H.2 Euler-Maclaur in Sum Formula
5.3 Euler Theorem of Homogeneous Functions
A functionf(x,y,z)is said to be a homogeneous function of degreenwith respect to the variablesx,y, andzif
f(λx,λy,λz)=λnf(x,y,z), (5.38) whereλis some parameter. Note that Eq. (5.38) requires a very special type of function and that many functions are not homogeneous. For a homogeneous function of degreen, the Euler theorem states that
x ∂f
∂x
y,z+y ∂f
∂y
z,x
+z ∂f
∂z
x,y=nf(x,y,z). (5.39) This theorem, illustrated for three dependent variables, holds for any number of depen- dent variables.
Proof. We differentiate Eq. (5.38) partially with respect toλto obtain
∂f(λx,λy,λz)
∂(λx)
∂(λx)
∂λ +∂f(λx,λy,λz)
∂(λy)
∂(λy)
∂λ +∂f(λx,λy,λz)
∂(λz)
∂(λz)
∂λ =nλn−1f(x,y,z) (5.40) and note that∂(λx)/∂λ=x,∂(λy)/∂λ=yand∂(λz)/∂λ=z. After the differentiation is done, we setλ=1 in Eqs. (5.40) and (5.39) results. Note especially that if the functionfdepends on additional variables, sayuandv, such that
f(λx,λy,λz,u,v)=λnf(x,y,z,u,v), (5.41)
6This state function is actually the Helmholtz free energy, a useful thermodynamic potential that we shall define later. For now, it is just a convenient state function that will allow us to get the desired result.
60 THERMAL PHYSICS
Eq. (5.39) still holds, with no corresponding terms foruandv. In other words, Eq. (5.41) should be interpreted to mean that f is homogeneous in the variablesx, y, and z; the variablesuandvare held constant during differentiation and are simply irrelevant insofar as homogeneity with respect tox,y, andzis concerned.
Examples:The functionφ(x,y,z):=x2+y2+z4/(x2+y2)is a homogeneous function of degree 2 inx,y, andz. We have∂φ/∂x=2x−2xz4/(x2+y2)2,∂φ/∂y=2y−2yz4/(x2+y2)2, and∂φ/∂z=4z3/(x2+y2). Thus
x∂φ
∂x+y∂φ
∂y +z∂φ
∂z =2x2− 2x2z4
(x2+y2)2 +2y2− 2y2z4
(x2+y2)2 + 4z3
x2+y2 =2φ.
The functionψ(x,y,z) := sin(x/y)+z2/x2is a homogeneous function of degree zero inx,y, andz. We have∂ψ/∂x = (1/y)cos(x/y)−2z2/x3,∂ψ/∂y = −(x/y2)cos(x/y)and
∂ψ/∂z=2z/x2, which yieldsx∂ψ/∂x+y∂ψ/∂y+z∂ψ/∂z=0.
The functionη(x,y,z):=x3+y3+z2is not a homogeneous function with respect to the variablesx,y, andz. The functionζ(x,y,z):=x3z+y3z2is not a homogeneous function with respect to the variablesx,y, andz, but it is a homogeneous function of degree 3 inx andywithzheld constant. Thenx(∂ζ/∂x)y,z+y ∂ζ/∂y
x,z=3ζ.
Note that it is not necessary fornto be an integer, and thatncan even be negative.
Thus, the functionφ(x,y,z) := (x/yz)1/3+(1/x)1/3is a homogeneous function of degree n= −1/3 inx,y, andzand Eq. (5.39) holds, as the reader may verify.
