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How can phase transitions be physical equilibriums?

  1. Dec 16, 2015 #1
    I have been told that phase transitions=physical equilibriums. How can phase transitions be physical equilibriums? For example, if you have liquid water going into gas phase, doesn't delta G have to be greater than 0 for the phase change to occur? Is physical equilibrium an assumption in phase transitions because the transition is so slow?
     
  2. jcsd
  3. Dec 16, 2015 #2
    Suppose there is liquid water. You will have to put the kettle on (heat it) to convert the water into its gas phase. That is a practicality of everyday life. The hot water kettle however is not a closed system. You will loose some water vapor in the boiling process and that signifies an energy loss to the environment around the kettle. If you stop the energy supply boiling will stop and cooldown begins But suppose there is a perfect volume that is shielded perfectly from the enviroment around it: there is zero energy loss to the environment around the volume. This is normally called a closed system. This volume will remain at a certain constant temperature and vapor pressure. The end result is an equilibrium between the two phase transition rates.
    1 from liquid to gas which needs energy
    2 from gas to liquid: this frees energy.
    You can of course study the closed system at other temperatures and pressures and make phase diagrams under these conditions. You can fill it with al kinds of substances: But the system is always closed and in equilibrium.
     
  4. Dec 16, 2015 #3
    Thank you! But does this mean that open system phase transitions are not equilibriums (delta G is not 0)?
     
  5. Dec 16, 2015 #4
    Can you please specify a specific problem, including an initial and final thermodynamic equilibrium state of a system for which you wish to establish the change in free energy?
     
  6. Dec 17, 2015 #5
    A closed system is always in equilibrium because no energy moves in or out of it. That is a definition and we use it as such.
    An open system may change its energy state because it interacts with its surroundings. The hot water kettle is an open system. You use heat to reach a higher temperature and you loose heat by boiling off. This I think you may agree to, is not compatible with the definition of a closed system and does not resemble anything like a proces in equilibrium.
     
  7. Dec 17, 2015 #6
    This terminology is not consistent with the usual definitions of closed and open systems employed in Thermodynamics. In conventional Thermodynamics, the definitions are as follows:

    Isolated System: No mass or heat can enter or leave the system

    Closed System: No mass can enter or leave the system, but heat can transfer in or out at the boundary

    Open System: Both heat and mass can transfer in and out of the system at the boundary
     
  8. Dec 17, 2015 #7
    That is of course a most usefull addition, to add mass transfer.
    I also quite like the Wikipedia definition of a closed system where they speak, maybe in even more general terms, of "transfer in and out" of the system.

    A closed system is a physical system that does not allow certain types of transfers (such as transfer of mass) in or out of the system. The specification of what types of transfers are excluded varies in the closed systems of physics, chemistry or engineering.

    https://en.wikipedia.org/wiki/Closed_system

    It makes me realize again how necessary it is for understanding to define, in a most exact way, excatly what one is contemplating about
     
  9. Dec 17, 2015 #8
    My latest post did not adress the original question of this thread, How can phase transitions be physical equilibriums?,
    as well as the post of the moderator concerning the exact definitions of the different sytems. So, here I go

    Suppose I am interested in studying phase transitions of water and do an experiment. I have a volume filled with an amount of water and it is perfectly isolated from its surroundings. It will decouple from its environment if left alone, no mass or energy will be exchanged. This corresponds with the moderator definition of an isolated system. However for the study I can add or remove energy from the volume but not any mass of water inside the volume. So this corresponds with the moderator definition of a closed system I think.
    I can look in the experimental set up at the phase transitions of water by adding heat or by cooling. Let's say I am interested in vapour pressure at different temperatures.
    If an accurate measurement of vapor pressure is perfomed I need the system to be in equilibrium and decoupled from the environment following the definition of an isolated system. During the measurement no heat or mass will be removed or added. When I like to do a vapor pressure measurement at another temperature heat is removed or added. The system is now a closed system but not in equilibrium, because it goes to another different temperature. That is where I go wrong in my previous post: A closed system is always in equilibrium because no energy moves in or out of it. So, a closed system will reach an equilibrium once no mass or heat is removed or added.

