Vapor pressure -- How does water still boil at 100°C in an open pot?

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  • #1
pisluca99
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Vapor pressure is the pressure of the vapor when it is in equilibrium with its liquid. This only happens when the container where the liquid is present is closed. Indeed, when the container is open, this liquid-vapour equilibrium is never reached, because the partial pressure of the vapor (at a given temperature) never reaches the vapor pressure. That said, how does water still boil at 100°C in an open pot? For what has been said, if the pot is open at 100 °C, the partial pressure of the vapor does not reach the vapor pressure (and therefore the atmospheric pressure) and boiling should not take place. Can you clarify? Thanks
 

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  • #2
Tom.G
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Please tell us where you found that statement, it sounds like it is talking about a slightly different condition.

In any case, boiling starts when the vapor pressure exceeds the ambient pressure of the 'immediate surroundings'. The 'immediate surroundings' of course include the weight of the water above the spot in question.

As the temperature is increased, more and more water molecules gain enough energy to separate from the remaining liquid, forming bubbles. The bubbles, being less dense than the water, then rise to the top of the water.

Since the contents of the bubbles have higher energy than the liquid molecules, they still have enough energy to escape the water surface, just as they did when they formed. The end result is the water 'boils away' as steam.

When water is not quite as hot as talked about above, it is said to 'simmer'. This is a condition where small bubbles form but when they rise thru the surrounding water, they cool enough that they condense back into liquid.

Hope this helps!
Tom
 
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  • #3
Hornbein
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if the pot is open at 100 °C, the partial pressure of the vapor does not reach the vapor pressure (and therefore the atmospheric pressure)

The vapor pressure is not the same as the atmospheric pressure. The H2O vapor pressure is a part of the atmospheric pressure, which is why you correctly call it a partial pressure.
 
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  • #4
256bits
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Vapor pressure is the pressure of the vapor when it is in equilibrium with its liquid. This only happens when the container where the liquid is present is closed. Indeed, when the container is open, this liquid-vapour equilibrium is never reached, because the partial pressure of the vapor (at a given temperature) never reaches the vapor pressure
Not exactly correct about the open container.
It is a more or less general statement that seems to be completely true but is not quite actually, considering that from our everyday experience water does evaporate. Sometimes we should be wary of such general statements often repeated, and re-analyze if it considers all conditions.

The air has a percentage of water vapour within it.
At any given day, we listen to the weather report and most of the time they give a temperature and the relative humidity. The relative humidity is the ratio of actual amount of water vapour within the air to that which could be within the air at that temperature. On a day with 100% relative humidity ( which doesn't happen all that often ), the water vapour is at the saturation pressure, meaning that the liquid-vapour equilibrium is reached ( as you have put it ) for any container, or any body of water, open to the atmosphere, where the body of water is at the same temperature as the air.

Dew will form on surfaces when the temperature of the air falls near or below the saturation temperature near the surface. So even if completely open to the 'rest' of the atmosphere, condensation onto a surface is an example of the vapour-liquid moving in the opposite direction - ie from the vapour state to the liquid state.
 
  • #5
Tom.G
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Just adding a bit to the above post by @256bits:

And when the temperature of the air falls well below the saturation temperature we get fog, or a cloud if it is well above the local ground level.
 
  • #6
pisluca99
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Just adding a bit to the above post by @256bits:

And when the temperature of the air falls well below the saturation temperature we get fog, or a cloud if it is well above the local ground level.

Thanks a lot for your replies, but I still have my doubt.
Originally, the vapor pressure experiment was performed in a closed container: in this case it is observed how, over time and at a constant temperature, the partial pressure of the vapor on the liquid increases progressively, until it reaches a constant value, which would be the saturated vapor pressure. Once this condition is reached, the number of water particles that condense is equal to the number of water particles that vaporize and there is no more mass change in the liquid.
If we now open the container, always at the same temperature, it happens that the water molecules in the vapor phase tend to escape into the surrounding environment, and the previously established liquid-vapor equilibrium no longer exists, so much so that the mass of The water evaporates until it disappears completely: this should mean that, always at that specific temperature, the saturated vapor pressure is not reached, as in the case of a closed container.
Therefore, at a certain temperature, if the vessel is open, a lower vapor pressure will be reached than with saturated steam. Consequently, this also happens at 100 °C and the water should not boil, unless the container is closed in such a way that the saturated vapor pressure, as well as atmospheric pressure, can be reached. What am I doing wrong?
Sorry for the boredom.
 
