Thermodynamical conjugate variables

Question

In thermodynamics the potentials are typically only a function of 2 variables, say $$U=U(S,V)$$ with entropy $S$ and volume $V$. I see that conjugate pairs $S,T$ or $p,V$ always have the unit of energy when multiplied. But what is the reason that for example $S$ and $T$ can not be independent leading to the potentials only depending on 2 of the 5 variables.

Answer 1

Indeed it is possible to express the internal energy of a thermodynamic system as a function of two conjugate variables. However the resulting function is not as useful as the thermodynamic potential expressed as a function of the independent variables $S,V,N$(in general don't forget N for fluid systems).

First of all, let me remark that it is not by chance that the product of two conjugate variables has the physical dimension of energy. This is a consequence of the energy being a homogeneous function of its extensive variables $S,V,N$. Indeed, Euler's theorem ensures that $$ U= TS - PV + \mu N $$

Let's use now the ideal gas as a simple example to show that it is possible to use as two independent variables the two conjugate quantities $P$ and $V$.

$$ U = \frac{3}{2}N k_BT = \frac{3}{2} PV $$

where use has been done of the equation of state.

Although this expression of the internal energy as a function of $P,V$ provides the correct value of the internal energy for any thermodynamic state, just using the corresponding values of $P$ and $V$, it cannot be defined a thermodynamic potential. A reason for this name is the possibility of obtaining explicitly all the thermodynamic quantities (by partial derivatives). This is true only if a specific set of variables is used (the so called natural variables for that potential). From the function $U=\frac{3}{2}PV$ is not possible to achieve such a result. The reason is evident in the present case because in order to obtain, say the dependence of $U$ on entropy, one should be able to get it from $$ P=-\left( \frac{\partial U}{\partial{V}}\right)_{S,N}= \frac{2}{3}\frac{U}{V}. $$ However, the integration with respect to $V$ of the previous equation provides $$ U=V^{-\frac{2}{3}}\phi(S,N) $$ where $\phi(S,N)$ is an arbitrary function of $S$ and $N$. Spoiling this result from any practical use.