and dV1 are each independent and arbitrary, we have

1T1

= 1T2

and P1T1

= P2T2

T1 = T2

P1 = P2

Let T1 = T2 and assume P1 > P2 initially


9/16/98 8

D S = 1T

(P1 P2) D V1 D S, D V small >0 >0 >0

D V1 must be D V1 > 0. That is, the volume of 1 gets bigger due to its higher initial pressure P1.The volume of 2 gets smaller.

Equilibrium with respect to Matter Flow

Permeable, fixed wall: dV1 = 0

fixed wall

N1, V1, E1 N2, V2, E2

adiabatic wall

fixed, impermeable wall

N1, V1, E1 N2, V2, E2

adiabatic wall

dS = 1T1

1T2

dU1 + m 1T1

+ m 2T2

dN1

dU1 and dN1 are independent. Thus

1T1

1T2

= 0 T1 = T2and

m 1T1

+ m 2T2

m 1 = m 2 chemical potentials are equal

Assume initially what T1 = T2 = T and m 1 > m 2

D S = m 1T

+ m 2T

dN1 = 1T ( m 2 m 1) dN1

> 0 < 0

D N1 < 0 if D S > 0Matter flows out of region 1. Matter flows from high m to low m .

Chemical Equilibrium


9/16/98 9

U, V are fixed.

Let separated species mix in aclosed, adiabatic container.

Turn on chemical reaction.

U = constant and V = constantLet l be a variable that measures the extent of the reaction. Given some small change dl , thestoichiometric coefficients provide the change in the number (of moles) of each type of species:

dN3 = n 3 dl then dN2 = +n 2 dl and dN1 = +n 1 dl

The other temperature and pressure equilibria still hold. We then have

dS = m jT

dNjj = 1

r

Let r be the number of different chemical species

= d lT

m j n jj = 1

r

dS = 0 m j n jj = 1

r

= 0

Energy Minimum Principle

Formal structure

U = U (S, V, N) S = S (U, V, N) *we can also write

T = T (S, V, N), P = P (S, V, N) .....but the functions U and S contain more information since they dictate the equilibrium state of thesystem.

Minimizing the energy is equivalent to maximizing the entropy

Let X denote other internal extensive variables of the system, such as the volume V.

The entropy maximization principle states that for fixed total internal energy U, the values ofany unconstrained internal parameters are those that maximize the entropy S.

Similarly, the energy minimization principle states that for constant entropy S, the values ofany unconstrained internal parameters are those that minimize the energy.

2 H2 + 1 02 2 H2O

fl fl fl

-n 1 -n 2 n 3 stoichiometric coefficients (reactants: n < 0) (products n > 0)


9/16/98 10

Ener

gy

Energy

other extensive variables

other extensive variables

AA

BB

plane of constant S

for process A: S = 0, E < 0for process B: S = q/T = E > 0

What if we didn't minimize the internal energy? Let's say U Umin for some point that isconsistent with a particular value of the entropy. If this were the case then, we could extractenergy as work without changing the entropy (process A in the figure) and then put that sameamount of energy back into the system as heat (process B) thereby increasing S without changingthe overall energy. S would not be at a maximum!

The equilibrium state must satisfy both extremum conditions. Thus we can work or either energyon entropy representations and we have conditions (equalities) to guide us to equilibrium. Is therea general way to transform to other useful representations for cases where other extensive orintensive variables naturally describe the state of the system, e.g., T = constant.?

Legendre Transforms

Legendre transforms provide a useful way to convert a function of many variables f(X1 ..., Xn) toone that is a function of some of the original variables and the derivatives of f with respect to theothers

f(X1 ..., Xn) fi g(X1,...,Xi,Ui+1,...,Un)

where Ui = f

Xi Xj Xi

Ui is said to be the conjugate variable of Xi.f = f (X1 ..., Xn) (f is a natural function of X1 ..., Xn)

df U dXi ii

n

=

=

1

Let


9/16/98 11

g f U Xi ii r

n

=

= +

1

dg df U dX X dUi i i ii r

n

= +( )= +

1

dg U dX X dUi ii

r

i ii r

n

= += = +

1 1

g is the Legendre Transform of f.

g contains all the same information as f which we can recover by back-transforming the conjugatevariables.

Example I. Helmholtz free energy.Given that we have a function (the internal energy) U(S, V, ni) that is a natural function of S,V, ni , find a function A that is a natural function of T, V, ni

A = U (S, V, ni) ST = A (T, V, n)A = U ST

dA = dU SdT TdS = (TdS PdV +

im i dni) SdT TdS

dA = SdT PdV +

im i dni

A is known as the Helmholtz free energy . For a reversible process, it is equal to themaximum possible work done by the system.

