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Thermodynamic properties of ideal gases

  1. Jul 15, 2010 #1
    Here are some general questions regarding my current reading. I am looking in my text at 2 equations for specific energy and specific enthalpy:

    [itex]u = u(T,v)\qquad(1)[/itex]

    [itex]h = h(T,p)\qquad(2)[/itex]

    Question 1: Are not the properties fixed by any 2 independent priorities? Why have we chosen to speak of u as u(T, v) in lieu of u(T, p) and the same for h ? Is it more convenient to out them in these terms for some reason?

    Now, if we put (1) and (2) in differential form, we have:

    [tex]du = \left(\frac{\partial{u}}{\partial{T}}\right)_v dT + \left(\frac{\partial{u}}{\partial{v}}\right)_T dv\qquad(3)[/tex]

    [tex]dh = \left(\frac{\partial{h}}{\partial{T}}\right)_p dT + \left(\frac{\partial{h}}{\partial{p}}\right)_T dp\qquad(4)[/tex]

    Question 2:

    It says that for an ideal gas:

    [tex]\left(\frac{\partial{u}}{\partial{v}}\right)_T \text{ and }\left(\frac{\partial{h}}{\partial{p}}\right)_T [/tex]

    are equal to zero. Can someone clarify this? Is there some mathematical reasoning behind this? Or is this simply something that we have observed? Or both?

    Question 3:

    Going along with assumptions above (i.e., dh/dp = 0 and du/dv = 0) we can assert that for an ideal gas, specific energy and specific enthalpy are both functions of temperature alone, correct?
  2. jcsd
  3. Jul 15, 2010 #2
    The thermodynamic properties of ideal gases were originally derived as the limit case of results obtained from experiences with real gases. Later on, Boltzmann showed they could be computed from the statistical description of an assemble of non-interacting, point-like particles
  4. Jul 16, 2010 #3


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    Q1: The state is fixed by any two independent, intensive, properties. As far as I know the choice is arbitrary (but I'm not 100% certain off the top of my head).

    Q2: There is both a mathematical reason and it is something that can be observed.

    Q3: Yes. In fact, that's what the math will show when you solve equations [3] and [4] above. I'm sure there should be a proof in your Thermo book. You'll probably see the more useful form of du = CvdT and dh = CpdT which also shows that for an Ideal Gas the internal energy and enthalpy are functions of temperature alone.

    Hope this helps.

  5. Jul 16, 2010 #4
    Q1: Right. I am thinking that the choices made here are more useful then others in some regard.

    Q2: OK.

    Q3: Right, however integrating (3) and (4) armed with the knowledge that dh/dp = 0 and du/dv = 0 is hardly a "proof" (and by the way, this is how my thermo book does it. Keep in mind this is "engineering thermodynamics.")

    I was trying to come up with an intuitive way to show why dh/dp = 0 and du/dv = 0 (if one exists) for ideal gases. I was thinking along the lines that since an ideal gas is very compressible that when we change the pressure or specific volume "differentially" the effect is negligible.

    Thanks again!
  6. Jul 16, 2010 #5


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    Well that's surprising that they don't go into detail! I'll see if I can help. First we have the total differential for u = u(T,v):

    [tex] du = \left(\frac{\partial{u}}{\partial{T}}\right)_v dT + \left(\frac{\partial{u}}{\partial{v}}\right)_T dv [/tex]

    Also using the fundamental differential: du = Tds - pdv and dividing by dv while holding T constant we get:

    [tex] \left(\frac{\partial{u}}{\partial{v}}\right)_T = T \left(\frac{\partial{s}}{\partial{v}}\right)_T - p [/tex]

    This next part is the part you probably haven't gotten to yet in your book which you would have figured out for yourself I'm sure once you read the associated chapter:

    Use one of Maxwell's relations:

    [tex] \left(\frac{\partial{s}}{\partial{v}}\right)_T = \left(\frac{\partial{p}}{\partial{T}}\right)_v [/tex]

    Substituting that relation into the previous equation gives:

    [tex] \left(\frac{\partial{u}}{\partial{v}}\right)_T = T \left(\frac{\partial{p}}{\partial{T}}\right)_v - p [/tex]

    Now for an Ideal Gas we have the equation of state: pv = RT

    Differentiating while holding v constant gives:

    [tex] \left(\frac{\partial{p}}{\partial{T}}\right)_v = \frac{R}{v} [/tex]

    Now substitute that into the previous equation gives:

    [tex] \left(\frac{\partial{u}}{\partial{v}}\right)_T = T \frac{R}{v} - p [/tex]

    and since the Ideal Gas equation of state can be arrange as p = RT/v we have the proof you are looking for:

    [tex] \left(\frac{\partial{u}}{\partial{v}}\right)_T = T \frac{R}{v} - p = p - p = 0[/tex]

    This shows that internal energy is not dependent on the specific volume (since it is 0 just above).

    Now from the very first equation we are only left with the first term on the RHS (since the second term was just shown to be zero):

    [tex] du = \left(\frac{\partial{u}}{\partial{T}}\right)_v dT [/tex]

    Which confirms your intuition that the internal energy of an Ideal Gas is dependent on temperature alone.

    Note that the coefficient in the last equation is defined as Cv (i.e. the specific heat capacity at constant volume). Or as I said before, the more familiar form is du = CvdT.

    Hope that helps.

  7. Jul 16, 2010 #6


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    BTW, the same can be shown true for the enthalpy (I was just too lazy to type anymore)! :rolleyes:

  8. Jul 16, 2010 #7
    No! This is more than enough! I will go through this in a little bit and make sure I understand it. And then I will return and post the solution to the enthalpy part now that I know where the starting point is.

    Thanks for all of your time CS! :smile:

  9. Jul 16, 2010 #8
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