Question about water/ethanol concentration.. 1

[a70duster]

This question comes from another forum

http://www.nastyz28.com/forum/showthread.php?t=85418

"I filled a glass with 40% ethanol solution and left it open on the lab bench. When I came back next morning to the lab, I found that the ethanol concentration in the glass was still 40%."

The other 60% is water and no tricks. Can anyone expand on what is going on here? Thanks

 

[25362]

Does anyone know the azeotropic (fixed V/L) composition of the water-ethanol azeotrope at the conditions given by a70duster ?

 

[dcasto]

the azeotrop is 190 proof. he started with 80 proof wiskey.

the solution is in steady state, not equalibrium.

 

[katmar]

I would be surprised if the concentration did not change (assuming this is not a trick question). If the initial strength of 40% is on a volume basis (the usual basis for potable spirits) then the mixture is definitely not at its azeotropic concentration and any vapor coming off the glass would be at a strength of higher than 40%. A rough calc indicates that the vapor would be at about 80% ethanol by volume if it was condensed to a liquid. The exact strength of the vapor is not important - if it is anything above 40% it will leave the liquid in the glass partially depleted of alcohol and the strength will decrease overnight.

I agree with dcasto that this is not an equilibrium situation, but I do not see it as a steady state either. In my opinion the vapor being evaporated would become progressively weaker in ethanol and this is not a steady state. Dcasto, in what aspect do you consider this steady state?

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[quark]

liar... ish eez foty percen water in the mornin...hick

 

[25362]

A question to a70duster, was the mix 40% ethanol by vol or by wt ?

The vapor pressures of ethanol and water over their solutions at 20^oC, in torr, taken from a book:

w% ethanol VP water VP ethanol

0 17.5 0
20 15.9 12.6
40 14.7 20.7
60 14.1 25.6
80 11.3 31.2
100 0 43.6

What comes to tell us that there is a possibility that ethanol has a lower VP than water. If the lab was at a lower temperature, this fact may have been more pronounced.

 

[katmar]

For ethanol/water a volume % of 40.0 is equivalent to a mass % of 33.3, so it still fits in with the data given by 25362. The VP will be proportional to the molar concentration, which is rather different from the volume or mass strength because ethanol has a much higher molecular weight than water.

Working with the above data for 40% (mass) the total VP will be 35.4 (= 14.7 + 20.7), and the molar fraction of ethanol in the vapor will be 20.7/35.4 = 0.585. Converting this to mass % gives 78.3% which also indicates that the vapor carries off a higher fraction of ethanol than exists in the liquid. This is why I said earlier that I would be surprised if the concentration did not change overnight.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[25362]

In order for the ethanol:water mass ratio in the vapor to be 4:6 (i.e., 40%) as in the liquid, the vapor pressure of water should be 3.8 times greater than that for ethanol.

As follows:

mass of ethanol in vapor = 46.07[×]1[÷]4.8 = 9.6 units
mass of water in vapor = 18.02[×]3.8[÷]4.8 = 14.3 units.

Total mass = 9.6+14.3 = 23.9 units

Ethanol in the vapor mix: 9.6[÷]23.9 = 40.2% mass

Am I wrong ?

 

[a70duster]

25362 - It was never stated in the problem but I assume volume.

 

[BenThayer]

i am with quark on this one. i don't understand why any of it was there.

 

[dcasto]

if the room were cold enough, the VP of the water and ethanol would become equal,like at -5C maybe.

in equalibrium means that the fluid may recieve enough energy to "vapor" off a few molecules, but the atmosphere removes the energy and the molecules become liquid..... no change in anything, mass or energy. In steady state, we can't control all the energy in or out as equal all the time, an open container so the equalibrium changes all the time (because of energy). In dynamic state, we can't contol anything because molecules are random, isn't that quantum theory stuff?

 

[katmar]

25362 - I agree with your calculation. In order to have the 3.8x higher water VP the liquid strength would have to be significantly less than 20% according to the data in your earlier post. To me, this just confirms that the glass would not remain at 40% overnight.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[quark]

Did extensive search and thought whole yesterday. Didn't find any reason for this and so agreeing with katmar et al.

The process seems to be not steady state either. The rate of evaporation reduces due to increased surface tension of the solution.

However, I thought of few things.

1. The surface tension of the solution was so high that evaporation was not possible.
2. Somebody was topping up the glass whole night.
3. The professor might be expecting an affirmative statement against this phenomenon.
4. A watchman might have consumed it in the night and refilled it in the morning.
5. Measuring device was erratic.

Cheers,

 

[25362]

If the glass with the 40% ethanol-solution was immersed in, or located very near, another open container with an ethanol-rich solution that evaporates keeping the dew point of the glass in question constant, is there a chance the original ethanol concentration remains unchanged ?

 

[katmar]

Yes, if the air in the lab is saturated with ethanol then no ethanol will escape from the glass.

If the lab is 10m x 10m x 2.5m then it has a volume of 250m³ and would contain roughly 250 kg of air. This is roughly 8.6 moles of air. From the data above pure ethanol has a vapor pressure of 43.6mmHg at 20C. The saturation level of ethanol in air is therefore 43.6/760 = 0.057 mole fraction.

Now 0.057 x 8.6 = 0.49 moles ethanol would be in the air. The molecular weight of ethanol is 46 so there would be 0.49 x 46 = 22.5 kg of ethanol in the air. So I suppose if you had a big enough container of ethanol to saturate the air, and no air was lost from the room you could prevent ethanol from leaving the glass.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[25362]

Just 10.3 kg ethanol and about 3 kg water vapors would probably do the job in such a room.

 

[quark]

Just 10.3 kg ethanol and about 3 kg water vapors would probably do the job in such a room.

I know what you are upto. Now you want CO<sub>2[/sup] gassing in the room.

Let us see. With 1:3 dilution (40% is too hard), you require approximately 6.56 kgs of ethanol and 3.24 kgs of water vapor and then you can sit in the room as often as you want</sub>

 

[dcasto]

Why does it always lead to ethanol curing the CO2 caused global warming?