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.)

Well, if we do some calculations, we find that the vapor pressure of propanol at 25°C is 10.7 torr. Now, let's calculate the number of moles of propanol in the room using the ideal gas law.

Using the formula n = PV/RT, where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature in Kelvin, we can find that n = (10.7 torr * 67.5 L) / (0.0821 L*atm/mol*K * 298.15 K) ≈ 3.013 moles.

So, in this case, not all of the propanol will evaporate. There will be approximately 3.013 moles remaining in the liquid form. Just be careful not to spill it or the room may become a slip-and-slide!

To determine whether all the propanol will evaporate or if some liquid will remain, we need to compare the vapor pressure of propanol with the partial pressure of propanol in the room.

1. Convert the volume of propanol from liters to milliliters:
10.0 L * 1000 mL/L = 10,000 mL

2. Calculate the moles of propanol using the ideal gas law equation:
n = PV/RT
n = (10.7 torr * 67.5 L) / (0.804 g/mL * 298.15 K)
n ≈ 0.352 mol

3. Convert the moles of propanol to grams:
0.352 mol * 60.1 g/mol = 21.2152 g

4. Calculate the mass of the remaining liquid propanol if not all of it evaporates:
Initial mass - mass of propanol vapor = remaining liquid mass
Initial mass = volume * density
Initial mass = 10,000 mL * 0.804 g/mL = 8,040 g
Mass of propanol vapor = n * molar mass
Mass of propanol vapor = 0.352 mol * 60.1 g/mol = 21.1352 g
Remaining liquid mass = 8,040 g - 21.1352 g ≈ 8,018.8648 g or 8,019 g

Therefore, if the initial volume of propanol is 10.0 L and the room dimensions are 7.5 m x 3 m x 3 m, not all of the propanol will evaporate. There will be approximately 8,019 grams of propanol remaining in liquid form.

To determine whether all the propanol will evaporate or if some liquid will remain, we can use the ideal gas law equation:

n = PV/RT

where:
n = number of moles
P = pressure
V = volume
R = ideal gas constant
T = temperature

First, we need to calculate the number of moles of propanol using the given information:

Given:
V = 10.0 L (volume of propanol)
P = 10.7 torr (vapor pressure of propanol)
T = 25 °C = 298.15 K (temperature)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)

n = (P * V) / (R * T)
= (10.7 torr * 10.0 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
≈ 3.013 moles

Now, we know that one mole of propanol occupies a volume of 0.804 g/mL.

To determine the amount of liquid remaining, we need to subtract the volume of the vapor from the total volume of the room.

The room dimensions are given as:
Length = 7.5 m
Width = 3 m
Height = 3 m

Total room volume = Length * Width * Height
= 7.5 m * 3 m * 3 m
= 67.5 m^3 (since all dimensions are exact numbers)

The volume of the vapor can be calculated using the ideal gas law equation for the number of moles:

Vapor volume = n * (R * T) / P
= 3.013 moles * (0.0821 L·atm/(mol·K) * 298.15 K) / (10.7 torr)
≈ 0.0666 L

Finally, we can calculate the amount of liquid remaining:

Liquid remaining = Total volume of propanol - Vapor volume
= 10.0 L - 0.0666 L
≈ 9.9334 L

Therefore, approximately 9.9334 L of propanol would remain as a liquid in the sealed room.