  • #8
pisluca99
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Boiling means that bubbles consisting of water vapor can form in the bulk of liquid water and rise. That happens at the boiling point, where the saturated vapor pressure is equal to atmospheric pressure.
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/vappre.html#c3
but in the link it says that water boils when the saturated vapor pressure reaches atmospheric pressure. But the saturated vapor pressure is reached only when the container is closed, as shown in the image. So how does water boil if the container is open and therefore, in theory, the vapor cannot become saturated?
 
  • #10
russ_watters
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The definition is theoretical/ideal/simplified. The practical use strays from the ideal and is more complicated. There's no problem/nothing wrong with that.
 
  • #11
256bits
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but in the link it says that water boils when the saturated vapor pressure reaches atmospheric pressure. But the saturated vapor pressure is reached only when the container is closed, as shown in the image. So how does water boil if the container is open and therefore, in theory, the vapor cannot become saturated?
@Tom.G did have an explanation.
Did you not understand it?

You have to understand that the saturation vapour pressure is that for a liquid, where equals amounts of vapour escape from the liquid surface into atmosphere, and the same amount return from the atmosphere back into the liquid. Hence equilibrium.
It so happens that when in a closed container the actual vapour pressure will reach the saturation vapour pressure of the liquid at that temperature.

So when you heat water to 100C, the liquid water's saturation vapour pressure becomes that of the atmosphere, which, when the liquid temperature was below 100C, had enough pressure to not allow bubbles to form within the liquid ( since the liquid's saturation vapour pressure was below atm and there is 1 atm pressure 'pushing down' on the liquid. )
At 100C the bubbles can form anywhere within the liquid and boiling occurs.
The bubbles have a vapour pressure which is the saturation vapour pressure of liquid water at 100C.

Note the distinction I am trying to make between saturation vapour pressure of the liquid and the actual vapour pressure.
 
  • #12
Lnewqban
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But the saturated vapor pressure is reached only when the container is closed, as shown in the image. So how does water boil if the container is open and therefore, in theory, the vapor cannot become saturated?
The hot water does not know what gas is above its surface, it only feels the pressure.
You can make water in a bottle boil at ambient temperature, by pulling the surface up with your thumb.

 
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  • #13
pisluca99
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@Tom.G did have an explanation.
Did you not understand it?

You have to understand that the saturation vapour pressure is that for a liquid, where equals amounts of vapour escape from the liquid surface into atmosphere, and the same amount return from the atmosphere back into the liquid. Hence equilibrium.
It so happens that when in a closed container the actual vapour pressure will reach the saturation vapour pressure of the liquid at that temperature.

So when you heat water to 100C, the liquid water's saturation vapour pressure becomes that of the atmosphere, which, when the liquid temperature was below 100C, had enough pressure to not allow bubbles to form within the liquid ( since the liquid's saturation vapour pressure was below atm and there is 1 atm pressure 'pushing down' on the liquid. )
At 100C the bubbles can form anywhere within the liquid and boiling occurs.
The bubbles have a vapour pressure which is the saturation vapour pressure of liquid water at 100C.

Note the distinction I am trying to make between saturation vapour pressure of the liquid and the actual vapour pressure.

So you mean: at a given temperature lower than the boiling point and considering an open system, there is evaporation at the surface level of the liquid, and this generates a vapor pressure which is not the saturated vapor pressure, but it is a pressure which simply contributes to the external atmospheric pressure (this is because liquid-vapour equilibrium can never be established under these conditions, since the water molecules in the vapor phase escape into the atmosphere). At the same time, it is in the mass of liquid that the liquid-vapour equilibrium is established, and therefore the vapor pressure is generated.
If we go up to 100°C, still considering the open system, the vapor pressure in the mass of liquid will be such as to overcome the atmospheric pressure, therefore the 'bubbles' of vapor present in the mass of water are released into the atmosphere, and this it's boiling. Contextually, at 100 °C evaporation continues to take place, but it does not contribute to the increase in atmospheric pressure, given that the mass of water is reduced.

Considering instead a hermetically closed system:
- Below 100 °C, the liquid-vapour equilibrium is ALSO reached in the environment above the water, therefore the vapor pressure is manifested both in the mass of water and above.

