0
$\begingroup$

The explanation for the boiling point of water is that at 100C, the vapour pressure becomes greater than atmospheric pressure. But say you had a a jar of water sealed in argon at 1atm, which is larger than the vapour pressure of water, wouldn't some water evaporate? I'm guessing yes, because there are no water molecules in the argon initially, so wouldn't some randomly leave the liquid, making the partial pressure on average non-zero?

Another reason I can think of it being based on partial pressure is that an open jar of water in a windless dark, dry room will still eventually evaporate. So why is it that at 100C there is a sudden dramatic change? Isn't the vapour pressure at e.g. 95C more than the partial pressure of water in the atmosphere as well?

$\endgroup$
2
$\begingroup$

But say you had a a jar of water sealed in argon at 1atm, which is larger than the vapour pressure of water, wouldn't some water evaporate?

Yes, but there will not be boiling, only evaporation from the surface.

there are no water molecules in the argon initially, so wouldn't some randomly leave the liquid, making the partial pressure on average non-zero?

Yes.

So why is it that at 100C there is a sudden dramatic change?

At this temperature the vapor pressure inside the bubble begins to be high enough to overcome the pressure in the liquid, so the bubble survives and rises.

Isn't the vapour pressure at e.g. 95C more than the partial pressure of water in the atmosphere as well?

Yes, and the water will evaporate to the atmosphere, but only on the surface. There will be no boiling, since the pressure in the liquid is higher than the pressure in the bubble could be, so any small bubble is crushed and the vapor turns back into liquid.

$\endgroup$
2
$\begingroup$

You are mixing up two different things.

  1. Evaporation. This has to do with the ratio of evaporating/condensing molecules only on the surface of the liquid. This happens all the time and (with an enclosed volume above the liquid) will eventually lead to a dynamic equilibrium (both rates being equal).

  2. Boiling: Whenever a vapor bubble develops internally a new "surface" is generated as well. But inside this bubble only molecules of the liquid are present. So inside the bubbles the equilibrium is not influenced by any other gas molecules. Consequence: when the vapor pressure rises above the external pressure (be it from the atmosphere or from some other device - like a pressure pot (higher) or a vacuum bell (lower)) the bubbles can grow -- the liquid is boiling.

$\endgroup$
1
$\begingroup$

There is some subtle terminology that may help in answering this question. First, the saturated vapor pressure is the pressure at which the number of water molecules evaporating from the surface of a liquid equals the number of water vapor particles condensing back to the liquid phase. The saturated vapor pressure is also non-linearly dependent on the temperature of the liquid. So, as the temperature of the water increases to a point where the saturated vapor pressure exceeds the total pressure (aka more water molecules are leaving the liquid phase than the number of water molecules condensing to the liquid phase) the liquid begins to boil. This means that you could change the boiling point by changing the total pressure on the system. If you put water in a vacuum chamber, by the way this would be a terrible idea, it would begin to boil at temperatures well below 100C as the total pressure is much less. An interesting area related to this is PVD, which you can learn more about on Wikipedia.

$\endgroup$
0
$\begingroup$

If you start with a volume of gaseous argon, all outside of a standing container of liquid water, and then seal those two volumes, you will then "see" that some of the water will evaporate and the pressure in the gaseous phase will probably increase, but at the same time some of the argon will dissolve into the water. You will get a new equilibrium where both substances are in the liquid and gaseous phases and where the pressure is determined by the pressure and the relative molar quantities of water and argon.

The partial pressures of argon and water in the gaseous phase will sum to the total pressure. And just to complexify things a bit, most solutes raise the boiling point of water.

$\endgroup$
  • $\begingroup$ "The partial pressures of argon and water in the gaseous phase will sum to the total pressure" Hi, in that case would the total pressure still be 1ATM? $\endgroup$ – Hilkjh Apr 7 '18 at 23:13
  • $\begingroup$ The water vapor from the liquid would add to the pressure but some of the argon would dissolve which would lower the gas pressure. I'm guessing the final pressure would be near 1 atm but I'm not exactly sure how close. $\endgroup$ – DWin Apr 7 '18 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.