Homogeneous Functions

5. Homogeneous Functions

Homogeneous functions possess certain mathematical characteristics and the term homogeneous must not be confused with the thermodynamic definition of homogeneity. The total energy U in the air contained in a vessel is readily determined if the state (say, number of moles, temperature, and pressure) is known. If three identical vessels containing air at the same conditions are combined, these will contain three times as many moles, and, therefore, three

times as much energy (since U is an extensive property). The combined internal energy U c = U c (T,P, 3 ×0.3 moles O 2 ,3 ×1.2 moles N 2 ) = 3 U(T, P, 0.3 moles O 2 , 1.2 moles N 2 ). Mathemati-

cally, U(a,b,

λN 1

O 2 , λN N 2 )= λ U(a,b, N O 2 ,N N 2 ).

The function U is called a homogeneous function of degree 1, λ is a multiplier, and a and b are constants (which in this case are fixed values of T and P). If the three vessels are combined in

an equilibrium state, the density, which is an intensive property, does not change. Therefore, ρ(T,P, λ N 0

O 2 , λN N 2 )= λ ρ (a,b, N O 2 ,N N 2 ),

where ρ is a homogeneous function of degree zero. The definition of homogeneous function can be generalized as follows:

In general, a function F(a,b,x 1 ,x 2 ,x 3 ,…,x k ) is a homogeneous function of degree m if

(32) where a and b are constants. Homogeneous functions for which m = 1 describe extensive prop-

F(a,b, λx , λx , λx ,…, λx )= λ m 1 2 3 k F(a,b,x 1 ,x 2 ,x 3 ,…,x k ),

erties, and those with m = 0 describe intensive properties. For instance, consider the function

F(a,b,x

Assuming x 1,new = λx 1 ,x 2,new = λx 2 ,...

F(a,b,x 1,new ,x 2,new ,,…) = F(a,b,

2 3 λx 2 1 , λx 2 ,…) = ax

1 , new 1 x 2 , new /( bx 3 , new ) = ax 2 23 3 2 4 2 3 2 λ 4 1 λ x 2 /( bx λ 3 ) = axx λ 1 2 /( bx 3 )= λ F(a,b,x 1,new ,x 2,new ,,…).

Therefore, F is a homogeneous function of degree 4. If a=b=1, x 1 =1, x 2 =2, and x 3 =1, F(1,2,1) =

8. Furthermore, if λ = 2, then F(2x 1 , 2x 2 ,...) = F(2,4,2), and using Eq. (32)

F (2,4,2) = 24 F(1,2,1) = 16 ×8 = 128. This result may be checked using the above values for the variables in Eq. (33) so that x 1,new =

λx 1 = 2x 1 = 2, x 2,new = λx 2 = 2x 2 = 4, and x 3,new = λx 3 = 2x 3 = 2. In that case as well, the func- tion F = 128. Consider the following homogeneous functions: F 1 (x,y) = sin 2 (x/y) is a function of degree 0, since its phase is unchanged by λ; F 2 (x,y) = x –π sin(x/y) + xy –π–1 ln(y/x) is one of de-

gree m = – π, and F 3 (x,y) = 3x 3 /y 2 of degree m = 1.

A necessary and sufficient condition for F to be homogeneous and of degree m, is that the Euler equation

(34) holds, the proof for which is contained in the Appendix to this chapter.

k = 0 x k ( ∂∂ F / x k ) = mF .

g. Example 7 Prove Euler’s equation with the function

(A) Solution

Z (x,y) = ax 2 y + 2bxy 2 .

F = Z(x,y) (B) is a homogeneous function of degree m = 3.

We must prove that xZ ( ∂∂ / x ) + yZ ( ∂∂ / y ) =3. Z Differentiating Eq. (A) with respect to x and then y, the resultant expression is = x(2axy + 2by 2 ) + y(ax 2 + 4bxy) = 2ax 2 y + 2bxy 2 + ax 2 y + 4bxy 2

= 3ax 2 y + 6bxy 2 = 3(ax 2 y + 2bxy 2 ) = 3Z.

A function F is oftentimes not homogeneous with respect to all of the variables. If F is partly homogeneous in terms of j among k variables so that

F(a,b,x 1 ,x 2 ,x 3 ,…,x k ) = F(a,b, λx 1 , λx 2 , λx 3 ,…, λx j ,x j+1 ,…,x k ), (35) the Euler equation Eq. (34) assumes the form

i = 0 x i ( ∂∂ F / x i ) = mF . (36)

h. Example 8 Is the function

F(a,b,x,y,t) = ax 3 y/t + x 2 y 2 /t 3 + bxy 3 /t 7

a homogeneous function, a and b being constants. What is the Euler equation? Solution The function F is not fully homogeneous, since F(a,b,x,y,t) ≠λ m F(a,b,x,y,t). If the powers of x, y, and t are added, the first term on the RHS of the expression yields 3, the second term 1, and the third term –3. However, if t is excluded, the sum of the powers of x and y

for each term is 4. Therefore, the function is partly homogeneous (with respect to x, and y) so that F(a,b,

λx,λy,t) = λ 4 F(a,b,x,y,t). The Euler equation assumes the form x ∂F/∂x + y∂F/∂y = 4F.

