Vapor Pressure, Open Container, and Boiling

[brookssp]

I've been confused about this for a while now and haven't found anything that directly addresses this problem. Please correct any faults you see in my reasoning.

My understanding of vapor pressure is that it is the pressure exerted by the vapor phase of a substance above the liquid phase onto the liquid phase (and container walls in a closed system). Molecules are always leaving the liquid phase due to their kinetic energy and entering the vapor phase. In a closed container, these molecules eventually enter an equilibrium state where the number entering the liquid phase equals the number entering the vapor phase. However, in an open container (e.g. pot on a stove) equilibrium cannot be reached due to the vapor being dispersed into the atmosphere which is not the case in a closed container.

Now, the definition of boiling is when vapor pressure reaches and/or exceeds the atmospheric pressure. If equilibrium cannot be reached in an open container, how is vapor pressure maintained above the liquid such that boiling can occur? Boiling in open containers occurs all the time so there must be some way that the vapor pressure above the gas can exceed atmospheric pressure even if it's being dispersed otherwise boiling wouldn't occur.

Side note: If the vapor is above the gas, does this not make it the atmospheric pressure above the liquid phase. Thus, the Earth's atmospheric pressure is intertwined with the vapor pressure making the atmospheric pressure above the liquid phase now a combination of vapor pressure and the Earth's atmospheric pressure? This is part of where my confusion is coming from because it seems that overcoming atmospheric pressure is a redundancy because vapor pressure is part of the atmosphere above the liquid thus vapor pressure would have to be overcoming itself. My only explanation for this is that the vapor pressure is also within the liquid phase and is exceeding the vapor pressure due to the gaseous phase above. Not sure if this is correct though.

 

[jbriggs444]

When the term "vapor pressure" is used, it usually refers to "saturated vapor pressure". That is the partial pressure of vapor that would be required to attain an equilibrium state. The actual partial pressure of vapor in the ambient atmosphere may be more or less than this.

The boiling temperature is the temperature where the saturated vapor pressure is equal to the total ambient atmospheric pressure. At this temperature there cannot possibly be a high enough partial pressure in the atmosphere to reach an equilibrium, even if it was 100% vapor. If the temperature exceeds boiling by a slight margin and if any bubbles (or nucleation sites) exist within the body of the liquid, the liquid will evaporate into these and cause the bubble to grow (or form). The resulting phenomenon is known as "boiling". At any lower temperature, a bubble of pure vapor would tend to shrink and boiling no longer occurs.

In the absence of nucleation sites it is possible for a liquid to exceed its boiling temperature without actually boiling.

 

[brookssp]

That is where I'm confused at is in reference to saturated vapor pressure with regards to an open container. It seems impossible to reach saturated vapor pressure when there is an open container yet boiling still occurs within open containers. If boiling occurs when saturated vapor pressure reaches total ambient atmospheric pressure then how does it occur in open containers if saturated vapor pressure is impossible in such a scenario?

Let me explain my thinking in an alternative way. A liquid has a certain partial pressure of vapor. The ambient atmosphere has a certain partial pressure of vapor. Increasing temperature increases both partial pressures. Is the partial pressure within the liquid overcoming the combination of ambient atmospheric pressure + ambient vapor pressure increasing the rate of nucleation and escape of vapor? I'm having trouble making the connection between saturated vapor pressure and boiling in an open container.

 

[russ_watters]

The vapor pressure ISN'T maintained under either boiling or evaporating. The vapor pressure of the liquid/gas mix is an equilibrium number and you get boiling or evaporation when the vapor pressure potential of the liquid is higher than the actual vapor pressure of the gas.

What is different about boiling is that the vapor pressure potential of the water is higher than the pressure in the water, enabling bubbles of saturated vapor to form.

 

[brookssp]

So, the definition for boiling commonly used by textbooks saying " when vapor pressure exceeds atmospheric pressure" is a little ambiguous? A better definition would be "potential for vaporization within the liquid exceeds the vapor pressure of the ambient atmosphere above the liquid and the thermodynamic/hydrostatic pressure of the liquid"? Correct me if I'm still misunderstanding.

 

[jbriggs444]

So, the definition for boiling commonly used by textbooks saying " when vapor pressure exceeds atmospheric pressure" is a little ambiguous? A better definition would be "potential for vaporization within the liquid exceeds the vapor pressure of the ambient atmosphere above the liquid and the thermodynamic/hydrostatic pressure of the liquid"? Correct me if I'm still misunderstanding.

...exceeds the pressure of the ambient atmosphere above the liquid. Not vapor pressure, just pressure. For shallow depths, this will be equal to the hydrostatic pressure within the liquid.

 

[russ_watters]

So, the definition for boiling commonly used by textbooks saying " when vapor pressure exceeds atmospheric pressure" is a little ambiguous?

If that's all it says -- but that would be surprising. Even the wiki is better than that:

Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the temperature at which the vapor pressure of the liquid is equal to the pressure exerted on the liquid by the surrounding environmental pressure.

http://en.wikipedia.org/wiki/Boiling

A better definition would be "potential for vaporization within the liquid exceeds the [STRIKE]vapor[/STRIKE] pressure of the ambient atmosphere above the liquid and the thermodynamic/hydrostatic pressure of the liquid"? Correct me if I'm still misunderstanding.

With the correction made in the previous post, that would then be correct.

 

[Chestermiller]

It doesn't matter whether the liquid water is in vapor-liquid equilibrium with water vapor in the air above. It won't be at equilibrium except at the interface with the air, where the partial pressure of the water vapor will be 1 atm. Above that, if the container is open, the partial pressure of the water vapor will drop off substantially with distance from the interface. But none of this matters to the issue of boiling.

The point is that, even with the open container, the gas pressure above the liquid water is 1 atm. To have boiling, you need to be able to form vapor bubbles well below the water surface. For this to happen, the added volume of the bubbles will be causing the pool of liquid to expand against the atmosphere above. The water vapor pressure within the pure water vapor bubbles forming below the surface has to be 1 atm in order to push back the atmosphere above and accommodate the added volume occupied by the bubbles. This is what we call boiling.