Here is the question: 5.You have two l-liter containers connected to each other by a valve which is closed. In one container, you have liquid water in equilibrium with water vapor at 25°C. The other container contains a vacuum. Suddenly, you open the valve. Discuss the changes that take place, assuming temperature is constant with regard to
(a) the vapor pressure,
(b) the concentration of the water molecules in the vapor,
(c) the number of molecules in the vapor state.
Answer: Vapour pressure and concentration of the gaseous molecules of the system remains essentially unchanged. Number of molecules doubled.
For (a) I agree that the vapour pressure remains the same. However, for (b) this was my thought process: Since P=F/A, and since the volume was doubled, the containers being of the same shape would have the same surface area. Hence the the surface area is doubled when the valve is opened. Since pressure remains the same, the force is also doubled. Hence the number of collisions per unit time would be the same. This would mean that the concentration is the same.
But thinking about it more deeply, if I double the volume it is not necessary that the surface area is doubled right? Because the piston adds to some surface area before removing it. So taking it away doesn't cause the surface area to double exactly right?
Then if it is not doubled then the frequency of collision would not be doubled and hence the concentration of the gas wouldn't be doubled too right?
So why can we simply say that the concentration would double? Are we assuming that the piston has a negligible surface area? Or is my concept here wrong?
Thanks