Is there an easy way to remember Maxwell relation in thermodynamics? Maxwell relations are very important in thermodynamics, is there a way to easily remember it once and for all?

 A: I would never recommend anyone memorize the Maxwell relations, as anyone who needs  them should already be intimately familiar with the information needed to derive any one of them in twenty seconds (maybe sixty seconds the first few times):

*

*The fundamental relation is $dU=T\,dS-P\,dV$. That is, there's two ways to add internal energy to a system (namely, heating it and doing work on it); $S$ and $V$ are the extensive variables that shift, driven by gradients in the intensive variables $T$ and $P$; and the unusual minus sign arises because pressure tends to decrease volume.


*Other useful potentials are the enthalpy $H\equiv U+PV$ ($U$ plus the $P$–$V$ work needed to move the atmosphere out of the way of the system), the Helmholtz energy $F\equiv U-TS$ ($U$ minus the $T$–$S$ heating available to bring the system up to temperature), and the combination of these for constant surrounding pressure and temperature, the Gibbs free energy $G\equiv U+PV-TS$.


*The corresponding differential forms of course switch variables and signs in the corresponding terms: $dH=T\,dS+V\,dP$, $dF=-S\,dT-P\,dV$, $dG=-S\,dT+V\,dP$.


*When we see a target for a Maxwell relation (e.g., $\left(\frac{\partial S}{\partial P}\right)_T$), we immediately remember that whatever is on top (here, $S$) arose from differentiating a potential earlier (here, $S=-\left(\frac{\partial F}{\partial T}\right)_V=-\left(\frac{\partial G}{\partial T}\right)_P$). Now that coefficient is being differentiated again, with respect to something else. (If this isn't the case, then no Maxwell relation is applicable; an example is the constant-pressure heat capacity $C_P=-T\left(\frac{\partial^2 G}{\partial T^2}\right)_P$.) The Maxwell relations arise from noticing that $$\frac{\partial^2\Phi}{\partial X\partial Y}=\frac{\partial^2\Phi}{\partial Y\partial X}.$$
for a potential $\Phi$ and state variables $X$ and $Y$.


*Thus, we immediately write down the thermodynamic relation where $S$ is the coefficient and something else is being differentiated with respect to $P$:
$$dG=-S\,dT+V\,dP$$
which we recall really means the expansion
$$dG=\left(\frac{\partial G}{\partial T}\right)_PdT+\left(\frac{\partial G}{\partial P}\right)_TdP,$$
so that
$$\left(\frac{\partial (-S)}{\partial P}\right)_T=\frac{\partial G}{\partial T\partial P}=\frac{\partial G}{\partial P\partial T}=\left(\frac{\partial V}{\partial T}\right)_P;$$
$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$
We confirm that the sign switches when only one minus sign appears in the thermodynamic relation.
Note the complete lack of mnemonics or tricks or any information outside what the Maxwell relation implementer must already know as a thermodynamics framework. The key parts of the technique are shown in bold.
This strategy is furthermore easily extendable to an infinite number of other Maxwell relations that one wouldn't wish to take the time to memorize;  $\left(\frac{\partial S}{\partial A}\right)_{V,T}=-\left(\frac{\partial \sigma}{\partial T}\right)_{V,A}$ for surface tension $\sigma$ and surface area $A$, for example, or $\left(\frac{\partial V}{\partial N}\right)_{T,P}=\left(\frac{\partial \mu}{\partial P}\right)_{T,N}$ for chemical potential $\mu$ and particle number $N$. One simply writes these down after a few seconds of thought.
A: You noted that you found my derivation-style answer unhelpful—which is fine—so here's a mnemonic technique for obtaining Maxwell relations: the thermodynamic square, as discussed, for instance, by Callen in his Section 7.2:

and by Brannon in her comprehensive collection of recursion tables:

A: There are four variables $(T,V,P,S)$ here. Let's abstract them to $(A,B,C,D)$ for now.
Notice that for
$$\left(\frac{\partial A}{\partial B}\right)_C \tag{1}$$
the corresponding Maxwell relation is
$$\left(\frac{\partial D}{\partial C}\right)_B$$
The pattern here is that the two elements on the "denominator" switch places, and the "numerator" switches (or circulated) to the element being "left out" in the original expression eqn. (1), in this case, it is $D$.
In terms of $(T,V,P,S)$, if $S$ and $P$ are both in the "fraction", a negative sign is appended to the front. To memorize this, keep checking for SuPer.
Verify this with the Maxwell relation.
A: There is! I call it “upside down and inside out.” You can even hum along to an OK Go song of that same name these days, which is nicer than what I had at college!
So the key is that you have to be able to determine from memory whether this manipulation is even likely to help. And then you can always derive it and look up any other details—“oh YEAH it was about commuting partial derivatives.” But the need for a mnemonic here, is just to have that initial spark where something looks familiar.
Let me explain the “inside out” thing first, this has to do with intensive and extensive thermodynamic variables that are conjugates of each other:

*

*Temperature (intensive) and entropy (extensive)

*Pressure and volume

*Chemical potential and amount

A Maxwell relation will always take two of these and turn them both inside out. A derivative of pressure with respect to temperature: this can also relate volume (pressure’s conjugate) to entropy (temperature's conjugate).
The upside down thing is that the symbols on the derivative flip sides from numerator to denominator, and you can kind of see that it has to flip if you take an intensive and an extensive variable in the derivative... You are going to have to have, say, both numerators intensive and both denominators extensive if you want the same behavior if I add “more of the same,” to a system (that is, grow the extensive quantities without changing the intensive quantities)... I guess that's based on treating a derivative as a ratio of deltas which isn't quite so easy with partial derivatives. But still, that's how it works in practice.
Once you know that Maxwell Relations are about turning things upside down and inside out, you can immediately look at a problem, “how is the pressure changing with the amount of heat that we put into it?”, recognize a partial, $\partial P/\partial S$ in this case, and do the inside out and upside down transform in your head to say “that is related to the temperature changes when you change the volume.” How exactly? That requires the full derivation, it turns out that they are negatives of each other. But at least you know that they are related—and you can immediately ask questions like “will they Maxwell relation help here? No...” to save yourself time and effort.
