# Psychrometry -- Ethanol in dry air

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1. Feb 16, 2015

### MexChemE

1. The problem statement, all variables and given/known data
Ethanol evaporates in a 10 x 20 x 50 ft3 room. Air is initially dry at a temperature of 30 °C and 765 mmHg of pressure.
a) If the partial pressure of ethanol after vaporization is 40 mmHg, how many pounds of ethanol evaporated?
b) The same air is compressed until it reaches a pressure of 25 atm and then cools down to 20 °C. What percentage of ethanol condensed?
Assume ideal behavior.

2. Relevant equations
PV = nRT
Pi = PYi; where Yi is the mole fraction of component "i"

3. The attempt at a solution
The procedure for part a) is pretty straight-forward. We know the partial pressure of ethanol, which is the pressure ethanol would have if it ocuppied the 10,000 ft3 room by itself. We also have temperature so we can clear "n" from the Ideal Gas Law and get n = 1.32 lb mol of ethanol. We multiply this value times the molar mass of ethanol in lb/lb-mol and get 60.72 lb, which is the amount of ethanol that evaporated in the room.

Now, part b) is where I'm having trouble. I can either assume a constant volume or not and calculate a new partial pressure for ethanol using either Gay-Lussac's Law or the Combined Gas Law, but these calculations assume we still have 1.32 lb mol of ethanol, which is not the case. How do I actually account for the moles of condensed ethanol?

Thanks in advance for any input!

2. Feb 16, 2015

### Bystander

What's the vapor pressure of ethanol at 20 C?

3. Feb 16, 2015

### MexChemE

0.05861 bar or ≈ 44 mmHg. The partial pressure of ethanol must be higher than its vapor pressure in order for it to condense, as far as I know.

4. Feb 16, 2015

### Bystander

And final pressure of air plus ethanol vapor pressure is 25 atm. (surprised they didn't throw gauge pressure at you --- this is a mixed unit sorting problem). How much air did they start with?

5. Feb 16, 2015

### MexChemE

Are you suggesting using the vapor pressure of ethanol as its partial pressure? Makes sense, if ethanol is condensing it is at least at saturation pressure. After ethanol evaporated we have 1.32 lb mol of ethanol vapor and 23.943 lb mol of dry air, for a total of 25.263 lb mol.

If I use 44 mmHg as ethanol's partial pressure I can calculate a new number of moles, should I make the assumption of constant volume? I think I should, because I can't calculate a new total volume because we don't have 25.263 lb mol of mixture.

If I calculate the number of moles of ethanol using 44 mmHg of partial pressure, 10,000 ft3 of volume and 20 °C of temperature I get 0.0601 lb mol. Which make up for 2.76 lb. That means 95% of ethanol condensed. Are these assumptions and calculations correct?

6. Feb 16, 2015

### Bystander

There are enough missing steps that it's tough to tell where/when the assumptions were applied, and I see no calculations.

7. Feb 17, 2015

### MexChemE

Sorry, I was in a hurry. The assuptions I made were for the partial pressure of ethanol and the volume being kept constant. And the only calculations I made were to get the number of moles from the Ideal Gas Law. I will try to post all the details tomorrow.

8. Feb 17, 2015

### MexChemE

I realized today I made a mistake in my assuptions and calculations (which I didn't post), but coincidentally arrived at the correct percentage of condensed ethanol. Using the vapor pressure of ethanol as its partial pressure in the room is right, but assuming the air keeps the same volume is wrong. You can't compress the air up to 25 atm and also cool it down to 20 °C while keeping the same original volume. But we don't need to include the volume in the calculations. I accidentally got the 0.0601 lb mol of ethanol (which coincidentally is a very close result to the one you get by following the correct procedure) using the Ideal Gas Law, keeping the 10,000 ft3 of volume, but instead of using the partial pressure of ethanol (44 mmHg), I used its mole fraction (calculated from 44 mmHg / 19,000 mmHg). Here's the right procedure:

We know from part a) that we have 23.943 lb mol of dry air. Now, we have the partial pressure of ethanol and the total pressure of the mix. We calculate:
$$Y_{C_2H_6O} = \frac{44 \ mmHg}{19000 \ mmHg} = 2.3158 \times 10^{-3}$$
Now we can also get the mole fraction of dry air, which is YAir = 0.9976842. We know the moles of dry air and know we know its mole fraction so we can calculate the amount of moles in the mix:
$$n_T = \frac{n_{Air}}{Y_{Air}} = \frac{23.943 \ lbmol}{0.9976842} = 23.99858 \ lbmol$$
Total moles times the mole fraction of ethanol result in 0.05558 lb mol of ethanol (close to the 0.0601 I calculated before), which make up for 2.557 lb of ethanol. Finally, we calculate the percentage of condensed ethanol:
$$\% \ Condensation = \frac{60.72 \ lbmol - 2.557 \ lbmol}{60.72 \ lbmol} \times 100\% = 95.8\%$$

Thanks, Bystander!