Isolated System

1. Isolated System

An example of an isolated system is one constrained by rigid adiabatic and impermeable walls, i.e., with specified values of U, V, and m. An isolated system is at equilibrium when its entropy reaches a maximum value so that δS = 0. At this state, if the system is perturbed such

that the values of U, V, and m of the system are unchanged (e.g., by increasing the temperature or internal energy by an infinitesimal amount), the perturbations are dampened since the sys- tem is stable. The perturbations at fixed values of U, V, and m actually decrease the entropy, i.e., δS < 0 at stable equilibrium. (e.g., consider adiabatic chemical reactions, Chapter 11).

a. Single Component In Chapters 6 and 7 we employed the real gas state equation and evaluated various thermodynamic properties. Let us consider the compression of water in the context of the RK equation of state is, say, at 593 K and 1 bar. The volume of water decreases (or its intermo- lecular spacing decreases) and the pressure first increases, then decreases with a further de- crease in the volume, and again increases. Since the intermolecular spacing continuously de-

First, let us consider an isolated system with specified values of U, V and m and ex- press S = S(U, V, m). From Chapter 7,

s = s (T,v) and u = u(T,v), i.e., s = s(u,v), (1) which is the entropy fundamental equation.

a. Example 1 Consider the Van der Waals equation of state. Obtain an expression for s = s(u,v) for

water assuming the specific heat to be constant and equal to the ideal gas value c –1 v0 =

28 kJ kmole . Select the reference condition such that s=c v0 ln (u/c v0 + a/(v c v0 )) + R ln (v–b). Plot the entropy vs. volume for an internal energy value of 7000 kJ kmole –1 .

Solution Since

du = c v dT + (T ( ∂P/∂T) v – P) dv, du T = (T( ∂P/∂T) v – P) dv.

(A) Using

P = RT/(v–b) – a/v 2 (B) in Eq. (A),

2 du 2 T = T(R/(v–b)) – (RT/(v–b) – a/v ) = (a/v ) dv Integrating at constant temperature, u(T,v) = –a/v + f(T).

(C) In case a = 0, u 0 (T) = f(T).

(D) Eliminating f(T) between Eqs. (C) and (D), u(T,v) = u 0 (T) – a/v.

(E) Assuming constant (ideal gas) specific heats and u = u ref,0 at T = T ref , u 0 (T) – u ref,0 =c v0 (T – T ref ),

(F) Eq. (E) assumes the form u=c v0 (T–T ref )+u ref,0 – a/v, i.e.,

(G) T = ((u–u ref,0 ) + a/v)/c v0 +T ref .

(H) Similarly, we can integrate the expression ds = c v dT/T + ∂P/∂T dv

at constant temperature to obtain the relation s(T,v) = s 0 (T,v) – R ln (v/(v–b)).

(I)

Using the ideal gas relation for s 0 (T,v),

s 0 (T,v) – s ref,0 (T Ref ,v ref )=c v0 ln (T/T Ref ) + R ln (v/v ref ), s 0 (T,v) – s ref,0 (T Ref ,v ref )=c v0 ln (T/T Ref ) + R ln (v/v ref ),

(K) Setting , s ref,0 (T ref ,v ref ) =c v0 ln(c v0 T ref ) + R ln v ref , u ref,0 =c v0 T ref

we obtain the expression s=c v0 ln (u + a/v) + R ln ((v–b)).

(L) For ideal gases, a = b =0, and Eq. (K) leads to the relation s 0 =c v0 ln (u/(c v0 T Ref )) + R ln (v/v Ref )

(M) This expression leads to a plot of s vs. both u and v for a real or ideal gas.

Using the values ¯a = 5.3 bar m 6 kmole –2 , ¯b = 0.0305 m 3 kmole –1 , u = 7000 kJ kmole –1 ,T = 1 K, v =1m 3 kmole –1 , c –1 –1

v0 = 28 kJ kmole K , u ref,0 = c v0 T ref = 28 kJ kmole in Eq. (K), a plot of s vs. v at u = 7500 kJ kmole –1 is presented in Figure 3 .

–1 ref

ref

Remarks The relation s =s (u,v) is the entropy fundamental equation. It is somewhat more diffi- cult to manipulate the RK equation and obtain an explicit expression for s = s(u,v)

using Eq. (L).

b. Example 2 Consider a rigid and insulated 0.4 m 3 volume tank filled with 4 kmole (24 ×10 26 mole-

cules) of water, and an internal energy of 7500 kJ kmole –1 . Divide the tank into two equal 0.2 m 3 parts. Assume that water follows the VW state equation. What is the en-

tropy of each section? Assume that section 1 is slightly compressed to 0.14 m 3 while the section 2 is expanded to 0.26 m 3 , but maintaining the same total volume and total

internal energy. What is the entropy change during the process? (Use Figure 3 .)

Solution The fluid molecules are distributed uniformly in the tank and initially, v 1 = v 2 = v =

0.4 ÷4 = 0.1 m 3 kmole –1 . Using Figure 3 or Eq. (K) of Example 1, we note that ¯s = 149.83 kJ kmole –1 . The state is represented by point D in the figure, and the extensive

K 149 B 148.5

s, kJ/ kmole K 147.5 147 L

Liquid

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 v, m 3 / kmole

Figure 3: Variation in the entropy vs. volume for water modeled as a VW fluid at u = 7500 kJ kmole –1 .

entropy of each section is initially

S D = (S 1 =N 1 ¯s 1 ) = (S 2 =N 2 ¯s 2 )=2 ×149.83 =299.7 kJ.

(A) The total entropy of the tank is S = 2S D = 299.7 + 299.7 = 599.4 kJ. After the pertur-

bation, v

= 0.14 ÷2 = 0.07 m 3 kmole 1 –1 , u 1 = 7500 kJ kmole –1 , and

v 2 = 0. 26 ÷2 = 0.13 m 3 kmole –1 , u 2 = 7500 kJ kmole –1 . The corresponding states are represented by points M and N in Figure 3 , i.e.,

S M =S 1 =N 1 ¯s 1 =2 ×149.8 = 299.6 kJ, and S N =S 2 =N 2 ¯s 2 =2 ×149.9 = 299.9 kJ.

The total entropy after the disturbance

S=S 1 +S 2 =S M +S N = 599.5 kJ.

Remarks We note that the entropy (at specified values of U, V and m) increases after the dis- turbance, i.e., S M +S N > 2 S D . Therefore, the initial state is unstable and changes in the direction of increasing entropy. At states B and E disturbances no longer cause a further increase in the entropy.