Atmospheric pressure inside a rigid vessel

Question

On a macroscopic scale one can explain the atmospheric pressure by the weight of the column of air over a given small area.

If you enclose this air in rigid vessel (for example by pressing together two Magdeburger hemispheres without evacuating them), the pressure of the air inside of the vessel remains the same as outside the vessel though the air column is not present anymore.

How can one explain this sticking only to the macroscopic/phenemological scale, i.e. avoiding microscopic kinetic gas theory?

Answer 1

The best you can do without reference to kinetic theory is apply the ideal gas law $P = \frac{nRT}{V}$ where $P$ is the pressure, $n$ is the number of moles of gas, $R$ is the ideal gas constant and $V$ is the volume.

Once the container is closed the pressure will obey this law (assuming and forces acting on the gas such as gravity are negligible) It may change if there are changes in the volume or temperature, so it is independent of the atmospheric pressure outside from the moment it is sealed.

If you want an explanation of the ideal gas law you need kinetic theory.

Answer 2

The macroscopic explanation is still a kinetic gas theory.

The weight of a column of air really only makes sense if the air molecules at the bottom are being pressed down by bouncing off the ones above them, and the ones above those and so on. Otherwise how does the weight of a molecule 30,000ft up have any effect?