Say you would have a plastic bottle partially containing water that is laying on its side. How would applying external pressure through a weight on the side affect the internal pressure. In other words, how well does the pressure transfer from outside to inside the system?
[Physics] Flow of pressure in water bottle
classical-mechanicsfluid dynamicspressure
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It took me quite some time to clearly understand the experiment you're describing.
Actually, pouring a full bottle in a container is a quite intriguing thing.
Consider the following starting configuration :
This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric pressure since the height of water in the bottle is higher than the level in the container.
So we should quickly get to this configuration instead :
You'll agree that along the red line, the pressure is $P_0$, so what is the pressure $P_1$ ?
Using simple hydrostatics, $ P_1 = P_0 - \rho \, g \, H$
Notice that in the picture as well as in this calculation, we consider the height $H$ to not have changed, i.e. very little water has moved out of the bottle into the container. We'll see why now.
What is now the volume of air in the bottle ?
Using the law of perfect gases $P_0 * V_0 = P_1 * V_1$, hence $$\frac{V_1-V_0}{V_0} = \frac{1}{\frac{P_0}{\rho \, g \, H} - 1} = \frac{1}{\frac{10^5}{10^3 \, 10 \, 10^{-1}} - 1} \approx 1 \% $$
For this numerical estimation I took a water height in the bottle of $10 \, cm$. The variation in volume is so small, it will be hardly noticeable !
The reason why pouring the bottle is intriguing is that it empties itself in bursts. A bubble of air gets in, and water gets out at once. But if you do it in a controlled way, you will end up in the initial configuration I described, and from that point onwards, no air can get in. The variation of volume of the air in the bottle we just obtained obviously corresponds to a volume of water that gets out of the bottle, but again, it is small, and hardly noticeable.
What if you take a longer bottle ?
gigacyan is right, something will happen after a while. Recall that I did the calculation assuming the amount of water exiting the bottle is very small, this assumption is now false. If you have a significant height of water, the pressure will be enough to push out quite a bit of water out of the bottle, in which case the pressure of the air in the bottle will go down, and the level of water in the container go up.
If you consider a very wide container, its level will stay roughly the same, but the level of water in your bottle will go down. A simple calculation leads to: $$P_0-\rho \, g \, h_{final}+\rho \, g \, (H-h_{final})=P_0$$ Hence $h_{final} = H/2$, which is the point when the low pressure in the air is able to lift the weight of the water underneath, down to the free surface.
Several interesting remarks can now be made.
To begin with, the pressure in the air keeps on dropping, $P_1 = P_0 - \rho \, g \, h_{final}$. Nothing prevents it from going to negative values, which happens when $h_{final} = \frac{P_0}{\rho \, g} = 10 \, m$. That's where this famous value of 10 meters comes from.
Now, if you think about trees, at first you may imagine they rely on capillary action to carry sap to their leaves, but that can't be the case, as the pressure drops too much after 10 vertical meters against gravity. Any presence of air would make the wood crumple under its own applied pressure.
Which means there is absolutely no air whatsoever in the sap canals of a tree (a.k.a. xylem).
The trees rely principally on another mechanism to pump up sap, known as evaporation. This easily produces (highly) negative pressures in the sap, and the actual limit to the size of a tree is the point when this pressure is small enough that a cavity of water vapor spontaneously appears in its canals, through cavitation. Pull hard enough on water, and you will create two interfaces and evaporate some of the liquid ! This cavitation pressure is around $-120 \, MPa$.
This catastrophic failure is know as embolism, and is also a bad health condition for humans (a gas bubble in a blood vessel).
Since you know the force on the bottle (roughly 200N), you would have to get an approximate area over which this force is distributed. You could try by spreading some ink on either the bottle or the weight to estimate the contact area. While this is not completely correct (the wall of the bottle does redistribute the pressure on the outside to a larger area on the inside), it should give a reasonable estimate to maybe a factor of two or so (but that's a guess of mine, of course). If you want to do it right, connect a long plastic pipe to the bottle cap and measure the height of the water column that gets squeezed out. That's a direct and fairly reliable pressure measurement. Lacking such a measurement, you could estimate from the squirt height. I am sue there are better ideas, still, but these are the easy ones that I can think of of the top of my head.
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Based on your comments I'd say you're on the right path on your own.
You already mentioned using the (ideal) gas laws. Beyond that there isn't a whole lot to consider in this problem. (assuming you want to do a cursory analysis of this situation, and not find exact pressure values or anything like that)
It's worth noting that water is approximately incompressible, so it doesn't build pressure quite like the gas does, but as long as the gas is pressurized the water will be at that pressure (and a bit higher the deeper you go, see hydrostatic pressure).
So the basic concept is apply a weight > deform the container > reduced air volume means increased pressure in the bottle due to ideal gas law.
Your rewording of the question is a bit more complicated. The effectiveness of transferring pressure between outside and inside depends quite a bit on the material of the bottle. Depending on the material, it will deform more or less when you apply a force to it.
A weak material, like a balloon very easily deforms, so it's very easy to apply a force to the outside and increase the pressure inside. Something like a steel container doesn't work as well. When you apply a force it does deform a bit, which would increase the pressure; but it barely deforms, so the volume barely changes and pressure barely increases.