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Reversibility and Entropy questions

  1. Feb 4, 2010 #1

    I am preparing to Thermal Physics exam. , and I think there are some basic subjects which I don't understand:

    1.Reversibility - As I understand, a reversible proccess, is one that can spontaneously happen in either direction, am I right?

    2.If this defenition is right, then, for example a glass of cold water, which is in a room with air in room temperature, heat flows from the air to the glass, so the glass become in room temp. too. apparently, this proccess will not happen spontaneously in the reverse direction - i.e the water in the glass won't spontaneously give heat to the air and become colder, so this proccess is irrevesible. Am I right?

    3.Another proccess - ice in a glass in 0[tex]^{o}[/tex]C gets heat (latent heat) from the surrounding air and melts. apparently, according to Clausius, the cold ice will not give it's heat to the hotter air, so this proccess is irrevesible too . Am I right?

    4.Entropy - if I understood well, entropy is defined as
    [tex]\Delta[/tex]S = [tex]\int dQ/T[/tex] and dS = dQ/T
    only in a reversible path/proccess.
    so in the preceding examples (2-3), I can't just calculate the entropy change by :
    [tex]\Delta[/tex]S = [tex]\int Cv\cdot dT /T[/tex] when Cv is the heat capacity of water
    in case 2.
    and [tex]\Delta[/tex]S = Q/273.15 when Q is the required latent heat, in case 3. (and 273.15 = 0[tex]^{o}[/tex]C).
    But I need to find some reversible path which have same initial and final state as this proccess. Am I right?

    5.Is there a general way to find a reversible path for a given proccess?

    Thank you
  2. jcsd
  3. Feb 4, 2010 #2


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    4) I don't think you need the process to be reversible to use that equation. After all, Entropy only ever increases. If there is delta S, the process can not be reversible (since, to reverse it, you need to lower entropy which violates the second law of thermodynamics).
  4. Feb 4, 2010 #3
    the argument seems to be right, but here is for example a quote from Wikipedia (Entropy) :

    the change in entropy, ΔS, is given by the equation

    [tex]\Delta S = \frac{Q}{T}[/tex], where Q is the amount of heat absorbed by the system in an isothermal and reversible process in which the system goes from one state to another, and T is the absolute temperature at which the process is occurring.
    If the temperature of the system is not constant, then the relationship becomes a differential equation:

    [tex]dS = \frac{\delta q}{T}[/tex].

    Then the total change in entropy for a transformation is:

    [tex]\Delta S = \int \frac{ \delta q }{T}[/tex].
  5. Feb 4, 2010 #4


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    This is not right. We can still exchange entropy in reversible processes (consider ice forming or melting in a water bath of 0°C), but the entropy lost by one phase must exactly equal the entropy gained by another (thus, the total entropy increase of the universe is zero, which is the definition of reversibility).

    naftali, I agree with all your points. To find reversible paths, look for processes that don't involve any imbalance of temperature, pressure, charge, etc. These imbalances would lead to entropy creation and thus irreversibility.
  6. Feb 4, 2010 #5
    two points :

    1.According to the the thermal identity : dU = Tds - Pdv, in the mentioned processes with good approximation dV = 0, so dU = dQ. so apperantly, we can still say :[tex]\Delta S[/tex] = [tex]\int \frac {\delta q}{T} [/tex] .
    If this argumet is correct, so is to any process with dV = 0.

    2.I'll try to find reversible pathes :
    In the melting ice : attach the ice (with [tex]0^{o}C[/tex] ) to a reservoir with infinitesimal more temperature that will provide the latent heat.
    In the glass of cold water which becomes to room temeperature : attach the water to a reservoir with infinitesimal more temperature, then it's tempeature rises a little, than attach it to another reservoir with infinitesimal more temperature so it's temperature rises again, and so on.

    Are these arguments correct?

  7. Feb 4, 2010 #6


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    Any reversible process.

    Sounds good.
  8. Feb 4, 2010 #7
    Why does it need to be reversible? Isn't the thermodynamic identity true for any process? so [tex] \Delta S = \int \frac {\delta q}{T} [/tex] will be true to any process with dV = 0?
  9. Feb 4, 2010 #8


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    How about a gas in a closed container with non-uniform temperature? For the system, [itex]\Delta V=0[/itex] and [itex]Q=0[/itex], but [itex]\Delta S\neq 0[/itex] as the gas temperature becomes uniform. But if you add the additional constraints that the system is closed, always at equilibrium, and with no other work sources, then I agree.
  10. Feb 5, 2010 #9
    Thank you, it helped me a lot
  11. Feb 5, 2010 #10


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    No, for example, you can stirr a container containing water and ice. Then, even if you neglect volume change, there is work done on the system which increases its internal energy due to friction. Hence W is not equal to p Delta V and Delta S is not equal to Q/T.
    Nevertheless Delta U=T Delta S - p Delta V and Delta U =W +Q both are true in that process.
  12. Feb 5, 2010 #11
    Hmm... , so I guess I can tell it's true in any quasi-static process, in which no work of any kind is done?
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