At constant T, dA = dU TdS = dW (reversible, quasi-static process)Thus, at constant temperature, the Helmholtz free energy is the maximum amount of energywe can usefully extract from the system. Note that we can also have bound energy that cannot be usefully extracted due to any entropy changes that might occur.

Example II. Gibbs free energy.Transform A(T, V, ni) to a natural function G(T, P, ni), where P is the pressure of thesystem (fixed pressure).

-P and V are conjugate are conjugate variables.G = A (PV)

G = A + PV A V T,ni

= P


9/16/98 12

dG = dA + PdV + VdP

= SdT PdV+

im i dni + PdV + VdP

= SdT +

im i dni + VdP

G is the Gibbs free energy G(T,P,ni)Example III (Transform back)

T, V, ni fi V, ni, S

dA = SdT PdV+

im i dni

S = A T V,ni

U = A (S) T = A + STdU = dA + TdS + SdT

= SdT PdV+

im i dni + TdS + SdT

dU = TdS PdV+

im i dni

U = U (S, V, ni)Note also that

dS = dUT

+ PT

dV m iT

dnii

, S = S (U, V, ni)

S = A UT

Recall energy entropy representations are equivalent.

Extrema Principles in Legendre Transformed RepresentationsWe already have conditions that specify equilibrium in the entropy and energy representations.

(1) For constant entropy, we need to minimize the internal energy UdU = 0, d2U > 0 (Hessian is positive)

U(S, V, Ni) fi minimize energy at equilibrium


9/16/98 13

(2) For constant energy, we need to maximize the entropy S

dS = 0, d2 S < 0 at equilibrium

S(U, V, Ni) fi maximize entropy at equilibrium

Note that for any arbitrary function f (Xi, ..., Xn) the total derivative and Hessian are

df = f Xi

dXii = 1

n

d2f =

i = 1

n

j = 1

n

2f Xi Xj

dXi dXj

Generally dn f = dXii = 1

n

Xi

n

f

From vector calculus we know we're at a minimum if d2f > 0, df = 0 and it max if df = 0,d2f < 0.

Helmholtz Potential

Consider a system contained in a very large heat bath or reservoir. The system can exchange heatwith the reservoir. The reservoir is so large that for any energy exchange with the system, thetemperature of the reservoir remains constant T = TR; this property defines a thermal reservoir.The system is subdivided in to two compartments 1 and 2. The value of any extensivethermodynamic quantity is the sum for the two compartments:

S(U,V,N) = S1(U1,V1,N1) + S2(U2,V2,N2)The total values of the internal energy and the entropy are the sum of those for both the system andthe reservoir.

UTOT = U + UR, STOT = S + SR

(Quantities without subscripts refer to those of the composite system.)


9/16/98 14

N1, V1, U1 N2, V2, U2

System

Bath or ReservoirT=TR

At equilibrium dUtot = 0, dStot = 0

d2Utot > 0, d2Stot < 0

dS1 + dS2 + dSR = 0

dUtot = 0 implies that

T1dS1 + T2dS2 + TRdSR = T1dS1 + T2dS2 TR (dS1 + dS2) = 0

T1 = TR and T2 = TR

We already showed that at equilibrium the system and the reservoir have same temperature:T = TR.

dUtot = dU + (T dSR PR dVR + m iR dNiR i

) = dU + T dSR

but sincedStot = 0 = dS + dSR,

we have that dS = dSR

dUtot = dU TdS = 0 , at equilibrium

= d (U TS) = 0Similarly,

d2Utot = d (dUtot) = d (dU TdS)

= d2U Td2S , since T = constant.

Since we must minimize UTOT,


9/16/98 15

d2Utot > 0 d2 (U TS) > 0Thus we can define a new function whose natural variables are T, V, N, that is at a minimum forequilibrium values of these quantities.

A U TS, dA = 0, d2A > 0

Helmholtz Potential Minimum Principle: For a system at constant temperature T (e.g, incontact with heat reservoir), the internal (unconstrained) variables are those that minimize A, theHelmholtz free energy.

Note that reservoir fixes T. N and V are also fixed.

Enthalpy minimum principle

For fixed P, N, VH U + PV

dH = 0, d2H > 0 at equilibrium

For systems in contact with a pressure reservoir, the values of unconstrained internalparameters are those that minimize the enthalpy.

Gibbs Potential Minimum PrincipleN, P, T are fixed

G U TS + PV

dG = 0 d2G > 0 at equilibrium

At constant pressure and temperature, the values of unconstrained internal parameters arethose that minimize the Gibbs free energy.

