Generalized Derivation for a Single Phase

2. Generalized Derivation for a Single Phase

We have thus far obtained the conditions for thermal (Example 33) and chemical (Ex- ample 37) equilibrium. Consider, once again, hot coffee contained in a rigid cup with a firm lid as in Example 33. If the coffee is replaced with warm nitrogen at 350 K, heat transfer will oc- cur from the cup. As the room air warms, the room pressure increases, while that of the gas in the cup decreases. Thermal equilibrium will be reached at the maximum entropy state (S max ) TE , but with a mechanical constraint in place. If the mechanical constraint is removed, i.e., the lid is replaced with a nonpermeable and moveable piston, then equilibrium will be achieved at another maximum entropy state (S max ) TM . Finally, if the impermeable piston is replaced with a permeable piston, the system pressure and temperature may not change, but the nitrogen mole fraction in the room will change until chemical or species equilibrium is reached at yet another maximum entropy state (S max ) TMC. In this case (S max ) TMC > (S max ) TM > (S max ) Thermal Equil , i.e., “equilibrium” is reached when the entropy attains the highest possible value when all the con- We have thus far obtained the conditions for thermal (Example 33) and chemical (Ex- ample 37) equilibrium. Consider, once again, hot coffee contained in a rigid cup with a firm lid as in Example 33. If the coffee is replaced with warm nitrogen at 350 K, heat transfer will oc- cur from the cup. As the room air warms, the room pressure increases, while that of the gas in the cup decreases. Thermal equilibrium will be reached at the maximum entropy state (S max ) TE , but with a mechanical constraint in place. If the mechanical constraint is removed, i.e., the lid is replaced with a nonpermeable and moveable piston, then equilibrium will be achieved at another maximum entropy state (S max ) TM . Finally, if the impermeable piston is replaced with a permeable piston, the system pressure and temperature may not change, but the nitrogen mole fraction in the room will change until chemical or species equilibrium is reached at yet another maximum entropy state (S max ) TMC. In this case (S max ) TMC > (S max ) TM > (S max ) Thermal Equil , i.e., “equilibrium” is reached when the entropy attains the highest possible value when all the con-

Assume that two subsystems A and B contain two species, namely, species 1 and 2, as illustrated in Figure 52(a) . Assume that subsystem A has a slightly higher pressure, a slightly higher temperature, and contains a slightly larger number of moles of species 1 as compared to subsystem B so that the two subsystems are infinitesimally apart from equilibrium with one another. There are three initial constraints: a rigid plate that is nonporous, is a good thermal energy conductor, and serves as a chemical constraint; a porous rigid insulation which serves as a thermal constraint allowing only mass transfer when subsystems A and B are at same tem- perature; and a pin holding the rigid plate firmly in place (serving as a mechanical constraint) which, when removed, allows work transfer. When all of the constraints are removed, assum-

ing the combined system to be insulated, rigid, and impermeable, changes in U, V, N 1 and N 2 occur only in each of the subsystems. Therefore, the entropy of each subsystem changes sub-

ject to the condition U = U A +U B , or

dU = 0 = dU A + dU B (112) Similarly,

(115) Since each subsystem is initially in a state of equilibrium, then

(117) the entropy of subsystems A and B changes as soon as the constraints are removed. Employing

S B =S B (U B ,V B ,N B1 ,N B2 ),

a Taylor series expansion for the relevant expressions around each initial subsystem state, S A + dS A =S A (U A ,V A ,N A1 ,N A2 ) + ((∂S/∂U) A dU A + (∂S/∂V) A dV A +

(118) Similarly,

∂S A /∂N A1 dN A1 + ∂S A /∂N A2 dN A2 ) + .... .

S B + dS B =S B (U B ,V B ,N B1 ,N B2 )+ ((∂S/∂U) B dU B + (∂S/∂V) B dV B +

(119) Since

(∂S/∂N B1 ) dN B1 + ∂S B /∂N B2 dN B2 ) + ... .

