Vapour pressure and volume of container Assume that a liquid is in equilibrium with its vapour in a container which can change volume. If we increase volume LE Chatelier principle states that pressure should increase but final pressure cant be equal to the initial pressure so final pressure will be less than initial pressure?
 A: Assuming you have a closed container to begin with, and you increase the volume via some mechanism to have more number of particles in the liquid, then you effectively are reducing the vapor pressure. Because the particles that were in the vapor phase have been forced to come down to the liquid. 
A: From what I gather, it seems you are asking confirmation for the question you have posed, which is "Is the final pressure of a system, when its pressure is decreased, less than the pressure of the initial system?". If this is not the case, then your question is unclear.
Let's get the assumptions clear. 


*

*Le Chatelier's principle states that a system, when one of it's properties is changed, will act in such a way as to counteract this change.

*The volume of a hypothetical system ($A_{(l)}⇆A_{(g)}$ for simplicity) is increased.


Increasing the volume of the system does not change the volume of the system in liquid state. It does, however, confer a decrease in the pressure of the gas, per Boyle's Law ($P \varpropto \frac{1}{V}$, given constant temperature). Le Chatelier's principle predicts that our hypothetical system counteracts this change by increasing the pressure of the gas to a final pressure. But this final pressure, as you have correctly claimed, is less than the initial pressure. As much as the system would like to return to the initial pressure, it cannot, because the ratio of gas particles to liquid particles must remain constant. 
A decrease in the general pressure of the gas must also mean a decrease in the vapour pressure of the system. An increase in volume means that less gas particles per unit area are pushing on the surface of the liquid, which means that it is easier for liquid particles to escape into the gaseous phase. As the system stabilizes into equilibrium, the final vapour pressure would be lower than the initial vapour pressure.
A: Reading again some basic thermodynamics and the useful comments/answers already provided, I came up with my own answer.
Vapor pressure is only function of temperature. There is a dynamic equilibrium between the liquid and the vapor:
$$A \, (\mathrm{l}) \longleftrightarrow A \, (\mathrm{g}) $$
with equilibrium constant $K = p_\mathrm{vapor}$.
Lets suppose that the volume of the container is increased. Momentarily, the vapor pressure is decreased, ($T$ and $n$ remain constant, $p = \frac{nrT}{V}$).
Pressure must increase according to the Le Chatelier's principle, which means that some molecules from liquid phase must pass to the gas phase. At the new equilibrium, the vapor pressure obtains again the value it had before the volume increase (is only function of the temperature).
A: In the example you have cited, and assuming that the temperature is constant, an increase in volume would potentially decrease the vapour pressure and the system reacts by increasing the number of molecules in the vapour phase by decreasing the number of molecules in the liquid phase until the vapour pressure returns back to its original magnitude.
At a given temperature there is a dynamic equilibrium between the rate at which molecules in the liquid phase move into the vapour phase and the rate at which molecules in the vapour phase move into the liquid phase.
If the density of vapour molecules is reduced when the volume of the system is increased the dynamic equilibrium no longer exists as the rate at which molecules in the liquid phase moving into the vapour phase is greater than the rate at which molecules in the vapour phase moving into the liquid phase.  This continues until both rates are the same and the vapour pressure becomes the same as it was before.
