Third Maxwell Relation
3. Third Maxwell Relation
The Helmholtz function is defined as
a = u – Ts, and da = du – d (Ts). (19) The entropy is a measure of how energy is distributed. The larger the number of quantum
states at a specified value of the internal energy, the larger the value of the entropy. Therefore, if two systems that exist at the same temperature and internal energy, the Helmholtz function is lower for the system that has a larger specific volume. Substituting from Eq. (9) for du in Eq. (19),
da = –P dv – s dT, and a = a(v,T). (20)
a. Remarks Equation (20) implies that
da = –P(v,T) dv – s(v,T) dT, where ( ∂a/∂T) v = – s(T,v) and ( ∂a/∂v) T = – P(T,v).
(21) Using the differentials of Eq.(21)
–ds = a TT dT + a Tv dv, and –dP = a vT dT + a vv dv. Using Eq. (20), the third relation is derived as ( ∂P/∂T) v =( ∂s/∂v) T .
(22) Equation (22) provides a relation for s in terms of the measurable properties P, v, and T.
(The value of the LHS of the equation is measurable while the RHS value is nonmeas-
v = 0.6 m 3 /kmole
{dP/dT} ‘
v=0.6, T=500
=13.9 kPa/K
6000 P, kPa
Figure 2: Illustration of relation in terms of P and T.
urable.) relations are illustrated in Figure 1 and Figure 2 . Since a = u – Ts and s = –( ∂a/∂T) v , a/T = u/T + ( ∂a/∂T) v . Therefore,
∂((a/T)/∂T) = (1/T) ∂u/∂T – u/T 2 –( ∂s/∂T) v =c v /T – u/T 2 –( ∂s/∂T) v . From the fundamental relation in entropy form, T( ∂s/∂T) v =( ∂u/∂T) v =c v , so that ∂((a/T)/∂T) = – u/T 2 or ∂((a/T)/∂(1/T)) = u.
(23) Furthermore, since
da T = P dv T , the area under an isotherm on a P–v diagram represents the Helmholtz function. The work
transfer during an isothermal process results in a change in the Helmholtz function. Recall that “a” is a measure of the availability in a closed system. Knowing P=P (v,T), one can obtain Helmholtz function “a”. The Massieu function j is defined as
j = –a/T = s – u/T, i.e.,
dj = –da/T + a/T 2 dT = (1/T 2 )(PT dv – u dT) = j(T,v).
b. Example 2
The fundamental relation for the entropy of an electron gas can be approximated as
(A) B=2 3/2 4/3
S(U,V,N) = B N 1/6 V 1/3 U 1/2 , where
B m N avag /(3 h P ).
(B)
Here, k B denotes the Boltzmann constant that has a value of R /N Avag = 1.3804 ×10 –26 kJ K –1 ,h
P is the Planck constant that has a value of 6.62517 ×10 kJ s, m denotes the electron mass of 9.1086
×10 –31 kg, N the number of kmoles of the gas, V its volume in
m –1 , and U its energy in kJ. Determine , T, and P when = 4000 kJ k mole , and v = 1.2 m 3 kmole –1 .
Solution The value of B = 5.21442 kg 1/2 k mole 1/6 sK –1 . From Eq. (A),
s = S/N = (B/N) N 1/6 ( v N) 1/3 ( u N) 1/2 =B v 1/3 u 1/2 , i.e., (C) s = 5.21442 (kg 1/2 K –1 Kmole 1/6 s)(1.2 m 3 k mole 2 –1 ) 1/3 (4000 kJ kmole –1 ) .
Recalling that the units kg (m/s 2 )m ≡ J. s = 350 kg 1/2 m kJ 1/2 kmole –1 K –1 . = 350.45 kJ kmole –1 K –1 .
From the entropy fundamental equation 1/T = ( ∂ s / ∂ u ) ¯v . Differentiating Eq. (C) with respect to u and using this relation,
1/T = (1/2) B 1/2 v / u = 0.04381 or T = 22.8 K. (D) Similarly, since P/T = ( ∂ s / ∂ v ) u ,
Upon differentiating Eq. (C) and using the above relation, P/T = (1/3) B u 1/2 / v 2/3 = 94.35 kPa K –1 .
(E) Using the value for T = 22.83 K, the pressure P = 2222.4 kPa.. The enthalpy
h = u +P v = 4000 + 2222.4
× 1.2 = 6666.9 kJ kmole –1 .
Remarks Eq. (C) can be expressed in the form
(F) Equation (F) is referred to as the energy representation of the fundamental equation
u ( s , v )= s 2 /(B 2 v 2/3 ).
