The mechanical equilibrium is a state in which the pressure (and elastic stress) inside a gas is evenly distributed. Do we make such an assumption while deriving the equation of state in thermodynamics?

I know that for ideal gas equation and van der Waal gas equation, mechanical equilibrium is assumed. But is this true for all the other more accurate equations of states?

If so, then is this not a very unrealistic way of modelling real gases? I mean, for a general gas, the force between particles can be assumed to be anything, in which case, the pressure of the gas away from the center will drop. Moreover, for a general gas, even the temperature will tend to decrease with increase in the distance away from the center, like in a star.

So if mechanical equilibrium is assumed while deriving the equations of state, then how can such assumption be physically justified?


To be precise about what you mean you should call the traditional thermodynamics "equilibrium thermodynamics" (as contrasted with non-equilibrium thermodynamics).

Because a thermodynamics state is described by a few macroscopically measurable variable and represents everything you need to know about that system it must be in internal equilibrium.1 Not just mechanical equilibrium, but also thermal and chemical equilibrium.

It is the tension between this basic property and the desire to study systems that are changing that necessitates the introduction of the "quasi-static" process (which implies quasi-equilibrium internal states though this is rarely stated explicitly).

In statistical mechanics we find that, in fact, internal equilibrium is not always present, but that fluctuation of macroscopic system away from internal equilibrium tend to be both small and brief. Which brings us full circle, because if the quasi-static driver of a change is slow enough, then the variation from equilibrium due to that driver will be similar to the size of statistical fluctuations of the system—giving us a way to characterize what we mean by "quasi-static" more precisely than just "slow".

1 For historical reasons a few non-equilibrium situations—such as the isothermal free expansion of a gas--are exampled in detail in the traditional study, but they represent boundary cases.


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