Vapor pressure, its decrease and subsequent lowering of boiling point


by ajassat
Tags: boiling point, pressure, temperature, thermodynamics, water
ajassat
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#1
Dec20-09, 08:51 AM
P: 55
I gather the following:

- A decrease in atmospheric pressure leads to a decrease in boiling point
- Thermodynamics explains this using mathematics
- A kinetic theory model can also be used to explain

I need to know:

- The science behind this occuring. Why does a lower atmospheric pressure result in a lower boiling point?
- How can we borrow concepts (including mathematical) from thermodynamics to explain why a lower atmospheric pressure results in a lower boiling point?

I have looked at:

- The Clausius-Clapeyron equation. When plotted as a function of pressure and temperature, it is seen that temperature is inversely proportional to pressure. Is this a sufficient explanation?
- Very brief explanations suggesting that molecules of a liquid have to do some sort of 'work' to push air molecules out of the way. A thinner atmosphere leads to less molecules of air to push away...so the energy required is greatly decreased (boiling point decreases). This is qualitative. How can we quantify this using mathematics?

I have asked this question before in a similar format on other forums. Replies don't usually answer my question. I would be grateful if the knowledgeable members of PF could answer my questions.

Regards,
Adam
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Mapes
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#2
Dec20-09, 10:31 AM
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Simply put, boiling occurs when the vapor pressure of a liquid exceeds atmospheric pressure. And the vapor pressure increases with temperature, because higher temperatures always promote the higher-entropy phase (in this case, gas, as compared to liquid). So the boiling temperature decreases with decreasing atmospheric pressure.

But this is just another qualitative description. The Clausius-Clapeyron equation explains the effect mathematically. It arises from the Gibbs free energy being equal between phases for phase transformations at constant temperature and pressure. Does this answer your question?
ajassat
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#3
Dec20-09, 11:31 AM
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Quote Quote by Mapes View Post
Simply put, boiling occurs when the vapor pressure of a liquid exceeds atmospheric pressure. And the vapor pressure increases with temperature, because higher temperatures always promote the higher-entropy phase (in this case, gas, as compared to liquid). So the boiling temperature decreases with decreasing atmospheric pressure.

But this is just another qualitative description. The Clausius-Clapeyron equation explains the effect mathematically. It arises from the Gibbs free energy being equal between phases for phase transformations at constant temperature and pressure. Does this answer your question?
This answers the qualitative part of my question very well.
I would be happy if you could show me some mathematics behind Clausius-Clapeyron and Gibbs free energy i.e a full mathematical description.

Thanks,
Adam

Mapes
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#4
Dec20-09, 11:44 AM
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Vapor pressure, its decrease and subsequent lowering of boiling point


1. For phase transitions at constant temperature and pressure, the Gibbs free energy [itex]G=U+PV-TS[/itex] of each phase is equal.

2. The differential form for phase [itex]i[/itex] is [itex]dG_i=-S_i\,dT+V_i\,dP[/itex].

3. Subtract [itex]dG_\mathrm{gas}[/itex] from [itex]dG_\mathrm{liquid}[/itex] to get [itex]0=-\Delta S\,dT_\mathrm{transition}+\Delta V\,dP_\mathrm{transition}[/itex], or [itex]\Delta S/\Delta V=dP_\mathrm{transition}/dT_\mathrm{transition}[/itex].

4. Since the higher-volume phase (gas) is the higher-entropy phase, the left side is positive and a decrease in the boiling pressure corresponds to a decrease in boiling temperature.
ajassat
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#5
Dec20-09, 11:47 AM
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Quote Quote by Mapes View Post
1. For phase transitions at constant temperature and pressure, the Gibbs free energy [itex]G=U+PV-TS[/itex] of each phase is equal.

2. The differential form for phase [itex]i[/itex] is [itex]dG_i=-S_i\,dT+V_i\,dP[/itex].

3. Subtract [itex]dG_\mathrm{gas}[/itex] from [itex]dG_\mathrm{liquid}[/itex] to get [itex]0=-\Delta S\,dT_\mathrm{transition}+\Delta V\,dP_\mathrm{transition}[/itex], or [itex]\Delta S/\Delta V=dP_\mathrm{transition}/dT_\mathrm{transition}[/itex].

