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
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