- At 100 °C, evaporation continues to take place, and as the system is hermetically closed, the vapor pressure that is generated contributes to increasing the atmospheric pressure which presses on the mass of water. This means that boiling no longer takes place at this temperature, as the vapor pressure developed in the mass of water cannot overcome the atmospheric pressure (principle of the pressure cooker).

I hope everything is correct and I await confirmation, thanks for your patience!
 
  • #14
256bits
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At the same time, it is in the mass of liquid that the liquid-vapour equilibrium is established, and therefore the vapor pressure is generated
Not the mass of liquid but the liquid phase.
A drop of water and a large swimming pool of water at the same temperature have the same saturation vapour pressure.

If we go up to 100°C, still considering the open system, the vapor pressure in the mass of liquid will be such as to overcome the atmospheric pressure, therefore the 'bubbles' of vapor present in the mass of water are released into the atmosphere, and this it's boiling. Contextually, at 100 °C evaporation continues to take place, but it does not contribute to the increase in atmospheric pressure, given that the mass of water is reduced.
Again liquid phase is the term to be used.
Below boiling evaporation can only occur at the water surface.
With boiling, the molecules with enough energy to form vapour can form anywhere within the liquid, rise to the surface and escape to the atmosphere. One can still call this evaporation, but we normally call it boiling.
Not sure what the 'Contextually...' sentence is getting at, but for this discussion for a pot of boiling water in a stove with Psaturation = Patm, there is no change in atmospheric pressure.

Considering instead a hermetically closed system:
- Below 100 °C, the liquid-vapour equilibrium is ALSO reached in the environment above the water, therefore the vapor pressure is manifested both in the mass of water and above.
In a closed container, Psaturation increases lockstep with the temperature.
As long as there is liquid water present, the vapour pressure will be Psaturation at that liquid temperature.

- At 100 °C, evaporation continues to take place, and as the system is hermetically closed, the vapor pressure that is generated contributes to increasing the atmospheric pressure which presses on the mass of water. This means that boiling no longer takes place at this temperature, as the vapor pressure developed in the mass of water cannot overcome the atmospheric pressure (principle of the pressure cooker).
There you go - pressure cooker. And why they have a safety relief valve. If you continue adding heat to the water raising its temperature, the liquid Psaturation does increase, and so will the vapour pressure above the liquid increase lockstep to stay in equilibrium.

Give it a day or more to sink in and you should understand it quite well.
A difficult subject nontheless.
 
  • #15
pisluca99
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Not the mass of liquid but the liquid phase.
A drop of water and a large swimming pool of water at the same temperature have the same saturation vapour pressure.


Again liquid phase is the term to be used.
Below boiling evaporation can only occur at the water surface.
With boiling, the molecules with enough energy to form vapour can form anywhere within the liquid, rise to the surface and escape to the atmosphere. One can still call this evaporation, but we normally call it boiling.
Not sure what the 'Contextually...' sentence is getting at, but for this discussion for a pot of boiling water in a stove with Psaturation = Patm, there is no change in atmospheric pressure.


In a closed container, Psaturation increases lockstep with the temperature.
As long as there is liquid water present, the vapour pressure will be Psaturation at that liquid temperature.


There you go - pressure cooker. And why they have a safety relief valve. If you continue adding heat to the water raising its temperature, the liquid Psaturation does increase, and so will the vapour pressure above the liquid increase lockstep to stay in equilibrium.

Give it a day or more to sink in and you should understand it quite well.
A difficult subject nontheless.


yes, by 'mass of water' I meant any quantity of water contained in a container. It is clear that any quantity of water always has the same vapor pressure at a constant temperature.

I might seem repetitive, but what doesn't convince me and which I probably have a hard time imagining is the situation in which the container is open.
By definition, boiling occurs when the *SATURATED* vapor pressure equals atmospheric pressure. But if the container is open (and excluding cases in which the humidity is 100%) the vapor can never become saturated, therefore the vapor pressure will be lower than in the case of saturation. All this is also true at 100 degrees: no saturated vapor pressure, no boiling. You also wrote before that, only in a closed container, the actual vapor pressure reaches saturation vapor pressure. So, in an open container this doesn't happen: no saturate vapor pressure at 100° C, no boiling. I don't understand what is incorrect. I probably have to distinguish what happens on the surface from what happens inside the water?
 
