Minimization of Potentials

1. Minimization of Potentials

a. Helmholtz Free Energy A at specified T, V and m In Chapters 1 and 3 we have discussed the phenomena of evaporation, condensation, and phase equilibrium. Evaporation occurs as a result of the chemical potential difference be- tween the liquid and vapor phases of a fluid. If an evacuated rigid vessel of volume V is in- jected with a liquid and then immersed in a constant temperature bath at conditions conducive to evaporation, µ l is initially higher, which is why evaporation occurs. As the vapor fills the

space within the vessel, the pressure increases, thereby increasing µ g . Evaporation stops at a saturation pressure P sat that is characteristic of the bath temperature T at which µ

l = µ g . Recall from Chapter 3 that

dA = –P dV – S dT – T δσ.

If we consider the evaporation to be an irreversible process that occurs in a rigid closed sys- tem, δσ > 0 at the specified temperature, volume, and mass, i.e.,

dA = – T δσ so that dA < 0.

dd. Example 30

A rigid container that has a volume of 0.35 m 3 is completely evacuated and then it is filled with 0.1 kmole of liquid water ( Figure 13a ). It is then immersed in an isother- mal bath at a temperature of 50ºC. The liquid evaporates to form vapor, and the vapor pressure is measured. We will refer to the liquid water as subsystem A and the va- por–filled space above the liquid as subsystem B. Phase equilibrium is reached when the vapor reaches a saturation pressure, i.e., when there is no net change in the mass of either the liquid or the vapor. This occurs when the net evaporation ceases. Deter-

mine the change in Helmholtz function with respect to the vapor pressure P v (= P B ) and determine the value of that pressure when Helmholtz function reaches a minimum value. Assume that at T = 273.15 K h f =s f = 0, h fg = 2501.3 kJ kg –1 and that s (323 K,

P –1

B ) ≈s g (273 K, P TP )+c p,o,v ln(T/T TP ) – R ln P B /P TP ,c p,o,v = 1.8 kJ kg K , R = 0.46 kJ kg –1 K –1 , c = 4.184 kJ kg –1 K –1 . The constant volume reactor is typically adopted to measure Reid’s vapor pressure.

Solution The Helmholtz energy of systems liquid (A) and vapor (B) (see Figure 13a )

A=A A +A B , where

(A)

(B) The value of A A changes, since the liquid mass decreases during vaporization. The

A A =N A a A , and A B =N B a B (or A v =N v a v ).

value of a v changes since P v changes. For the liquid phase, u

f (323 K) ≈h f (323 K) = c w (T – 273) = 4.184(323 – 273) = 209.2 kJ kg , s

A = c ln(T/273.15) = 0.7033 kJ kg K –1 ,

A =u A – Ts A = 209.2 – 323.15 × 0.7033 = –18.075 kJ kg , so that initially

A A =N A a A = 0.1 × (–17.97 × 18.02) = –32.57 kJ (and A v = 0), and A=A A +A v = –32.57 + 0 = – 32.57 kJ.

For an arbitrary amount of vapor accumulation, say 0.0002 kmole, since the total number of moles of water N is unchanged,

N A =N–N v = 0.1 – 0.0002 = 0.0998 kmole, i.e.,

A A = 0.0998 × (–18.075 × 18.02)= –32.506 kJ. The vapor pressure

P v =N v R T/V B , where

B =V–V A =V–N A W A /v A = 0.35 – 0.0998 × 18.02 ÷ 1000 ≈ 0.35 m , i.e., P v = 0.0002 × 0.08314 × 323.15 ÷ 0.35 = 0.01535 bar. Furthermore,

u v ≈u vo =h vo – RT = (h g,ref +c p,o (T – 273.15)) – RT = (2501.3 + 1.8 –1 × (323.15-273.15)) – (8.314 ÷ 18.02) × 323.15 = 2442.2 kJ kg .

s v (T, P v )=s v (323 K, P v ) ≈s g (273 K, P TP )+c p,o ln (323/273) – R ln p v /P TP = 2501.3 ÷273.15 + 1.8 ln(323.15÷273.15)– 0.461 ln(0.015÷0.0061)

= 9.046 kJ kg –1 K , and

v =u v (T) – Ts v (T, P v ) = 2442.2 – 323.15 × 9.046 =– 477.36 kJ kg , i.e.,

(a)