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Problem from Fermi's Thermodynamics

  1. Oct 1, 2009 #1
    Hi all, I am getting stuck on this problem from chapter 4 of Fermi's Thermodynamics:

    1. The problem statement, all variables and given/known data

    "A body obeys the equation of state:

    pV[tex]^{1.2}[/tex] = 10[tex]^{9}[/tex]T[tex]^{1.1}[/tex]

    A measurement of its thermal capacity inside a container having the constant volume of 100 L shows that under these conditions, the thermal capacity (heat) capacity is constant and equal to 0.1 calories / K. Express the energy of the system as a function of T and V."


    2. Relevant equations

    The first law: dQ = dU + dW.

    Since we want U=U(T,V), we can express the first law as:

    [tex]
    \left(\frac{\partial Q}{\partial T}\right)_V dT + \left(\frac{\partial Q}{\partial V}\right)_T dV = \left(\frac{\partial U}{\partial T}\right)_V dT + [(\left(\frac{\partial U}{\partial V}\right)_T + (\frac{10^9*T^{1.1}}{V^{1.2}})] dV
    [/tex]

    We also know:

    [tex]
    \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p
    [/tex]

    where again p can be substituted from the defined equation of state.

    3. The attempt at a solution

    I'm tempted to say that [tex]\left(\frac{\partial Q}{\partial T}\right)_V = 0.1 cal / K [/tex] with the corresponding V = 100 L. If I do this and assume all dV = 0 for constant volume I get what appears to be a trivial result. I feel like I need to get a perfect differential to solve for U(T,V), but I don't see how. Other than that I haven't done much more than toy with the above equations. Thanks for your time.
     
  2. jcsd
  3. Oct 7, 2009 #2
    I have looked at the problem again recently but still don't have much more insight. I see that, by the first law, since no work is being done (constant volume):

    [tex]

    \left(\frac{\partial Q}{\partial T}\right)_V = C_v = \left(\frac{\partial U}{\partial T}\right)_V

    [/tex]

    Using [tex]

    \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p

    [/tex], setting up a perfect differential and taking the derivatives, we can get:

    [tex] dU = (0.1 cal / K) dT + \frac{10^{8}T^{1.1}}{V^{1.2}}dV [/tex], but I don't see how to go further. Any ideas?
     
    Last edited: Oct 8, 2009
  4. Oct 8, 2009 #3

    Mapes

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    Gold Member

    Try expressing the constant-volume heat capacity as

    [tex]C_V=T\left(\frac{\partial S}{\partial T}\right)_V[/tex]

    Then determine S by integration (note that a function of V appears that needs to be evaluated via a Maxwell relation), and use

    [tex]U=TS-PV+\mu N[/tex]

    where [itex]\mu[/itex] needs to be evaluated by using the heat capacity condition.

    This was pretty difficult to work through. There may be an easier way. I eventually got [itex]U=(0.1\,\mathrm{cal}/\mathrm{K})T[/itex] plus a function of T and V that vanished when [itex]V=100\,\mathrm{L}[/itex].
     
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