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Phase changes, change of entropy is temperature independent

  1. Dec 6, 2015 #1
    Good morning everyone!
    My chemistry final is approaching and I have a few difficulties in the thermodynamics chapter.
    One of the things that are bothering me is the calculation of the temperature of phase change.

    We know that if a mole of ice at 273K melts, the entropy change would be ΔS=ΔHf/T=6030/273=22.1J/K, we compare this to ΔSsurr and we find that their sum (ΔSuni) is 0.
    But when we want to calculate ΔS at other temperatures to compare them with ΔSsurr, and see if the process is spontaneous, we always assume that ΔS is equal to 22.1J/K and that the heat needed to melt 1 mole is 6030J; as if when 1 mole of ice at 263K (-10 degrees Celsius) hypothetically melts its change of entropy would be the same as if it melted at 273K, and the heat it would need to change phases is not temperature dependent.
    To quote the book : "In these calculations we are assuming ΔS and ΔH to be temperature independent".
    To what extent is this true? Why would melting of ice at whatever temperature always cause an entropy change of the system by 22.1J/K, the same goes for ΔH.
    Is there a formula that would give the hypothetical change in entropy of ice when it melts at diffrent temperatures so that I could see that temperature dependence is low?
  2. jcsd
  3. Dec 6, 2015 #2
    This is the given table. I first assumed that the temperatures indicated where those of the surroundings and assumed that the the mole of ice was at 0 degrees Celcius, and this would be quite logical, because in this case ΔSuniv would give a measure of the driving force in this process and would tell us that at 10 degrees Celcius, ice cubes melt faster than, say at 1 degree Celsius. But then in the formula for ΔG, isn't the temperature T, the temperature of the system, and in ΔSsurr isn't T the temperature of the surroundings*. So I suppose the correct assumption is that Tsystem=Tsurroundings. And we are given successively an ice cube ( 1mole) at -10 degrees C in a -10 degrees C environment, and then at 0 and 10 degrees C.
    *For this I am not so sure because for it not to have a subscript, T must designate the temperature of the system, and in this case we have no idea about T surroundings
    Last edited: Dec 6, 2015
  4. Dec 6, 2015 #3
    Try using Hess' law to calculate the change in H at 263, based on the heat capacities of water ice and liquid water and on the change in H at 273. For liquid water, if there is temperature dependence of the heat capacity (which I doubt is significant), you can extrapolate to the lower temperatures. Do the same for 283, this time extrapolating the heat capacity of ice to higher temperatures. How much of a difference is there in delta H?
  5. Dec 6, 2015 #4

    James Pelezo

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    One consideration is that temperature in ∆H =T∆S is a ‘positional temperature’ chosen by the observer. At very low temperatures close to 0 Kelvin, ‘changes’ in entropy ( say +10K) are very noticeable, but at much higher temperatures around 25oC (298 K) ± 10oC/K, ‘changes’ in entropy are, for practical purposes, undetectable and considered constant and temperature independent. Consider the following hypothetical:


    Remember ‘Entropy (S)’ is a measure of disorder relative to zero Kelvin typically showing a positive value, whereas ∆S is a change in Entropy [∆SoRxn = Σn·So(products) - Σn·So(reactants)] of the system and can be either positive or negative depending upon the reaction system being considered.
    See reference: https://www.chem.wisc.edu/deptfiles/genchem/netorial/modules/thermodynamics/entropy/entropy04.htm for further support; or, for a more mathematical approach see https://www.quora.com/How-are-standard-entropy-values-determined [Broken].
    Last edited by a moderator: May 7, 2017
  6. Dec 6, 2015 #5

    James Pelezo

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    During phase change Ice => Water the system temperature is constant (see heating curve of ice => water => steam), so the reference -10 to 0 to +10 degrees would refer to the temperature values of the surroundings chosen by the observer to examine the effect that such choices have on the system.
  7. Dec 13, 2015 #6
    Sorry for answering a week late, it's very rude I know, and thank you for answering so promptly yourselves, but I have been incredibly busy studying the incredible ton of math and physics material I need to know for my exams. I am doomed to fail my finals seemingly.
    This is what I thought to be the most logical explanation of the table myself at the beginning, since solid water does not exist at 10 degrees C (obviously). But my instructor told me that in the formula for ΔSsurroundings=-ΔH/T, T designates the temperature of the system. The aim of this table is to make a hypothetical assumption that solid water exists at all three temperatures, and measure the driving force behind its melting.

