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Hello I have a question related to fluid parcels in hydrodynamics.

Say in Lagrange fluid dynamics a parcel is a part of the system that is macroscopically small, but still a thermodynamical system (enough particles). This parcel may be represented with $s,\rho$, the specific entropy and the density.

Say I would additionally assume the volume of the parcel to be constant. May I reduce the number of variables to one? Say I want the (extensive entropy $S$) be the only thermodynamical variable ?

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  • $\begingroup$ Complete specification of the thermodynamic state of a single-phase single-component fluid parcel needs two intensive variables. Knowing that volume (an extensive variable) is fixed doesn't help. $\endgroup$
    – Deep
    Commented Apr 9, 2019 at 5:19
  • $\begingroup$ I don't see why. I mean a fixed volume reduces the number of possible thermodynamical changes. So I should be able to represent this reduced number of possibilities with a reduced number of variables. In particular, if the volume is fixed, rho ist additionally constant and I don't see any difference between $s$ and $S$. $\endgroup$
    – Q.stion
    Commented Apr 10, 2019 at 20:59
  • $\begingroup$ If density is fixed then yes, you have fixed one intensive variable, and then only one more intensive variable is needed to determine its thermodynamic state. $\endgroup$
    – Deep
    Commented Apr 11, 2019 at 5:23
  • $\begingroup$ Okay thanks. But I have the feeling, fixing the density is equal to fixing the volume. As stated by (Massimo Materassi in Lagrangian Hydrodynamics, Entropy and Dissipation) parcels do not exchange matter (parcel identity conservation). So the mass is fixed and additionally fixing the volume fixes the density, right? $\endgroup$
    – Q.stion
    Commented Apr 11, 2019 at 19:55
  • $\begingroup$ Yes that's right. $\endgroup$
    – Deep
    Commented Apr 12, 2019 at 4:51

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