5.3.1 Euler Theorem Applied to Extensive Functions
We note that U, which is extensive, is a homogeneous function of degree one in the extensive variablesS,V,N1,N2,. . .,Nκ. Thus,
U(λS,λV,λN1,λN2,. . .,λNκ)=λU(S,V,N1,N2,. . .,Nκ). (5.42) For example, if we double all of the extensive variables on whichUdepends, we will obtain a system that is twice as large but whoseintensivevariables are unchanged,7so we will have twice as much of the same thermodynamic state. Applying the Euler theorem for n=1, we obtain
U=TS−pV+ κ i=1
μiNi. (5.43)
We call this theEuler equationforU. By taking its differential, we obtain dU=TdS+SdT−pdV−Vdp+
κ i=1
μidNi+ κ i=1
Nidμi. (5.44)
7This follows because the intensive variables are partial derivatives of the extensive variableUwith respect to the extensive variables on whichUdepends, so any constant multiple such as 2 will cancel.
By comparing with Eq. (5.10), we deduce that 0=SdT−Vdp+
κ i=1
Nidμi. (5.45)
Eq. (5.45) is the Gibbs-Duhem equation for a multicomponent system. It shows that T, p, and theμi are not independent intensive variables because changes in them are related. Thus for a κ component system, there are only κ + 1 independent intensive variables.
For a monocomponent system, there are two independent intensive variables, sayp andT, andμ(p,T)can be regarded to be a function of them.8In that case, Eq. (5.45) can be divided byNand written in the form
dμ= −sdT+vdp, (5.46)
wheres:=S/Nis the entropy per mole andv :=V/N is volume per mole (molar volume).
If the functions s(T,p) and v(T,p) are known (these are the two equations of state of the system), Eq. (5.46) can be integrated to determineμup to an additive constant that results from the arbitrary zero of energy. For a two component system, there would be three independent intensive variables, sayp,T, andμ1, and thenμ2(p,T,μ1). For a three component system, there would be four independent intensive variables, etc.
Example Problem 5.3. By using the chemical potential of an ideal gas given by Eqs. (5.5) and (5.46), determine its equations of state. From these results, calculate the enthalpy per moleh and the internal energy per moleuand comment on their dependence on pressure. Deduce the relationship between the molar heat capacitiescV andcpat constant volume and constant pressure and compare with Eq. (2.13).
Solution 5.3. We havev= ∂μ/∂p
T =RT/p, which just reproduces the ideal gas law, which is one equation of state. Also,s= −(∂μ/∂T)p= −dμ∗(T)/dT−Rlnp, which is the other equation of state. Thus, by dividing the Euler equation Eq. (5.43) for a single component byN, we obtain u=Ts−pv+μ, so the molar enthalpy
h=u+pv =μ+Ts=μ∗(T)−Tdμ∗(T)
dT , (5.47)
which is a function of only the temperature, independent of pressure. Moreover, u=h−pv =μ∗(T)−Tdμ∗(T)
dT −RT, (5.48)
which is also a function of only the temperature, independent ofv. We readily computecp = dh/dT = −Td2μ∗(T)/dT2andcv = du/dT = cp−Rin agreement with Eq. (2.13), even ifcp andcvdepend onT.
8We could make other choices, such as regardingT andμas the independent variables, and then writing p(T,μ).
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Composition:For multicomponent systems, one often regards the independent intensive variables as beingp,T, andcomposition, where composition9is designated by theκ−1 independentmole fractions
Xi:=Ni/N, (5.49)
whereN =κ
i=1Niis the total number of moles. Note that theXiare intensive variables, because they are ratios of extensive variables. Moreover, we have
κ i=1
Xi=1 (5.50)
so onlyκ−1 of them are independent, as already stated. Taking the differential of Eq. (5.50) gives
κ i=1
dXi=0 (5.51)
so we note that it isimpossibleto take a partial derivative with respect to one of theXiwhile holdingallof the others constant. In particular, we cannot calculate chemical potentials by taking a single partial derivative with respect to anXi, that is,
μi= ∂U
∂Xi
S,V,{Xi} (5.52)
because the right-hand side is meaningless.
For a system with two components, we could take a set of the independent inten- sive variables to be p, T, andX1, in which case μ1(p,T,X1)and μ2(p,T,X1). For three components, independent intensive variables could bep, T,X1, andX2, in which case μ1(p,T,X1,X2),μ2(p,T,X1,X2), andμ3(p,T,X1,X2). To recover an extensive description, we could add to these variable setsNor any one of theNi.