    Back to the question of phase transition using the same experiment. Now I am interested in the transition of water to ice or gas. I will remove/add heat but no water from the system. Then the system is left alone and another equilibrium will be reached, either in ice or gas phase. The phase transition itself will thus be not an equilibrium.

    Hope to hear from you all, share your thoughts and correct.
     
  10. Dec 17, 2015 #9
    We know that G is a function of state, independent of the path between the initial and final equilibrium states of a system. So, if we follow a reversible path between the initial and final equilibrium states at constant pressure (meaning very small differences in temperature between the surroundings and the system), the change in enthalpy has to be ΔH=Q, and the change in entropy has to be ΔS=Q/T, where T is constant (since, for a phase change of a pure substance, temperature is constant). Therefore, ΔG is equal to zero. But, since G is a function of state, ΔG is zero irrespective of the path between the initial and final states. Using the reversible path has just been a convenience in evaluating the ΔG.

    Chet
     
    Last edited: Dec 17, 2015
  11. Dec 17, 2015 #10
    Thank you! Therefore, the system must be at equilibrium then--since delta G is 0, right? However, my question is: is delta G actually 0 or very close to 0? We are making the assumption that T is constant, but in reality, T is not completely constant. I am wondering whether we are also making the assumption that delta G is 0 when in fact is only very close to 0 (perhaps it is slightly positive, in order for the phase change to occur)?
     
  12. Dec 17, 2015 #11
    A perfectly reversible path is an idealization which can be approached as closely as we desire in practice. In the limit of a perfectly reversible path ΔG = 0 for a phase change. Therefore, it must be zero for all paths, including irreversible paths.

    In geometry, there is no such thing as a perfect circle, rectangle, or triangle, but that doesn't stop us from using the equation for the area of a circle A = πr2.
     
    Last edited: Dec 17, 2015
  13. Dec 17, 2015 #12
    Thank you! I like to think about it as delta G=0 (phases in equilibrium) at the boundaries of the two phases (liquid/gas). However, the gas molecules diffuse away and thus slowly, the liquid turns into gas.

    I now have another question: My teacher told me that when phases are in equilibrium, DG° = 0 since DG=DG°+RTlnK, in which DG would be set to 0 because of the assumption of equilibrium and K would be equal to 0 because of standard state conditions and activities for solids and liquids taken to be 1. However, if DG°=0, doesn't that imply that at 298K, phase equilibria occurs? However, shouldn't phase equilibriums occur only at boiling points/melting points?
     
  14. Dec 17, 2015 #13
    This is not a good to think of it. Whenever there is a change in a thermodynamic function, it is important to think of it in terms of two thermodynamic equilibrium states, an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state. In your case, you should be thinking of it as

    STATE 1: 1 mole of saturated liquid water at temperature T
    STATE 2: 1 mole of saturated water vapor at temperature T

    This precisely defines the initial and final equilibrium states. The change in free energy for this transition is zero.
    G0(298) is the free energy of formation at 298. What is the definition of the free energy of formation of a liquid or a vapor at 298 K? Is the free energy of formation of a liquid equal to the free energy of formation of a vapor at 298 (if the equilibrium vapor pressure is not 1 atm)?
     
  15. Dec 18, 2015 #14
    By the way, from the thermo book that I have, the heats of formation of liquid water and water vapor at 298 K are -237129 J/mole and -228572 J/mole, respectively. So ΔG0 for the change from liquid water to water vapor at 298 is not equal to zero. It is equal to +8557 J/mole. Do you know why it is not equal to zero, and how to calculate this difference using only the specific volume of liquid water and the equilibrium vapor pressure of liquid water at 298K.
     
  16. Dec 18, 2015 #15
    Hi! Thank you for your explanation. Unfortunately, now I am confused about these words in my notes (are they wrong?):
    "At a given pressure, phase transitions occur at a specific/constant temperature. At this temperature, the phases of a substance are in equilibrium, at which point there is no change in Gibbs free energy: DG° = 0 and there is no driving-force entropic or otherwise. This applies for physical equilibriums."

