As we recall, the entropy of vapor is much
greater than the entropy of the corresponding amount of liquid. A
look back at Table 1 shows that,
at 25°C, the entropy of one mole of liquid water is
69.9JK69.9JK,
whereas the entropy of one mole of water vapor is
188.8JK188.8JK.
Our first thought, based on our understanding of spontaneous
processes and entropy, might well be that a mole of liquid water at
25°C should spontaneously convert into a mole of water
vapor, since this process would greatly increase the entropy of the
water. We know, however, that this does not happen. Liquid water
will exist in a closed container at 25°C without
spontaneously converting entirely to vapor. What have we left
out?

The answer, based on our discussion of free
energy, is the energy associated with evaporation. The conversion
of one mole of liquid water into one mole of water vapor results in
absorption of
44.0kJ44.0kJ
of energy from the surroundings. Recall that this loss of energy
from the surroundings results in a significant decrease in entropy
of the surroundings. We can calculate the amount of entropy
decrease in the surroundings from
ΔSsurr=−ΔHTΔSsurrΔHT.
At 25°C, this gives
ΔSsurr=-44.0kJ298.15K=-147.57JKΔSsurr-44.0kJ298.15K-147.57JK
for a single mole. This entropy decrease is greater than the
entropy increase of the water,
188.8JK−69.9JK=118.9JK188.8JK69.9JK118.9JK.
Therefore, the entropy of the universe
*decreases* when one mole of liquid water
converts to one mole of water vapor at 25°C.

We can repeat this calculation in terms of the
free energy change: ΔG=ΔH−TΔSΔGΔHTΔS ΔG=44000Jmol−(298.15K)(118.9JKmol)ΔG44000Jmol298.15K118.9JKmol ΔG=8.55kJmol>0ΔG8.55kJmol0
Since the free energy increases in the
transformation of one mole of liquid water to one mole of water
vapor, we predict that the transformation will not occur
spontaneously. This is something of a relief, because we have
correctly predicted that the mole of liquid water is stable at
25°C relative to the mole of water vapor.

We are still faced with our perplexing
question, however. Why does any water evaporate at 25°C? How
can this be a spontaneous process?

The answer is that we have to be careful about
interpreting our prediction. The entropy of one mole of water vapor
at 25°C
*and 1.00 atm pressure* is
188.8JK188.8JK. We should clarify our prediction to say that one mole of
liquid water will not spontaneously evaporate to form one mole of
water vapor at 25°C and 1.00 atm pressure. This prediction
is in agreement with our observation, because we have found that
the water vapor formed spontaneously above liquid water at
25°C has pressure 23.8 torr, well below 1.00 atm.

Assuming that our reasoning is correct, then
the spontaneous evaporation of water at 25°C when
*no* water vapor is present initially must have
ΔG<0ΔG0.
And, indeed, as water vapor forms and the pressure of the water
vapor increases, evaporation must continue as long as
ΔG<0ΔG0.
Eventually, evaporation stops in a closed system when we reach the
vapor pressure, so we must reach a point where
ΔGΔG
is no longer less than zero, that is, evaporation stops when
ΔG=0ΔG0.
This is the point where we have equilibrium between liquid and
vapor.

We can actually determine the conditions under
which this is true. Since
ΔG=ΔH−TΔSΔGΔHTΔS,
then when
ΔG=0ΔG0,
ΔH=TΔSΔHTΔS.
We already know that
ΔH=44.0kJΔH44.0kJ
for the evaporation of one mole of water. Therefore, the pressure
of water vapor at which
ΔG=0ΔG0
at 25°C is the pressure at which
ΔS=ΔHT=147.6JKΔSΔHT147.6JK
for a single mole of water evaporating. This is larger than the
value of
ΔSΔS
for one mole and 1.00 atm pressure of water vapor, which as we
calculated was
118.9JK118.9JK.
Evidently,
ΔSΔS
for evaporation changes as the pressure of the water vapor changes.
We therefore need to understand why the entropy of the water vapor
depends on the pressure of the water vapor.

Recall that 1 mole of water vapor occupies a
much smaller volume at 1.00 atm of pressure than it does at the
considerably lower vapor pressure of 23.8 torr. In the larger
volume at lower pressure, the water molecules have a much larger
space to move in, and therefore the number of microstates for the
water molecules must be larger in a larger volume. Therefore, the
entropy of one mole of water vapor is larger in a larger volume at
lower pressure. The entropy change for evaporation of one mole of
water is thus greater when the evaporation occurs to a lower
pressure. With a greater entropy change to offset the entropy loss
of the surroundings, it is possible for the evaporation to be
spontaneous at lower pressure. And this is exactly what we
observe.

To find out how much the entropy of a gas
changes as we decrease the pressure, we assume that the number of
microstates WW
for the gas molecule is proportional to the volume
VV. This would
make sense, because the larger the volume, the more places there
are for the molecules to be. Since the entropy is given by
S=klnWSkW,
then SS must
also be proportional to
lnVV.
Therefore, we can say that

SV2−SV1=RlnV2−RlnV1=RlnV2V1SV2SV1RV2RV1RV2V1

(4)We are interested in the variation of
SS with
pressure, and we remember from Boyle's law that, for a fixed
temperature, volume is inversely related to pressure. Thus, we find
that

SP2−SP1=RlnP1P2=−(RlnP2P1)SP2SP1RP1P2
RP2P1

(5)For water vapor, we know that the entropy at
1.00 atm pressure is
188.8JK188.8JK
for one mole. We can use this and the equation above to determine
the entropy at any other pressure. For a pressure of
23.8torr=0.0313atm23.8torr0.0313atm,
this equation gives that S23.8torrS23.8torr)
is
217.6JK217.6JK
for one mole of water vapor. Therefore, at this pressure, the
ΔSΔS for evaporation of one mole of water vapor is
217.6JK−69.9JK=147.6JK217.6JK69.9JK147.6JK.
We can use this to calculate that for evaporation of one mole of
water at 25°C and water vapor pressure of 23.8 torr is
ΔG=ΔH−TΔS=44.0kJ−(298.15K)(147.6JK)=0.00kJΔGΔHTΔS44.0kJ298.15K147.6JK0.00kJ.
This is the condition we expected for equilibrium.

We can conclude that the evaporation of water
when no vapor is present initially is a spontaneous process with
ΔG<0ΔG0,
and the evaporation continues until the water vapor has reached its
the equilibrium vapor pressure, at which point
ΔG=0ΔG0.

Comments:"Reviewer's Comments: 'I recommend this book. It is suitable as a primary text for first-year community college students. It is a very well-written introductory general chemistry textbook. This […]"