    One more thing but this is slightly off topic.
    In my book, ΔG is calculated for reactions so that we could have an idea of the driving force behind the reaction and how far it will go to the right.
    Because we know, that when T and P are constant ΔG<0 ⇔ reaction is spontaneous to the right.
    However we sometimes use this rule, even when P is not constant throughout the process.
    Take this reaction for example,
    forwhich we calculated ΔG for different reaction conditions to measure spontaneity although the pressure is not constant throughout this process assuming the gases are mixed in an incompressible flask.
  8. Dec 13, 2015 #7
    This is something that has caused students unending confusion. Using the value of ΔG0 as a measure of spontaneity or non-spontaneity is only a rule of thumb. For any reaction, if you combine reactants (and there are no products present), the reaction will proceed spontaneously until equilibrium is reached. But if ΔG0 is negative, the reaction will tend to go further toward products, and if ΔG0 is positive, the reaction will tend to go further toward reactants.

    All the development at constant T and P using ΔG and ΔG0 is designed to do is to obtain a value for the equilibrium constant for the reaction at any given temperature and pressure. That equilibrium constant can then be applied to any process system in which the reaction has equilibrated.

  9. Dec 13, 2015 #8
    We actually used the T constant, P constant assumptions, to be able to say that ΔSuniverse=-ΔG/T and thus if ΔSuni>0 the reaction is spontaneous and ΔG<0. But if P is not constant this delta(Suniverse)=-delta(G)/T does not hold true anymore. Are you saying that even if P is not constant this formula:
    ΔG° = -RT ln(K) informs us that if ΔG°<0, K>1 and the reaction is product favored and if ΔG°>0, K<1 and the reaction is reactant favored; and thus that ΔG° is still useful even if P is not constant?
  10. Dec 13, 2015 #9
    I'm saying that that equation gives you the equilibrium constant value at whatever temperature ΔG0 is evaluated, irrespective of the pressure. However, the sign of ΔG0 can only be used as a rough rule of thumb for telling you what would happen. If you start out with all reactants, products will always form, and if you start out with all products, reactants will always form. But, in either case, depending on the sign of ΔG0, the equilibrium mixture will tend to be more toward products or more toward reactants.

  11. Dec 13, 2015 #10
    And then when exactly could we need that P is constant? Sorry I am giving you a hard time
  12. Dec 13, 2015 #11
    We wouldn't. The pressure effect is embodied in the K term. As I said, the development involving constant T and total P is only used to provide the equation for the equilibrium constant at a specified temperature and total pressure of reactants and products.

    See if you can go back and review the development, preferably for the approach that involves the use of the van't Hoff equilibrium box.

  13. Dec 13, 2015 #12
    Thank you, I understand now!
  14. Dec 13, 2015 #13

    James Pelezo

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    Here's a note on determination of change of entropy of surroundings and it specifies Temperature of Surroundings not System...

    Entropy Change of Surroundings (and Total)
    The second law depends on the entropy change of everything, not just the system. It is possible for a process to occur that lowers the energy of the system, but raises the entropy of the surroundings. As long as the surrounding increase more than the system goes down, then this process will occur. Water freezing in a constant temperature surroundings at -5 °C is an example of this. The system (the water) will decrease in entropy first because its temperature will go down and second because it will go from a liquid to a solid. The entropy of the surroundings will increase since energy (heat) is flowing into the surroundings from the system.

    ∆Ssurroundings = qsurroundgs /Tsurroundings = -qsystem /Tsurroundings

    How do we calculate entropy changes for the surroundings? In most cases, the surroundings will be at a constant temperature. Therefore the entropy change will simply be related to the amount of energy that enters the surroundings in the form of heat divided by the temperature of the surroundings. Since we typically give heat a sign based on the system, the heat from the perspective of the surroundings is equal to the heat of the system but opposite in sign. Heat flowing out of the system is flowing into the surroundings. Then entropy change of the surroundings is

    ΔSsurroundings =qsurroundings/Tsurroundings=−qsystem/Tsurroundings

    Reference link: http://ch301.cm.utexas.edu/section2.php?target=thermo/second-law/entropy-surroundings.html

    As heat is generally defined from the perspective of the system, the subscript "system" is often left off of the heat in the last version of this equation.
    If we have the entropy changes of the system and surroundings, we can calculate total entropy change. The total entropy change is simply the sum of the system and the surroundings.