  • #16
Lnewqban
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I might seem repetitive, but what doesn't convince me and which I probably have a hard time imagining is the situation in which the container is open.
By definition, boiling occurs when the *SATURATED* vapor pressure equals atmospheric pressure. But if the container is open (and excluding cases in which the humidity is 100%) the vapor can never become saturated, therefore the vapor pressure will be lower than in the case of saturation.

In a closed container, boiling stops when saturation pressure is reached, as long as the liquid does not receive additional thermal energy.

In an open container, boiling continues until all liquid is depleted, as long as it keeps receiving additional thermal energy.

 
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  • #17
jbriggs444
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but in the link it says that water boils when the saturated vapor pressure reaches atmospheric pressure.
Saturated vapor pressure is a function of temperature. It is a number that can be computed and measured for ideal circumstances. i.e. closed pot.

Water boils when the temperature within the bulk of the water reaches the temperature at which this saturated vapor pressure would be equal to atmospheric pressure. Technically, the relationship is between temperature at a point and the pressure within the fluid at the same point where temperature is measured. If the two are equal then the fluid at that point is "at the boiling point".

As has been explained, when the fluid is above the boiling point, small bubbles can expand. [Too small and surface tension can make tinier bubbles deflate despite local saturated vapor pressure exceeding local pressure. Hence the notion of "nucleation points", "superheated water" and water in a clean glass sometimes spontaneously boiling when removed from the microwave]

It does not matter whether the kettle is sealed or open at the top. It does not matter whether the surface of the fluid is exposed to dry air or to one atmosphere of saturated steam. It is the temperature and pressure within the bulk of the fluid that matters.

With dry air, one has evaporation on the surface. So one does not have equilibrium. One needs a heat source. Like a stove. Or a microwave. Or a Bunsen burner. Water heated on a stove will normally have a temperature gradient. Hotter near the bottom, cooler near the top. Boiling tends to initiate at the hotter, deeper portions of the pot where the temperature first exceeds the boiling point at that depth.
 
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  • #18
pisluca99
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With dry air, one has evaporation on the surface. So one does not have equilibrium.

this is exactly the point: equilibrium is not established in an open container, therefore vapor pressure is not generated, because vapor pressure is the pressure that is observed when liquid-vapour equilibrium is established. If the vapor pressure is not generated, it will never reach the external pressure for boiling, because all the evaporated water molecules escape into the surrounding 'open' atmosphere!

It is a different matter if the equilibrium is not established on the surface of the liquid, but is established inside the liquid, inside the mass of water. Maybe that's what I didn't understand?
 
  • #19
jbriggs444
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this is exactly the point: equilibrium is not established in an open container, therefore vapor pressure is not generated, because vapor pressure is the pressure that is observed when liquid-vapour equilibrium is established. If the vapor pressure is not generated, it will never reach the external pressure for boiling, because all the evaporated water molecules escape into the surrounding 'open' atmosphere!

It is a different matter if the equilibrium is not established on the surface of the liquid, but is established inside the liquid, inside the mass of water. Maybe that's what I didn't understand?
My point is that the "saturated vapor pressure" that is used in the definition of boiling point is not the vapor pressure that actually exists anywhere around the fluid that is possibly boiling. Instead, it is a defined pressure associated with a temperature and a particular fluid.

If you have water and you have a temperature of 100 degrees then the associated "saturated vapor pressure" is one atmosphere (approximately 100 kpa). We can measure this with a closed pot and verify that it is so.

With this value in hand and written down in a table somewhere, we apply the definition of "boiling temperature" and find that for water under atmospheric pressure, the boiling temperature is 100 degrees.

We observe water under these conditions and see that small water vapor bubbles in the fluid are stable, neither expanding nor contracting. We observe water under higher temperatures and observe that small water vapor bubbles expand and float to the surface. More rapidly at higher temperatures such as a "rolling boil".

We also notice that water on a stove actually does boil. So any reasoning that it cannot boil must be mistaken somehow.
 
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  • #20
pisluca99
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My point is that the "saturated vapor pressure" that is used in the definition of boiling point is not the vapor pressure that actually exists anywhere around the fluid that is possibly boiling. Instead, it is a defined pressure associated with a temperature and a particular fluid.

If you have water and you have a temperature of 100 degrees then the associated "saturated vapor pressure" is one atmosphere (approximately 100 kpa). We can measure this with a closed pot and verify that it is so.

