Suppose I put water in a closed vessel and heat it, so the vessel becomes pressurized above ambient pressure. Now, there should be liquid and gaseous water inside the vessel. Since it's under pressure, when I open the vessel at the bottom, the liquid starts to flow out. If I wait until all liquid is expelled, and close the opening before the water vapor exits too, what state do I have in the vessel? Specifically:

  • Releasing pressure from a system usually means it cools down. What does that mean for the vessel's inside? If I want the vessel to stay at the temperature it was, do I have to reheat (if we ignore ordinary heat dissipation for now)?
  • The pressure inside the vessel now is lower, does that mean water vapor will condense and the system moves toward a new equilibrium? If I keep the temperature constant, will it be the same pressure as before?
  • Does it matter for this question to what extent the vessel is filled at the beginning? If I understood vapor pressure correctly, the contents will reach an equilibrium of liquid and gaseous such that the pressure reaches a point which is solely dependent on temperature.
  • Does it matter for the purpose of this question how quickly I expel the liquid phase?
  • I'm not sure what influence the ambient pressure has on this question.

If you need sample values, assume 100 kPa (i. e., normal atmosphere) for the ambient pressure, 200 °C for the water and 1.555 MPa inside the vessel when I open the valve.

  • $\begingroup$ In case the context is helpful: I'm wondering if the remaining pressure inside a batch operated reactor can be used to both expel the reaction product and then do some useful work when the reaction is complete, instead of just venting or cooling down the reactor. $\endgroup$ Commented Dec 2, 2012 at 18:41
  • $\begingroup$ Vapor is more considered the liquid phase than the gaseous phase. Vapor by itself in a container should (partially) condense into water. $\endgroup$
    – resgh
    Commented Dec 3, 2012 at 11:04

2 Answers 2


It is not clear from your question whether the gas in the bottle starts out as air, or the only substance in the bottle is water, either in liquid or gas form. I'll assume the latter since it's easier to answer.

Once the bottle is closed, ambient temperature doesn't matter. The pressure of the gas part in the bottle will be stricly a function of temperature, with that function being solely a property of water. You can find what that function is in what is commonly referred to as a steam table.

As you heat the bottle, and presumably everything inside gets to the same temperature, the vapor pressure increases. This causes a little more of the liquid to boil and thereby make less liquid and more gas. Eventually for any one temperature, a new equilibrium is reached where the gas is at the pressure listed in the steam table for that temperature.

If you let the system reach equillibrium at 200°C, for example, then open a valve at the bottom, water at the bottom will be forced out under pressure. As the volume of water in the bottle decreases, the volume available to the gas increases, which decreases its pressure. The liquid that was at equillibrium now no longer is, since its pressure is reduced but its temperature (for the short term) remains the same. This will cause the liquid to boil and make more gas. This cools the liquid, and if done slowly enough the liquid will track the ever decreasing boiling temperature for the ever decreasing pressure.

Meanwhile, the gas is expanding, so it will cool. It won't condense because there is no place for the heat to go. We are making the assumption that the bottle is insulated at this point. Boiling liquid will keep the gas "topped off" at whatever temperature it is according to the steam table.

Once all the liquid is gone, the steam will still be at pressure above ambient. If the valve stays open, the steam will vent until its pressure is the same as ambient. If the bottle is truly insulated, nothing more should happen. If heat is lost thru the bottle, then the temperature will decrease and the pressure of the steam decrease, again according to the steam table. Outside air is then sucked into the bottle. If this air is at 20°C, for example, it will significantly cool the steam, which will cause much of it to condense, which creates lower pressure, which sucks in even more cool air, etc. The proceeds until most of the water is condensed to liquid, with the only remaining water gas being whatever maximum partial pressure of water that air at that temperature and pressure can support.

If the valve is opened so that the 200°C water is expelled "quickly", then it gets a lot more complicated because there is not enough time for the system to track equillibrium as pressure is abruptly release. In that case, I think all we can say is that a bunch of super-heated water will be violently released and quickly undergo decompression, which will cause some of it to boil quickly, making a lot of steam, with the left over liquid being at boiling temperature for ambient pressure, which would be 100°C. After the water is gone from the bottle, a rush of steam will come out, condensing in the (relatively) cold ambient air and adding to the already large saturated water vapor cloud. Once enough steam is expelled so that ambient pressure is reached the remaining steam acts as above.

  • $\begingroup$ "It won't condense because there is no place for the heat to go." -- what exactly do you mean here? $\endgroup$ Commented Jul 12, 2014 at 4:57
  • $\begingroup$ Also, I'd add that, in the quick expelling case, if the valve is closed once all the water is expelled, the remaining gas will expand to full container volume (not much bigger than initial volume). Work effected here will reduce overall enthalpy, thus decreasing temperature (and pressure). Correct? $\endgroup$ Commented Jul 12, 2014 at 5:13

I believe that one can look at this another way. Assuming all materials "strong enough" and perfectly insulated, then imagine two containers separated by a capillary tube with a stopcock in it. Now imagine that the liquid is totally contained in the lower container and the upper container holds all the vapor and no liquid.

You now close the valve and discard the lower container. What happens in the upper container? In particular how does the temperature and pressure change? Indeed one should be able to answer all your questions easily.

  • 2
    $\begingroup$ I don't think it's quite the same thing -- in the setup as described by the OP, the gas will expand to also occupy the space formerly occupied by the liquid. $\endgroup$ Commented Jan 2, 2013 at 5:19

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