Phase Equilibrium

Author: John Hutchinson

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 69.9 J/K, whereas the entropy of one mole of water vapor is 188.8 J/K. 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 44.0kJ 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 ΔS_surr = -(ΔH/T). At 25°C, this gives ΔS_surr = -44.0 kJ / 298.15K = -147.57 J/K for a single mole. This entropy decrease is greater than the entropy increase of the water, 188.8 J/K - 69.9 J/K = 118.9 J/K. 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-TΔS

ΔG = 44000 J/mol - (298.15K) (118.9 J/Kmol)

ΔG=8.55 kJ/mol > 0

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 188.8 J/K. 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<0. And, indeed, as water vapor forms and the pressure of the water vapor increases, evaporation must continue as long as ΔG<0. Eventually, evaporation stops in a closed system when we reach the vapor pressure, so we must reach a point where ΔG is no longer less than zero, that is, evaporation stops when ΔG=0. 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-TΔS, then when ΔG=0, ΔH=TΔS. We already know that ΔH=44.0kJ for the evaporation of one mole of water. Therefore, the pressure of water vapor at which ΔG=0 at 25°C is the pressure at which ΔS=ΔH/T = 147.6 J/K for a single mole of water evaporating. This is larger than the value of ΔS for one mole and 1.00 atm pressure of water vapor, which as we calculated was 118.9 J/K. Evidently, ΔS 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 W for the gas molecule is proportional to the volume V. 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=klnW, then S must also be proportional to lnV. Therefore, we can say that

S(V2) - S(V1) = RlnV2 - RlnV1 = Rln(V2/V1)

[4]

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

S(P2) - S(P1) = Rln(P1/P2) = -Rln(P2/P1)

[5]

For water vapor, we know that the entropy at 1.00 atm pressure is 188.8 J/K for one mole. We can use this and the equation above to determine the entropy at any other pressure. For a pressure of 23.8 torr = 0.0313 atm, this equation gives that S(23.8 torr)) is 217.6 J/K for one mole of water vapor. Therefore, at this pressure, the ΔS for evaporation of one mole of water vapor is 217.6 J/K - 69.9 J/K =147.6 J/K. 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 = 44.0 kJ - (298.15K)(147.6 J/K) = 0.00 kJ. 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<0, and the evaporation continues until the water vapor has reached its the equilibrium vapor pressure, at which point ΔG=0.