Simple Normal Boiling Point question

[iceman132]

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I have a really simple question that just doesn't seem intuitive to me at all.

When

Pressure of Boiling point = Vapor pressure

Shouldn't there just be no activity since it's at equilibrium? Is there evaporation?

I understand that if the vapor pressure is even a tiny bit HIGHER then evaporation will occur..... But AT equilibrium?

 

[riddler]

Yup, there's still evaporation, I think. How do you think there is any vapor pressure anyway? Even below the boiling point, there is some evaporation which causes the vapor pressure to be what it is.

From my understanding, at this equilibrium, rate of evaporation = rate of condensation.

 

[MD Odyssey]

I have a really simple question that just doesn't seem intuitive to me at all.

When

Pressure of Boiling point = Vapor pressure

Shouldn't there just be no activity since it's at equilibrium? Is there evaporation?

I understand that if the vapor pressure is even a tiny bit HIGHER then evaporation will occur..... But AT equilibrium?

I think you may be missing the fact that boiling occurs when the vapor pressure of the liquid (which depends upon temperature) equals the atmospheric pressure surrounding it.

 

[iceman132]

Yup, there's still evaporation, I think. How do you think there is any vapor pressure anyway? Even below the boiling point, there is some evaporation which causes the vapor pressure to be what it is.

From my understanding, at this equilibrium, rate of evaporation = rate of condensation.

Okay, so the rate of evaporation=rate of condensation that makes sense.

I think you may be missing the fact that boiling occurs when the vapor pressure of the liquid (which depends upon temperature) equals the atmospheric pressure surrounding it.

Pvapor = Patm. I understand this point. It's just that when I think of a boiling liquid I imagine a lot of vaporization going on, but I think I'm just wrong in thinking that way.

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So the only difference between boiling (Pvapor=Patm) and being more liquid (Pvapor< Patm) is the RATE. (There is always some vaporization going on)

When you are at the boiling you don't lose any liquid (Because you will have liquid condensing as soon as you evaporate) correct ?

 

[riddler]

Let me clarify.

There can be equilibrium (rate of evap = rate of condense) before boiling point. It just means that the relative humidity is 100% (easier to accomplish in a closed container, I think).

Also, I'm not so sure there is evaporation at or beyond the boiling point. Instead, I believe the proper term is vaporization (I could be wrong about this). Evaporation is just random surface molecules gaining enough energy to escape the liquid phase, while after the boiling point the liquid molecules themselves begin to vaporize (bubbles).

If there is ever an equilibrium, the vapor pressure then remains constant, and is called the saturated vapor pressure.

 

[iceman132]

The vapor pressure of any substance increases non-linearly with temperature. The atmospheric pressure boiling point of a liquid (also known as the normal boiling point) is the temperature at which the vapor pressure equals the atmospheric pressure. With any incremental increase in that temperature, the vapor pressure becomes sufficient to overcome atmospheric pressure and lift the liquid to form vapor bubbles inside the bulk of the substance. Bubble formation deeper in the liquid requires a higher pressure, and therefore higher temperature, because the fluid pressure increases above the atmospheric pressure as the depth increases.
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I found this off of ask.com and I wanted to see what anyone here thought.

This makes sense to me since the vapor pressure must become LARGER then atmospheric pressure to start vaporizing/losing liquid.


The problem I'm having to visualize is what happens at Pvapor=Patm. (Also, I'm not 100% understanding why it boils anyways since Patm and Pvapor both point toward the liquid)

 

[mikexima]

at equilibrium the rate at which water molecules jump out of the liquid phase and become a gas molecule is THE SAME as the rate at which the atmosphere forces a gas molecule back into the liquid

 

[circulus vitios]

I have a really simple question that just doesn't seem intuitive to me at all.

When

Pressure of Boiling point = Vapor pressure

Shouldn't there just be no activity since it's at equilibrium? Is there evaporation?

I understand that if the vapor pressure is even a tiny bit HIGHER then evaporation will occur..... But AT equilibrium?

At equilibrium the net change is zero. That doesn't necessarily mean the forward and reverse reactions (or condensation and evaporation) are zero.