User:Tim Starling/Derivation of hydrostatic equilibrium from kinetic theory

Take a cuboid container with height ${\displaystyle h}$ and horizontal area ${\displaystyle A}$, as in Hydrostatic equilibrium § Derivation from force summation but instead of a continuous fluid we will have ${\displaystyle N}$ particles of a gas, each with mass ${\displaystyle m}$, following the kinetic theory of gases.

Each particle is of negligible size in the sense that they collide only with the walls of the container, the particles do not collide with each other. In other words, ${\displaystyle h}$ is much less than the mean free path.

Typically, kinetic theory ignores gravity, but here we include it, giving each particle a constant downwards acceleration of ${\displaystyle g}$.

A particle leaves the bottom of the container with a vertical velocity of ${\displaystyle v_{0}}$. By the time it reaches the top of the container, the velocity has been reduced to ${\displaystyle v_{1}=v_{0}-gt}$, where ${\displaystyle t}$ is the time taken for the particle to go from the bottom to the top.

There is a solution for ${\displaystyle t}$ if ${\displaystyle v_{0}^{2}/2\geq gh}$. We ignore particles that fail to reach the top of the container.

The particle collides elastically at the top and returns to the bottom, where the velocity immediately before impact is ${\displaystyle -v_{0}}$.

The change in particle momentum imparted by the collision at the bottom of the container is ${\displaystyle \Delta p_{0}=2mv_{0}}$, twice the particle momentum, since it elastically returns to an upwards trajectory.

The magnitude of the average force imparted on the bottom of the container by the whole ensemble of ${\displaystyle N}$ particles is

${\displaystyle F_{0}={\frac {N\Delta p_{0}}{2t}}}$,

since each particle collides with the bottom with a period of ${\displaystyle 2t}$.

The pressure at the bottom is

${\displaystyle P_{0}={\frac {F_{0}}{A}}={\frac {Nmv_{0}}{At}}}$,

and similarly, the pressure at the top is

${\displaystyle P_{1}={\frac {Nmv_{1}}{At}}={\frac {Nm(v_{0}-gt)}{At}}=P_{0}-{\frac {Nmg}{A}}}$.

The density of the contained gas is ${\displaystyle \rho ={\frac {Nm}{Ah}}}$, and so

${\displaystyle P_{1}=P_{0}-\rho gh}$.

Thus we have recovered the standard result of hydrostatic equilibrium.

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