I am currently reading a textbook called Engineering and Chemical Thermodynamics but I am stuck in one part about the State Postulate and the Gibbs Phase Rule. The Gibbs Phase rule says that the number of independent intensive properties that one needs to specify to constrain the properties of a given phase is:

n = Number of Chemical Species - Number of Phases + 2

Therefore, if there is a two-phase system with one chemical species, such as boiling water, one needs to specify one property in each phase to constrain all the other properties of that phase. However, the author states that we can take advantage of the fact that "the properties Temperature and Pressure are equal in both phases, thus, if we know either T or P of the system, we constrain the properties in each of the phases". This statement confused me a lot because the author didn't mention if this applies only for a saturated system. For me, it is logical that the temperature of both of the phases should be the same even if the system is not saturated, however, I am struggling to understand why the pressure of the system is the same for both phases always? The water could be boiling at 100 ºC, which means that the vapor pressure of the liquid water should be 1 atm, but the system could not be saturated and therefore the pressure of the vapor would not be 1 atm.

I assume that the author says that this applies even for unsaturated systems because of the following lines:

To illustrate this concept, consider a pure system of boiling water where we have both a liquid and a vapor phase. The phase rule tells us that for the liquid phase of water, we need only one property to constrain the state of the phase. If we know the system pressure, P, all the other properties (T, vl , ul...) of the liquid are constrained. The subscript “l” refers to the liquid phase. It is omitted on T since the temperatures of both the liquid and vapor phases are equal. For example, for a pressure of 1 atm, the temperature is 100 °C. We can also determine that the volume of the liquid is 1.04x10-3 m3/kg, the internal energy is 418.94 kJ/kg, and so on. The system pressure of 1 atm also constrains the properties of the vapor phase. The temperature remains the same as for the liquid, 100 °C; however, the values for the volume of the vapor (1.63 m3/kg), the internal energy (2,506.5 kJ/kg), and so on are different from those of the liquid.

Then he said:

The pressure (and temperature) in each phase of a two-phase system is equal; hence, if we know P (or T), we know the values of all the intensive properties in both phases. However, we have not yet constrained the state of the system. To do so, we need to know the proportion of matter in each phase. Thus, a second independent intensive property that is related to the mass fraction in each phase is required. Specifying that a system of boiling water is at 1 atm does not tell us how much liquid and how much vapor are present. We could have all liquid with just one bubble of vapor, all vapor with just one drop of liquid, or anything in between.

If my question isn't clear enough: if we have "all liquid with just one bubble of vapor", how is it possible that the pressure in each phase is equal?, this is confusing me a lot. The only explanation that I can find is that the vapor pressure of the liquid is the same as the vapor pressure of the vapor, however I don't know if the concept of vapor pressure applies for a vapor.

I would appreciate if someone can clear my doubts on this question. For more details, the chapter of the textbook is 1.5 of the 2nd edition. If I didn't explain myself enough, please let me know. Thanks in advance.


1 Answer 1


A key assumption underlying all of this is that we’re talking about equilibrium states.

Briefly, when regions are adjacent, barring any constraints, there tends to be heat transfer to equalize the temperature. There tends to be volume shifting to equalize the pressure. And there tends to be mass transfer to equalize the chemical potential.

All three can be rigorously derived from entropy maximization within the system.

There are three factors here because we’ve implicitly assumed three ways to change the system energy: heat it, do mechanical work on it, or add mass to it. (If we wished to include electrical work, for instance, then equilibrium would also involve charge transfer to equalize the electric field.)

In each case, note that an extensive parameter shifts (entropy, volume, mass) to remove gradients in the corresponding conjugate intensive parameter (temperature, pressure, chemical potential, respectively).

If imagining a small amount of gas having the same pressure as a large amount of liquid (here, we’re ignoring hydrostatic and Laplace pressures, among other nuances) is proving tricky, try an analogy: a small bit of one material adjacent to and having the same temperature as a large amount of another material. Hopefully this seems like a natural, everyday condition. The pressure can be thought of in essentially in the same way.

Does this make sense?

  • $\begingroup$ Thanks, I think it makes sense. It is still complex to think about though!, that volume shifts to equalize the pressure of the system with the surroundings is the key I think. If there is a piston-cylinder with a block of mass 'm', with saturated liquid, then pressure is that exerted by the block. Then if heat is added, vapor will be produced. For the pressure to stay constant, the volume must increase due to the increase in moles of vapor. I think this explanation makes things clear to me. $\endgroup$
    – prado5083
    Aug 3 at 17:09
  • $\begingroup$ Adding to my last comment, of course the fact that the pressure all the way throughout the system is not exactly the same due to hydrostatic pressure still confuses me, but I guess that it is negligible. I suppose that in saying that the pressure P is the same for both phases, we are neglecting the effect of gravity, is this correct? $\endgroup$
    – prado5083
    Aug 3 at 17:11
  • 1
    $\begingroup$ Exactly; for simplicity, we’re ignoring gravity and the energy implications of creating more surface area at phase interfaces, among other effects. Something to look forward to when you’re comfortable with the present aspects! $\endgroup$ Aug 3 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.