posted by Sarah .
If you place 10.0 L of propanol (C3H8O) in a sealed room that is 7.5 m long, 3 m wide, and 3 m high,
will all the propanol evaporate? If some liquid remains, how much will there be? The vapor pressure of
propanol is 10.7 torr at 25 °C, and the density of the liquid at this temperature is 0.804 g/mL. Treat the
room dimensions as exact numbers
n = PV/RT
(10.7)(67.5)/(0.804)(298.15) = 3.013
I would convert the room dimensions to cm because it's easier to convert cc to L.
(750 cm x 300 cm x 300 cm)/1000 = ? L and I get 67,500 L.
Then grams propanol = 10,000 mL x 0.804 Lg/m: = 8040 grams and mols propanol = 8040g x (1 mol/60 g) = 134 mols.
Plug in 10.7 torr (converted to atm) into PV = nRT and solve for n. I get something like 38 mols. You have 134 mols initially; therefore not all of it will vaporize and you will have 134-38 = ? mols remaining.
I think the spirit of the problem expects you to ignore the volume in the room occupied by the 10 L propanol; however if you wish you may correct for that by 67,500-10 = ? but that doesn't change the volume by much and has very little effect on the calculations. Actually, subtracting 10 isn't quite correct; technically its the amount of the liquid remaining. The easiest way to handle that is to do nothing but you can subtract 10, calculate n remaining, convert that to L and subtract that amount. This is continued until it makes no difference in the answer. That is called using iteration to solve the problem. (In this problem it makes almost no difference anyway.)