Free Energy, Vapor Pressure and the Equilibrium Between
a Vapor and Condensed Phase
(Entirely plagiarized from Physics 5719, “The Physics of Nuts and
Bolts”, aka “Lab Survival Skills”)
References:
Thermodynamics, G. N. Lewis and M. Randall,
rev. K. Pitzer and L. Brewer, McGraw Hill, New York (1961).
Chemical Thermodynamics, I. Klotz, Benjamin, New York
(1964).
The Gibbs Free Energy is generally agreed to be the “weapon of
choice” for describing (a) chemical reactions and (b) equilibria
between phases. It is defined as:
G = H – TS = E + PV – TS (1)
Where
H = Enthalpy
E = Total internal energy
T = [Absolute] Temperature
S = Entropy
Obviously dG = dE + PdV +VdP – TdS – SdT
But from the first Law of Thermodynamics
dE = TdS – PdV
Hence:
since dS = δQ/T and the mechanical work
done on a system when it expands is –PdV.
dG = -SdT + VdP
Let us apply this to a closed system containing a pure substance
consisting of its vapor and a condensed phase co-existing in
equilibrium. The objective will be to see how the pressure of the
vapor depends on the temperature of the system. Clearly we may
write the above equation twice, once for each phase:
dGc = -ScdT + VcdP where c refers to the condensed phase
dGv = -SvdT + VvdP where v refers to the vapor phase
The objective of the exercise is to change the temperature of the
system (obviously the vapor and condensed phase are in thermal
equilibrium, at the same temperature) and see how the pressure of
the vapor changes. The system remains in thermal equilibrium if
the molar free energy of the system, also known as the chemical
potential, remains constant.
The definition of chemical equilibrium between two phases is that
the free energy is the same in both phases: Gc = Gv,.
Hence:
dGc = dGv Changes in free energy when some independent
variable is changed must be the same if they are
to remain in equilibrium.
-ScdT + VcdP = -SvdT + VvdP
(Sv - Sc )dT = (Vv- Vc)dP
dP/dT = (Sv – Sc)/(Vv – Vc) = ΔHv/(TΔV)
Here we are talking about the entropy change associated with the
reversible, isothermal evaporation of a small quantity of material
from a condensed phase to the vapor phase. The entropy change in
such a process is simply the heat of vaporization, ΔHv, divided by
the temperature at which the vaporization took place, the transition
temperature.
This is the Clapeyron equation (E. Clapeyron, J. ẻcole polytech.
(Paris) 14(23), 153 (1834). It relates the change in pressure of a
vapor to the temperature in a closed, mono-component system to
the heat of vaporization, system temperature and molar volume
change of the material on vaporization.
For lack of a better model, we treat most vapors as ideal gases,
whose molar volume is given by:
V/n = RT/P
Under most circumstances, the molar volume of the vapor is about
three orders of magnitude larger than the molar volume of the
condensed phase. Hence the required volume change can be
approximated as the volume of the vapor alone, which is a constant
for the isothermal transfer of mass at equilibrium we are discussing
and is given by the above expression when calculating equations of
state.
dP = (ΔHv/Vv)dT/T = (PΔHv/RT)dT/T
dP/P = ΔHv/R)dT/T2
ln(P/ P0) = -(ΔHv/R)(1/T – 1/T0)
P = P0 exp(-ΔHv/R(1/T – 1/T0))
The vapor pressure in equilibrium with a condensed phase
increases exponentially (sort of: exp(-1/T) isn’t exactly an
exponential!) with temperature from zero up to the critical
temperature. This is borne out in the vapor pressure charts and
generating functions for them. Deviations from linearity on the
log-log plot reflect the temperature dependence of the heat of
vaporization and the fact that exp (-1/T) isn’t really linear in the
exponent.
Vapor pressure of some elements; stolen from VEECO website.