With this value in hand and written down in a table somewhere, we apply the definition of "boiling temperature" and find that for water under atmospheric pressure, the boiling temperature is 100 degrees.

We observe water under these conditions and see that small water vapor bubbles in the fluid are stable, neither expanding nor contracting. We observe water under higher temperatures and observe that small water vapor bubbles expand and float to the surface. More rapidly at higher temperatures such as a "rolling boil".

We also notice that water on a stove actually does boil. So any reasoning that it cannot boil must be mistaken somehow.
Maybe I understood: the vapor pressure is always the same for a fluid, at a fixed temperature, regardless of whether the liquid-vapour equilibrium is formed. The fact that liquid-vapour equilibrium can be formed in a closed environment simply allows us to obtain a particular condition to be able to measure this vapor pressure. But boiling occurs regardless of whether equilibrium is formed or not.
Let me know if this Is correct.
 
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  • #21
Lnewqban
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As your intuition correctly noted, the volume of the steam "cavity" tends to infinite for an open container, reason for which no molecules remain close enough to the interphase as to jump back into the liquid (reason for which the liquid eventually disappears from the open container).

Nevertheless, the liquid remains pressurized at 1 atmosphere.
Any formation of steam still must "fight" against that pressure (which remains constant).

This is all about the kinetic energy that the molecules of liquid and vapor have to jump back and forth through the liquid-vapor interphase.

We have only two practical ways to measure that molecular energy: temperature and pressure (at constant volume or closed fixed walls container).
Vapor pressure is an indirect value that is proportional to the internal kinetic energy of the vaporizing liquid, nothing else.

Rather than a cause, it is a consequence of the heat that has been absorbed or yield by the substance, which is what modify the thermodynamic characteristics.

For our closed container:
More external thermal energy being transferred into it increases the kinetic energy of each molecule of liquid: measured temperature increases.

As a consequence, higher number of molecules, which have higher velocity, jump out of the liquid and into the cavity containing steam or gas: measured pressure increases.

We then say that the vaporization phase is initiated and ended at a constant pressure and within a range of fixed temperature values.

When balance is reached, exactly the same number of molecules jump back into the liquid (reason for which the level of the liquid remains constant within the closed container).

More external thermal energy being transferred into the liquid-vapor mix, or subtracted from it, naturally changes the molecular energy, as well as those previous values of pressure and temperature.

Step by step, by modifying the internal energy of the liquid-vapor mix, a curve and a table can be created.
Those serve to predict exactly how that mix will behave regarding vaporization and condensation for that substance.

Please, see:
https://www.ohio.edu/mechanical/thermo/Intro/Chapt.1_6/Chapter2a.html

https://thermopedia.com/content/1150/


WATER_PROPERTIES_FIG1.gif
 
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  • #22
Tom.G
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this is exactly the point: equilibrium is not established in an open container,
But it IS established within the liquid where the temperature is high enough. This is evidenced by the bubbles forming at and near the bottom of a pan being heated on a stove.

Above the liquid surface in an open container both the temperature and the pressure is lower with the end result that the steam starts condensing (which is what you can see) and it also dissipates into the environment.
 
  • #23
Drakkith
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Maybe I understood: the vapor pressure is always the same for a fluid, at a fixed temperature, regardless of whether the liquid-vapour equilibrium is formed.
Per wiki, vapor pressure: is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

So vapor pressure isn't something you measure, it is something you define. What you actually measure is the partial pressure of a gas, which in water's case is water vapor. So when you boil a pot of water, the vapor pressure of that water gradually rises because it is getting heated, but you wouldn't measure the vapor pressure since it is a defined quantity, not a measured one.

A similar thing happens with the boiling point of a substance. If we changed the pressure around the pot of water instead of changing its temperature we would see that the boiling point changes as pressure falls or rises. And similar to vapor pressure, the boiling point isn't what you measure. You'd measure the temperature or pressure and then define the boiling point to be a certain temperature at a certain pressure, just like how vapor pressure is defined to be a certain pressure at a certain temperature.

Note that by 'measure' I mean physically place a device to record some quantity, such as placing a pressure gauge or thermometer near the surface of the pot of water.
 
  • #24
pisluca99
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But it IS established within the liquid where the temperature is high enough. This is evidenced by the bubbles forming at and near the bottom of a pan being heated on a stove.

Above the liquid surface in an open container both the temperature and the pressure is lower with the end result that the steam starts condensing (which is what you can see) and it also dissipates into the environment.