General thermodynamic potential extremum principleFor system in contact with reservoirs that fix intensive variables (f 1R, f 2R, ...), the values ofinternal unconstrained parameters are those that minimize the thermodynamic potential U (f 1R,f 2R, ...) at constant. ( f 1R, f 2R ...).In other words, to determine the equilibrium values of unconstrained internal variables, minimizethe thermodynamic potential that is a natural function of those thermodynamic variables we wantfixed.

Masseiu FunctionsUp till now we've just looked at Legendre transforms of the energy U. We could also Legendretransforms of the entropy. These are known as Masseiu functions

For given N, V, we know that we need to maximize S at constant. U. What about at constant 1T(which is the conjugate variable of U). Let B be some new Legendre transform of the entropy.


9/16/98 16

B = S S U NV

U = S 1T

U = - AT

(this is the Masseiu function)

dSTot = 0 d(- AT ) = 0 fi dA = 0T = constant due to reservoir

d2STot < 0fi d2S 1

T d2U < 0

or

d2(- AT

) < 0 d2A > 0 T = constant

Thus transforming the entropy gives us the Helmholtz minimization principle.

The S-representation at constant U can be transformed to the A-representation at constant T.

Maxwell Relations

Many different possible partial derivatives in thermodynamics, but they're not all independent. Wecan develop relations among them. For the Maxwell relations, we use the simple equality amongmixed second partial derivatives.

dU = U S NV

dS + U V SN

dV + U N SV,S

dN = TdS PdV + m dN

( )S UV = ( )V SU

S P

V,N =

V (T)

N,S

=

PS

TVV N N S, ,

Similarly 2U

S N=

S m =

=

S S

=

2U N S

=

N

U S

=

T

N

Masseiu function Representation Thermodynamic Potential

S (U, N, V) Microcanonical EntropyA (T, N,V) or F (T, N, V) Canonical Helmholtz free energy

W (T, m , V) or U (T, m , V) orJ(T, m , V)

Grand Canonical Grand potential


9/16/98 17

So =

TN SS V N V, ,

Similarly, we can also obtain

P

N V,S =

m

V S,N

We get a different set of Maxwell relations for each thermodynamic potential.

For example, if T, V, and N are the natural variables, thendA = SdT PdV + m dN

and S V N,T

=

P

T V,N ;

P

N T,V =

m

V N,T;

S N TV

=

m

T N,V

Maxwell relations allow us to transform derivatives that involve A, U, S to those involving V, Pand T.

Relation of Derivatives to Properties of Physical Interest(N=constant.)Coefficient of Thermal Expansion

a = 1V

V T P

Isothermal Compressibilityk

T

= 1V

V P T

Adiabatic Compressibility (constant. heat, no heat flow) TdS = dQ = 0 k s =

1V

V P S

Heat Capacity at Constant Pressure C TST

dQdTP P P

= dQ is quasi-static, reversible change in heat

Heat Capacity at Constant Volume C TST

dQdTV V V

=


9/16/98 18

We can show that CV is also equal to a derivative of the internal energy. Vary the internal energyin such a way that the total volume is kept fixed: dUV

dUV = TdSV PdV = TdSV since the volume is fixed, dV=0

dUVdTV

= T dSVdTV

Thus,

=

=UT

T ST

CV V

V for N=constant

We can also show this another way.

CV = T S T V

= T

T S (U (T, V),V)

Note that CV applies when T, V are the "natural" variables.

S T V

=

T S (U (T, V),V)

we can write each thermodynamic function in terms of the new variables

U = U (T, V), S = S (U (T, V), V) = S (T, V). (Change of variables)

0 S T V

=

S U V

U T V

= 1T

U T V

CV = U T V

In general, we can use the Maxwell relations to

(a) "simplify" derivatives so that they are expressed in easily determined quantities, e.g., a ,CV, k T. (Quantities such as G, A, and S are difficult to measure directly)

(b) Find relationships among physical parameters


9/16/98 19

Example:

C C TVP VT

=

2 for solids and liquids CP is easier tomeasure than CV. This gives easy wayto measure CV.

(Show this as a homework problem.)

Stability Revisited

At equilibrium S is at extremum and S is at a maximum dS = 0, d2S < 0.

Stability also has physical consequences, just as dS = 0 did.Now we will show that S must be a concave-down function of the internal energy U. Let us dothis by considering a case where S is at somepoint a concave-up function of U. We will show thatfor a homogeneous system (not in the vicinity of a phase transition) that this is not physicallyreasonable.

