If you place 10.0 L of propanol (C3H8O) in a sealed room that is 3 m long, 2.75 m wide, and 2.5 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.
I would convert the measurements to cm; therefore, 300 cm x 275 cm x 250 cm = 2.06E7 cc.
Convert cc to L = 2.05E7 cc x (1L/1000cc) = approx 2.06E4 L but you need to confirm all of these numbers and if needed with more accuracy.
Then PV = nRT.
Substitute 10.7 torr(converted to atm) \ 10.7/760 = ?
Use V from above in the room.
You know R and T (in kelvin), solve for n = mols and convert to grams. Compare with 10.0L propanol converted to grams with density. Post your work if you get stuck.
To determine if all the propanol will evaporate in the given room, we need to calculate the total volume of the room and compare it to the volume of the propanol.
Step 1: Calculate the volume of the room.
Volume of the room = length × width × height.
Volume of the room = 3 m × 2.75 m × 2.5 m = 20.625 m^3.
Step 2: Convert the volume of the room to liters.
1 m^3 = 1000 L.
Volume of the room = 20.625 m^3 × 1000 L/m^3 = 20,625 L.
Step 3: Calculate the weight of the propanol using its density.
Weight of propanol = volume × density.
Weight of propanol = 10.0 L × 0.804 g/mL = 8.04 kg.
Step 4: Convert the weight of the propanol to grams.
1 kg = 1000 g.
Weight of propanol = 8.04 kg × 1000 g/kg = 8040 g.
Step 5: Convert grams to moles.
Molar mass of propanol (C3H8O) = (3 × atomic mass of carbon) + (8 × atomic mass of hydrogen) + (1 × atomic mass of oxygen).
Molar mass of propanol = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (1 × 16.00 g/mol) = 60.10 g/mol.
Moles of propanol = weight of propanol / molar mass of propanol.
Moles of propanol = 8040 g / 60.10 g/mol = 133.94 mol.
Step 6: Calculate the volume of the propanol vapors at 25 °C using the ideal gas law.
P = nRT/V, where P is the pressure, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, and V is the volume.
Rearranging the formula: V = nRT/P.
Convert the pressure to atm:
1 torr = 0.00131579 atm.
Vapor pressure = 10.7 torr × 0.00131579 atm/torr = 0.014 atm.
Convert the temperature to Kelvin:
25 °C = 298 K.
Volume of propanol vapors = moles of propanol × R × T / P.
Volume of propanol vapors = 133.94 mol × 0.0821 L·atm/(mol·K) × 298 K / 0.014 atm = 16788.62 L.
Step 7: Compare the volume of the propanol vapors to the volume of the room.
Since the volume of the propanol vapors (16788.62 L) is larger than the volume of the room (20,625 L), not all the propanol will evaporate.
Step 8: Calculate the remaining liquid propanol.
Remaining liquid propanol = volume of the room - volume of propanol vapors.
Remaining liquid propanol = 20,625 L - 16788.62 L = 3836.38 L.
So, there will be approximately 3836.38 L of liquid propanol remaining in the sealed room.
To determine whether all the propanol will evaporate or if some liquid will remain, we need to compare the vapor pressure of propanol to the partial pressure of propanol in the sealed room.
First, let's calculate the volume of the room:
Room Volume = length × width × height = 3 m × 2.75 m × 2.5 m = 20.625 m^3
Next, we need to find the number of moles of propanol in 10.0 L:
Number of moles of propanol = volume of propanol (Liters) × density of propanol (g/mL) ÷ molar mass of propanol (g/mol)
Molar mass of propanol (C3H8O) = (3 × molar mass of carbon) + (8 × molar mass of hydrogen) + (1 × molar mass of oxygen)
Molar mass of carbon = 12.01 g/mol
Molar mass of hydrogen = 1.01 g/mol
Molar mass of oxygen = 16.00 g/mol
Now we can calculate the molar mass of propanol:
Molar mass of propanol = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (1 × 16.00 g/mol)
Using the given data, the density of propanol at 25 °C is 0.804 g/mL. Let's convert that to g/L:
Density of propanol (g/L) = 0.804 g/mL × 1000 mL/L
With this information, we can calculate the number of moles of propanol:
Number of moles of propanol = 10.0 L × (0.804 g/mL × 1000 mL/L) ÷ (molar mass of propanol)
Now we have the number of moles of propanol. To find the partial pressure of propanol, we can use the ideal gas law equation:
Pressure × Volume = n × R × Temperature
We know the volume of the room and the temperature is 25 °C, which we need to convert to Kelvin:
Temperature (Kelvin) = 25 °C + 273.15
The ideal gas law equation becomes:
Pressure × 20.625 m^3 = (number of moles of propanol) × (R constant) × (temperature in Kelvin)
We can rearrange the equation to solve for pressure:
Pressure = (number of moles of propanol) × (R constant) × (temperature in Kelvin) ÷ 20.625 m^3
Given that the vapor pressure of propanol is 10.7 torr, we need to convert it to atm:
Vapor pressure of propanol (atm) = 10.7 torr ÷ 760 torr/atm
Now, we can compare the vapor pressure of propanol to the partial pressure of propanol in the room. If the partial pressure is less than the vapor pressure, then all the propanol will evaporate; otherwise, if the partial pressure is greater than or equal to the vapor pressure, some liquid will remain.
Let's calculate the partial pressure of propanol and compare it to the vapor pressure to determine whether all the propanol will evaporate or if some liquid will remain.