Equalization of pressure in heat exchanger I was solving numericals on Kinetic Theory of Gases when I came across this question


  Two closed vessel of equal volume contain air at 105 kPa, 300 K and are connected through narrow tube. If one of the vessel is now maintained at 300 K and other at 400 K what will be the pressure in the vessel?


I don't want the solution of this question but I want to know why the pressure will be same in both the vessel when one is maintained 300 K and other 400 K. The molecules in  the vessel at 400 K will have greater kinetic energy and will exert greater pressure on the walls than the molecules of the other vessel.

Now one will say that some of the molecules of vessel at higher temperature will move to the other vessel so that the pressure remains same in both the vessels, but why is it so. Why can't the pressure in both the vessels be different?
Macroscopics

Imagine the tube is initially closed. Then one vessel is heated up. As $V$ and $m$ are constant:
$$p=k\,T$$

So if $T$ is increased, $p$ is increased. As $p$ is increased, there will be a net flow of mass, until they are balanced. That's what we observe. Now, we can think of it microscopically.

Microscopics

Initially, there are the same number of particles in each vessel. When heating up there is an increase in velocity. So, in the high-temperature vessel the same number of molecules are with higher velocity. Imagine a plane in the middle of the narrow tube, the number of collisions per time, because of velocity, will be higher in $400\ \mathrm K \to 300\ \mathrm K$ direction. It indicates a net flow of mass. This flow of mass will equal pressures on both sides.