    ΔStotal =ΔSsystem+ΔSsurrounding

    ∆Ssrndgs = - q / Tsrndgs = -∆H/Tsrndgs where Temperature (T) relates the Surroundings. How does your prof come to the T being the system? Maybe thermal equilibrium between system and surroundings perhaps? Just curious.
  15. Dec 14, 2015 #14
    Good morning,
    Well this question that you are raising has literally been tormenting me and has been the source of quite some anxiety on my side. In physics class we were taught that:
    ΔSsomething=heat flowing into or out of this "something"/Temperature of this same exact something
    The teacher seemed to imply this relationship was correct because she said like you mentionned in your ealier message that when the tempertaure is low, heat creates more chaos, so change in entropy in inversely proportional to temperature. This is why in the beginning this was the equation I adopted ΔSsurroundings=-ΔH/Tsurroundings, which perfectly made sense to me.
    But then we studied free energy, this is what we wrote:
    G=H-TS all of which pertaining to the system (this will always be the case when there is no subscript)
    ΔG=ΔH-TΔS supposing T constant

    When I read this, I thought to myself they are equating ΔSsurroundings=-ΔH/Tsurroundings=-ΔH/T with T temperature of the system.
    First thing I supposed was that Tsurroundings=Tsystem
    But then I came back to the table of phase changes at different temperatures, and I said to myself if Tsystem=Tsurroundings then no heat flow could ever occur and no phase changes either. So I asked my instructor who specifically told me T in ΔSsurroundings=-ΔH/T designates T of the system which is not necessarily equal to T surroundings.
    The only way that I could thus explain why ΔSsurroundings=-ΔH/Tsystem was the explanation given in khan academy's video* on the matter, where they said that if T system is high the system could take the heat (or energy) necessary for the reaction from its own internal energy and thus the change in entropy iof the surroundings is inversely proportional to T system.
    *https://fr.khanacademy.org/science/...ee-energy/v/gibbs-free-energy-and-spontaneity (which is in french sorry)
  16. Dec 14, 2015 #15
    I somewhat disagree with this assertion. If it possible (and advantageous) to express the Second Law solely with in terms of the system, by including the effects of the surroundings only at the interface with the system. This is covered in my Physics Forums Insights article at the following link:


    This article has assisted untold numbers of confused Physics Forums members and visitors in understanding the 2nd law and applying it successfully to their homework and real world problems.
    In most cases, in homework problems, the surroundings will be specified at constant temperature. However, this is not typically the case in the real world, where there will be temperature gradients within the surroundings as well as in the system, such that the entropy change in the surroundings is not Q at the boundary divided by T of the surroundings (since T of the surroundings is not even constant). Even if one uses the temperature at the interface between the system and the surroundings to calculate the change in entropy of the surroundings, this will give the wrong value for the entropy change since the surroundings are experiencing an irreversible process. Furthermore, there can be additional irreversibilities occurring in the surroundings, such as viscous dissipation of mechanical energy and species diffusion. This is another good reason for not including the surroundings is the mathematical specification and application of the 2nd law.

  17. Dec 14, 2015 #16
    Tsystem is always equal to Tsurroudings at the interface between the system and surroundings, because temperature is continuous at the interface even in an irreversible process. That does not mean that there are no temperature gradients in the system to provide heat transfer, particularly if the temperature of the surroundings is different from the average temperature of the system. Furthermore, even though we regard a reservoir (representing the surroundings) as having a very high thermal conductivity and a very high heat capacity (so that temperature gradients in the reservoir and the temperature change of the reservoir is extremely small), Qsurr/Tsurr is only an idealization that applies in the limit. But, for an ideal reservoir with these characteristics, the entropy change would Qsurr/Tsurr.
    I don't know what your instructor was thinking, but this is definitely not correct, particularly if T is supposed to represent the average temperature of the system in an irreversible process. For an ideal infinite reservoir, ΔS=Qres/Tres.
  18. Dec 14, 2015 #17

    James Pelezo

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    Let’s try this … Go back to the table at the beginning of this thread and note the ∆G values and (yes) let ‘T’ be the system temperatures -10oC =>∆G = +220 J/mol, 0.0oC => ∆G = 0.00 J/mol and +10ooC =>∆G = -220 J/mol. What I’m interpreting the table to say is if T(system) = -10oC and ∆G = +220 J/mol., there is very little tendency toward spontaneous phase change because ∆G >> 0.00 J/mole; if T(system) = 0.0oC and ∆G = 0.00 J/mol., phase change is spontaneous and occurring at a thermodynamic equilibrium with constant temperature of both system and surroundings (which, in this case = 0.00o C) and if T(system) = +10oC and ∆G = -220 J/mol, the tendency toward phase change would be (hypothetically) 220 times more probable than at 0.00oC and 440 times more probable than at -10oC and would be an indication of the driving force present to initiate and sustain the phase change. Just a different way of looking at the problem as it relates to giving the system the temperature values defined in the table, but it sure seems more plausible to relate the temperature to surroundings that would be influencing the phase change tendency.
  19. Dec 15, 2015 #18
    I think that I finally made some sense out of the table in post #2 that may be helpful to you. The columns involving the surroundings seem superfluous to the entire situation, since, in the end, we are interested in determining the change in free energy of the system. So we should be focusing primarily on the system. And, since free energy is a function of state, we only need to know the initial and final states of the system. However, since you do have some interest in what is implied for the surroundings in this table, let's talk a little about that.

