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I read that the slope of the pressure, ($P$), versus volume ($V$) graph should always be negative for thermodynamic stability, but there was no further explanation in the text. Could someone give an explanation?

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    $\begingroup$ Can you write the pressure from the ideal gas law as a function of volume? $\endgroup$ – Kyle Kanos May 16 '15 at 13:45
  • $\begingroup$ @KyleKanos $p=RT/v$. But my question is for any general phase, not necessarily obeying ideal gas law $\endgroup$ – Bosnia May 16 '15 at 14:05
  • $\begingroup$ That's fine, you can get the understanding of the general case from the ideal gas law just as well. $\endgroup$ – Kyle Kanos May 16 '15 at 15:10
  • $\begingroup$ Ideal gas law is of little help here as it is indeed a very specific case. A more natural thing to consider is the van der Waals equation of state which does infact predict positive slopes. What is the assumption that goes wrong? That the system is uniform. In reality the system will immediately phase separate (and by definition then be the very opposite of uniform indeed) to achieve a lower free energy state as per Maxwell equal area construction (except for very small systems or special experimental conditions where the metastability can be maintained for prolonged times). $\endgroup$ – alarge May 16 '15 at 21:51
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A state is thermodynamically stable when its Gibbs free energy is at a minimum.

$$G=U-TS+PV$$

Holding all variables but $P$ and $V$ fixed, it means that:

$$dG=dP V+PdV=0 \implies {dP\over dV}=-{P\over V}$$

Since neither of $P$ nor $V$ could be negative ${dP\over dV}$ must be negative.

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