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Enthalpy vs. Internal Energy (Thermodynamics)

  1. Aug 2, 2010 #1
    Hello!

    First post here :-) Hope this isn't too simple for this forum...if so, by all means give me a pointer...

    I'm wondering about some basic thermod. concepts. Could somebody please shed a little light on these?

    Internal Energy and Enthalpy (both from macroscopic viewpoint)

    Does the internal energy U of a gas include kinetic energy from pressure? Since enthalpy is defined as U + PV, where P is pressure and V is volume, one might think that internal energy does not include pressure.

    The thermodynamics text I have here makes it clear: all energy of a thermodynamic system other than 1) kinetic energy of the whole system in reference to an external frame, and 2) potential gravitational energy of the whole system in reference to an external frame, is labeled as internal energy U.

    So, basically, if internal energy includes kinetic energy from pressure, what is the significance of enthalpy?


    Thanks for your thoughts,

    Kubus



    ( ( ( I should note that I think that pressure is not as straightforward as kinetic energy. As a simple explanation would have it, pressure is the kinetic energy of molecules as they move in a gas, but does not include the microscopic vibrations associated with thermal energy (temperature). However, thermal energy contributes to pressure, so my previous sentence cannot be completely accurate. Maybe classical thermodynamics is not enough here, and statistical mechanics / etc. is needed to explain the relation between pressure and temperature ? ) ) )
     
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  3. Aug 2, 2010 #2

    Andy Resnick

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  4. Aug 2, 2010 #3
    I don't understand what the "kinetic energy from pressure" is.
     
  5. Aug 3, 2010 #4
  6. Aug 3, 2010 #5

    Andy Resnick

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    I just cut-n-paste. Super lo-tech.
     
  7. Aug 4, 2010 #6
    Thank you! The link at the bottom of the older thread gave good insight.

    I should've searched the forum first though. Sorry about that, newb mistake haha :-D.
     
  8. Aug 24, 2010 #7
    This might help conceptually as well:

    Think about creating a system out of thin air. The amount of total energy one would need to do this is the energy associated with the system itself (energy stored in bonds, thermal energy, rest energy of the mass itself, etc. -> internal energy, U) as well as the work done to the on the medium to place the system there (the PV term).
     
  9. Aug 25, 2010 #8
    Austen_G, thanks for replying to my thread. That is a new way to look at it, and it makes sense to me - tracing where each component of the system's energy (enthalpy) came from.

    But now that I look at this topic again after a while, I'm realizing that a lot of my original confusion stemmed from ascertaining whether or not internal energy is affected by pressure. That was the "kinetic energy from pressure" I talked about and which was unclear in my first post.

    The simple answer I should have come to is: Assuming ideal gas behavior, NO, internal energy is a function of ONLY temperature.

    Realizing this as I looked through my thermodynamics book, I stumbled across this statement:

    "Similarly, for a gas obeying the ideal gas model, the specific enthalpy depends only on temperature, ..."

    Specific enthalpy is, of course, h = u + pv , where u and v are internal energy per unit mass and volume per unit mass.


    Which completely scrambled my thoughts. With that definition of specific enthalpy, how could specific enthalpy ever (ideal gas or not) be a function of ONLY temperature ???!


    Thanks in advance if you dare to untangle my latest mess. :-)
     
  10. Aug 25, 2010 #9
    There is no such thing like "kinetic energy from pressure" .
    The internal energy of a system is U .
    U is supposed to included kinetic and potential energy, and this includes everything.
    However, when analysing open systems, it is often more convenient to work with the enthalpy.
    This is because of the mechanical work involved on the input or output streams.
     
  11. Aug 25, 2010 #10
    Austen G indicates the right direction for this discussion. However his apporach, which can also be found in Levenspiel 1996 (Understanding Engineering Thermo), implies that p in H=U+pV is the external (environmental) pressure not the internal pressure as is commonly believed, and this causes the confusion for Kubus and the rest of us. In 'equilibrium thermodynamics' the internal and external pressure are assumed to be equal, and then there seems to be no need to make the distinction, but in all real world thermodynamics there is.
     