So, for example, at 25°C, equilibrium does not occur on the surface, because the vaporized water is dispersed in the atmosphere, but it manifests itself 'inside' the water: in fact, small bubbles form. The pressure inside these bubbles in equilibrium with the water will be precisely the vapor pressure and it can be measured if we close the container, because, in this condition, this equilibrium will also be established 'above' the water.
Obviously this is true at any temperature, even at 100°C, but in this case, a tension develops in the bubbles in equilibrium with the water such as to overcome the atmospheric pressure and boiling occurs.
I have simplified, but the important thing is that it is correct in broad terms.
 
  • #25
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I agree with @Lord Jestocost.

OP: What is your understanding of the contents of a bubble beneath the surface? (a) a mixture of air and water vapor at 1 atm. or (b) pure water vapor at 1 atm

Boiling has very little to do with what is happening above the surface (other than setting the total pressure on the liquid below). Bubbles will form when the equilibrium vapor pressure is high enough to physically push back the surrounding liquid at the total pressure that exists a little below the free surface.
 
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  • #26
pisluca99
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I agree with @Lord Jestocost.

OP: What is your understanding of the contents of a bubble beneath the surface? (a) a mixture of air and water vapor at 1 atm. or (b) pure water vapor at 1 atm

Boiling has very little to do with what is happening above the surface (other than setting the total pressure on the liquid below). Bubbles will form when the equilibrium vapor pressure is high enough to physically push back the surrounding liquid at the total pressure that exists a little below the free surface.
From what I understood, at any constant temperature, an equilibrium is established between vapor and liquid, therefore this means that bubbles form in the 'mass' of water, which however cannot rise because the atmospheric pressure does not allow it. However, at 100 degrees, the vapor that makes up the bubbles, always in equilibrium with the liquid, reaches such a pressure as to overcome the atmospheric pressure and the bubbles can rise.
If the container is closed, this equilibrium is also established on the surface of the liquid, while if the container is open, this equilibrium cannot occur on the surface because the vaporized water molecules escape into the atmosphere.

Having said that, namely that the vapor of the bubbles is in equilibrium with the liquid, I would say that the correct answer is b, i.e. there are vapor bubbles in equilibrium with the surrounding liquid.
 
  • #27
jbriggs444
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which however cannot rise because the atmospheric pressure does not allow it
A bubble in a pot of water on the stove will rise to the top of the water and burst there. This happens even though atmospheric pressure is one atmosphere.

What cannot happen due to atmospheric pressure is that the bubble will not expand further. It is in evaporation/condensation/thermal equilibrium with the surrounding water.

It is not, however, in buoyant equilibrium.
 
  • #28
pisluca99
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A bubble in a pot of water on the stove will rise to the top of the water and burst there. This happens even though atmospheric pressure is one atmosphere.

What cannot happen due to atmospheric pressure is that the bubble will not expand further. It is in evaporation/condensation/thermal equilibrium with the surrounding water.

It is not, however, in buoyant equilibrium.
Ok, so at any constant temperature, vapor in the 'mass' of water is in equilibrium with the liquid water itself and appears in the form of bubbles, which however are able to rise (I suppose because the vapor is less dense than the liquid water). The vapor in these bubbles has a pressure which is the vapor pressure, at that particular temperature and once equilibrium has been established.
At 100°C, the vapor pressure of the bubbles reaches atmospheric pressure and these become able to expand precisely because they overcome the atmospheric pressure pressing on them. These 'big' bubbles rise, generating the phenomenon of boiling.
So essentially boiling is a 'rising of bubbles' which is much more visible and grosser than when the temperature is lower than 100 °C.

I hope I finally found the solution.
 
  • #29
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Bubbles will not form under the liquid surface unless the liquid temperature is such that its equilibrium vapor pressure equals or exceeds the total static pressure within the liquid.
 
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  • #30
pisluca99
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Bubbles will not form under the liquid surface unless the liquid temperature is such that its equilibrium vapor pressure equals or exceeds the total static pressure within the liquid.
So Bubbles form only during boiling and not during a simple evaporation (under 100 °C)?
 
  • #31
russ_watters
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So Bubbles form only during boiling and not during a simple evaporation (under 100 °C)?
Right. And why would they? The static pressure under the surface exceeds (or even far exceeds) the vapor pressure.
 
