How to calc. amount of liquid going to vapor in closed container

[dwalter68]

I am sure this is a simple question for someone with chemical background. I need to make sure I have solved this problem correctly.

I take an open container and fill it partially with ethanol. Then I seal it up and assume no leaks. Note that the container is not evacuated of air beforehand and temperature is ambient. Container sits until full vapor pressure of ethanol is reached inside. There is enough ethanol so that some stays in liquid form.

I need to calculate the mass of ethanol that has gone into vapor inside the container.

I have made these assumptions:

- Press.(absolute) inside container = P(atmos.)+P(vp of ethanol)

-ideal gas law can be applied to ethanol vapor

-partial pressure of ethanol in this case is equal to vapor pressure of ethanol at ambient temp.

-because of Dalton's law I use the partial pressure of ethanol vapor (vapor pressure) in the ideal gas law instead of using the total absolute pressure (?)

So to find the mass I solve this:

m=[(vap. pres.of ethanol)x(volume of headspace)]/[(R)x(ambient temp)]

Are my assumptions correct?

 

[Torricelli]

I agree with your conclusion:
quantity of ethanol in a saturated mixture is
m = (PVP.V) / R.T

but my assumption would be TP=PVP+PAP
TP (total pressure) = PVP (partial vapour pressure of EtOH) + PAP (partial air pressure).

TP = Patm. = const. as long as Tamb. const.

 

[btrueblood]

He specified a closed (sealed) box, no leaks. The vapor pressure of the added ethanol will then add to the existing 1 atm of air in the box.

 

[mbeychok]

dwalter68:

Unless I have completely misunderstood what you have done, the universal gas law is PV=nRT where n is the number of mols (not the mass).

Are you using the universal gas law constant R or are you using the specific R_s for ethanol? Keep in mind that R_s = R / (Molecular Weight). If you are using R_s, then PV=mR_sT where m is the mass of ethanol.



Milton Beychok
(Contact me at www.air-dispersion.com)

 

[25362]

btrueblood, dwalter68 indicated the container was open to the atmosphere when ethyl alcohol was added. The unanswered question was at what temperature was the alcohol added and if equilibrium was reached prior to closing and sealing the container.

Humid air may have allowed some moisture to dissolve in ethanol and slightly change the vapor pressure. Anyhow, all contributors appear to think the total pressure in the closed container is atmospheric, the alcohol is pure 200^o-proof stuff.

mbeychok is absolutely right. If R=0.08206 L.atm/(K.mol), pressures are given in atm, volume in liters, and absolute temperatures in K, then n is number of moles, not mass.
Thus n multiplied by 46.07 g/mol would give the wanted result.

 

[dwalter68]

Yes, when I wrote R in my formula, I was actually referring to the specific Rs for ethanol.

Also the ethanol is pure 200 proof and it is at ambient temp.

My thanks to all for your help.

 

[btrueblood]

Yes, good point and good thinking 25362 -- "when is equilibrium reached?". If unknown, calculate both ways and assume reality is somewhere in between.