# Phase changes in relation to pressure changes

## Main Question or Discussion Point

I understand the basic connections between vapor pressure of substances and how this effects their boiling points. However, I am a bit hazy on applying it to some real life situations so I want to give some examples to check where I might be going wrong in my thinking.

1. I have 70F water in a beaker. I put it into a tube where the pressure is at 0.739 in.Hg so the water is boiling and has a temperature/pressure relationship. I now reduce the pressure in the tube so it is at .180 in Hg corresponding to the boiling point of 32F of water. Since the pressure in the tube is reduced, what changes here? I mean, the water temperature of 70F is well over 32F so the water would still be boiling so would this just all change to vapor or would the temperature decrease? I am also confused as to what point the boiling would stop if a decrease occurs.

2. Let's say I have a superheated vapor of water where the absolute pressure is .180 in.Hg and the temperature is let's say 40F (so 8 degrees F above boiling point). The vapor then goes through a .180 in.Hg tube and into a piston which compresses and decreases the volume of the vapor. Since the molecules are being compressed, they increase in their KE which results in an increase in their pressure and an increase in temperature of the superheated vapor as a result correct? Thus, if they exited in a different tube, the pressure/temperature would be higher but it would still remain as a superheated vapor.

3. I have a subcooled liquid where the temperature/pressure relationship is 125F/278 psig for a substance so the substance is a liquid-vapor mixture. I now take away some thermal energy so the liquid-vapor mixture becomes a subcooled liquid. The subcooled liquid now immediately is forced into another tube where the pressure is 69 psig. The temperature/pressure relationship is 69 psig/-41F. In terms of pressure/temperature relationship, since the temperature of the subcooled liquid is 110F and is above the boiling point of -41F, the liquid should start boiling. However, I also know that in general, reducing the pressure will result in molecules moving slower. Thus, I am having issues connecting these concepts in this context. In any case, if the subcooled liquid boils, how much is it boiling? Does it boil to -41.0F and then stop since this is at equilibrium?

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mfb
Mentor
1. The water will keep boiling until
a) the pressure went up (from the additional vapor)
and/or
b) the temperature went down (as boiling needs energy)
or
c) no liquid water is left

Since the molecules are being compressed, they increase in their KE which results in an increase in their pressure
That is not the dominant effect, the increased density is more important. Some fraction might condensate, if the boiling point rises quicker than the temperature.
Thus, if they exited in a different tube, the pressure/temperature would be higher but it would still remain as a superheated vapor.
If you keep that pressure: Probably.

I don't think you mean superheated/subcooled in the way you use the words.

1. The water will keep boiling until
a) the pressure went up (from the additional vapor)
The rest was helpful but can you clarify this part? Isn't the boiling going to reduce the temperature which will reduce the vapor pressure?

mfb
Mentor
Isn't the boiling going to reduce the temperature which will reduce the vapor pressure?
The vapor pressure in equilibrium, which you do not have. Until the real pressure is equal to the vapor pressure, the water boils, and that vapor can increase the (real) pressure.

The vapor pressure in equilibrium, which you do not have. Until the real pressure is equal to the vapor pressure, the water boils, and that vapor can increase the (real) pressure.
So pretty much, the liquid will keep boiling. In doing so, the vapor pressure will increase since more of the molecules will spring up into the closed container and increase the vapor pressure. At the same time though, the temperature of the liquid will be reduced due to a reduction in the KE since molecules are leaving. Since it is a closed container, at some point, the amount of vapor leaving will match the amount of vapor coming back and turning into water. This however will not match a temperature/pressure relationship since the pressure might be higher than the temperature/pressure relationship for that temperature. Is this correct?

So, at this point, if I wanted to reduce the pressure to match the temperature of the liquid, would I manually just open a valve slightly (for example) until I am satisfied?

mfb
Mentor
So pretty much, the liquid will keep boiling. In doing so, the vapor pressure will increase since more of the molecules will spring up into the closed container and increase the vapor pressure. At the same time though, the temperature of the liquid will be reduced due to a reduction in the KE since molecules are leaving. Since it is a closed container, at some point, the amount of vapor leaving will match the amount of vapor coming back and turning into water.
Right
This however will not match a temperature/pressure relationship since the pressure might be higher than the temperature/pressure relationship for that temperature. Is this correct?
It will happen at the phase transition curve. In equilibrium, this is the only way to get two phases at the same time.