Would it be possible to construct a thermodynamic potential with all the intensive variables ($T$, $p$, and $\mu$—the temperature, pressure, and chemical potential, respectively) fixed?
This is one of my homework questions. I thought it was mathematically possible, using Legendre transforms like you would transform any other potential, but my professor mentioned later in one of the lectures that it's a "big no no". So, am I missing something here? And if it's not possible then why?
Your professor should remember that in Physics, it is a good idea to keep in mind the motto "never say never".
The "big no no" requires some qualification. It is true that if one is dealing with a usual extensive macroscopic system, the triple Legendre transform of $U(S,V,N)$: $$ {\mathbb Z}(T,P,\mu)= U-TS+PV-\mu N $$ is identically zero for every temperature, pressure, and chemical potential. Therefore it looks useless. Notice that here I am using a symbol like ${\mathbb Z}$ because there is no established convention for notations and names. Sometimes I saw it called "zero-thermodynamic-potential," although I do not have a reference for this name.
However, ${\mathbb Z}=0$ is a consequence (through Euler's theorem) of the homogeneity of degree $1$ (extensiveness) of the internal energy $U(S,V,N)$. In turn, the extensiveness is a consequence of the continuity and the additivity of the energy.
The latter property is true only for large systems and in the absence of long-range interactions.
Therefore, for small systems where sub-extensive terms are not negligible, the ${\mathbb Z}$ potential may play a role because it allows focusing on the sub-extensive thermodynamics of nanoparticles.
Actually, with a different name and symbol, subdivision energy, ${\bf E}$, such a thermodynamic potential was introduced years ago by TL Hill. A recent paper by Hill on such thermodynamic potential is Hill, T. L. (2001). A different approach to nanothermodynamics. Nano Letters, 1(5), 273-275.
my professor mentioned later in one of the lectures that it's a "big no no"
You can certainly construct a thermodynamic potential $\Phi$ for cases of fixed temperature $T$, pressure $P$, and chemical potential $\mu$. Recall that the internal energy $E$ is
$$E\equiv TS-PV+\mu N+\sum_i X_iY_i,$$
with entropy $S$, volume $V$, particle number $N$, and $X$ and $Y$ as intensive and extensive conjugate variables, respectively, corresponding to all types of work that aren't pressure–volume work (e.g., magnetic field–magnetization work or surface tension–area work).
Based on the fundamental relation
$$dE=T\,dS-P\,dV+\mu \,dN+\sum_i X_i\,dY_i,$$
we define our potential $\Phi$ as usual through a Legendre transform:
$$\Phi\equiv E-TS+PV-\mu N,$$
yielding a new relation
$$d\Phi=-S\,dT+V\,dP-N\,d\mu+\sum_i X_i\,dY_i,$$
which is just $d\Phi=\sum_i X_i\,dY_i$ at constant $T$, $P$, and $\mu$. Now, if you never considered any type of work other than pressure–volume work, then this potential wouldn't be too useful to you, as it would be identically zero! Perhaps this is what your professor is referring to.
Answer by Chemomechanics is right. I thought I would mention a related point.
In a simple model of a paramagnet, we can write $$ dU = T dS - m dB $$ where $m$ is total dipole moment and $B$ is magnetic field. So then you might form the combination $\phi = U + mB$ and get $$ d\phi = T dS + B dm. $$ The trouble with this is that in this model we also have $$ U = - m B $$ so $$ \phi = 0. $$
Temperature (T) and pressure (P) are inherently intensive (independent of mass) properties. (Not sure what your $\mu$ is.)
The thermodynamic potentials Helmholtz free energy (F), Enthalpy (H), Gibbs free energy (G), and internal energy (U) are extensive properties, i.e., they depend on the amount of mass. The mass dependent thermodynamic potential equations are
$F=U-TS$
$H=U+PV$
$G=U+PV-TS$
Dividing the above equations by the mass of the system makes all the extensive properties intensive, called specific properties (lower case by convention).
$f=u-Ts$
$h=u+Pv$
$g=u+Pv-Ts$
Hope this helps.