(120) considering only the first-order derivatives, dS A =( ∂S/∂U) A dU A +( ∂S/∂V) A dV A +( ∂S A / ∂N A1 ) dN A1 +( ∂S A / ∂N A2 ) dN A2 , (121)

dS = dS A + dS B = δσ

dS B =( ∂S/∂U) B dU B +( ∂S/∂V) A dV B +( ∂S B / ∂N B1 ) dN B1 +( ∂S B / ∂N B2 ) dN B2 . (122) Recall that for a closed system dS= dU/T + P/T dV – Σ( µ /T) dN k k . Therefore,

dS A = dU A /T A +P A /T A dV A – Σ(µ k /T) A dN k,A ,

B B B B Porous

Copper

sponge plate

Copper

plate

A A A A roller Insulation

(d)

Figure 52 a. Initial state of a composite system; b. thermal equilibration with a copper plate in con- tact; c. mechanical equilibration with a movable partition; d. chemical equilibration with a porous

partition.

(∂S A /∂U A ) VA,NA1,NA2 = 1/T A , and

(124) We can also define

(∂S/∂V) = P A /T A .

(125) The process within the adiabatic and isolated composite system is irreversible so that dS = δσ.

∂S A /∂N A1 =–µ A1 /T, and ∂S/∂N A2 =–µ A2 /T.

Therefore, the entropy increases due to heat, work, and mass transfer, and dS > 0. Using Eqs. (121) to (125) we can express Eq. (120) in the form

dS = δσ = (1/T A – 1/T B ) dU A + (P A /T A –P B /T B ) dV A +

µ B1 /T B – ( µ A1 /T A ) dN A1 +( µ B2 /T B – µ A2 /T A ) dN A2 ≥0. (126) This relation can be expressed in rate form that is valid during the entropy transfer, i.e.,

δσ /dt = (1/T A – 1/T B ) dU A /dt + (P A /T A –P B /T B ) dV A /dt +

( µ B1 /T B – µ A1 /T A ) dN A1 /dt +( µ B2 /T B – µ A2 /T A ) dN A2 /dt ≥0.

a. Special Cases iv. No Thermal Constraint

In this case, consider the rigid impermeable plate and pin in place, but with the porous insulation removed (cf. Figure 52(b) ) so that the two subsystems are each rigid and imperme-

able, i.e., dV A = 0, dV B = 0, dN 1 = 0. dN 2 = 0. Using Eq. (126), δσ = (1/T A – 1/T B ) dU A , or

δσ/dt = (1/T A – 1/T B ) dU A /dt.

(128) We can also express (128) in terms of heat flux. From the First law δW A /dt = 0, since the pin

is in place. Therefore, the rate of change in internal energy dU A /dt equals the conduction heat flux Q A (= J Q ) from subsystem A, and the relation

δσ/dt = J Q (1/T A - 1/T B ).

If dU A /dt < 0 (e.g., heat loss from A and J Q < 0) and δσ/δt > 0, T A >T B . At thermal equilib- rium δσ = 0. Then, using Eq. (128)

T A =T B . v. No Mechanical Constraint

Now assume that, initially, both temperatures are equal. Once the pin and insulation are removed, heat and work transfer is possible, but mass transfer is not (cf. Figure 52(c) ). Thereupon, as subsystem A expands, subsystem B will be compressed. The temperature in both sections will equilibrate with the consequence that the first term on the RHS of Eq. (124)

will equal zero, since T A =T B . In this case

dS/dt = δσ/dt = (P A /T A –P B /T B ) dV A /dt = ((P A –P B )/T) J V ,

(129) where J V is the deformation rate flux as a result of the generalized deformation force (P A –

P B )/T. If dV A > 0 and δσ > 0, P A >P B . At the mechanical equilibrium condition, δσ = 0 so that P A =P B .

vi. No Chemical Constraint If the initial temperatures and pressures are equal in the subsystems (as illustrated in Figure 52(d) ), but the boundary is permeable Eq. (127) takes the form

( µ B1 /T B – µ A1 /T A )dN A1 /dt + ( µ B2 /T B – µ A2 /T A ) dN A2 /dt ≥0.