Vapor pressure data are given in the CRC Handbook in the form
Log10p(Torr) = -0.2185*A/T + B
For water, CRC gives: A = 10999.4, B = 9.183837
Plotting this gives:
Vapor Pressure of Water
Vapor Pressure (Torr)
10000
"Normal boiling point"
1000
100
10
1
0.1
-20
0
20
40
60
80
100
Temperature (C)
The first thing we see is that the generating function is imperfect:
the normal boiling point of water on the plot is about 108.2 C. (I
cheated; I looked at the data!)
120
Kubaschewski, Evans and Alcock, Metallurgical
Thermochemistry, Pergammon, Oxford (1967), use the higher
order expression:
Log10p(Torr) = A/T + BlogT + CT + D
Vapor Pressure of Water
Vapor Pressure (Torr)
10000
"Normal boiling point"
1000
100
10
Kubaschewski et al.
1
0.1
-20
0
20
40
60
80
100
120
Temperature (C)
In this case the vapor pressure at 100 C is 758 Torr, a much better
approximation!
Caveat emptor!
Great resource tool: the “RCA Charts”, which live in 327 JFB.
Treat them with great respect: they are much older than you are
and irreplaceable.
Phase diagram of water.
So what does all this mean?
 Vapor pressure is a continuous function of temperature for all
temperatures from absolute zero to the critical point, above
which only a single phase is defined.
 The melting point on a vapor pressure curve conveys no
significance whatsoever. The vapor pressure of water at 0 C
is 4.6 Torr; it’s just another point on an otherwise smooth
curve. Actually this isn’t quite true: because the solid and
liquid have different heat capacities, the vapor pressure curve
changes slope at the melting point. The vapor pressure still
changes continuously across the melting point and is of no
particular significance there. Homework: Find or calculate
the vapor pressures of CO2 and gallium at their melting
points.
 The Clapeyron Equation can be used equally well to
determine the decrease in the freezing point of water with
increasing pressure. This is an interesting calculation that
apparently gives the wrong answer to the question: “what
makes ice skates work?”. Note that this is not the normal
case: the freezing point decreases with increasing pressure
only because water expands on freezing; the solid is less
dense than the liquid! (You know this: ice floats on water.)
 Conversely, the freezing point of water increases with
decreasing pressure. Hence the Triple Point, at which all
three phases, solid, liquid and vapor, coexist is 0.01 C.
(Actually this is now a “defined temperature”.)
 “Normal boiling point”. At 95 C, the generating function
says the vapor pressure of water is 633 Torr; at 97 C the
generating function says 657 Torr. Ambient pressure in SLC
is about 640 Torr, obviously depending on the weather. A
pan on a stove is heated from below. Convection in the water
distributes heat and establishes a reasonable facsimile of
thermal equilibrium. At 95 C a vapor nucleus in the liquid,
having an internal pressure of only 633 Torr, will be
compressed by the surrounding isostatic pressure; at 97 C a
vapor nucleus will have an internal pressure greater than the
ambient isostatic pressure and will expand and rise due to a
net buoyant force. This is called “boiling”. Attempting to
raise the temperature further is futile: any heat which flows
into the system, which is at a constant pressure of 640 Torr,
is consumed by the latent heat of vaporization and simply
converts more material to the vapor phase. Because raising
the temperature above the boiling point would create an
imbalance in mechanical equilibrium between the internally
generated vapor and the externally imposed ambient
pressure, it is impossible to raise the temperature above the
boiling point as long as there is condensed phase in the
system. This statement is independent of ambient pressure;
conversely, the temperature at which this “boiling”
phenomenon takes place depends strongly on pressure.
Homework: what is the boiling point of water in a vacuum
system at 10-6 Torr?
So what does all this have to do with an ESEM????
As you know, the NovaNano is a variable pressure machine that
bleeds in water to passivate the accumulated charge.
The Quanta has a thermoelectric cold stage that reduces sample
temperatures to 0 +/- 1- C.
The Quanta bleeds in even higher pressures of water than the
NovaNano so that the sample is in thermodynamic equilibrium
with its environment at temperatures of 0 +/- 10 C!