    I was also told that this concept applies because of the following derivation:

    dG=dG°+RTlnQ
    dG=0, Q=K for equilibriums, giving:

    dG°=-RTlnK
    Then, assuming we are talking about water vapor and liquid water…
    K=(activity of water vapor)/(activity of liquid water);
    The activity of water vapor can be taken to be the partial pressure of water vapor, which is taken to be 1atm at standard state.
    The activity of liquid water is taken to be 1.

    Doesn't this give K=1, therefore dG°=-RTlnK=0?

    Thanks for all your help.
     
  17. Dec 18, 2015 #16
    With all due respect to your professors and your notes, if this was such a good way of thinking about it, you wouldn't be so confused. Like I said, the approach I recommend is the think in terms two thermodynamic equilibrium states, and initial state and a final state. Define these two states precisely, and then determine the change in whatever thermodynamic function you are interested in between the two states.
    Let's look at this in detail to see what we are talking about here. We are going to determine ΔG0 (298), OK?

    Initial State: One mole of liquid water at 1 bar pressure and 25C

    Final State: One mole of water vapor in the hypothetical ideal gas state of 1 bar and 25C (This is how the standard state is to be specified for a gaseous species - check your textbook)

    Let p* be the equilibrium vapor pressure of water at 25 C (0.0313 bar), and let G* be the free energy of water at p*= 0.0313 bar and 25 C. This value of the free energy is, of course, the same for both the saturated liquid and the saturated vapor because it is the equilibrium condition at 25 C. Next, let's determine the free energy of liquid water in the initial state, ##G_l^0##, and the free energy of water vapor in the final state, ##G_v^0##. For the vapor,
    $$G_v^0=G^*-RT\ln p^*\tag{1}$$
    where p* is in bars. The second term represents the change in free energy in going from the saturation vapor pressure to p* to 1 bar (the standard state) for the vapor. Similarly, for the liquid water,
    $$G_l^0=G^*+V(1-p^*)\tag{2}$$
    where V is the molar specific volume of liquid water (0.018 liters/mole). The second term represents the change in free energy in going from the saturation vapor pressure p* to 1 bar for the liquid. If we subtract Eqn. 2 from Eqn. 1, we obtain:
    $$ΔG^0=-RT\ln p^*-V(1-p^*)\tag{3}$$
    The magnitude of the first term on the right hand side of this equation is +8583 Joules/mole. The magnitude of the second term is 2 Joules/mole. This compares with the value of 8557 J/mole that I mentioned in post #14 (which was obtained by taking the difference of two 6 digit numbers, which may have featured some roundoff error). So, if we neglect the second term on the right hand side of Eqn. 3, we obtain:
    $$ΔG^0=8583=-RT\ln{K}=-RT\ln{p^*}$$
    So, K=p*
     
  18. Dec 19, 2015 #17
    Thanks! I have a few questions about your explanation though:
    1. How can you have 1 mole of liquid water at 1 bar pressure? Is 1 bar pressure referring to the atmospheric pressure?
    2. As for the path you analyzed, are you assuming that: Initially, there was 1 mole of liquid water. Then, the liquid water was in equilibrium with a vapor pressure of 0.0313 bar, and then at the end, there is water vapor at 1 bar?

    Thanks again.
     
  19. Dec 19, 2015 #18
    Yes. I bar is approximately 1 atm.
    If you want to think of it in terms of a process path, I start out with 1 mole of liquid water at 1 bar, then I drop the pressure on the liquid water to 0.0313 bar, then I add heat to evaporate the liquid water to one mole of vapor at the constant temperature of 25 C and the constant pressure of 0.0313 bar (there is no free energy change for the mole of water in this process step), then I compress the vapor isothermally to the hypothetical ideal gas state of 1 bar.

    Chet
     
  20. Dec 19, 2015 #19
    Thank you for all your help! This makes sense to me now. I have one more question:
    Are both boiling and evaporation equilibriums, with dG=0? (I imagine that with boiling, the partial pressure of liquid vapor would be much higher though).
     
  21. Dec 19, 2015 #20
    Yes.
     
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