if u place 10 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 eveporate?if some liquid remains, how much will there be?the vaopr pressure of propanol is 10.7 torr at 25 C, and the density if the liquid at this temperature is 0.804 g/mol.treat the room dimensions as exact number..thnks guys..
Did you make a typo? Should the density be 0.804 g/mL?
I have assumed you made a typo and the density of propanol is 0.804 g/mL.
Volume room is 7.5 x 3 x 3 cubic meters. Convert to liters for volume.
Then use PV = nRT and solve for n. That will give you the moles of the gas at the conditions listed. (10.7 torr is what you use for pressure).
Then use density of the propanol to solve for grams, convert to moles. If moles of the liquid exceed moles of the gas, subtract to find moles liquid remaining after vaporization has ceased; i.e., reached equilibrium. If moles gas exceeds moles liquid at start, then all of the liquid will evaporate.
yeah ur right..it should be 0.0804 g/mL..sorry bout that..
dont i have to use 10.7 torr and convert it to atm,then it gives me o.01407..otherwise.my mol is a very big number..
Yes, P must be in atmospheres in PV = nRT. V must be in L. However, n still is a relatively large number, considering you had 10L (10,000 cc) propanol.
To determine if all the propanol will evaporate in the sealed room, we need to compare the partial pressure of propanol vapor to its vapor pressure at the given temperature.
1. Calculate the volume of the sealed room:
Volume = length × width × height
Volume = 7.5 m × 3 m × 3 m
Volume = 67.5 m³
2. Convert the volume of the room to liters:
1 m³ = 1000 L
Volume = 67.5 m³ × 1000 L/m³
Volume = 67500 L
3. Convert the mass of propanol to moles:
Given that the density of the liquid propanol is 0.804 g/mL, and we have 10 L of propanol:
Mass = density × volume
Mass = 0.804 g/mL × 10 L
Mass = 8.04 g
Now, convert the mass of propanol to moles using the molar mass of propanol (C3H8O):
Molar mass of propanol = (3 × atomic mass of carbon) + (8 × atomic mass of hydrogen) + atomic mass of oxygen
Molar mass of propanol = (3 × 12.01 g/mol) + (8 × 1.008 g/mol) + 16.00 g/mol
Molar mass of propanol = 60.10 g/mol
Moles = mass / molar mass
Moles = 8.04 g / 60.10 g/mol
Moles = 0.134 mol
4. Calculate the partial pressure of propanol:
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
Rearrange the equation to solve for pressure (P):
P = (nRT) / V
Given that the temperature is 25°C, convert it to Kelvin:
T(°C) + 273.15 = T(K)
25°C + 273.15 = 298.15 K
Convert the pressure of propanol vapor from torr to atm:
1 atm = 760 torr
Vapor pressure of propanol = 10.7 torr / 760 torr/atm
Vapor pressure of propanol = 0.0141 atm
Now, calculate the partial pressure of propanol in the room:
P = (0.134 mol × 0.0821 atm⋅L/mol⋅K × 298.15 K) / 67500 L
P = 0.0142 atm
5. Compare the partial pressure of propanol to its vapor pressure:
If the partial pressure of propanol is less than its vapor pressure, then not all the propanol will evaporate. Otherwise, if the partial pressure is greater than or equal to the vapor pressure, then all the propanol will evaporate.
Since the partial pressure of propanol (0.0142 atm) is less than its vapor pressure (0.0141 atm), not all the propanol will evaporate in the sealed room.
6. Calculate the amount of remaining liquid propanol:
To find the amount of remaining liquid, subtract the amount of propanol that evaporated from the initial amount.
Remaining liquid = Initial liquid - Evaporated liquid
Remaining liquid = 0.134 mol - 0 mol (since all propanol did not evaporate)
Remaining liquid = 0.134 mol
Convert the remaining liquid to mass:
Mass = moles × molar mass
Mass = 0.134 mol × 60.10 g/mol
Mass = 8.07 g
Therefore, approximately 8.07 grams of propanol will remain in the sealed room.