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A A question on absolute kelvin temperature scale.

  1. Mar 10, 2017 #1

    MathematicalPhysicist

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    I was asked in an exam the following question:

    Q: How is the absolute Kelvin Thermodynamic temperature scale defined ?

    My answer is the following:

    A: The absolute zero kelvin thermodynamic temperature is defined as the temperature for which the entropy of the system is zero, i.e ##S|_{T=0} = 0 ##.
    Which happens when we have only one state, in the ground state.

    Am I right, how would you answer this question?
     
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  3. Mar 10, 2017 #2

    BvU

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    That's only one point on the scale !

    And I don't think the ground state is low enough (but I could be wrong) -- zero K is hypothetical.
     
  4. Mar 10, 2017 #3

    MathematicalPhysicist

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    I didn't know how else to answer this question to tell you the truth.

    I know of negative temperatures which are hotter than ordinary positive absolute temperatures, but I think you still need to find the point where the entropy is zero, in which case we would have zero temperature, and since entropy is non-negative all other temperatures will follow; i.e you need to locate the absolute zero kelvin in this scale and the rest will follow; but I don't know really what's the answer.
     
  5. Mar 10, 2017 #4

    hilbert2

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    Absolute zero temperature doesn't necessarily mean zero entropy, if you have a solid material that has crystal defects (Schottky, etc.).

    If you have something that obeys the Boltzmann distribution where probability of a state is proportional to ##e^{-E/k_B T}##, then the system being certainly in the ground state is pretty much equivalent to the statement that T = 0.
     
  6. Mar 10, 2017 #5

    MathematicalPhysicist

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    @hilbert , but the entropy is zero at the ground state of energies, is it not?
     
  7. Mar 10, 2017 #6

    hilbert2

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    Yes, but a system that's at absolute zero can be frozen in a metastable state that is neither the ground state or has zero entropy.
     
  8. Mar 10, 2017 #7

    MathematicalPhysicist

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    So @hilbert2 how would you answer this question?

    Thanks.
     
  9. Mar 10, 2017 #8

    hilbert2

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    You do it by filling a rigid container with very low-density gas, measuring pressures at different temperatures and extrapolating to zero pressure, which corresponds to the absolute zero. The size of one degree is the same in Celcius and Kelvin scales.
     
  10. Mar 10, 2017 #9

    jbriggs444

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    Zero temperature is not defined in terms of zero entropy. The thermodynamic definition of temperature is more nuanced.

    Consider a sample of an ideal gas in an insulated container. A real gas under low pressure is a good approximation. Hold volume constant and allow the sample to reach thermal equilibrium. Now introduce (or drain) some energy from the system and let it equilibriate again. The thermodynamic temperature of the system is the ratio between how much energy you put in and how much entropy increased as a result.

    That is, it is the partial derivative of entropy with respect to energy, ##\frac{\partial S}{\partial U}##

    [Hopefully, I have not butchered the definition -- I've never taken a course in thermodynamics]

    A natural consequence of this definition is that if you have two objects at different temperatures and allow [heat] energy to flow from one to another, entropy is increased when energy flows from from higher temperature to lower temperature.

    A negative temperature results when a system is in a state where adding energy actually reduces entropy. If such a system is brought into contact with a system with a positive temperature, entropy is increased when energy flows from negative temperature to positive temperature. This means that negative temperatures are "hotter" than any positive temperature.

    It is possible to invert the definition of temperature and define a scale based on ##\frac{\partial U}{\partial S}##. That gets rid of the strange behavior at absolute zero.
     
  11. Mar 10, 2017 #10

    f95toli

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    Zero temperature is not actually defined in terms of anything "physical" at the moment.
    One kelvin is defined as a fraction (1/273.16)of of the triple point of water (which would be the answer to the question), and the current ITS-90 temperature scale only goes down to something like 0.65K right now. Hence, it is not possible nor necessary to discuss absolute zero in order to answer the question.

    Now, the SI temperature will be re-defined from 2019 and the Kelvin will then be defined by giving an exact value to the Boltzmann constant; however there will still not -as far as I am aware- be a direct definition of what absolute zero is.
     
  12. Mar 10, 2017 #11

    Philip Wood

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    I thought the thermodynamic temperature scale was defined by [tex] \frac{T_1}{T_2}\ =\ \frac{Q_1}{Q_2}[/tex] in which [itex]Q_1[/itex] and [itex]Q_2[/itex] are the heat taken in by a Carnot engine on the high temperature isothermal and the heat given out on the low temperature isothermal, the equation defining the ratio of thermodynamic temperatures of these isothermals.

    If we define 273.16 K as the temperature of the triple point of water, then in theory, using the above equation, we have a means of assigning a temperature to any other source or sink of heat. A Carnot engine is, of course, and ideal engine, and the equation can't be used directly, but its consequences lead to methods that can be used. For example, it's not difficult to show that the ideal gas scale temperature, pV/nR for a gas as its density approaches zero, is identical to the thermodynamic scale.
     
  13. Mar 11, 2017 #12
    Ok.
    Somewhat of a wordy question, and that tricked up the answer.

    Note that the following are all the same thing, as perceived in this day and age:
    The explanations after the "-" my input so probably debatable as actual definitions.
    Absolute temperature scale - ideal gas PV=T ( constant ) ; if P=0, then T=0 see post 8 ( Experiments on gases gave indications of an absolute scale )
    Kelvin temperature scale - see post 10 for the size of a Kelvin
    Thermodynamic scale - Carnot engine , see post 11

    Zero entropy: S=0 at T=0 for a perfect crystal. Otherwise it is not.
    But then you will be asking if something such as a diamond crystal and graphite crystal could both have, at absolute zero, an S=0.
    Good question.
    Some answer yes, others no.
    I guess that is related to post 2.
     
  14. Mar 11, 2017 #13

    Vanadium 50

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    This is a first - B and I answers for an A-level question. Usually it's the other way round.

    1/T = dS/dE, so the condition is not S = 0. The condition is that for any energy input that the system can absorb, the number of available states increases. Effectively, this means you are in the ground state, but the ground state may be degenerate. An atom with angular momentum J has a ground state that is (2j+1) degenerate, so at absolute zero its entropy is k ln (2j+1).
     
  15. Mar 12, 2017 #14

    MathematicalPhysicist

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    I wonder what is the official answer to this question my lecturer had in mind... :-)
     
  16. Mar 12, 2017 #15

    Philip Wood

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    It depends how your teacher has approached the subject. If you've been taught about Carnot Cycles and suchlike, then I'd back my answer (reply 11). If you've been introduced to entropy without first studying Carnot cycles, then the equation at the beginning of the second paragraph of post 13 can be used as a definition of T. Note that the dS/dE is actually a partial derivative; in the case of a gas it's the volume which is kept constant.
     
  17. Mar 13, 2017 #16

    f95toli

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    See my post #10.

    I can of course not guarantee that this is what your lecturer had in mind; but it is certainly the correct answer (refer your lecturer to the BIPM webpage with the official definitions of all SI units if he/she disagrees).

    Note that there are lots of common misconceptions when it comes to units; most of the time because people assume they are always defined in some "clever" way that somehow refers to fundamental physics. However, the point of the SI -and all other systems of units- is that both base- and derived units have to be useful and possible to realize , i.e. you have to be able calibrate measurement equipment in some sensible way. Defining the temperature scale using entropy would not work, simply because there is good direct way to directly measure entropy.
     
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