1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the thermodynamics temperature scale?

  1. Apr 3, 2010 #1
    What is the thermodynamics temperature scale, both in theory and in practice?
    Is it really independent of physical substance? Why?
     
  2. jcsd
  3. Apr 3, 2010 #2
    I believe theoretically the (thermodynamic) temperature is found by assuming you have an ideal gas with
    [tex]pV\propto T[/tex]
    So basically you keep on variable constant and thus find the relative change of the other variables.

    Experimentally
    https://www.physicsforums.com/showpost.php?p=2270246&postcount=10

    The ideal gas law is valid whenever the energy and the density of states are power laws of the momentum.

    There might be better experts who can answer this...
     
  4. Apr 3, 2010 #3

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    I'm not sure what temperature 'is', but thermodynamics uses 'temperature' as a real, positive, number that (somehow) relates to the energy of a system. The specifics of that relationship have not yet been worked out for the general case, only for equilibrium conditions.

    Kelvin and Rankine scales are 'absolute' in the sense that the measured temperature is independent of the thermometer- something that was not the case in the early days of thermometry.

    In practice, the temperature of an object tells you how hot it is.
     
  5. Apr 3, 2010 #4
    What you say is the ideal gas temperature scale.It is equivalent to the thermodynamical scale in practice,but NOT in the fundamentals.The ideal gas scale depends on the universal behavior of gas,but the thermodynamical scale is independent of all material.

    In theory,the thermodynamical scale is based on the efficiency of Carnot engine because according to the Second Law of Thermodynamics,all reversible cycles between two reservoirs have the same efficiency.Then we choose a standard,in practice the triple point of water as 273.16K.

    In practice,it is realized using different ways in different range of temperature.See http://www.bipm.org/utils/common/pdf/its-90/SUPChapter1.pdf
     
    Last edited: Apr 3, 2010
  6. Apr 3, 2010 #5
    Even thermodynamic temperature can be negative
    See http://en.wikipedia.org/wiki/Negative_temperature
     
  7. Apr 4, 2010 #6
    Could you please describe the way with which they get a temperature reading, other than using the ideal gas law? But don't forget that it should be a fundamental principle not relying on data parameters which is collected by ideal gas law thermometers.
    I believe it's impossible to practically apply the definition of temperature you used.

    Also theoretically the reversible cycle definition seems impractical. I actually went through that definition in detail and found that you need to be able to identify
    - reversible processes and also
    - isothermal processes
    Both of which you cannot do in a strictly rigorous way.
     
  8. Apr 4, 2010 #7

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    I assume you are referring to the statistical-mechanical description of a two-level system. While that's an interesting result, it cannot honestly represent a thermodynamic temperature- after all, then there would be a large flow of *heat*, which is not observed.
     
  9. Apr 5, 2010 #8
    I think the ideal has scale is the same as the thermodynamic scale because of the way ideal has is defined. PV= nRT , where T is already the thermodynamic scale.
     
  10. Apr 5, 2010 #9
    I believe the usual approach to defining temperature is as follows: Considering two compartments which are divided by a diathermal wall (and isolated as a whole), one can define a temperature if the compartments are in thermal equilibrium. This occurs when
    [tex]\frac{\partial S_1}{\partial E_1}=\frac{\partial S_2}{\partial E_2}[/tex],
    where the volume of each compartment and the number of particles in each compartment is kept constant. S is the entropy.

    Since the compartments are in thermal equilibrium, they have the same temperature. So
    [tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex].
    That is, the partial derivatives define a function of the temperature. When we choose f(T)=1/T, this definition becomes identical with the perfect gas scale.

    That said, I'm just a recent student of the subject. There might be a better definition. But this one is pretty general, since it makes no assumptions about specific properties of the system. It basically only uses the second law of thermodynamics.

    About negative temperatures: Defining temperature this way, negative temperatures arise quite naturally. If the partial derivative is negative, so is the temperature. This can happen, for example, in a paramagnetic solid in a magnetic field. Consider the interaction between the dipoles and the magnetic field. The system has a minimal energy if all the dipoles in the solid are parallel with the field and maximal energy if they are anti-parallel with the field. Both the minimum and maximum energy correspond to an entropy of 0 (that is, the arrangement of the dipoles is completely known).
    If we'd plot the entropy against the energy, we would see a curve that goes up from zero at the minimal energy and goes down from some point to reach zero again at maximum energy. When the curve goes down, we have a negative slope and therefore a negative temperature.

    The key to negative temperature is the maximum energy, which forces the slope to become negative at some point. When we consider the complete system (not just the interaction between the dipoles and the magnetic field) there can be no maximum energy, since the kinetic energy of particles is unbounded. But we can still talk about these separate aspects of the system if we assume that the different aspects interact very weakly with each other, which is the case in this situation.
     
    Last edited: Apr 5, 2010
  11. Apr 5, 2010 #10

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    This topic comes up a lot.

    The truth is that there IS no good general definition of temperature, which is why the international temperature scale (ITS-90) uses many different "definitions" (depending onf which fixed point is used for a particular range); although there is e.g. no fixed point at very low temperatures (ITS-90 starts at 0.65K).
    While this might not be very satisfactory from a philosophical point of view it does have the advantange that it works...

    Also, some primary thermometers (e.g. nuclear orientation thermometers and noise thermomenters) are based on statistical principles that are quite difficult to connect to more classical thermodynamics.
     
  12. Apr 5, 2010 #11
    Sure, the scale is arbitrary to some degree. But temperature can be related to entropy and energy. By doing so, we see that f(T)=1/T (in the derivative in my previous post) is a reasonable choice for the scale: Defined this way, heat goes from hot bodies to cold bodies, which is consistent with our idea of hot and cold. f(T)=T would imply heat goes from cold bodies to hot bodies. So, in my opinion, not all scales are created equally. Perhaps like reference frames in mechanics, some are more useful than others.

