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

[russ_watters]

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.

 

[pisluca99]

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.

 

[Drakkith]

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.

 

[pisluca99]

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...

 

[Drakkith]

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.

 

[russ_watters]

But bubbles are made of vapor, and they are generated from the bottom of the container...

When boiling. Not when not boiling.