# boiling/condensing water in closed, expandable system

Assume you have a closed, expandable container (such as a balloon or piston) filled with water (no air). The water is initially at standard conditions (25C and 1 atm). The water is then heated until it begins to boil, creating water vapor within the system (assume no compression in container, so it is always at 1 atm). The system is then allowed cool back down to room temperature and pressure. Will all of the water vapor condense, or will some of it remain as a vapor?

• What do you think will happen? And why (from a physics standpoint)? – Jon Custer Nov 13 '17 at 21:02
• Note: If you try this experiment at home, you'll need to account for the fact that tap water isn't pure water. Among other things, it contains small quantities of dissolved gasses. – Solomon Slow Nov 13 '17 at 21:04
• I assume at least most of it will condense, but since water has a non-zero vapor pressure, then some amount of water will remain as a vapor. – KoldBeans Nov 13 '17 at 21:06
• Indeed, assume the water is degassed. Perhaps the water was boiled first to remove the dissolved water. – KoldBeans Nov 13 '17 at 21:07
• On a related note, if the degassed water was allowed to sit for a very long time in a closed, freely expanding system, would a water vapor bubble form? – KoldBeans Nov 13 '17 at 21:13

## 2 Answers

Let's assume that there is no dissolved air within the water to begin with. The air pressure outside the balloon initially is 1 atm., and the pressure within the contained water initially is slightly higher than 1 atm., because the balloon is stretched slightly (and exerts a radial compressive stress on the water). So in the initial state of the system, the water is in a "compressed liquid" equilibrium state at slightly higher than 1 atm. (and no water vapor is present).

After the water is heated to a temperature at a little over 100 C (such that its vapor pressure at this temperature matches the initial pressure exerted by the balloon), bubbles containing water vapor begin forming inside the balloon, and the balloon rubber membrane begins to stretch. Boiling is occurring. The stretching of the rubber membrane causes the pressure exerted on the balloon contents to increase a little, and the boiling temperature must also rise a little to match the new membrane compressive stress (normal to the membrane).

Now, the addition of heat is discontinued. The temperature inside the membrane decreases, and, as it does, the vapor in the balloon begins to condense. The amount that the balloon is stretched decreases, and, this allows the pressure on the balloon contents to decrease a little. Once the temperature drops below a value close to 100 C, all the vapor will have condensed, and the pressure on the balloon contents (exerted by the surrounding rubber membrane) will again be a little above one atm. There will be no vapor present, and the contents will be fully liquid. Further cooling of the liquid will bring down its temperature to room temperature again, but with no vapor present. The final state will be identical to the initial state before any heating was introduced, and we again have a "compressed liquid" equilibrium state.

Even if dissolved air were initially present within the water, all the air would have redissolved in the liquid water at this point.

For a closed container that is initially full of liquid and is set to keep a constant pressure, the equilibrium depends on two factors: the pressure $P$ of the container, and the temperature $T$ of the liquid. The liquid has a vapor pressure $P_v(T)$, a solid-gas transition pressure $P_g(T)$, a solid-liquid transition pressure $P_s(T)$, a triple point $(P_t,T_t)$, a critical pressure $P_c$, and a critical temperature $T_c$.

There are a few scenarios. Let's start with things that aren't water (i.e. those materials that contract when frozen, so that $P_s(T)$ slopes upward):

• $T<T_t$:
• $P>P_g(T)$: Final state is completely solid.
• $P<P_g(T)$: Final state is solid-vapor mixture. Lower $P/P_g$ means more vapor. At $P=0$, final state is entirely vapor.
• $T_t\leq T<T_c$, $P\neq P_t$:
• $P>P_s(T)$: Final state is completely solid.
• $P=P_s(T)$: Final state is solid-liquid mixture. Unstable.
• $P_v(T)<P<P_s(T)$: Final state is completely liquid.
• $P<P_v(T)$: Final state is liquid-vapor mixture. Lower $P/P_v$ means more vapor. At $P=0$, final state is entirely vapor.
• $T=T_t$,$P=P_t$: Final state is solid-liquid-gas mixture. Unstable.
• $T>T_c$:
• $P<P_c$: Final state is entirely vapor.
• $P>P_c$: Final state is entirely supercritical fluid.

Water is different. It expands when freezing, so $P_s(T)$ slopes downward. This changes our final states a bit.

• $T<T_t$:
• $P>P_s(T)$: Final state is completely liquid.
• $P=P_s(T)$: Final state is liquid-solid mixture. Unstable.
• $P_g(T)<P<P_s(T)$: Final state is completely solid.
• $P<P_g(T)$: Final state is solid-vapor mixture. Lower $P/P_g$ means more vapor. At $P=0$, final state is entirely vapor.
• $T_t\leq T<T_c$, $P\neq P_t$:
• $P>P_v(T)$: Final state is completely liquid.
• $P<P_v(T)$: Final state is liquid-vapor mixture. Lower $P/P_v$ means more vapor. At $P=0$, final state is entirely vapor.
• $T=T_t$,$P=P_t$: Final state is solid-liquid-gas mixture. Unstable.
• $T>T_c$:
• $P<P_c$: Final state is entirely vapor.
• $P>P_c$: Final state is entirely supercritical fluid.

The equilibrium state of the liquid in the container is a state function, and as such, depends only on the final temperature and pressure of the system. Since the pressure is always kept constant, and the system is heated, then cooled back down to its initial temperature, both of these state variables are the same as in the initial state. So really, equally heating and cooling the system doesn't do anything at all to the system, from an equilibrium perspective.

To address the problem at hand, room temperature is above $T_t=.01$ Celsius for water, and $P_v$ at room temperature is $23.8$ torr, which is less than atmospheric pressure, so the system begins and ends fully liquid. In the middle, if the water is heated to 100 Celsius, $P_v$ exceeds 1 atm, so a liquid-vapor mixture forms. The more you heat it, the less liquid remains. But once the container is allowed to cool, it returns to its initial state, which was entirely liquid.

• probably_someone, it would be a good idea to do some research on pure-component vapor-liquid equilibrium. Your current concept of vapor pressure is substantially different from what happens in nature. – David White Nov 14 '17 at 1:23
• @Pieter I didn't account for containers to be able to exert a significant amount of pressure - I had erroneously made the assumption that the container was in a vacuum and could not itself exert a ton of pressure on the liquid. Edited to address all possible situations. – probably_someone Nov 14 '17 at 4:39
• @DavidWhite Edited to fix numerous problems. – probably_someone Nov 14 '17 at 4:39