idlexp
What pressure build up is there in a 5000ml solvent bottle that was stoppered at room temperature (20C) and atmospheric pressure with a 50 ml air gap above the solvent - the bottle is in now a hot store room (30C)?

Does the filling level make a difference?

I was told this would be atmospheric pressure because the solvent vapour would be in equilibrium with the liquid - I'm not so sure,
1. aren't there pressure increases with temperature in the trapped air?
2. How does the solvent vapour pressure figure - does it reach saturation or does it reduce so that in combination vapour and air match the pressure as calculated in (1)?
3. Is there a formula I could use for pressure in a stoppered bottle at elevated temperatures (approximate will do eg ideal gas, minimal gas solubility in the solvent, Raoult's law followed, pragmatic vapour pressure equation)?

Apologies if this is muddled, I haven't done chemistry for 30 years. Reasoned explanations appreciated

boit
This is a good place to start. http://www.yorks.karoo.net/aerosol/link7.htm
Sorry if it doesn't begin to explain anything. I too am rusty in my chemistry. Anyway pressure increases with temperature when volume is held constant. This may not happen in the solvent's case cause it will quickly liquify, taking up less volume hence restoring pressure (an equilibrium as you correctly noted).
Kino
You are correct that raising the temperature will raise the pressure inside the bottle. The air pressure will rise per the ideal gas law; the vapour pressure will rise per Clausius-Clapeyron because there is more heat energy available to cause molecules of the liquid to evaporate.

When the bottle was stoppered, the gas mixture in the neck would have been at atmospheric pressure (100kPa). Dalton's Law tells you that the total pressure is the sum of the partial pressure of air and the partial pressure of vapourised solvent. If you can find the vapour pressure of the solvent at 20C (let's call this P_v) then you can work out the partial pressure of air at 20C (let's call this P_a, and it is given by 100kPa-P_v).

Now raise the temperature to 30C. The ideal gas law says that pressure is proportional to temperature. If we are happy to treat air as an ideal gas, and assume that any change in the solubility of air in the solvent is negligible, then we can write P_a'=P_a.(T'/T)=P_a.(303/293). This is a rise of about 3.5%.

According to Wikipedia (a skim read of their source seems sound), you can relate vapour pressure P1 at temperature T1 to vapour pressure P2 at temperature T2 by

ln(P2/P1)=(DH/R)(1/T1-1/T2)

where R is the gas constant and DH is the molar enthalpy of vapourisation, which you will have to look up for your solvent. For this particular case, then:

ln(P_v')=ln(P_v)+(DH/R)(1/293-1/303)

which is approximately

ln(P_v')=ln(P_v)+1.355x10^-5.DH

Per Dalton's Law again, the final pressure is P'=P_v'+P_a'.

As an example, if your solvent is ethanol (cheers) the vapour pressure at 20C is 5.83kPa and the enthalpy of vapourisation of 38.6kJ/mol. Numbers are also from Wikipedia, and health warnings abound because it quotes a vapour pressure of 5.95kPa elsewhere.

P_v=5.83kPa
P_a=100kPa-P_v=94.17kPa

That makes the values at 30C:

P_v'=P_v.exp(1.335x10^-5 x 38.6x10^3)=9.85kPa
P_a'=P_a.(303/293)=97.38kPa

That makes the total pressure at 30C:
P'=P_a'+P_v'=107kPa

This is a rise of about 7% compared to 20C. I very much doubt that that would threaten your bottle integrity, but you might want to calculate the area of your stopper and make a quick estimate of how much force is needed to pop off the top.

Random notes:
You will get different answers for different solvents, obviously.
If the vapour pressure at 30C exceeds atmospheric pressure (100kPa), it boiled.
The filling level does not make a difference, assuming that there's more than a tiny drop of liquid in the bottom.
idlexp
Kino,
this is magic very full, clear convincing and to the point. Thank you indeed.
idlexp
idlexp
Further observations. From the reasoning, the vapour pressure rises with the temperature and adds to the ideal gas law pressure. Thereby there are more molecules in the space.

One thing that puzzled me - is there is not a question of Avogadro's law being violated - ie there are more molecules in the space at 30C than the ideal gas law?

idlexp
Kino
QUOTE (idlexp+Sep 19 2011, 08:31 AM)
Further observations. From the reasoning, the vapour pressure rises with the temperature and adds to the ideal gas law pressure. Thereby there are more molecules in the space.

One thing that puzzled me - is there is not a question of Avogadro's law being violated - ie there are more molecules in the space at 30C than the ideal gas law?

Yes - but that's fine because Avogadro's law assumes that there are only gases. Here, you have a liquid phase which acts as a source-or-sink of gas molecules.

If you somehow placed a rigid seal over the surface of the liquid solvent so that it could not evaporate into the neck of the bottle then the pressure inside the bottle would follow the ideal gas law and hence Avogadro's law (give or take the actual ideality of the air-plus-vapourised-solvent mixture). However, without this seal what happens is that when you raise the temperature, extra molecules leave the liquid phase and enter the gas phase. That is a situation that Avogadro excludes in his assumptions (he speaks only of ideal gases which the liquid solvent is not, however ideal the vapourised solvent); his law does not apply.

Does that make sense?
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