    That, and we may at least say that temperature IS independent of the physical substance, which was one of the OP's questions. It just depends on the second law of entropy (and conservation of energy).
     
  13. Apr 5, 2010 #12

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    Nobody here (IIRC) will disagree that temperature is related to energy. Or that temperature and entropy are related.

    That's different from claiming that the conceptual foundations of temperature are known and completely elucidated. I would disagree with that statement.
     
  14. Apr 5, 2010 #13
    I don't disagree with that. All I tried to say is one can devise a definition of temperature that is "good" and general. I don't see how a definition like dS/dE=1/T isn't good or general. This may very well be the result of my ignorance, which is why I'm participating in the discussion. Perhaps we have a different notion of what goodness and generality is in this case?
     
  15. Apr 5, 2010 #14
    Actually that is the most general definition that I support. But the question back to you: So you are given a piece of wood. How would you measure the temperature in practice, now?
    Your definition only shifts the problem to a new (unknown) quantity, the entropy.


    Could you mention the method they use to find a temperature (which is not based on ideal gas laws)? How can you guarantee that any other method doesn't cause inconsistencies? For example with a flawed method I could measure the temperature of one body, bring it on contact with another body, measure the other body and suddenly get a different temperature reading.

    What do you mean by related? Of course if you change one quantity of a system, all others will most likely also change. But note that energy is a function of temperature alone (no p or V dependence) only in the special case that [itex]p\propto T[/itex] at constant V. That special case doesn't necessarily have to hold.
     
  16. Apr 5, 2010 #15
    Well, by sticking a thermometer in it. When the thermometer is in equilibrium with the wood, it gives a certain reading. Then calibrate the thermometer in a controlled experiment, where the value of the partial derivative is known. A gas that can be described as a perfect classical gas would do.

    But now I'm really curious as to what the objections to this are, since I assume you've considered this as well. I can't find any hidden assumption in this process.
     
  17. Apr 5, 2010 #16
    The hidden assumption is, that you need someone who has actually solved the problem of measuring temperature before you. And he must have used another method than you, because with your method alone you wouldn't be able to measure temperature.

    OK, so in the end you are using the ideal gas law definition all the time? You have a definition of temperature with entropy, but you never use it. Instead you bring all you objects in contact with an ideal gas and read off the temperature?

    Now say you measure temperature of body A. Then you bring body A in contact with body B. Now you measure temperature of body B. Why are you sure that temperature B will give the same reading?
     
  18. Apr 5, 2010 #17
    I'm not really sure that I understand the objection.

    I claim that
    [tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex]
    for compartiments in thermal equilibrium and I choose
    [tex]f(T_i)=\frac{1}{T_i}[/tex].
    When considering an ideal classical gas, this statement happens to be equivalent to pV=NkT. This is based on the second law, not on the ideal gas law. I don't see how that undermines the generality of the first equation.

    The particular choice for the function is made, so that the ideal gas law is a special case in which the microscopic variables may be replaced by macroscopic ones. Again, the ideal gas law is be derived from this definition, not the other way around.

    About the last part, I don't see why the second measurement would give the same reading. Could you elaborate on this situation? I don't get the issue you're trying put forward.

    correction: I shouldn't say the ideal gas law is derived from the second law. It just corresponds to the definition of temperature that I gave.
     
    Last edited: Apr 5, 2010
  19. Apr 5, 2010 #18
    I'm not sure what you mean in your last post. I try express my objection clearer:

    Imagine you are the first person in the world to measure a temperature. No such thing as a thermometer exists, yet. What would you do experimentally to get a temperature reading? Please describe what you do step by step.



    For the second part:
    1) you've somehow managed to build a thermometer
    2) you stick it in body A and read off T_a
    3) you remove the thermometer
    4) you bring body A and body B in contact
    5) you stick the thermometer in body B and read off T_b
    6) can you guarantee with your definition of temperature that T_a=T_b as it should be?
     
  20. Apr 5, 2010 #19
    Hypothetically speaking: I prepare an array of test chambers in which gases reside with different pressures and equal volume and amount of gas. All the test chambers can act as heat baths for a mini-chamber that can be thermally connected to the test chambers.
    I look for a device that, when put in the mini-chamber, reacts predictably to the different pressures (and thus temperatures, by the ideal gas law). I then assign values to the states of the device.

    Then, since the ideal gas law is compatible with the definition of temperature as given before, the device could now be used to measure other things. These other things don't have to follow the ideal gas law, just the second law of entropy.

    I still don't see why measurement of body B needs to give the same reading as body A. It seems to me that this should only true if body B does actually have the same temperature as body A. If this is not the case, there is no thermal equilibrium when the two are put in contact and everything changes. I can guarantee that when the bodies are in contact and thermal equilibrium has been achieved, the readings will be the same (when both measurements are made after equilibrium). I can also guarantee that different readings will be made when the bodies are not in thermal equilibrium.
     
  21. Apr 5, 2010 #20
    How do you pick which number exactly you want to assign? Who is telling you which test chamber pressure corresponds to which temperature?

    Why do you think that if you take two different devices, you won't get contradictory results when cross-checking the temperature of the same test object?

    When do you ever need your quoted definition of temperature? You might as well have said [itex]\partial \sqrt{S}/\partial E=T[/itex] and nothing would be different?

    Step 4) was to bring both bodies in thermal contact (for a long time)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the thermodynamics temperature scale?
  1. What is temperature (Replies: 8)

  2. What is temperature? (Replies: 9)

  3. What is temperature? (Replies: 4)

Loading...