Mathematical Criterion for Stability

2. Mathematical Criterion for Stability

a. Perturbation of Volume

i. Geometrical Criterion Consider the state B illustrated in Figure 3 which undergoes a small disturbance ∆V at

a specified value of U. Due to the disturbance, Section 1 of the system in Example 2 reaches state L while Section 2 reaches state M ( Figure 3 ) . With respect to stability

(2) δS = (S L (U,V– ∆V,N) – S B (U,V,N)) + (S M (U,V+ ∆V,N) – S B (U,V,N)) < 0. Since

δS = δS 1 + δS 2 < 0, i.e.,

δS = S L (U,V– ∆V,N) + S M (U,V– ∆V,N) – 2S B (U,V,N) < 0, i.e., (S L (U,V– ∆V,N) + S M (U,V– ∆V,N))/2 < (S B (U,V,N).

(3) The entropy after a disturbance decreases in order that the initial state of the system is stable.

In the context of Figure 3 , the ordinate of the midpoint C of the chord LCM that connects the points L and M is represented by the LHS of Eq. (3) while the RHS represents the ordinate of the point B. Therefore, the chord LCM must lie below the curve LBM for the system to be stable. The curve LKBHM satisfying the criteria given by Eq. (3), is a concave curve with re- spect to the chord LCM. On the other hand, the midpoint of the chord MAN lies above the convex curve MDN, thereby violating this stability criterion. We find that for a system to be stable the fundamental relation for S = S (U, V, N) one must satisfy the concave condition, which is established by Eq. (3).

ii. Differential Criterion The discussion so far pertains to disturbances, which are of large magnitude ∆V. Con- sider a disturbance is in the neighborhood of state B (cf. Figure 3 ) that extends from state K at (U, V– ∆V, N) to state H at (U, V + ∆V, N). In that case

( δS = (S K (U,V– ∆V,N)–S B (U,V,N)) + (S H (U,V+ ∆V,N)–S B (U,V,N))) < 0, (4) where δS denotes the entropy change due to the disturbance. The entropy should decrease fol-

lowing disturbance at the stable points A, B and E illustrated in Figure 4 . Expanding S K and S H in a Taylor series

S K (U, V– ∆V, N) = S B +( ∂S/∂V) B (–dV) + (1/2!)( ∂ 2 S/ ∂V 2 ) 2 B (–dV) + ..... , and

H (U, V+dV, N) = S B +( ∂S/∂V) B (dV) + (1/2!) ( ∂ S/ ∂V ) B (dV ) + ... , i.e., δS = S +( ∂S/∂V) (–dV) + (1/2!)( ∂ 2 S/ ∂V B 2 B ) B (–dV) 2 + ... +

B +( S ∂S/∂V) B (dV) + (1/2!)( ∂ 2 S/ ∂V 2 ) B (dV) 2 + ... – 2 S B (U, V, N) < 0. We will represent the contribution to the disturbance by the first derivatives in the form

dS B =( ∂S/∂V) B (–dV) + (( ∂S/∂V) B (dV) = 0,

and by the higher–order derivatives as

d 2 S B = (1/2!)( ∂ 2 S/ ∂V 2 ) B (–dV) 2 + (1/2!)( ∂ 2 S/ ∂V 2 )

B (dV) + ….

So that Eq. (4) assumes the form (δS = (dS)

B + (d S) B ) < 0.

Since (dS) B = 0,

(δS = (d 2 S) B =( ∂ 2 S/ ∂V 2 ) B (dV) 2 ) < 0.

Omitting the subscript B, the general criteria for stability at any given state that

(5) which are the same as the conditions for entropy being maximized at specified values of U, V

(dS) = 0, and (d 2 S) < 0,

and N (stable points A, B and E and unstable point D in Figure 4 ). Since (dV) 2 > 0, the rela- tion

( ∂ 2 S/ ∂V 2 )<0 (6) must be satisfied in the neighborhood of an equilibrium state for it to be stable. If case the sec-

ond derivatives are zero, the third derivatives in Taylor series are included to obtain the stabil- ity condition d 3 S < 0, and so on.

c. Example 3 Assume that water follows the ideal gas state equation and show that for a unit mass, s vv ∂ 2 = s/ ∂v 2 < 0 at all volumes.

Solution We will employ the relations

Tds – P dv = du and ds = du/T + P/T dv. At constant internal energy

( ∂s/∂v) u = P/T, and since for ideal gases, P/T = R/v,

s v =( ∂s/∂v) u = R/v = f(v) alone, and s

vv =( ∂ s/ ∂v ) u = –R/v <0 Remark

This illustrates that ideal gases are stable at all states. Figure 5 contains plots of ¯s vs. ¯v at various values of ¯u . At very high values of

¯u (e.g., at temperatures larger than the critical temperature), the fluid is stable for any vol- ume, i.e., there is no increase in entropy in the presence of a disturbance at any state. For a solid there are certain regimes in which stability criteria are not satisfied and disturbances re- sult in phase transitions. e.g., in case of water from ice–phase I, to ice–phase II.

