From the expression $$U=TS - PV + \mu N$$ $U$ can be described as a function of three of these six variables: $$U=U(S,V,N),$$ implying that the other variables are also functions of these; i.e: $$T=T(S,V,N), P=P(S,V,N), \mu = \mu(S,V,N)$$ In particular, $$T=\partial U/ \partial S, P= \partial U/ \partial V, \mu = \partial U/ \partial N.$$

Now one would naively expect the follow differential relation to hold: $$dU = TdS + SdT - PdV -VdP +\mu dN + N d\mu.$$ However, the thermodynamic identity is written: $$dU = TdS - PdV +\mu dN.$$ What is the meaning of this? It seems to necessitate: $$SdT-VdP+ Nd\mu =0$$ which may seem clear when viewed from the point of view of extensive vs. intensive variables? Expanding the above, $$Sd(\partial U/ \partial S)-Vd(\partial U/ \partial V)+ Nd(\partial U/ \partial N) =0$$ $$S(\frac{\partial^2 U}{\partial S \partial S}dS+ \frac{\partial^2 U}{\partial S \partial V }dV + \frac{\partial^2 U}{\partial S \partial N}dN) - V(...)+ N (...)=0$$ Which is not very illuminating.

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    $\begingroup$ The Gibbs–Duhem identity resolves this, I think. $\endgroup$ Commented Jul 26, 2023 at 5:17
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    $\begingroup$ I'm afraid there is not much more to be understood apart from you have already stated. It means that a system can't be described by intensive variables only, because they are constrained by the Gibbs-Duhem idendity (hence one fewer degree of freedom less). $\endgroup$
    – Abezhiko
    Commented Jul 26, 2023 at 9:05

1 Answer 1


I disagree with the comment that there is not much more to be said about the Gibbs-Duhem equality $$SdT+\sum_{k=1}^N X_kdY_k=0 \tag{1}$$ as that follows directly from $dU=TdS+\sum_kY_kdX_k.$ It is true that the conventional interpretation of (1) just being a differential constraint among the thermostatic variables existing between two equilibrium states infinitesimally close to each other. This is the thermo-static interpretation and is used to derive Gibbs's phase rule, see.

But there is also another interpretation, namely a process oriented dynamic interpretation. In this view, take the thermodynamic system, the engine, and sandwich it between reservoir pairs, specifically a pair of thermal reservoirs at temperatures $T^1$ and $T^2$, a pair pressure reservoirs at pressures $-p^1$ and $-p^2$, a pair of gravitational energy reservoirs (two fixed heights) at potentials $\phi^1$ and $\phi^2$, a pair of electric voltages $\psi^1$ and $\psi^2$, a pair of chemical potentials $\mu^1$ and $\mu^2$,... etc. These represent the intensives $Y_0=T$, $Y_1=-p, Y_2=\phi, Y_3=\psi, Y^4 =\mu, ..$ and their corresponding extensive quantities are entropy $X_0=S$, volume $X_1=V$, gravitational mass $X_2=m$, electric charge $X_3=e$, moles of chemical species $X_4 = c$, etc.

The thermo-dynamic system, the engine, between these reservoirs acts as a transport mechanism by which the reservoir pairs exchange a certain amount of quantity, let us denote it by $\Delta X_k$, as it moves from one potential level to another, that is, from $Y_k^1$ to $Y_k^2$.

The partial work done by moving a quantity $\Delta X_k$ between potentials $Y_k^1$ and $_kY^2$ is of course $(Y^1-Y^2)\Delta X$. It is not assumed that $Y_k^1-Y_k^2$ be infinitesimal, instead, it is only assumed that the intensive values $Y_k^1,Y_k^2$ of the reservoirs do not change by absorbing or releasing the corresponding quantities $\Delta X_k$. Then for any reversible process the total work is zero and is expressed by the work conservation principle:

$$\sum_{k=0}^N (Y_k^1-Y_k^2)\Delta X_k= (T^1-T^2)\Delta S + \sum_{k=1}^N (Y_k^1-Y_k^2)\Delta X_k = 0 \tag{2}.$$

When the intensives of the reservoir pairs are infinitesimally close in Eq. (2) you get the differential constraint resembling that of Gibbs-Duhem but now the extensives have different interpretations. You recover the conventional interpretation by noting that the Gibbs-Duhem equality is scale independent, and Eq. (2) must also hold for any amount of $\Delta X_k$ as long as the reservoirs do not change, in other words the size of the engine very small, insignificant, relative to the reservoirs.

One could legitimately view Eq. (2) as the operational definition of a reversible process: A process is reversible if and only if Eq. (2) holds. Any deviation from zero (ie. non-conservation) is, in fact, a measure of process irreversibility, and it represents the dissipated work in said process, so that $$\sum_{k=0}^N (Y_k^1-Y_k^2)\Delta X_k=\mathcal D \tag{3}$$ and the $2^{nd}$ law of thermodynamics is just the statement in two parts: $$\mathcal D \ge 0 \tag{4a}$$ and $$\mathcal D = T^*\delta S^* \tag{4b}$$ where $\delta S^*$ is the irreversible entropy produced inside the engine at temperature $T^*$.


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