If a function F = F (a,b,x,y) is homogeneous and of degree m, λ can be specified equal to 1/x so that

F(a,b, λx,λy)= F (a,b,1,y/x) = (1/x) m F(a,b,x,y). (37) Therefore,

F (a,b,x,y) = x m F(a,b,y/x) (38)

i. Example 9 Consider the function

Z(a,b,x,y) = ax 3 + bxy 2 ,

(A)

and show that Z (a,b,x,y) = x 3 Z (a,b,y/x).

Solution Equation (A) may be written in the form

(B) where the terms in the brackets correspond to Z(a,b,y/x). Therefore,

Z(a,b,x,y) = x 3 (a + b(y/x) 2 ),

Z (a,b,x,y) = x 3 Z(a,b,y/x).

We now summarize the properties of homogeneous functions: F(a,b,

λx m

1 , λx 2 , λx 3 ,…, λx k )= λ F(a,b,x 1 ,x 2 ,x 3 ,…,x k ),

k = 0 x k ( ∂∂ F / x k ) = mF .

F (a,b,x,y) = x m F(a,b,y/x)

a. Relevance of Homogeneous Functions to Thermodynamics i.

Extensive Property

A thermodynamic variable or property F is extensive if it is a homogeneous function of the first degree with respect to all of its extensive parameters in a functional relation. Mathematically, F is an extensive property if m = 1 in Eq. (32), namely

F( λx 1 , λx 2 ,...) = λF(x 1 , x 2 ,...),

where x 1 ,x 2 ,... are all extensive properties. If F is taken to represent the internal energy U = U(S,V, λN O 2 , λN N 2 , λN Ar ) of air

contained in a vessel of volume V and of entropy S, where N O 2 , N N 2 , and N Ar denote moles of oxygen, nitrogen and argon. The entropy is an extensive property (that is discussed in greater

detail in Chapter 3 which has the units of kJ K –1 . If λ identical vessels are combined into a system, the internal energy of the composite system is λU, and the volume is λV contains λN i moles of each of the species i. Therefore,

(39) For sake of illustration assume each vessel to be at S = 2 kJ K –1 , with volume of 5 m 3

U( λS, λV, λN 1 , λN 2 , λN 3 )= λU(S,V,N 1 ,N 2 ,N 3 ).

containing 1 kmole of argon (N 1 ), 78 kmole of nitrogen (N 2 ) and 21 kmole of oxygen (N 3 ). Assume that the internal energy in each vessel is 500 kJ. If λ = 3, three vessels have been combined and the volume and number of moles increases threefold. Using the notation of Eq. (39)

U(3 ×2,3×5,3×1,3×78,3×21) = 3×U(2,5,1,78,21). Therefore, m = 1, and we confirm once again that U is an extensive variable.

ii. Intensive Property

A thermodynamic variable or property F is said to be intensive if it is a homogeneous function of zero degree with respect to all of its extensive parameters. In mathematical terms F is intensive when m = 0 in Equation (32) or if

F( λx 1 , λx 2 ,...) = F(x 1 , x 2 ,...),

We can define T = ∂U/∂S. Since U is a function of S, V, and of the number of moles of various species, as discussed above, ∂U/∂S is also a function of those variables. Therefore,

(40) If the energy of each vessel in the above discussion is increased by, say, dU = 3 kJ and, for

T= ∂U/∂S = T(S,U, N 1 ,N 2 ,...).

sake of illustration, the corresponding change in dS = 0.01 kJ K –1 , the temperature inside the vessels must be

T= ∂U/∂S = 3 kJ/0.01 kJ K –1 = 300 K. If three vessels are combined, the volume, number of moles, energy, and entropy all triple, i.e.,

dU = 3 ×3, dS = 3×0.01. Therefore, the temperature of the combined system is still T = 9/0.03 = 300 K,

as expected. More rigorously,

T= ( ∂ U / ∂ S ) VN , , N ,... = T(S,V,N

1 2 1 ,N 2 ,...). Since for the combined system U c = λU, and S c = λS,

T c = ( ∂ U c / ∂ S cVNN ) ,

= ∂(λU)/∂(λS) = ∂U/∂S = T.

