Vacuum by condensation causes water to boil?

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SUMMARY

The discussion centers on the effects of creating a vacuum in a closed container with water and the implications for boiling. When steam condenses in a closed system, it creates a vacuum, but this does not necessarily lead to boiling unless the temperature of the water reaches its corresponding vapor pressure. The equilibrium vapor pressure of water is temperature-dependent, with 100°C corresponding to one atmosphere. The conversation also touches on practical experiments involving canning jars and the use of salt solutions in vacuum systems, specifically referencing the Icyball refrigeration method.

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askingask
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So basically if I have a closed container with a valve, and inside the container there is water. Now i heat the container and boil the water. The valve is open so steam escapes form there. I now close the valve and cool the container causing the steam to condense inside. Inside the container is a vacuum now.

Will this vacuum cause the water to boil? Or will only so much steam have condensed, to create the exact pressure which matches the evaporation point of the water to that of the ambient temperature?
 
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askingask said:
... Inside the container is a vacuum now.

Will this vacuum cause the water to boil?
It would, but then there will no longer be a vacuum. The temperature of the water in the container, will decide the vapour pressure of the steam in the container. The contents of the container are then in equilibrium.

askingask said:
Or will only so much steam have condensed, to create the exact pressure which matches the evaporation point of the water to that of the ambient temperature?
The "evaporation point" is confusing because evaporation and condensation take place continuously. Evaporation is equal to condensation when at equilibrium. The equilibrium vapour pressure is a function of temperature. At 100°C the vapour pressure of water is one atmosphere, so it boils. It is not zero at 0°C, it is 611 Pa.
Code:
'  Temp       pp_H2O        Atm
'   °C        pascal         %
'  100.0     101,331.      100.0
'   95.0      84,524.       83.4
'   90.0      70,113.       69.2
'   85.0      57,811.       57.1
'   80.0      47,368.       46.7
'   75.0      38,556.       38.1
'   70.0      31,168.       30.8
'   65.0      25,015.       24.7
'   60.0      19,926.       19.7
'   55.0      15,746.       15.5
'   50.0      12,340.       12.2
'   45.0       9,586.        9.5
'   40.0       7,378.        7.3
'   35.0       5,624.        5.6
'   30.0       4,243.        4.2
'   25.0       3,167.        3.1
'   20.0       2,337.        2.3
'   15.0       1,704.        1.7
'   10.0       1,227.        1.2
'    5.0         872.        0.9
'    0.0         611.        0.6
'   -5.0         421.        0.4
'  -10.0         286.        0.3
'  -15.0         191.        0.2
'  -20.0         125.        0.1
'  -25.0          81.        0.1
'  -30.0          51.        0.1
'  -35.0          31.        0.0
'  -40.0          19.        0.0
'  -45.0          11.        0.0
'  -50.0           6.        0.0
 
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askingask said:
So basically if I have a closed container with a valve, and inside the container there is water. Now i heat the container and boil the water. The valve is open so steam escapes form there. I now close the valve and cool the container causing the steam to condense inside. Inside the container is a vacuum now.

Will this vacuum cause the water to boil? Or will only so much steam have condensed, to create the exact pressure which matches the evaporation point of the water to that of the ambient temperature?
You can do this experiment with a canning jar filled halfway with water and heated to boiling in a microwave. Just make sure the lid is loose enough to let the steam out, and new enough to maintain a seal. The lid acts like a one way check valve, letting out steam and air while heating. But once the heat is removed, the lid should seal the jar. To work, all the air needs to be pushed out, so I would recommend heating to a violent boil.
If the experiment is successful, it will boil for as long as it takes the water in the jar to reach room temperature. And then if you put it in the refrigerator, or put an ice cube on the lid, it will start boiling again! If the experiment fails, which is when, post heating, air gets into the jar, then it won't do this, and you'll just have a very boring jar full of hot water.

You can read up on this effect in food canning forums.
Example:
I just canned up some ground turkey. It's been 3 hours and the jars are still bubbling!
...
ref: https://permies.com/t/138925/Jars-bubbling-hours

I learned of this effect about a year ago, when I delved into the art of food canning.
 
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OmCheeto said:
To work, all the air needs to be pushed out, so I would recommend heating to a violent boil.
Yes, boil it vigorously.
The molecular weight of water is 18 g/mole, while air averages 29 g/mole.
Water vapour is lighter than air, so it rises. You must therefore stir the gasses to progressively dilute and remove the air from the bottom of the ullage space.
 
OmCheeto said:
If the experiment is successful, it will boil for as long as it takes the water in the jar to reach room temperature.
Alright that’s really interesting thanks for sharing.


Baluncore said:
Yes, boil it vigorously.
The molecular weight of water is 18 g/mole, while air averages 29 g/mole.
Water vapour is lighter than air, so it rises. You must therefore stir the gasses to progressively dilute and remove the air from the bottom of the ullage space.
So if I create a vacuum using this method with water and place a salt solution in another container, which is connected ( both containers connected being evacuated), could I create something similar to the Icyball?
https://en.wikipedia.org/wiki/Icyball

I know that salt is really bad at absorbing moisture, but in the case of a vacuum how strongly would it absorb the water vapor formed?
 
askingask said:
Alright that’s really interesting thanks for sharing.



So if I create a vacuum using this method with water and place a salt solution in another container, which is connected ( both containers connected being evacuated), could I create something similar to the Icyball?
https://en.wikipedia.org/wiki/Icyball

I know that salt is really bad at absorbing moisture, but in the case of a vacuum how strongly would it absorb the water vapor formed?
That all depends on what you mean by salt. Lithium Bromide salts are commonly used in absorption refrigeration devices, which is what your Icyball is. As to whether or not Sodium Chloride would work, I can't say for sure, but I've never heard of such a device.
 
OmCheeto said:
That all depends on what you mean by salt. Lithium Bromide salts are commonly used in absorption refrigeration devices, which is what your Icyball is. As to whether or not Sodium Chloride would work, I can't say for sure, but I've never heard of such a device.
I know about lithium bromide. Thing is, sodium chloride is known to be very ineffective when it comes to being a desiccant. My question is: does that change if its operating in a vacuum, where the water evaporates and the system fills up with steam?
 
askingask said:
My question is: does that change if its operating in a vacuum, where the water evaporates and the system fills up with steam?
NaCl is a very poor desiccant.

The best way to remove water vapour, is physically with a vacuum pump, or in a chemical reaction with CaO, but that can get angry when it expands, and is difficult to reverse without an 825°C kiln.

The cheapest way would be with silica gel, sold in the supermarket as cat litter.
 
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