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School of Physics and Astronomy

Junior Honours Thermodynamics GJA 2013-2014

Lecture TOPIC 10 1 (Finn: 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, 6.8, 6.9)

Synopsis: The thermodynamic potentials: internal energy U, enthalpy H, Helmholtz functionFand the Gibbs functionG. Maxwells relations. dG= 0 as the condition for equilibrium.

Equilibrium and the thermodynamic potentials

The second Law tells us that entropy increases in the universe as a whole. So it must increasein any isolated system.

But for a system in contact with surrounds, the Second Law tell us only that the entropy of

system+surrounds must increase. A system can reduce its entropy, provided the entropy of thesurroundings increases by more.

Main postulate of thermodynamics tells us that systems tend towards equilibrium. How-ever, the equilibrium state of a non-isolated system depends on its boundary conditions.

We can define state functions, U,H,F and G which are useful for determining thermodynamicequilibrium under different external conditions, and the maximum work that can be extractedfrom a system when those boundary conditions are changed. They all involve state functionsand are independent of the process applied to the system.

Maxwell Relations

The four Maxwells relationsare general thermodynamic relations involvingP,V,TandSwhichestablish links between different measurable properties of substances; specific heat capacities,compressibilities etc. They are most conveniently derived from the requirement that the secondderivatives of U,H,F,G with respect to their natural variables do not depend on the order ofdifferentiation; this is required since the potentials are state functions.

dU = TdS -PdV implies that: 2U/V S= (T/V)s= 2U/SV = (P/S)v

To recall the Maxwell relations from their derivation it can be seen that:

1. The independent (natural) variables of the potential from which each Maxwell relation isderived appear in the denominators of the relation.

2. Cross multiplication of numerators and denominators yields products of pairs of conjugatevariables, S T and P V.

3. The sign can be deduced by recourse to the appropriate potential function.

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potential differential Maxwell relation natural variables

internal energy U dU=T dS P dVTV

S

= PS

V

S, V

enthalpy H

H=U+ P V dH =T dS+ V dPTP

S

=VS

P

S, P

Helmholtz free energy F

F =U T S dF = P dV SdTPT

V

=SV

T

T, V

the Gibbs free energy G

G= H T S dG= V dP SdTVT

P

= SP

T

T, P

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Internal EnergyU

From the central equation of thermody-namics.

dU=T dS P dV

suggests putting U=U(S, V).ThendU=

US

V

dS+UV

S

dVEquate coefficients to get:

1. T =US

V

2. P =UV

S

If volume constant & no other workis done, for ANY changes:

U=Q (V constant, no work)

note d qR = T dS for any REVERSIBLEchange with no restriction on dV

dU=T dS (|V, no work, REV)

Heat capacityFor reversible, isochoric (constant volume)heat flows: CV =

d QVdT

= dUVdT

=UT

V

andCV =TdSdTV

=TST

V

From partial derivatives forT andP.TV

S

=

2UVS

= 2U

SV

=

PS

V

,

a Maxwells relation.

Enthalpy H=U+ P V

dH=dU+ P dV + V dP =T dS+ V dPsuggests putting H=H(S, P).ThendH=

HS

P

dS+HP

S

dPEquate coefficients to get:

1. T =HS

P

2. V =HP

S

For changes in contact with constantPreservoir at pressureP0, if no workother than P V work is done:

H=Q (Pf =Pi = P0, no W except PV)

for isobaric (constant P) processes.

note d qR =T dSfor any REVERSIBLEchanges with no restriction on dP.

dH=T dS (|P, no W except PV, REV)

Heat capacityCP =

d QPdT

= dHPdT

=HT

P

and CP =TdSPdT

=TST

P

From partial derivatives for Tand V. TP

S

=VS

P

,another Maxwells relation.

Helmholtz free energyF =U T S

dF = dU T dS SdT = T dS P dVT dS SdT = P dV SdTsuggests putting F =F(V, T).ThendF =

FV

T

dV +FT

V

dTEquate coefficients to get:

1. P =FV

T

2. S=FT

V

Finally, from partial derivatives forP and S.PT

V

=SV

T


The Gibbs free energyG = H T S

dG = dH T dS SdT = dU+P dV +V dP T dS SdT =V dP SdTsuggests putting G = G(P, T).ThendG =

GP

T

dP+GT

P

dTEquate coefficients to get:

1. V =GP

T

2. S=GT

P

Finally, from partial derivatives forV and S.VT

P

= SP

T


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Equilibrium with different boundary conditions

(i) A system in contact with a T & Preservoir

Proof that the condition for equilib-rium is dG= 0

Certain processes occur in systems whichare constrained to operate or evolve whenthey are open to the atmosphere, and so aresubject to (very nearly) constant pressureand temperature at their boundaries.

adiabatic wall

reservoir T0P0

T PQ

system

free diathermal

wall

Suppose that the system is initially at the temperature (T0) and pressure (P0) of the heatand pressure reservoir. If the system evolves, possibly by chemical reaction, its temperature andpressure will, in general, both vary with time. If the system cools, heat Q (assumed measurable)will flow into it from the reservoir. The entropy of the reservoir will change by S0 = Q/T0.If the system expands, it will do work P0V on the reservoir.