Consider 2 identical systems separated by an adiabaticwall

In what follows, we will assume that the amount ofstuff N is fixed and that the adiabatic piston is fixed (itdoes not move). If the two subsystems are identical,then the total entropy of the system is

S = 2S1(U1, V1, N1)Let us suppose that the entropy of each subsystem was at some point a concave-up function of U.We could take D U out of one side and add it to the other. New entropy is

S' = S1 (U + D U) + S2 (U D U)

Si S

UU-U U+U

S'/2

S/2S1 or S2

N1, V1, U1 N2, V2, U2


9/16/98 20

Note that S' > S ; the entropy increases upon redistributing the energy.

If we removed the adiabatic wall, energy would spontaneously flow from one side to the other.Similar things would also happen within a homogeneous system. The system would be unstable.

[In some cases, e.g. near phase transition points, stability isn't satisfied and the system breaks upinto different phases or different types of matter (chemical reactions).]In order for the system to be stable, we must have that

S (U + D U) + S (U D U) 2S (U, V, N) V = constant.N = constant.

Or, if we consider infinitesimal displacements of the energy (D U fi 0), then

2S

U2 V, N

0 ,

Similarly 2S

V2 U, N

0

Otherwise, we could squeeze one side andincrease the entropy S.

Convexity of the entropyS (U, + D U, V + D V) + S (U D U, V D V) 2 (S (U, V)) N constant.

D V fi 0, D U fi 0

At a maximum, we must satisfy the above inequalities on the second derivatives of S as well as

2S

U2

2S

V2

2S U V

2

0

5

10

15

20

5

10

15

200

2

4

6

5

10

15

200

2S

UV


9/16/98 21

(See any text on vector or multivariable calculus)

2S

U2 V,N =

U

S U V,N

=

U

1T V,N

= 1T2

T

U N,V =

1T2 CV

0

CV 0

Internal Energy

Energy must be convex function of extensivevariables

2U

S2 V,N =

T

S VN 0;

2U

V2 N,S =

P

V 0

Helmholtz Free Energy

dA = SdT PdV

T = U S N,V

S = A T N,V

S T N,V

=

2A

T2 N,V

T

S N,V =

S

U S N,V

-1 = +

2U

S2

NV

-1

0

12

34

5

2

4

6

0

10

20

30

40

0

10

20

30

V

S

U


9/16/98 22

If U is a concave up function of S , A is a concave down function if T . Note also that since

= ST

CTV

V 0, A must be a concave down function of T. (Similarly with the other conjugatevariables.)

(T, V, N) are the natural variables

2A

T2 NV 0

2

2 0A

V T N,

(P, S, N) are the natural variables

2

2 0H

S P N,

2H

P2 SN 0

1020

30

510

150

-200

-150

-100

-50

0

1020

30

A

TV

24

6 810

2.55

7.510

0

200

400

600

24

6 810

H

PS


9/16/98 23

(P, T, N) are the natural variables

2G

T2 PN 0

2G

P2 TN 0

Thermodynamic potentials: concave down functions of intensive variables; concave up functionsof extensive variables .

Here by thermodynamic potential, we mean the appropriate Legendre transform of the internalenergy U.

Physical Consequences of Stability

=

= 2

21 0A

VPV VT T T

k

T

0

CP easier to measure than CV for solids.Condense with

C C TVP VT

=

2

CV 0, k T 0 CP CV 0

Le Chatelier's PrincipleOne way of stating Le Chateliers principle is to say that the system evolves so as to removeinhomogeneities.

From microscopic (molecular) "chaos" we know that fluctuations induce local inhomogenity inenergy density, entropy density, particle density, etc.

12345

24

68

10

-60

-40

-20

0

-60

-4

-2

G

T

P


9/16/98 24

Thermodynamic potentials provide a guiding force back to equilibrium (Le Chatelier).density fluctuation

The thermodynamic potentials act as a sort of restoring force against inhomogeneous fluctuationsin the system.

Thermodynamics summary

From the postulates we see that we must either maximize S on minimize U at equilibrium.

The concept of equilibrium leads to equalities among intensive variables

Using the Legendre Transform, we can obtain a thermodynamic potential for any representation orset of macroscopic intrinsic variables, e.g. G(T,P,N).The Maxwell Relations provide useful relationships among the derivatives of thermodynamicquantities.

From the concept of thermodynamic stability, we can obtain useful bounds and inequalities:CP > CV > 0, k T > k S > 0

References

[1] D. Chandler, Introduction to Modern Statistical Mechanics. New York: OxfordUniversity Press, 1987.[2] D. A. McQuarrie, Statistical Mechanics. New York: Harper & Row, 1976.[3] H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, Second ed.New York: John Wiley & Sons, 1985.

review of thermodynamics - by j g saven

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