I have studied that are 4 thermodynamical potentials wich are useful, which are the internal energy $U$, the helmholtz function $F$, the Gibbs free energy $G$ and the enthalpy $H$:

$$dU=TdS-PdV$$ $$dH=TdS+VdP$$ $$dG=VdP-SdT$$ $$dF=-SdT-PdV$$

where, $T$ is temperature, $S$ is entropy, $V$ volume and $P$ pressure. But when do we know that these processes are conserved? For example I know (and is somehow intuitive) that in an isolated system internal energy has to be conserved. But what about the others?

  • $\begingroup$ Entropy is conserved in reversible processes taking place in an isolated system. $\endgroup$ Dec 3, 2018 at 23:54

1 Answer 1


First, let's write the complete form of these differentials:

Energy $dU = T dS - P dV + \mu dN$: To get $dU = 0$ we must set $dS=0$, $dV=0$, $dN=0$. This means adiabatic+constant volume+closed.

Enthalpy $dH = T dS + V dP + \mu dN$: To get $dH=0$ set $dS=0$, $dP=0$, $dN=0$. This means adiabatic+isobaric+closed.

Gibbs energy $dG = -S dT+V dP + \mu dN$. To get $dG=0$ set $dT=0$, $dP=0$, $dN=0$. This means isothermal+isobaric+closed

Free energy $dF = - SdT - P dV + \mu dN$: The get $dF=0$ set $dT=0$, $dV=0$, $dN=0$. This means isothermal+constant-volume+closed.

Not much memorization needed if you just follow the rules of calculus.

  • $\begingroup$ I don't think I ever saw $\mu dN$. What is it? Other than that good answer $\endgroup$
    – Bidon
    Dec 4, 2018 at 10:45
  • 1
    $\begingroup$ $\mu$ is the chemical potential. The internal energy is a function of entropy, volume, and number of particles. When we write $dU = TdS - P dV$, we imply a closed system, i.e., $dN=0$. If we allow $N$ to change, then the full form of the differential is the one I gave above. Sam with all other potentials. $\endgroup$
    – Themis
    Dec 4, 2018 at 11:07

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