  12. Aug 25, 2010 #11
    Zeppos, what you are saying is a revelation to me. If I had paid more attention, I would have seen that this is what Austen G was pointing at.

    Trying to find more about how enthalpy relates to the distinction between internal and environmental pressure, I came to the wikipedia page about enthalpy, and it looks like somebody improved it since I first started this thread:


    I think that finally unravels the significance of enthalpy for me. The most important distinction is that the energy associated with enthalpy is 'contained' both inside and OUTSIDE the system boundary. Does that make sense?
     
  13. Aug 25, 2010 #12
    lalbatros, thanks for replying. Could you explain more why there is no such thing as kinetic energy from pressure, why I'm coming up with this idea?

    Without implying that this term of mine makes sense, I'd like to put out this example:

    A common engineering device is a compressor. Such a device clearly must be supplied with energy, and if it were completely efficient, the only effect it would produce would be an increase in pressure of the fluid (the thermodynamic system) which flows through it. Without academic rigor, a simple way to put it is that a compressor 'converts kinetic energy into pressure'.

    The first time I read that quoted statement on some website, I thought it sounded unlikely...but then I realized that, since energy must be conserved, where else would it go?

    If I use my NEW understanding of enthalpy (^^^ post above ^^^), I would say that the work done by the compressor becomes the work specified by enthalpy's pressure x volume, pV, term. This work is transferred further as the fluid (system) pushes against its surroundings on, say, the other end of the pipe.

    Am I on the right track?

    Thank you,
    Kubus
     
  14. Aug 25, 2010 #13

    presbyope

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    The work done by the compressor on the gas typically heats up the gas. So that is considered internal energy (the U term). I suppose it is kinetic in the sense that hot gas molecules move faster, but it's confusing to call it kinetic energy since the system as a whole isn't zooming off anywhere.
     
  15. Aug 26, 2010 #14
    I'd like to bring up a point about your compressor. In general, a compressor will decrease the volume of the container, which done through a quasi-static process (done "infinitely" slowly so that the system goes through infinitesimal changes) will increase the pressure of the system. To me this seems like a rather useless process. Kubis, is this really the formal definition of a compressor? Can I get the source please?

    Take for instance an air compressor (a device which is used to fill car tires and such): This device will actually increase pressure by adding addition gas to the system. If you increase the number of particles (again just looking at an ideal gas) then the pressure will increase. In addition to an increase in pressure, as presbyope has pointed out, the temperature will also increase. The kinetic energy as you have mentioned is simply the increase in the momenta of the molecules inside. Realistically, this system will come to equilibrium with the temperature outside of the compressor, in which case your kinetic energy has been released in the form of heat. This is the sort of compressor I'm used to thinking about.
     
  16. Sep 7, 2010 #15
    Kubus,

    I reacted to your sentense:

    "Does the internal energy U of a gas include kinetic energy from pressure?"

    In the same way, a compressor does not convert pressure into energy, but it eventually works on a gas and transfers energy to this gas.

    Moreover, there is no way to recognize how a part of the internal energy has be transfered to the gas. The same final state could have been reached without external work, but by heat transfer.

    I agree with you that this is more about "precise wording".
    However, in thermodynamics, even more than in any other parts of physics, wording is quite important.

    The best way to talk about enthalpy is to say that it is a state function.
    You can also add that this state function is useful because it is the best "word" to use when the first law is applied in a constant pressure transformation. This function is also useful to analyse open systems.

    As such the enthalpy cannot be said to have a physical meaning like mass or internal energy.
    Yet, it is a state function that makes the energy balance looking short in some circumstance.
    This is simply because it include work done on the system.

    Of course, you are on track!
    Enjoy thermodynamics!

