Vapor pressure, its decrease and subsequent lowering of boiling point

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SUMMARY

The discussion centers on the relationship between atmospheric pressure and boiling point, emphasizing that a decrease in atmospheric pressure results in a lower boiling point due to the principles of thermodynamics and kinetic theory. The Clausius-Clapeyron equation is highlighted as a mathematical framework that explains this phenomenon, illustrating that vapor pressure increases with temperature and that boiling occurs when vapor pressure exceeds atmospheric pressure. The Gibbs free energy concept is also discussed, providing a deeper understanding of phase transitions at constant temperature and pressure.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Familiarity with Gibbs free energy concepts
  • Basic knowledge of thermodynamics and phase transitions
  • Concepts of kinetic theory related to molecular behavior
NEXT STEPS
  • Study the mathematical derivation of the Clausius-Clapeyron equation
  • Explore Gibbs free energy calculations for phase transitions
  • Investigate the effects of temperature on vapor pressure in different liquids
  • Research practical applications of reducing boiling points in industrial processes
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Students and professionals in chemistry, physics, and engineering, particularly those interested in thermodynamics, phase transitions, and the behavior of substances under varying pressure conditions.

ajassat
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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 occurring. 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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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?
 
Mapes said:
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
 
1. For phase transitions at constant temperature and pressure, the Gibbs free energy G=U+PV-TS of each phase is equal.

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

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

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.
 
Mapes said:
1. For phase transitions at constant temperature and pressure, the Gibbs free energy G=U+PV-TS of each phase is equal.

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

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

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
 
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 celsius to be in the gaseous phase at room temperature?
 
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.
 
Another thing i would like to know is that if a gas having a boiling point of about 60 degrees celsius 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?
 
lnlohiya said:
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 celsius 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?
 
  • #10
ajassat said:
I gather the following:

- A decrease in atmospheric pressure leads to a decrease in boiling point
I need to know: The science behind this occurring.
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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