Suppose that the system eventually returns to a new equilibrium state with temperature T0and pressureP0, but with its composition somehow changed (possibly its chemical composition).Then, in general, the values of some of its state functions will have changed, possibly the Gibbsfree energy and the entropy, amongst others. The change in value of the Gibbs free energy canbe found using the principle of increasing entropy, and the 1st law of thermodynamics. In whatfollows, symbols with the subscript 0 denote properties of the reservoir and no subscript isused for properties of the system. We examine the ensemble of the system and a sufficientlylarge volume of reservoir such that the ensemble can be considered to be thermally isolated fromthe rest of the universe.

entropy increase First LawS+ S0 0 U=Q P0V

S+ QT0

0 Q= U+ P0V

T0SQ 0 Q T0S 0

Combining the above,U+ P0V T0S 0

U+ P0(Vf Vi) T0(Sf Si) 0SinceP0 = Pi= Pf and T0= Ti = Tf

U+ (PfVf PiVi) (TfSf TiSi) 0

U+ (P V) (T S) 0

(U+ P V T S) 0

G 0

A system in thermal and mechanical contact with a heat and pressure reservoir is

in equilibrium when the Gibbs free energy is a minimum

This means that spontaneous changes in a system in contact with a heat and pressure reservoirare accompanied by a decrease in the value of its Gibbs free energy. Further if any infinitesimalchanges are somehow induced in a system already in an equilibrium state, the changes arereversible anddG= 0.

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Equilibrium condition for different system boundaries

With reference to the boxed EQNs below, identify the natural variables for each potential.

U dU=T dS P dV U(S, V)H=U+ P V dH=T dS+ V dP H(S, P)

F =U T S dF = SdT P dV F(T, V)

G= U T S+ P V dG= SdT+ V dP G(T, P)

Be aware that if there are additional thermodynamic variables and conjugate forces morepotentials can be defined eg. for dU = T dS P dV +xdX it may be useful to define a newpotential G(T , P , x) =U T S+ P V xX

If the boundary constraints fix the natural variables for a particular potential, that po-

tential is at an extremum at equilibrium with respect to changes of all internal degrees offreedom (eg fixing P = P0 & T = T0 requires dG = 0 at equilm for all possible internalchanges of composition). For U, H, F, G the extremum is a minimum. For S(U, V) theextremum is a maximum. The proof in each case follows the same line of argument as wasgiven on the previous page for G: a generic proof will be given in topic 11.

Example Derivation of a Maxwell relation

eg. Suppose we need to re-expressSP

T

in terms of other derivatives and expect a Maxwell

relation might help.

1. The relevant independent variables can be identified as P, T. Since the Maxwell relationsjust reverse the order of differentiation, the above derivative can be related to a partialderivative with respect to T with P held constant. (We could also work with the reciprocalPS

T

and relate this to a partial derivative with respect to T with S held constant.)

2. The cross multiplication rule tells us that the derivative it is related to isVT

P

. (At

this stage it should be apparent whether the application of a Maxwell relation is a usefulstep in solving a particular problem or not).

3. We know thatG has natural variables (T, P) therefore look at dG = SdT+ V dP:

SP

T

=

2

GPT

=

2

GTP

=

VT

P

Thus a negative sign is required in this case:SP

T

= VT

P

.

This Maxwell relation relates a derivative that does not correspond to a common typeof measurement to the isobaric thermal expansivity, = (V/V T)P, which is easilymeasured or can be written down directly from the equation of state if the latter is known.

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Legendre Transformations - optional background material

Legendre transformation, like Fourier transformation acting onf(x) creates a new functionof a transformed variable. L[f(x)] fL(f/x) without loss of information: L[L[f(x)]] =f(x)

You have already used a Legendre Transformation in classical mechanics to change aLagrangian L(x, x) into a Hamiltonian, H(x, p) where p = (L/x).

In thermodynamics the transformation can be used to transform the internal energy ex-pressed as function of S and V to give different potentials that depend on (i) (U/S)V =Tinstead ofS(ii) (U/V)S= Pinstead ofV (iii) (U/x)S,V =X instead ofx (iv) Anycombination of these.

The transformations are achieved by writing fL = fx.(f/x) fX.x and replacingx by x[(f/x)] on the RHS to give the new function fL[(f/x)].

Geometrically for a given value of (f/x) the function fL[(f/x)] gives the value atx=0 of the tangent of f(x) taken at the point, x, at which (f/x) has the specified value.

It is necessary to assume that the internal energy is an increasing (convex) function ofentropy, volume, etc, for all this to work.

The new potentials generated are useful as a short cut for deriving different Maxwellrelations and because they are found to be minimized when equilibrium is achieved underdifferent external conditions.

Knowledge about the formal properties of Legendre transforms is not necessary for un-derstanding G,H,F in thermodynamics; we can derive all the useful properties of these

potentials more easily on a case by case basis.

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