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Consider a solid isolated cylinder containing water, in the earth's gravitational field as we know there is a pressure gradient inside the water like:

$\frac{dP}{dz}=-\rho g$ ; (1)

but from postulated thermodynamics,for a simple system in thermodynamic equilibrium the pressure is defined as Eq.(2):

$P\equiv -\left(\frac{\partial U}{\partial V}\right)$ ; (2)

according to Eq.(2), the pressure gradient must be zero in a thermodynamic equilibrium state (i.e., $\nabla P=0$) but in a gravitational field as Eq.(1) says, it is not!!

So, is the mentioned system in thermodynamic equilibrium?

By the way, I've seen this: (Could a gas be in thermodynamical equilibrium even with a gravitational field?) but this question has not pointed to the postulated thermodynamic definition of pressure (Eq.(2)).

GiorgioP-DoomsdayClockIsAt-90
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Danial Rezai
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  • Your thermodynamic relation defines pressure in terms of work done during compression or expansion. How have you turned this into a constraint on the pressure gradient? – rob Jul 07 '22 at 08:59

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No, the cylinder is not at equilibrium, or at least not a “global” equilibrium, since the pressure is not uniform in the system. However you can slice up the column into smaller subsystems which have nearly uniform pressure (and temperature and chemical potential) then treat each subsystem as if it were at equilibrium. That is, we assume each subsystem is in its own “local” equilibrium. This approach lets you use thermodynamic methods to analyze the whole system even though it is not at (global) equilibrium. See Section 8.1.4 in this book for an example of this approach in gas columns.

Look up the “local equilibrium hypothesis” for more on this topic.

C. Taylor
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