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I am confused about the independent variables in thermodynamics. I know that for a one-component, one-phase system, there are only two independent intensive variables that can be chosen and which values will determine all other ones. I would like to know which variables we can choose for the intensive variables. As I understand, it must not be pressure and temperature (the natural variables of the internal energy) but can be any intensive variable, so two from a pool of variables like:

$I = \{p, T, \rho = m/V, M = m/n, V_m, \dots\}$

Are we really free to choose any 2 of those? I can e.g. not imagine choosing the molar mass $M$ and $p$ to describe all properties of an ideal gas, as the ideal gas equation can be reformulated as:

$pV=nRT \Leftrightarrow pm/\rho =nRT \Leftrightarrow pM/\rho=RT$

which means that for any choice of $M$ and $p$, $T$ is not determined, since another variable $\rho$ needs to be chosen.

Edit:

I think my question was too specific on the molar mass. Actually, I wanted to quite generally know how we can choose them. For example, if we include also extensive variables, such as conjugates of the intensive ones, or even $U$, $H$, $G$ and $A$ can the independent variables still be chosen completely freely?

  • Could we e.g. take $U$ and $H$ as our two independent variables?
  • Could we take $p$ and $V$ as out two independent variables?
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In addition to p and T, the third variable is either molar density $\rho=\frac{m}{MV}=\frac{n}{V}$ or its reciprocal, molar volume $V_m=\frac{1}{\rho}=\frac{MV}{m}=\frac{V}{n}$.

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  • $\begingroup$ Thanks, this is great and answers a part of my question. This means that the molar mass is no suitable thermodynamic variable for this system right? Could you extend your answer in a more general way and also consider my new edit of the question? $\endgroup$ – Guiste Mar 7 at 13:19
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I think your confusion is here:

  • "I know that for a one-component, one-phase system, there are *only two independent intensive variables that can be chosen* and which values will determine all other ones," (italics are mine).

Having only two intensives is not what characterizes a simple one-component one-phase system, it is rather *your choice* to have the simplest possible characterization that way, i.e., with two intensives. It is perfectly possible that the molecules of your material have electric dipole moments but you keep the external bias field at some arbitrary but fixed value (it could be zero, too), and you do not explicitly care about the mutual electric dipole interaction and just take that as part of the over-all thermal/pressure response.

Similarly, if you wish so, you can ignore the total amount of mass/volume of the system and look only at its local thermal pressure/response but then you are ignoring, for example, surface effects. In general, you need two equations of state to describe a thermostatic system: (1) thermal equation of state that is a relationship among the directly measurable but not necessarily controllable variables including temperature $$f(T, p,\rho, V, \mu, \mathcal E,\mathcal M, \mathcal B,...)=0$$ (2) and a caloric equation of state that describes how energy/entropy/work are related: $$g_1(U, T, p,\rho, V, \mu, \mathcal E,\mathcal M, \mathcal B,...)=0$$ or $$g_2(S, T, p,\rho, V, \mu, \mathcal E,\mathcal M, \mathcal B,...)=0.$$ For the simplest case the equations can be $T=T(p,V)$ or $f=f(T,p,\rho), ...$ and $U=U(T,p)$, or $U=U(p, v)$, or $S=S(T,\rho)$, $g_1(U,p,v)=0$, etc.$

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  • $\begingroup$ Thanks. Of course, the choice of variables should be made regarding the system. But is it true that mathematically we are really free to choose any 2 variables from the set of conjugate thermodynamic variables (p/V, T/S, \dots) as the independent variables of a specific coordinate patch of the thermodynamic manifold? Let's say we have an open system, for which we need to define 3 independent variables and we choose $\rho=m/V$, $V$ and $m$, would these not be linearly dependent? $\endgroup$ – Guiste Mar 8 at 1:59

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