Auxiliary Functions Legendre Transforms · Euler’s theorem • Euler's homogeneous function theorem States that: Suppose that the function ƒ is continuously differentiable, then

Auxiliary Functions

Legendre Transforms

Min Huang

Chemical Engineering

Tongji University

Auxiliary Function

• The term "auxiliary function" usually refers to the functions created during the course of a proof in order to prove the result.

• In thermodynamics, quantities with dimensions of energy were introduced that have useful physical interpretations and simplify calculations in situations where controlled set of variables were used.

Work

• In general, work can be divided into two parts:

• work of expansion and contraction, and

• work of the sum of all other forms

• Therefore in the reversible case,

where i will be defined as the chemical

potential of species i, but not yet at this

moment.

i

iidnμ+pdV=Xdf

Euler’s theorem

• Euler's homogeneous function theorem

States that: Suppose that the function ƒ is continuously differentiable, then ƒ is positive homogeneous of degree n if and only if

• n= 1, f is a first-order homogeneous function

xfxf n

Euler’s theorem

• Let f(x1,…, xn) be a first-order homogeneous

function of x1,…, xn.

• Let ui = xi

• Then f(u1,…,un) = f(x1,…,xn)

• Differentiate with respect to ;

1 ...

...1

1n

ix

n x,,xf=λ

u,,uf

Euler’s theorem

• From calculus,

• and,

2 /...1

1 i

n

=ij

uin duuf=u,,udf

i

xii

n

=ij

uii

xi λuuf=λf 1

//

3 / 1

i

n

=ij

ui xuf=

Euler’s theorem

• Substitute back to the first equation,

• Take = 1,

• This is Euler’s theorem for first-order

homogeneous functions

(4) /...1

1 i

n

=ij

uin xuf=)x,,f(x

(5) /...1

1 i

n

=ij

xin xxf=)x,,f(x

Legendre Transform

• Recall the 2nd law of thermodynamics,

• and

• we arrive at,

• Thus, , is a natural function of S, V, and the ni’s.

XdfTdSdE

XdTfdETdS

)/()/1(

i

iidnμ+pdVTdSdE

i

iidnμ+pdV=Xdf

rnnnVSEE ,,,, 21

Legendre Transform

• However, experimentally, T is much more

convenient than S.

• Assume f = f(x1,…,xn) is a natural function of

x1,…,xn.

• Then,

• Let

j

xiii

n

=i

i

i

n

=ij

xin

xf=udxu=df

xxf=)x,,f(x

/

/...

1

1

1

i

n

+r=i

idxuf=g 1

Euler’s theorem for

first-order

homogeneous functions

Legendre Transform

• Then,

• Thus, g = g(x1,…,xr,ur+1,…,un) is a natural

function of x1,…,xr and the conjugate

variables to xr+1,…,xn, namely ur+1,…,un.

• The function g is called a Legendre transform

of f.

i

n

+r=i

i

r

=i

ii

n

+r=i

iiii

dux+dxu=

dux+dxudf=dg

11

1

Legendre Transform

• It transform away the dependence upon

xr+1,…,xn to a dependence upon ur+1,…,un.

• It is apparent that this type of transformation

allows one to introduce a natural function of T, V,

and n, since T is simply the conjugate variable to

S; so as to p to V.

Legendre Transform

• From the first and second law, we have

E = E(S, V, n)

• We construct a natural function of T, V and n, by subtract from the E(S, V, n) the quantity

S ╳ (variable conjugate to S) = ST.

• Let A(T, V, n) = E – TS called the Helmholtz free energy

• Therefore,

i

r

=i

idnμ+pdVSdT=dA 1

Legendre Transform

• Let G(T, p, n) be the Gibbs free energy

G = E – TS – (–pV)

• And H(S, p, n) be the Enthalpy

H = E – (–pV) = E + pV

• Therefore,

i

r

=i

i

i

r

=i

i

dnμ+Vdp+TdS=dH

dnμ+VdpSdT=dG

1

1

Think also about, volume to U pressure to H

Maxwell Relations

• Armed with the auxiliary, many types of different measurements can be interrelated.

• Consider,

• implies we are viewing S as function of the natural function of T, V and n.

nTVS ,/

Maxwell Relations

• If df = adx + bdy, from calculus,

• Recall

• Then we have

• and

yx xb=ya //

μdn+pdVSdT=dA

nV,nT, Tp=VS //

μdn+VdpSdT=dG

np,nT, TV=pS //

Example I

• Let

• then

nV,

nV,nV,

nV,nT,

nT,nV,nT,

v

T

pT=

T

p

TT=

V

S

TT=

T

S

VT=

V

C

2

2

nV,v TST=C /

Quiz (exercise 1.10)

• Derive an analogous form for (15 Mins)

nT

p

V

C

,

solution

np,

np,np,

np,nT,

nT,np,nT,

p

T

VT=

T

V

TT=

p

S

TT=

T

S

pT=

p

C

2

2

Example II

• Let

• Viewing S as a function of T, V and n

• We have

np,

pT

ST=C

pn,nT,nV,np,

n

nT,

n

nV,

n

T

dV

V

S+

T

S=

T

S

dVV

S+dT

T

S=dS

Maxwell Relations

• Hence

• Note that

• So

• Therefore

pn,nV,

vpT

V

T

p+C

T=C

T

11

xyzy

z

z

x=

y

x

np,nT,nV, TVVp=Tp ///

2// np,nT,vp TVVpT=CC

Euler’s chain rule

Euler’s theorem

• From the 2nd law of thermodynamics,

• the internal energy E is extensive, it depends

upon S and X, which are also extensive.

• Thus, E(S,X) is a first order homogeneous

function of S and X.

XS,E=E

XS,λE=XλE

Euler’s theorem

• Therefore, from Euler’s theorem, Eq.5,

where X is a vector means system volume

• And work is,

Xf+TS=

XXE+SSE=E sX

//

i

iidnμ+pdV=Xdf

Extensive Function

• This flow naturally as we gave earlier,

• That is, E = E(S,V, n1,…,nr)

• and Euler’s theorem yields,

r

=i

iidnμ+pdVTdS=dE1

r

=i

iinμ+pVTS=E1

Extensive Function

• Its total differential is

• Therefore,

r

=i

iiii dμn+dnμ+VdppdVSdT+TdS=dE1

r

=i

iidμn+VdpSdT=1

0

This is the Gibbs-Duhem Equation

Extensive Function

• Recall the definition of Gibbs free energy

G = E – TS – (–pV)

• Apply Euler’s theorem gives,

• For one component system= G/n, Gibbs

free energy per mole

i

r

=i

i

i

r

=i

i

dnμ=

pVTSdnμ+pVTS=dG

1

1

Quiz (exercise 1.14)

• Show that for a one component p-V-n system

• where v is the volume per mole. [Hint: show that , where s is the entropy per mole.

TT v

pv

v

vdpsdTd

solution

• The Gibbs-Duhem Equation,

• Implies, for one component,

• Hence,

r

=i

iidμn+VdpSdT=1

0

vdpsdT=d

TTT v

pv

v

Ts

v