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MexChemE
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Homework Statement
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.
Homework Equations
PV = nRT
Pi = PYi; where Yi is the mole fraction of component "i"
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!