Enthalpy of a multicomponent system:Recall that we defined the enthalpyH:=U+pV. For a multicomponent system
dH=dU+pdV+Vdp=TdS+Vdp+ κ i=1
μidNi, (5.53)
9For a mass based description, we can describe composition by the mass fractionsωi:=Mi/M, whereMiis the mass of theith component andM=κ
i=1Miis the total mass. The relationship of theωito theXiis nonlinear and depends on the molecular masses,mi. Specifically,ωi=miXi/
jmjXj.
where Eq. (5.10) has been used. Thus,His anatural function10ofS,p, and theNiand we have
T= ∂H
∂S
p,{Ni}
; V= ∂H
∂p
S,{Ni}
; μi= ∂H
∂Ni
S,p,{Ni}
. (5.54)
Furthermore, with regard to homogeneity, we see that
H(λS,p,λN1,λN2,. . .,λNκ)=λH(S,p,N1,N2,. . .,Nκ) (5.55) becausep, being intensive, does not participate. Thus, application of the Euler theorem gives
H=TS+ κ
i=1
μiNi (5.56)
which is in agreement with Eq. (5.43) once the definition ofHis used. So we actually get nothing new (except self-consistency) and Eq. (5.45) follows as well.
5.3.2 Euler Theorem Applied to Intensive Functions
Intensive functions are homogeneous functions of degree zero with respect to extensive variables. For example, the energyu:=U/N per mole or the energyuV :=U/V per unit volume are intensive. They are therefore homogeneous functions of degree zero in the variablesS,V,N1,N2,. . .,Nκ, which means that they can depend only onratiosof these variables, which ratios are themselves intensive. To see this formally, note that
u=U(S,V,N1,N2,. . .,Nκ)
N =U(λS,λV,λN1,λN2,. . .,λNκ)
λN (5.57)
and then chooseλ=1/N to deduce
u=U(s,v,X1,X2,. . .,Xκ), (5.58) wheres = S/N is the entropy per mole andv = V/N is the volume per mole. But since theXiare not all independent we can omit the last of them and write, in terms ofκ +1 independent variables,
u=u(s,v,X1,X2,. . .,Xκ−1) (5.59) whose differential turns out to be
du=Tds−pdv+
κ−1
i=1
(μi−μκ)dXi. (5.60)
10A natural function is a thermodynamic potential that contains information equivalent to a fundamental equation (forU orS) and whose independent variables are either members of the original complete set of extensive variables (on whichUorSdepends) or their conjugate variables (which are the partial derivatives ofUorSwith respect to their extensive variables).
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Equation (5.60) can be verified by taking the differential of Eq. (5.57) and using Eqs. (5.10) and (5.43) to simplify the result. Thus
du=dU N − U
N2dN = 1 N
TdS−pdV+ κ i=1
μidNi
− 1 N2
TS−pV+ κ i=1
μiNi
dN. (5.61) But ds=dS/N−(S/N2)dN, dv=dV/N−(V/N2)dNand dXi=dNi/N−(Ni/N2)dN, so Eq. (5.61) becomes
du=Tds−pdv+κ
i=1
μidXi, (5.62)
which reduces to Eq. (5.60) after dXκis eliminated. Note in this derivation that the total number of molesNwas treated as a variable, even though the result appears as if we just treated it as a constant and divided by it.
In a similar way, we can deduce that
uV=u(sV,c1,c2,. . .,cκ), (5.63) wheresV :=S/Vis the entropy per unit volume and theci:=Ni/Vare theconcentrations of each component (in moles per unit volume). The corresponding differential is
duV =TdsV+ κ i=1
μidci. (5.64)
Similar considerations apply to other intensive variables, such as the enthalpyh :=H/N per mole, which is a function ofs,p,X1,X2,. . .,Xκ−1and whose differential is
dh=Tds+vdp+
κ−1
i=1
(μi−μκ)dXi. (5.65)