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There is something I don't understand about thermodynamic potentials.

We know that the necessary condition for a system to evolve is an increase in entropy. For a system kept at constant volume and temperature for example, this means :

$$dU = \delta Q, \hspace{7mm} dS>\frac{\delta Q}{T} = \frac{dU}{T}$$ $$\Leftrightarrow TdS > dU$$ $$\hspace{2cm}\Leftrightarrow d(U-TS) = dF < 0$$

But if $V$ and $T$ are held constant, $dV=dT=0$.

So how is $\hspace{3mm}dF = -SdT - pdV \hspace{3mm}$ supposed to be anything else than zero ?

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Your thermodynamic system has 2 degrees of freedom. We can place all states of the system on a 2d plane, which we can parametrise by whichever 2 thermodynamic variable we choose, lets say $S$ and $V$ for definiteness.

If we impose a constraint upon the system, such as fixing the temperature, we reduce the number of degrees of freedom by 1 and the allowed states of the system now lie on a 1d curve; an isotherm in the previous example.

If we impose another constrain, such as fixing the volume, we reduce the number of degrees of freedom again, so our system is now limited to a point; that is there is only one state of the system with that volume and temperature. If you now try and force the system to change state, you find that the system is overconstrained and you encounter a problem. You cannot keep the temperature fixed and keep the volume fixed and change the entropy.

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  • $\begingroup$ Are you saying that a system with fixed temperature and volume cannot evolve ? I don't have a specific system in mind, I just used math of thermodynamics on an abstract system and it seems to lead to a contradiction. $\endgroup$ – mwa1 Oct 8 '15 at 18:45
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    $\begingroup$ For a system with only 2 degrees of freedom yes. Systems can have more than 2 degrees of freedom, but then the differentials of the thermodynamic potentials will have more than 2 terms. If you put in an example it is not unreasonable. what can you do to a gas that will change it while keeping the temperature and volume constant? $\endgroup$ – By Symmetry Oct 8 '15 at 18:57
  • $\begingroup$ I still don't understand what's wrong with the equations. Should $F$ have another parameter ? This isn't mentioned in my book or in my course, it just says $F(T,V) = U - TS$ and that evolution implies $dF<0$ and equilibrium implies $F$ is minimum. $\endgroup$ – mwa1 Oct 8 '15 at 19:33
  • $\begingroup$ The problem is that you are holding too many variables fixed. If you all the variables constant, then clearly nothing can change. The subtlety is that $T$, $S$, $p$ and $V$ are not independent, so if you fix any 2, you fix the others as well. explicitly. $p = \frac{\partial U}{\partial V}$ and $T = \frac{\partial U}{\partial S}$ and $U = U(S,V)$, so fixing 2 variables fixes everything else. $\endgroup$ – By Symmetry Oct 8 '15 at 19:48
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There must be unconstrained variables in addition to $T$ and $V$.

Suppose that the system consists of multiple species that can undergo a chemical reaction. At a fixed $T$ and $V$, the number of each species changes (via the chemical reaction) in such a way that $dF<0$.

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  • $\begingroup$ Then $F$ has to be $F(T,V,N_i)$ for example and not only $F(T,V)$, right ? $\endgroup$ – mwa1 Oct 8 '15 at 20:15
  • $\begingroup$ @mwa1 That is correct. $\endgroup$ – higgsss Oct 8 '15 at 20:17

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