# Chapter 14-part III

```Chapter 14
Part III- Equilibrium and Stability
A system with n components and m
phases
• Initially in a non-equilibrium state (mass
transfer and chemical reactions may take
place)
• T and P are uniform
• System is in thermal and mechanical
equilibrium with surroundings
• What changes may happen to the system?
• What will be the final state of the system?
Changes to the system translate to:
• Heat exchange
• Expansion work
dS surroundings
dQsurr  dQ


Tsurr
T
• By the second law, what happens to the
entropy?
dSsys + dSsurr &gt; 0
dSsys - dQ/T &gt; 0  dSsys &gt; dQ/T
dQ &lt; T dSsys
By the first law, dUsys =- PdVsys + T dSsys
dUsys + PdVsys &lt; T dSsys , or
dUsys + PdVsys -T dSsys &lt; 0
dUsys + PdVsys -T dSsys &lt; 0
• Valid for any closed system
• The inequality determines the direction of
change between non-equilibrium states
• The equality holds for changes between
equilibrium states (reversible)
dUsys + PdVsys -T dSsys &lt; 0
• Special cases:
– At V and S constant
(dUsys)SV &lt; 0
– At U and V constant
(dSsys)UV &gt; 0
Process at constant T and P
dUTP+ d(PV)TP –d(TS)TP &lt; 0
d(U+PV-TS)TP &lt; 0
(dG) TP &lt; 0
 All irreversible processes at constant T and P
tend to decrease the Gibbs free energy
Equilibrium criterion
• For a closed system at constant T and P, the
Gibbs free energy is a minimum
• Given an expression for G, we find the set of
composition values that minimize G
At equilibrium, differential changes
may occur
• The system is not static !!!
• At constant T and P changes may happen but
they do not change G.
• Therefore:
(dG) TP = 0
DGmixing = G – SxiGi
If the system is stable, G must decrease, therefore G &lt; SxiGi , G – SxiGi &lt; 0
For curve II, the system has a lower G by splitting into two phases than in a single phase
(at compositions between x1a and x1b)
Stability criterion for a single phase
binary system
• At constant T and P, DG and its first and
second derivatives must be continuous
functions of x1, and the second derivative
must everywhere be positive
d DG
 0 constant T and P
2
dx1
2
d (DG / RT )
 0 constant T and P
2
dx1
2
Alternative stability criterion:
Relation to GE
• Since
DG
GE
 x1 ln x1  x2 ln x2 
RT
RT
d 2 (DG / RT )
1
d 2 (G E / RT )


2
dx1
x1 x2
dx12
d 2 (G E / RT )
1

2
dx1
x1 x2
At constant T and P
Other alternative stability criteria
d 2 (G E / RT )
1

2
dx1
x1 x2
GE
 x1 ln  1  x2 ln  2
RT
d 2 (G E / RT ) 1 d ln  1

0
2
dx1
x2 dx1
Alternative criteria, at constant T and P, valid for each of
the components:
d ln  1
1

dx1
x1
dfˆ1
0
dx1
d1
0
dx1
How the stability criteria affect VLE?
dfˆ1 ˆ d ln fˆ1
 f1
0
dx1
dx1
but
fˆ  0
1
d ln fˆ1

0
dx1
How is the criterion for component 2?
d ln fˆ1 d ln fˆ2

0
dx1
dx1
For an ideal gas mixture, you can show that
d ln fˆ1 d ln fˆ2
1 dy1


dx1
dx1
y1 y2 dx1
Then the stability criterion is dy1/dx1 &gt; 0
What does it mean for a y-x diagram?
For the liquid phase, at constant T and P
Gibbs  Duhem
d ln fˆ1
d ln fˆ2
x1
 x2
0
dx1
dx1
Low pressure VLE, assume ideality of gas phase, you can show
1 dP ( y1  x1 ) dy1

P dx1
y1 y2 dx1
What can we say about the sign of dP/dx1?
dP dP / dx1

dy1 dy1 / dx1
Therefore, what is the sign of dP/dy1?
What happens at an azeotrope?
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