(130) If dN A1 < 0 (i.e., there is net transfer of species 1 from subsystem A), since δσ > 0, µ A1 /T A >

µ B1 /T B . Furthermore, since T A =T B ,

µ A1 > µ B1 , and Eq. (130) may be written in the form

δσ = ((µ B1 – µ A1 )/T) J NA1 + (( µ B2 – µ A2 )/T) J NA2 ≥0.

The term dN A1 /dt is the species flux J N1 crossing the boundary as a result of the generalized (species–1) flux force ( µ B1 – µ A1 )/T that is conjugate to J N1 . At chemical or species equilib-

rium µ A1 = µ B1 . Similarly, µ A2 = µ B2 .

Consider the example of pure liquid water at a temperature of 100ºC and pressure of 1 bar (subsystem A) and water vapor at the same temperature and pressure contained in subsys- tem B. When both subsystems are brought into equilibrium T A =T B ,P A =P B , and µ 1A = µ 1B .

This is also known as the phase equilibrium condition. vii. Other Cases

Oftentimes in a mixture, the higher the concentration of a species, the higher is its chemical potential (cf. Chapter 8) and hence the species transfers from regions of higher con- centration to those at lower concentration through molecular diffusion.

Since there are four independent variables dU A , dV A , dN A1 and dN A2 contained in Eq. (126), when δσ > 0 each term in the relation must be positive. Consequently T A >T B if dU A <

0, P A >P B if dV A > 0, and µ B1 > µ A1 if dN A1 > 0, i.e., heat flows from higher to lower tem- peratures, work flows from higher to lower pressures, and species likewise flow from higher to

lower chemical potentials. In case of multiphase and multicomponent mixtures, the derivation of the equilibrium condition is not simple. In this case the LaGrange multiplier scheme can be used to maximize the entropy subject to constraints, such as fixed internal energy, volume, and mass. This method is illustrated in the Appendix.

If a system initially in equilibrium (i.e., T A =T B ,P A =P B , µ A1 = µ B1 , µ A2 = µ B2 ) is disturbed, its entropy decreases, i.e., dS < 0. A Taylor series expansion around the equilibrium

state that includes second order derivatives reveals that dS = 0, and d 2 S < 0,

which are conditions for entropy maximization indicating the initial state to be stable. This will

be further discussed in Chapter 10.

A process (or effect) occurs only if a nonequilibrium state (or cause) exists. For in- stance, fluid flows in a pipe due to a pressure differential, and heat transfer only occurs if a temperature differential exists. A stable equilibrium state will not support the occurrence of any process.

kk. Example 38 Consider a mixture of O 2 and N 2 (in a volumetric ratio of 40:60) contained in chamber

A at a 10 bar pressure as shown in Figure 53 . Chamber B contains only O 2 at the same temperature, but at a pressure P B such that chemical equilibrium is maintained for O 2 between chambers A and B that are separated by a semipermeable membrane

that is permeable to only O 2 . Determine P B . (A semipermeable membrane is a device that allows the transfer of specific chemical specie.) Solution At chemical equilibrium, there is no net flow of molecular oxygen, i.e., µ O2,A = µ O2,B , and

0 h 0 O2 (T) – T (s O2,A (T) – R ln p O2,A /1) = h O2 (T) – T (s O2,B (T) – R ln p B /1). Upon simplification p B = 0.4 × 10 = 4 bars.

Remark Note that there is a mechanical imbalance of forces across the semipermeable mem- brane so that the membrane must be able to withstand a pressure difference of 10 – 4 = 6 bars.