The vapor pressure of CCl3F at 300 K is 856 torr. If 11.5 g of CCl3F is enclosed in a 1.0 L container, will any liquid be present? If so, what mass of liquid?
What would be the mass of the gas?
PV=nRT solve for N. Now what is the mass of n moles of CCl3F? The mass left would be liquid.
123
To determine if any liquid will be present when the CCl3F is enclosed in a 1.0 L container, we can compare the given vapor pressure to the pressure exerted by the gas at a given temperature.
First, we need to convert the mass of CCl3F to moles. To do this, we'll use the molar mass of CCl3F, which is:
CCl3F: 12.01 (C) + 1.01 x 3 (H) + 35.45 x 3 (Cl) + 18.99 (F) = 137.32 g/mol
Number of moles of CCl3F = mass of CCl3F / molar mass of CCl3F
Number of moles = 11.5 g / 137.32 g/mol ≈ 0.0837 mol
Next, we can use the ideal gas law to calculate the pressure exerted by the gas using moles, temperature, and volume:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature
Let's substitute the given values into the equation:
P = (nRT) / V
P = (0.0837 mol)(0.0821 L·atm/(mol·K))(300 K) / 1.0 L
P ≈ 2.47 atm
Now, we can compare the calculated pressure to the vapor pressure of CCl3F at 300 K, which is 856 torr. To convert the vapor pressure to atm, we divide by 760 (as 1 atm = 760 torr):
Vapor pressure of CCl3F = 856 torr / 760 torr/atm ≈ 1.13 atm
Since the calculated pressure (2.47 atm) is higher than the vapor pressure (1.13 atm), it implies that the CCl3F will be fully gaseous at these conditions. Therefore, there will not be any liquid present in the container.
In conclusion, no liquid will be present in the 1.0 L container.
To determine if any liquid will be present, we need to compare the vapor pressure of the substance to the partial pressure exerted by the substance in the given conditions. If the vapor pressure is equal to or greater than the partial pressure, then all the substance will be in the gas phase, and no liquid will be present. However, if the vapor pressure is less than the partial pressure, then some of the substance will be in the liquid phase.
To find the partial pressure, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the mass of CCl3F to number of moles by using its molar mass.
The molar mass of CCl3F can be calculated as follows:
Molar mass = atomic mass of C + (atomic mass of Cl × 3) + atomic mass of F
Molar mass = (12.01 g/mol) + (35.45 g/mol × 3) + 18.998 g/mol
= 12.01 g/mol + 106.35 g/mol + 18.998 g/mol
= 137.358 g/mol
Now, let's calculate the number of moles of CCl3F:
Number of moles = mass / molar mass
= 11.5 g / 137.358 g/mol
≈ 0.0838 mol
Next, let's calculate the partial pressure:
Partial pressure = (n/V) × RT
Where:
n/V = moles of CCl3F per liter = 0.0838 mol / 1.0 L
R = 0.0821 atm·L/(mol·K) (ideal gas constant)
T = 300 K
Partial pressure = (0.0838 mol / 1.0 L) × (0.0821 atm·L/(mol·K)) × 300 K
= 2.473 atm
Now, we can compare the vapor pressure (given as 856 torr) to the partial pressure (2.473 atm):
To convert the vapor pressure to atm:
1 atm = 760 torr
Vapor pressure = 856 torr / 760 torr/atm
= 1.126 atm
Since the vapor pressure (1.126 atm) is less than the partial pressure (2.473 atm), some of the CCl3F will be in the liquid phase.
To determine the mass of liquid present, we need to find the mass of CCl3F that corresponds to the difference between the total mass (11.5 g) and the mass in the gas phase.
Mass of liquid = total mass - mass in gas phase
To calculate the mass in the gas phase, we need to determine the number of moles of CCl3F that will be in the gas phase.
Using the ideal gas law:
PV = nRT
Rearranging the equation to solve for n:
n = PV / RT
The number of moles in the gas phase:
n = (partial pressure) × V / (R × T)
= (2.473 atm) × (1.0 L) / (0.0821 atm·L/(mol·K)) × (300 K)
≈ 0.100 mol
Now, let's calculate the mass of CCl3F in the gas phase:
Mass of CCl3F in the gas phase = number of moles × molar mass
= 0.100 mol × 137.358 g/mol
= 13.736 g
Finally, let's calculate the mass of liquid:
Mass of liquid = total mass - mass in gas phase
= 11.5 g - 13.736 g
≈ -2.236 g
The negative mass indicates that there is an excess amount of CCl3F in the gas phase. Therefore, no liquid will be present in the 1.0 L container.