  • #32
pisluca99
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Right. And why would they? The static pressure under the surface exceeds (or even far exceeds) the vapor pressure.
ok, to recap:
- below boiling temperature, only evaporation occurs, involving only the surface of the liquid. In particular, if the container is open, the liquid-vapour equilibrium is established only in the 'mass' of water and not above it, because the vaporised water molecules are dispersed in the surrounding atmosphere (it is as if they became an integral part of the gases in the air), therefore evaporation never stops.
If, on the other hand, the container is hermetically closed, the liquid-vapour equilibrium is also established 'above' the solution and this allows us to evaluate the vapor pressure of this saturated vapor at equilibrium. If so, evaporation stops.

- At boiling temperature, vapor pressure developed in the liquid-vapour equilibrium reaches and overcomes the atmospheric pressure, so the vapor is released in the form of bubbles which rise throughout the mass of water (if the container is open). Conversely, if the container is hermetically sealed, boiling will never occur, as the vapor pressure will never be able to reach the pressure that presses on the water (it is true that as the temperature increases, the vapor pressure increases in the water, but also increases the pressure pressing on the water itself, given by the sum of the pressure of the saturated steam developed + atmospheric pressure), unless there is a vent, as in the case of the pressure cooker.
 
  • #33
Drakkith
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In particular, if the container is open, the liquid-vapour equilibrium is established only in the 'mass' of water and not above it
How can there be a liquid-vapor equilibrium inside the liquid water if there's no vapor and no surface that liquid-gas or gas-liquid transitions can occur?

At boiling temperature, vapor pressure developed in the liquid-vapour equilibrium
This doesn't really make any sense. Vapor pressure isn't 'developed', and an equilibrium is something that is reached, not something in which you can develop something else.

Instead I'd say that at boiling temperature, the vapor pressure equals the surrounding pressure. Note that this does not mean that bubbles will immediately form, as the phase transition from liquid to gas itself takes energy. You can indeed have a mass of water at 100 c (or whatever its boiling point happens to be depending on the surrounding pressure) without it boiling.

Also, remember that when you boil a pot of water the bubbles form at the BOTTOM of the pot, not the top. This means that the bubbles are forming not at a boundary of liquid and gas, but within the liquid water itself. The vapor pressure of the water has to equal the combined atmospheric pressure plus an additional amount equal to the pressure exerted by the water at whatever depth the bottom of the pot is.

Conversely, if the container is hermetically sealed, boiling will never occur, as the vapor pressure will never be able to reach the pressure that presses on the water (it is true that as the temperature increases, the vapor pressure increases in the water, but also increases the pressure pressing on the water itself, given by the sum of the pressure of the saturated steam developed + atmospheric pressure), unless there is a vent, as in the case of the pressure cooker.
No, this is incorrect. Boiling can and does occur in a sealed container if you heat the water up fast enough or if the container is large enough. What needs to happen is that you need to reach the boiling temperature of the water before the maximum allowable pressure of the container is reached. Or, as I explain below, you seal the container after boiling begins.

Note that in a pressure cooker (at least the one that I have in my kitchen) the water boils BEFORE it pressurizes. There are two valves that allows for the release of pressure located in the lid. One is a manual valve that you open when the food is done to release the pressure completely. The other valve is smaller and is designed to close when a large enough pressure differential is developed between the inside and outside of the cooker. Being small, it cannot release the water vapor from the inside of the cooker faster than the cooker can generate it by heating the water. The water starts to boil, generating large amounts of water vapor, which pressurizes the inside of the cooker until the valve close completely. The water continues to boil until the rising pressure inside pushes the boiling point high enough to stop boiling.

So a pressure cooker itself is an example of boiling water in a closed system.
 
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How can there be a liquid-vapor equilibrium inside the liquid water if there's no vapor and no surface that liquid-gas or gas-liquid transitions can occur?
But bubbles are made of vapor, and they are generated from the bottom of the container, so vapor and liquid-vapour equilibrium must necessarily manifest 'inside' water...
 
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Drakkith
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But bubbles are made of vapor, and they are generated from the bottom of the container, so vapor and liquid-vapour equilibrium must necessarily manifest 'inside' water...
Prior to bubble formation there's no water vapor inside of liquid water. So there's nothing for the liquid to be in equilibrium with. The concept of liquid-vapor equilibrium only makes sense if there are both liquid and vapor phases present.
 
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