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Situation 1: Consider an uncapped partially filled bottle of water. What would be the pressure of air inside the bottle?

My answer: It is open to atmosphere. So it would be atmospheric pressure $P_0$

Situation 2: Now cap the above bottle. What would be the pressure of the air in the bottle?

My answer 1: It would still be $P_0$ because $PV=nRT$ and neither the temperature nor the no. of moles of gas in the container, nor the volume of gas available for these $n$ moles of gas changed. So pressure wouldn't change.

Experiment: Try pricking a small hole at the bottom of the bottle and water doesn't come out of the orifice when the bottle is capped but when you open the cap of the bottle, the water starts coming out. (I observed this when I tried this with a pen cap which has a hole at the bottom). Which implies that when the bottle is capped, the pressure of air inside the bottle is $<P_0$.

So my answer 1 is wrong. I correct my answer 1 to $<P_0$ And I explain this by saying, the atmospheric pressure $P_0$ is due to the weight of entire column of air from from the beginning of the atmosphere till the surface$(= \rho_{air}.g.H_{atmosphere})$. But when I capped the bottle the pressure will only be due to air column in the bottle.$(= \rho_{air}.g.h_{airinbottle})$

Which of my answers is wrong? What am I missing? What exactly changed among $V, n, T$ in which accounts for change in pressure in the equation $PV = nRT$? If I want to calculate this pressure of air inside a capped bottle, how to go about it?

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closed as off-topic by John Rennie, ZeroTheHero, sammy gerbil, Kyle Kanos, Yashas Mar 28 '17 at 12:40

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As soon as you try to take fluid out of the sealed bottle, consider what happens to the air pressure.

Removing water will require the volume of air to increase. This increased air volume will lower the pressure of that air sealed in the bottle. This pressure difference will help to keep the water sucked in, as long as the surface tension and similar effects prevent an air bubble from rising through the hole.

If the hole is big enough this is generally what happens. It builds negative pressure, then sends a bubble through to equalize it, builds up negative pressure, sends another bubble, etc. Same thing as trying to pour a whole bottle of water/pop out at once. You're better off to only half pour it so that the air can freely flow into the bottle as the liquid flows out.

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  • $\begingroup$ I don't think this addresses my question. You are explaining about, "As soon as you try to take fluid out of the sealed bottle, consider what happens to the air pressure." But the situation I am referring to is, 'consider an open bottle with some water in it, what is the pressure in it. Now cap the bottle, what will be the pressure inside'. Are these two pressures same or different? If different, why? If we consider air to be ideal gas, how would PV=nRT be in both cases? $\endgroup$ – claws Mar 28 '17 at 1:40
  • $\begingroup$ @claws You yourself said in your evaluation they should be the same. You then explained an experiment that "disproved" what you thought. What I'm saying is in your question you say " What am I missing? What exactly changed among V,n,T in which accounts for change in pressure in the equation PV=nRT? " and that is what I am addressing. I'm saying the attempt to increase volume of air in the container (water pouring out) will decrease the pressure. The system resists this if the hole is small and the forces at the hole can keep up with the force of water pressure (proportional to area). $\endgroup$ – JMac Mar 28 '17 at 1:45
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I think JMac is getting to the heart of the issue. As a side note, the air in the bottle will not necessarily be exactly atmospheric pressure. Any liquid forms an equilibrium with its vapor phase as the surface slowly evaporates. The air just above your water will be more populated with gaseous water molecules than the air outside the bottle (assuming there aren't any exposed bodies of water outside the bottle). Thus, when the bottle is closed, one may expect the internal pressure to be slightly higher than the external atmospheric pressure, because the vapor pressure of water contributes to the total internal pressure. The vapor pressure of water at room temperature is around 23.8 mmHg (iirc). This is small relative to 760 mmHg, but I figured you may want to be aware of this relevant phenomenon.

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  • $\begingroup$ Even that 23.8 mmHg wont be fully relevant. There will already be a partial vapour pressure in the air regardless, depending on how long you have it in the bottle before sealing it should be at near-equilibrium near the surface. So when you close it up the vapour pressure shouldn't be much greater than atmospheric if any. It's already at approximate equilibrium for the conditions. $\endgroup$ – JMac Mar 27 '17 at 16:56
  • $\begingroup$ I agree it is negligible. I just had a feeling OP might want to be aware of this, perhaps for other problems/thought experiments where it may play a role. $\endgroup$ – electronpusher Mar 27 '17 at 22:00

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