Since, in our illustrations, U, V and N (or m) were specified during a disturbance, the

U(D) plane

d 2 S<0

d 2 S>0 d 2 S<0

Disturbed Volume

Figure 4: Illustration of the entropy perturbation following a disturbance in the vol- ume of sections 1 and 2 of a rigid container.

151 s, kJ/ kmole K

u = 7500 kJ/kmole 149

v, m 3 / kmole

Figure 5: Variation of s vs. v in case of water according to the VW state equation.

relation

δW opt = – dU + T 0 δS,

with disturbance at a stable state yields δW opt =T 0 δS < 0. (7)

If the system is initially unstable (cf. State D in Figure 1 ), then δW opt > 0, i.e., the system could have performed work during the entropy increase in the isolated system.

b. Perturbation of Energy Repeating the procedure outlined in the previous section, but with the disturbance pa- rameter in terms of U, it is again possible to show that for stability

UU =( ∂ S/ ∂U ) < 0, (8) which is identical to the “concavity condition” described previously.

Consider the plot of entropy vs. internal energy at specified values of N (or m) illus- trated by the curve ABCDEFGH in Figure 6 for a fluid following a real gas equation of state. At some equilibrium states d 2 S >0 (i.e., these states are unstable) and at others, d 2 S < 0 (stable states). In the context of Figure 6 consider the fluid at state B at which the curve ABC is con- cave with respect to U. Here, the stability criterion is satisfied. On the other hand, curve DEF is convex, the criterion is violated.

A plot of ( ∂S/∂U) V = 1/T vs. U is presented in Figure 6b and that of T vs. U is illus- trated in Figure 6c . It is apparent that for certain ranges of U (U D to U F ), T decreases with in- creasing U. Mathematically, the stability condition ∂ 2 S/ ∂U 2 < 0 is equivalent to the condition ( 2 ∂/∂U)(∂S/∂U) < 0, i.e., ∂/∂U(1/T) < 0 or –1/T ( ∂T/∂U) < 0.

(9) In the context of Eq. (9), since ( ∂U/∂T) = mc v , then mc v > 0, and at a specified value of either

m or N,

c v >0 (10) c v >0 (10)

of energy as intermolecular potential energy and less as vibrational energy) with the distur- bance. The entropy generation is still positive for the adiabatic cup since δσ = δQ (1/T A

–1//T B ) > 0 as T A >T B . It still satisfies Second Law with entropy generation indicating that the initial state is thermally unstable. The specific heat c v > c vo for fluids following the RK equation, where c vo > 0. This implies that a curve representing variations in S vs. U must be concave (i.e., it must have a decreasing radius of curvature) with respect to U at a specified value of V. In other words, thermal stability is always satisfied for satisfying most of the real gas state equations. Read Example 17 for instances when this may be violated.

d. Example 4 Consider a VW gas and show that (s uu = ∂ 2 s/ ∂u 2 ) < 0 at all volumes. Solution G

Consider the relation T ds – P dv =

du. At fixed volume ∂s/∂u = 1/T.

Differentiating this expression,

2 2 2 ∂ G s/ ∂u = –1/T ( ∂T/∂u)

Since ( ∂u/∂T) v =c v ,

2 2 ( 2 ∂ s/ ∂u = –1/(c

v T )) < 0.

Remarks Figure 6: Qualitative illustration of the varia- For a real gas,

tion of a) entropy, b) ∂S/∂U=1/T and c) tem-

perature with respect to the internal energy. i.e., for a VW gas ( ∂c v / ∂v) T = 0, which implies that c v = f(T) = c v0 (T) >0

( ∂c v / ∂v) T = T( ∂P/∂T 2 ) v ,

A VW gas is thermally stable at all points.

c. Perturbation with Energy and Volume The discussion so far has pertained to disturbance in the internal energy at specified volume or in the volume at specified values of U. If the volume and energy are both perturbed, the following conditions must be satisfied, i.e.,

δS = S(U+dU, V+dV, N) + S(U–dU, V–dV, N) – 2 S(U, V, N) < 0, i.e., S(U+dU, V+dV, N) + S(U–dU, V–dV, N) < 2 S(U, V, N).