    Michel

    see also: http://www.mathpages.com/home/kmath184/kmath184.htm (for example)
     
  17. Sep 8, 2010 #16
    In my view:
    the best way to talk about enthalpy is to say that it is a form of energy, which implies that it is a state function. Enthalpy is the sum of two froms of energy: internal energy (which includes the kinetic energy of molecules) and pV-energy which goes by many names like energy of displacment, pressure-energy, flow-energy, but still has to be recognized as an official form of energy. Note however that p must be understood as external pressure, not internal pressure (the difference can be subtle).
    The LINK states: The justification for defining this variable (Enthalpy) is really only a matter of convenience, because we often find that the sum U + PV occurs in thermodynamic equations.
    Having a complete energy balance is indeed very convenient, but it is not only a matter of convenience: it is necessary.
     
  18. Sep 8, 2010 #17
    Zeppos10,

    I do not fully agree with you.
    The term PV has the dimension of an energy, of work, of heat.
    But calling it energy should mean more than that.
    The term PV can even not be called work, since work is PdV.

    If you consider a certain amount of a perfect gas, why would you consider H = U + PV as an energy related to this amout of gas?
    My understanding is that H is an amount of energy related to an energy balance.
    More precisely, I see it as an accounting artefact.
    In accounting, things can be accounted in various ways, and some ways are simpler.
    This is similar in thermodynamics.

    The internal energy, U, is directly related to the kinetic (K) and potential energy (P) of the molecules of the gas. In fact U=K+P. Therefore, U can be called an energy of this gas, why the more precise terminology the "internal energy" of the gas.

    Nothing similar can be said for the enthalpy.
    However, in a transformation, when an amount of gas interacts with the environment, then the energy balance will of course include the work done on the gas. It is no surprise then that the energy balance might include the variation of the PV term, specially at constant pressure since then d(PV) = pdV = work.

    Otherwise, the kinetic energy of the particles has not such flavor has "pressure".
    The kinetic energy can however be split in some flavors, like rotational kinetic energy, translational energy, kinetic energy in the x-direction, ...
    But the kinetic energy has no such flavor as "pressure-kinetic-energy".
    The same applies for the potential energy.

    This what I wanted to say.

    Michel
     
  19. Sep 9, 2010 #18
    Lalbatros:
    when you believe that enthalpy is an 'accounting artefact' you are in trouble. (The only 'accounting artefact' I know of was called ENRON.)
    If you try to ALWAYS (!!) distinguish between internal and external pressure (which most textbooks do not) you will find that enthalpy has physical meaning as a form of energy, but p in (U+pV) is the external pressure. Try to move a system with volume V from an environment of p' to an environment of p": tell me how much work you did and where this energy went.
    PS: I have never heard of "pressure-kinetic-energy" exept in this thread so I leave it alone.
     
  20. Sep 9, 2010 #19
    Zeppos10,

    My problem is that I don't know what you mean by:

    "... has physical meaning as a form of energy ..."

    By "form" do you mean kinetic, potential or what else?
    And by energy, what do you mean, the energy of 1kg of oxygen?

    I tried to solve these questions by explaining that the enthalpy appears naturally in an energy balance because of the work done on the system (eventually by a reversible transformation if you prefer so). That's why I used the language of accounting. Like ENRON, thermodynamics makes use of balance sheets, but thermodynamics doesn't intend to steal nature!

    Enthalpy is obviously a state function, with the dimension of energy.
    But you cannot related -I believe- enthalpy to a sum of energie of the different molecules in the gas.
    Am I wrong about that?

    Any way, as long as we use the laws of thermodynamics correctly, the names and meta-ideas don't really matter.

    Michel
     
  21. Sep 10, 2010 #20
    Lalbatros:
    if you google around a bit you can find under energy-wiki:
    Different forms of energy include kinetic, potential, thermal, gravitational, sound, elastic and electromagnetic energy.
    The wiki does not claim this is a complete list, nor do I. Note that gravitational energy is one form of potential energy, which in turn is part of mechanical energy.
    Also note that no reference is made to chemical energy (1 kg of oxygen ??), while our society runs on chemical energy, which is called enthalpy if you want to make any calculation.

    Enthalpy = (U+pV). The internal energy is related to the sum of energy of different molecules, the pV part is independent of them, but the pV part is always small (a few %) of the total enthalpy, and it is convention to attribute all enthalpy to the molecules.
    So strictly speaking you are right, but .....
     
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