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Thermodynamic state entirely determined by only two quantities?

  1. Jul 11, 2014 #1
    I am trying to understand why I can specify the specific volume [itex]v[/itex] of a fluid element as a function of the equilibrium pressure, [itex]p[/itex], and the equilibrium entropy, [itex]s[/itex]. This is for example done in this article http://www.sno.phy.queensu.ca/sno/str/SNO-STR-95-051.pdf , on this website: http://wind.mit.edu/~emanuel/geosys/node3.html , and in Landau & Lifshitz fluid mechanics, page 7.

    I have spent a while looking some source explaining this, and remembered from statistical physics (quote from Landau & Lifshitz, Statistical Physics, Page 42):
    I am trying to understand how the knowledge of [itex]p[/itex] and [itex]s[/itex] would help me to, for example, obtain [itex]T[/itex]? How can I get access to other quantities knowing only those two variables?
     
  2. jcsd
  3. Jul 11, 2014 #2

    WannabeNewton

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    The statement in Landau is only true if there is only one generalized force associated with the system or equivalently if there is only one generalized coordinate characterizing the system; in the case of Landau's discussion the generalized coordinate is the volume and the conjugate force is the pressure. In more general cases, wherein one has ##n## generalized coordinates or generalized forces, one needs ##n+1## variables in order to completely determine the thermodynamic properties of the system.

    With that being said, the enthalpy ##H## of a system is entirely characterized by ##p## and ##S## i.e. ##H = H(S,p)##. You can therefore determine the temperature using ##T = \frac{\partial H}{\partial S}|_p## so that ##T = T(S,p)##. The enthalpy also allows you to determine the volume as a function of these coordinates i.e. ##V = V(S,p)## using ##V = \frac{\partial H}{\partial p}|_S##. All you need to do is determine the enthalpy of the system as a function of the pressure and entropy.

    This can always be done in principle by putting the system in thermal and mechanical contact with a heat bath and using the Gibbs canonical ensemble to calculate the Gibbs partition function ##\mathcal{Z}## from which we get the enthalpy ##H = -\frac{\partial}{\partial \beta}\ln \mathcal{Z}##. This will tell us ##H(T,p)## since ##\mathcal{Z} = \mathcal{Z}(T,p)## on account of the system being in both thermal and mechanical contact with the heat bath. We can also get ##S = S(T,p)## from ##\mathcal{Z}## and in principle we can invert this to get ##T = T(S,p)## so that ##H = H(S,p)##.
     
    Last edited: Jul 11, 2014
  4. Jul 11, 2014 #3
    Alright, thanks for the explanation. This leaves me with two questions:
    • [itex]T[/itex] and [itex]S[/itex] are not viewed as a pair of generalized forces/coordinates in this context?
    • What is the argument for needing [itex]n+1[/itex] variables in order to completely determine all thermodynamic properties?
     
  5. Jul 11, 2014 #4

    WannabeNewton

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    This is really a matter of taste. Some authors will consider ##T## and ##S## as a pair of generalized forces/coordinates whereas others will reserve these qualifications only for coordinates and forces that are related through some form of work ##\delta W## and not through spontaneous heat flow ##\delta Q## as ##T## and ##S## are. I prefer the latter.

    Say we have ##n## generalized coordinates for a system. This can be for example volume ##V## and an external magnetic field ##B## if we have a sample of ferromagnetic material and wish to calculate its average magnetization and pressure. The Gibbs partition function will have dependence ##\mathcal{Z} = \mathcal{Z}(T,x_1,...,x_n)## i.e. it will depend on ##n+1## coordinates. As you know, all thermodynamic properties of a system can be determined entirely from ##\mathcal{Z}##. For example, going back to the example of the ferrmagnetic, we have ##p = \frac{1}{\beta}\partial_V \ln \mathcal{Z}## and ##\langle M \rangle = \frac{1}{\beta}\partial_B \ln\mathcal{Z}## as well as ##H = -\partial_{\beta}\ln\mathcal{Z}##.
     
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