(cf. Chapter 5). Rewriting Eq. (D)
(G) Differentiating this relation with respect to T we obtain the result
u (T, v ) = (1/4) B 2 v 2/3 T 2 .
c v =( ∂u/∂T) v = (1/2) B 2/3 v 2/3 T. (H) Dividing Eq. (E) by Eq. (D) we obtain the expression
u (P, v ) = (3/2) P v . (I) Likewise, using the entropy fundamental state equation (Eq. (A)), we can also tabu-
late other nonmeasurable thermodynamic properties such as a (= u – T s ) and g (=
h –T s ). Eliminating u in Eqs. (D) and (E) we obtain the state equation P = P(T, v ) for an electron gas in terms of measurable properties, i.e.,
(J) If this state equation (in terms of P, T and v ) is known, it does not imply that s , u ,
P (T, v )= (B 2 /6) T 2 / v 1/3 .
h , a , and g can be subsequently determined. This is illustrated by considering the temperature and pressure relations
T= ∂ s / ∂ u , and P/T = ∂ s / ∂ v . (K) One can use Eq. (J) in (K). These expressions indicate that Eqs. (K) are differential
equations in terms of s and, in order to integrate and obtain s =s(T, v), an integra- tion constant is required which is unknown. Therefore, a fundamental relation is that
relation from which all other properties at equilibrium (e.g., T, P, v , s , u , h , a , g ,
c p , and c v ) can be directly obtained by differentiation alone. While the Eq. (A) repre- sents a fundamental relation, we can see that the relation Eq. (J) does not.
c. Example 3 An electron gas follows the state equation
a (T, v ) = –(1/4) B 2 v 2/3 T 2 ,
(A)
where B = 5.21442 kg 1/6 K kmole s. Determine the functional relations for prop-
erties such as s , P, u , and h .
Solution Using Eq. (21), we obtain the relation
2 (d 2/3 /dT) v =– = –1/2 B T v . (B) The pressure is obtained from the expression
(C) Since u =a+T s , using Eqs. (A) and (B), we obtain
2 2 v (d 1/3 /d ) T = –P = –(1/6) B T /
2 2 u 2/3 = –(1/4) B T + (1/2) B T v
2 v 2/3 2 2 2 2/3
= 1/4 B T v .
(D) Differentiating Eq. (D), we obtain an expression for the constant volume specific
heat, i.e.,
c =( = (1/2) B v 2 ∂u/∂T) v T v 2/3 .
Furthermore, h =u+P v so that
h = (1/4) B 2 T 2 v 2/3 + (1/6) B 2 T 2 v 2/3 = (5/12) B 2 T 2 v 2/3 , and (E)
2 c –1/3 P =( ∂h/∂T) P = (5/12) B (2T v + (2/3) T v ( ∂ v / ∂T) P ). (F) The value of ( ∂ v / ∂T) P can be obtained from Eq. (C).
Remarks Alternatively, one can use Eq. (23) and get u shown in Eq. (D) directly. The Gibbs energy is a measure of the chemical potential, and
g = h –T s = (5/12) B 2 T 2 v 2/3 + (1/2) B 2 T 2 v 2/3 = (11/12) B 2 T 2 v 2/3 . The above relation suggests that the value of the chemical potential becomes larger
with an increase in the temperature. A temperature gradient results in a gradient in-
2 2 volving the chemical potential of electrons. The state equation, P = (1/6) B 1/3 T / v indicates that v increases (or the electron concentration decreases) as T increases at
fixed P. Hence, the warmer portion can have a lower electron concentration.
Figure 3: Illustration of the variation in some properties, e.g., h, s and g, with temperature.
Example 3 illustrates that the relation a = a (T,v) is a fundamental equation that contains all the relevant information to construct a table of properties for P,u, h, g, s, etc., (e.g. Tables A-4 for H 2 O, A-5 for R134a, etc.). One can plot the variation in h, g, and s with respect to temperature as illustrated in Figure 3 .
d. Example 4 Obtain an expression for the entropy change in an RK gas when the gas is isother- mally compressed. Determine the entropy change when superheated R–12 is isother-
3 mally compressed at 60ºC from 0.0194 m –1 kg (state 1) to 0.0126 m kg (state 2). Compare the result with the tabulated value of s 1 = 0.7259, s 2 = 0.6881.
Solution Consider the RK state equation
P = RT/(v–b) – a/(T 1/2 v(v+b)) (A) Note that the attractive force constant a is different from “a” Helmholtz function.
From the third relation Eq. (22) and Eq. (A), ( 3/2 ∂s/∂v)
T =( ∂P/∂T) v = R/(v–b) + (1/2) a/(T v(v+b)). (B) Integrating Eq. (B),
s 2 (T,v 2 ) –s 1 (T,v 1 )=
Rln((v
2 –b)/(v 1 –b)) +(1/2)(a/(T b)) ln(v 2 (v 1 +b)/(v 1 (v 2 +b))). (C)
From Table 1 for R–12, T c = 385 K, and P c = 41.2 bar. Therefore a =208.59 bar (m 3
kmole 3 ) K , and b = 0.06731 m –1 kmole . The molecular weight M = 120.92 kg kmole –1 , and
2 –1 a= 2 /M = 208.59 bar (m kmole ) K ÷120.92 (kg kmole )
a 2 3 –1 2 1/2
= 1.427 k Pa (m 3 kg –1 ) 2 K 1/2 , and
b= b /M = 0.557
×10 –3 m 3 kg –1 .
Since, R = 8.314
÷ 120.92 = 0.06876 kJ kg –1 K –1 , s 2 –s 1 = 0.06876 ln[(0.0126 – 0.000557) ÷ (0.0194 – 0.000557)]
+ (1/2){172.5 ÷ (333 1.5 0.000557)} ln [0.0126 (0.0194 + 0.000557) ÷ {0.0194 × (0.0126+ 0.000557)}].
= – 0.06876 × 0.448 – 0.211 × 0.01495 = –0.03396 kJ kg –1 K –1 .