We, therefore, conclude that intensive properties are unchanged upon addition of identical systems, i.e.,

T( λS, λU, λN 1 , λN 2 ,...).= 0 λ T( λS, λU, λN 1 , λN 2 ,...).

Additional applications will be discussed in Example 10 and Chapters 3, 5 and 8.

iii. Partly Homogeneous Function The volume given by the ideal gas law V = NRT/P where V = V(T,P,N) is a partly

homogeneous function of the number of moles N. Consider a vessel containing air at a tem- perature of 298 K and pressure of 1 bar. If three identical vessels are combined into another system, the values of V and N triple, although T and P are unaffected. Therefore,

V(T,P, λN O 2 , λN N 2 , λN Ar )= λV(T,P, N O 2 ,N N 2 ,N Ar ),

(42) which shows that V is a partly homogeneous function of degree 1 with respect to N O 2 , N N 2 ,

and N Ar . iv.

Conversion of Extensive Into Intensive Properties We have shown that U = U(S,V,N) is a homogeneous function of degree 1, namely,

U=U( λS, λV, λN) = λU(S,V,N) Using a value of λ = 1/N, U (S/N,V/N,1) = (1/N) U(S,V,N), or

Usv (,,) 1 = (/)(,,) 1 NUSVN or Nu s v (,) = USVN (,,) .

j. Example 10 Consider the following state equation for the entropy of an electron gas

(A) Show that S is a homogeneous function of degree 1 (i.e., it is extensive).

S(N,U,V) = C N 1/6 U 1/2 V 1/3,

Assuming T = ( ∂U/∂S) V,N , show that T is a homogeneous function of degree 0 (i.e., it is intensive).

Solution S(

λU) 1/6 = λCN V 1/3 U 1/2 = λS(N,U,V). (B) Therefore, S is homogeneous function of degree m = 1, S being an extensive property.

λN,λV,λU) = C(λN) 1/3 λV) (

From Eq. (A), dS

= CN 1/6 V 1/3 ((1/2)U V,N –1/2 dU V,N ) and T(N,U, V) =

(C) The temperature

/ = 2U 1/2 /(CN 1/6 V ∂U 1/3 V,N ∂S V,N ).

T( λN,λV,λU) = 2(λU) 1/2 /(C( λN) 1/6 ( λV) 1/3 )= λ 0 2U 1/2 /(CN 1/6 V 1/3 ) , that proves that T is a homogeneous function of degree 0 which cannot be altered by in-

creasing or decreasing the system size (or λ). Remarks

The entropy S is an extensive property (m = 1), whereas the temperature T is an in- tensive property (m = 0). Since m = 1, Euler’s equation for S(U, V, N) assumes the form

U( ∂S/∂U) + V(∂S/∂V) + N(∂S/∂N) = S. (D) We will show in Chapter 3 that

∂S/∂U = 1/T, ∂S/∂V = P/T, and ∂S/∂N = –µ/T. (E) where

µ is called the chemical potential. If S is expressed in units of J K –1 and U in J, ∂S/∂U is in units K –1 . (Similarly, you may verify that ∂S/∂V can be expressed in units of N m –2 K –1 .) Using Eqs. (D) and (E),

U/T + V(P/T) – µ/T = S, i.e., U + P V – T S = µN. k. Example 11

The internal energy U is an extensive property, since it is a homogeneous function of de- gree m = 1. In general, U = U(S,V,N 1 ,N 2 ,...,N k ) so that k+2 extensive properties are re- quired to determine U for a k–component simple compressible system. Show that u ,

which is an intensive property, is a function only of k+1 intensive variables. Solution

Select λ = 1/N, where N denotes the total number of moles in the system so that U(S/N,V/N,N 1 /N,N 2 /N,...,N k /N) = (1/N)U(S,V,N 1 ,N 2 ,...,N k ), or

U(S,V,N 1 ,N 2 ,...,N k )=N u ( s , v ,x 1 ,x 2 ,...,x k ),

where x i represents the mole fraction of the i–component in a gaseous system (we can re- place x i with x l,i for a system containing a liquid mixture). Therefore,

u = U/N = u ( s , v ,x 1 ,x 2 ,...,x k ).

Since N 1 +N + ... 2 +N k = N, then

N /N+N /N+ ... 1 2 +N k /N = 1, or x 1 +x 2 + ... +x k = 1.

Therefore, x k = 1–x 1 –x 2 –...–x k–1 , and u ( s , v ,x 1 ,x 2 ,...,x k–1 ) is an intensive property which is

a function of only k–1+2 = k+1 intensive variables.