4. Since the higher-volume phase (gas) is the higher-entropy phase, the left side is positive and a decrease in the boiling pressure corresponds to a decrease in boiling temperature.
This sort of logic represented by mathematical equations is exactly what I was looking for. Thank you very much. I have written some maths down for Clausius-Clapeyron but I am not sure it is correct.

Can you explore Clausius-Clapeyron mathematically and I'll see if it matches with mine?

Thanks again,
Adam
lnlohiya
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#6
Jan11-12, 01:59 PM
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I have a question. So is reducing the pressure the only way to reduce the boiling point of a gas? In short, what are the ways I can make a gas having a boiling point of around 60 degrees Celcius to be in the gaseous phase at room temperature?
lnlohiya
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#7
Jan11-12, 02:04 PM
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and yea...another thing to keep in mind in connection to my previous statement is that how can it be done on a large scale (like in a public place) where pressure cannot be controlled.
lnlohiya
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#8
Jan11-12, 02:18 PM
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Another thing i would like to know is that if a gas having a boiling point of about 60 degrees Celcius is introduced in room temperature scenario, will it continue to spread like a gas if it is initially compressed into a gas and later released into atmospheric pressure or will it turn into a liquid and not spread like a gas?
klimatos
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#9
Jan12-12, 12:34 PM
P: 403
Quote Quote by lnlohiya View Post
I have a question. So is reducing the pressure the only way to reduce the boiling point of a gas? In short, what are the ways I can make a gas having a boiling point of around 60 degrees Celcius to be in the gaseous phase at room temperature?
This is in response to INLOHIYA's posts #7 and #8:

A substance may be in the gaseous phase at temperatures far below its boiling point. Expose a liquid to the air at a temperature below its boiling point and it will vaporize until the ambient vapor pressure is equal to the equilibrium vapor pressure for that temperature. The boiling point is irrelevant.

Your mention of a public place makes me uneasy. Just what do you have in mind?
klimatos
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#10
Jan12-12, 02:43 PM
P: 403
Quote Quote by ajassat View Post
I gather the following:

- A decrease in atmospheric pressure leads to a decrease in boiling point
I need to know: The science behind this occuring.
Why does a lower atmospheric pressure result in a lower boiling point?

- Very brief explanations suggesting that molecules of a liquid have to do some sort of 'work' to push air molecules out of the way. A thinner atmosphere leads to less molecules of air to push away...so the energy required is greatly decreased (boiling point decreases).

Regards,
Adam
Adam,

I can’t answer all your questions, and I suspect that some are not answerable at our present understanding of the physics of the matter.

I can, however, address the phenomenon of boiling from a molecular perspective.

Let us postulate a surface of liquid water and an overlying humid atmosphere.
Within that liquid water, two opposing sets of forces are at work. There are the disruptive forces which are largely the thermal motions (kinetic energies) of the individual molecules as they translate, rotate, vibrate, and librate. These tend to force the molecules apart from one another. Since all of these kinetic energies are subject to quantum limitations, the mathematics involved are a bit tricky.

In opposition, there are the constrictive forces which keep the water in its liquid phase. These are of three kinds, the intermolecular bonding forces, surface tension, and the transmitted pressure of the overlying air. The relationships between these various forces are not well described mathematically.

Any time the disruptive forces become greater than the constrictive forces, the water will boil. You can bring this about by increasing the liquid temperature, decreasing the surface tension, or decreasing the transmitted air pressure.

Air pressure affects boiling by transferring impulses from the surface throughout the liquid. The average surface molecule (part of the surface tension net) will be hit by an air molecule some billion times per second. The impulses generated by these impacts will travel throughout the water and have a net effect of keeping the water molecules closer together. That is, the net effect is to counter intermolecular disruption (boiling).

On the average, a water molecule will travel some 500 molecular diameters away from the surface under normal laboratory conditions before it encounters an air molecule. Consequently, “pushing an air molecule away” is not a factor in establishing the boiling point.


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