Expanding this expression in a Taylor series, and retaining terms up to the second derivative,

δS = ( ∂S/∂U dU + ∂S/∂V dV + (∂/∂U + ∂/∂V) 2 S) +

∂S/∂U dU – ∂S/∂VdV +(∂/∂U + ∂/∂V) (– 2 S) < 0

Since,

dS = ( ∂S/∂U dU + ∂S/∂V (–dU)) + (∂S/∂V dV + ∂S/∂V (–dV)) = 0, then

( ∂/∂U + ∂/∂V) 2 S<0 Expanding this relation, we obtain the expression,

2 2 2 2 2 2 2 (d 2 S= ∂ S/ ∂U dU + 2( ∂ S/ ∂U∂V)dU dV + ∂ S/ ∂V dV ) < 0, i.e.,

(11) Multiplying Eq. (11) by S UU , since S UU < 0, Eq. (11) assumes the form

2 2 (d 2 S=S

UU dU +2S UV dU dV + S VV dV ) < 0.

2 2 2 (d 2 S=S UU dU +2S UU S UV dU dV + S UU S VV dV ) > 0, i.e., (12)

(13) Now, (S

(S UU dU + S UV dV) 2 + (S UU S VV –S 2 UV 2 ) dV >0

UU dU + S UV dV) > 0, and since dV > 0, it is apparent from the perspective of stability that

VV –S UV ) > 0, (14) which is the stability condition in the presence of volumetric and energetic fluctuations within

(S UU S

a system. The stability criterion at a given state can be summarized as

(15) In determinant form, the stability criterion is SS UU UV

D 1,U 2 =S UU = ∂ S/ 2 2 ∂U 2 < 0, and D 1,V =S VV = ∂ S/ ∂V < 0,

D 2 = > 0 (16) SS VU VV

where D 2 is determinant of second order. If D 1,U < 0, and D 2 > 0, then D 1,V < 0. It is noted that since ∂s/∂u = 1/T, then s

vu =( ∂ s/ ∂v ∂u)= –(∂T/∂v) u /T .

e. Example 5 For an ideal gas show that Eq. (16) applies for all volumes.

Solution From Example 1 we note that s vv < 0, and from Example 2 that s uu < 0.

Since ∂s/∂u = 1/T, then s vu = ∂ 2 s/ ∂v ∂u = –(∂T/∂v) u /T 2 .

If the energy of an ideal gas is specified, its temperature cannot change with respect to changes in the volume. For this reason ( ∂T/∂v) u = 0. Therefore,

s vu = 0, and using Eq. (16) s uu s uu – s vu s vu > 0, which satisfies the stability criterion.

f. Example 6 Apply the RK equation of state

P = RT/(v–b) – a/(T 1/2 v(v+b)) (A) to water at 593 K. (a) Plot the pressure with respect to volume; (b) determine s VV ,s UU

and s at v = 0.1 and 0.4 m 3 kmole UV –1 ; and (c) check whether the stability criteria are satisfied at these states. Solution

The plot of pressure vs. volume is contained in Figure 7 . Since, ∂s/∂v= P/T, and

(B) ∂s/∂u = 1/T, then

(C) ∂s/∂v= R/(v–b) – a/(T 3/2 v(v+b)).

(D) Differentiating this expression with respect to v at constant u, we obtain the relation

s =( 2 s/ 2 ) =– R/(v–b) 2 +a(2v+b)/(T 3/2 2 2 vv 5/2 ∂ ∂v u v (v+b) ) +(3/2)a( ∂T/∂v) u /(T v(v+b)). (E) Recall that du = c v dT + (T ∂P/∂T – P) dv, hence, (dT/ dv) u = - ((T ∂P/∂T – P))/ c v .

In the context of the RK equation of state (

= – ((3/2) a/(c v T ∂T/∂v) 1/2 u v(v+b))) (F)

150 F U D K G 1500 e

1000 g, k J

v, m 3 /kmole

Figure 7: Plot of pressure and Gibbs energy with respect to volume for water using the RK equation of state.

Similarly differentiating Eq. (C) with respect to u, we obtain the relation

(G) and differentiating Eq. (C) with respect to v, we have

s uu =∂ 2 s/ ∂u 2 =–( ∂T/∂u) v /T 2 = – 1/(c v T 2 ),

s = ∂ 2 s/ ∂v∂u = –(∂T/∂v) /T vu 2 u . (H) Thus, all of the differentials involved with the stability criterion represented by Eq.

(16) can be evaluated. For v = 0.4 m 3 kmole –1 ; and T = 593 K, P = 95.4 bar, c v = 30.09 kJ kmole –1 K, a =

142.64 bar m 3 kJ kmole –2 , b = 0.02110 m 3 kmole –1 ,( ∂T/∂v)u = –173.04 kmole K m –3 , ∂ 2 s/ ∂v 2 = –31.92 kJ kmole m –6 K –1 , ∂ 2 s/ ∂u 2 = –9.4 × 10 –8 kmole kJ –1 K –1 , and ∂ 2 s/( ∂v∂u) = 0.000492 kmole m –3 K –1 . For stability,