    The row in the table labeled -10 C refers to the following irreversible process:
    Initial State: One mole of Ice at -10 C
    Final State: One mole of liquid water at -10C
    In this change, it is assumed that the surroundings is being held at -10 C throughout the change because it is being modeled as an ideal infinite reservoir at -10 C throughout. During this change, the average temperature within the system may not be constant, and may drop below -10C temporarily, but in the end, it is again at -10 C. So there can be temperature gradients within the ice/water during the change. However, at the interface with the surroundings (ideal reservoir), the temperature remains -10C. This is how there is a heat transfer driving force within the system during the change. But, because the reservoir is ideal, its entropy change is Q/263, where 263 is both the temperature throughout the surroundings/reservoir and the temperature of the system at the interface with the reservoir.

    The change in free energy of the system (the ice/water) at -10 C for the initial and final states described above can be properly determined in either of three ways. The first way is to make use of the equation:
    But ΔG(273) = 0
    Therefore, $$ΔG(263)=\frac{10ΔH}{273}=+221\, J/mole$$
    Similarly, $$ΔG(283)=-\frac{10ΔH}{273}=-221\, J/mole$$
    The other method of getting this same result is to make use of
    So $$d(ΔG)=-ΔSdT$$
    where the Δ's refer to the change from ice to liquid.
    So, $$ΔG(263)-ΔG(273)=(-ΔS)(-10)$$
    So, $$ΔG(263)=+10ΔS=+221\, J/mole$$

    The third way is simply to write ##ΔG(263)=ΔH-263ΔS=+221\, J/mole##

    It still doesn't make sense to me how they arrived at the changes in free energy in the table using the change in entropy of the surroundings, or why invoking the surroundings was even necessary.

  20. Dec 15, 2015 #19
    Hi James,

    I like your development quite a bit. I have something additional that I think you may find interesting.

    Let G(T,P*) represent the free energy of water vapor at temperature T and pressure P*, where P* is the vapor pressure at the triple point. Then the free energy of water vapor at temperature T and pressure ##P_{liq}##, where ##P_{liq}(T)## is the equilibrium vapor pressure of liquid water at temperature T, is given by:
    Similarly, for water vapor at temperature T and pressure ##P_{ice}(T)##, where ##P_{ice}(T)## is the equilibrium vapor pressure of water ice at temperature T,
    At temperatures above 0 C, Eqn. 2 will involve extrapolation of the ice vapor pressure behavior, while, at temperatures below 0 C, Eqn. 1 will involve extrapolation of liquid water vapor pressure behavior (say, in either case using the Clausius Clapeyron equation).

    Since, at saturation, the free energy of liquid water and water ice will be equal to that of the corresponding saturated vapor, we can write $$G_{liq}(T,P)=G(T,P^*)+RT\ln{(P_{liq}(T)/P^*)}+V_{liq}(P-P_{liq}(T)\tag{3}$$
    and $$G_{ice}(T,P)=G(T,P^*)+RT\ln{(P_{ice}(T)/P^*)}+V_{ice}(P-P_{ice}(T)\tag{4}$$
    where the V's are the specific volumes of water and ice and P is the total pressure on the ice and liquid water system. So the change in free energy between ice and water at any given temperature and pressure P is given by:
    Typically, the terms involving the specific volumes are going to largely cancel, and the equation, to an excellent approximation in our region, can be reduced to:
    So we can not write:
    At 263 K, ΔG=221 J/mole, and this equation predicts that,
    At 283 K, ΔG=-221 J/mole, and this equation predicts that,
    So, at -10 C, if liquid water, ice, and water vapor are in contact, there is an identifiable pressure driving force for liquid water to form ice. And, at +10 C, if liquid water, ice, and water vapor are in contact, there is an identifiable pressure driving force for the ice to form liquid water. Actually, of course, the water vapor does not even have to be present for this same driving force to exist.

    Hope you find this interesting.

    Last edited: Dec 16, 2015
  21. Dec 16, 2015 #20

    James Pelezo

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    Yes, VERY interesting... but check me if I'm misinterpreting your logic... if, for a given temperature (T), Pliq > Pice the driving force (tendency) favors the solid; where as, if Pliq<Pice the driving force (tendency) favors the liquid. If that's right, then you have graciously given me a new tool for the toolbox. Oh, what happens when Pliq = Pice? I interpret that to be a thermodynamic equilibrium where the Pliq/Pice = 1.000 and the driving forces are equivalent but in opposite directions. Hope that's right, cause that's a very useful tool to have. Thanks much.
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