Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 4.50atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.
Use PV = nRT and solve for n = total mols
mols CH4 = grams/molar mass = ?
mols C2H6 = grams/molar mass = ?
mols unknown gas = total mols - mols CH4-mols C2H6.
mol fraction = XCH4 = mols CH4/total mols.
XC2H6 = mols C2H6/total mols
Xunknown = mols unkn/total mols
pCH4 = XCH4*Ptotal
pC2h6 = XC2H6*Ptotal
punk = Xunk*Ptotal
To calculate the partial pressure of each gas in the container, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Let's calculate the number of moles for each gas:
For methane (CH4):
molecular weight of methane (CH4) = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen) = 16.05 g/mol
moles of methane (CH4) = mass / molecular weight = 8.00 g / 16.05 g/mol = 0.498 mol
For ethane (C2H6):
molecular weight of ethane (C2H6) = 2 * 12.01 g/mol (carbon) + 6 * 1.01 g/mol (hydrogen) = 30.07 g/mol
moles of ethane (C2H6) = mass / molecular weight = 18.0 g / 30.07 g/mol = 0.598 mol
Now, let's calculate the number of moles of propane (C3H8):
To find the number of moles of propane, we need to know the mass of propane that was added to the container.
Assuming the mass of propane is x grams:
molecular weight of propane (C3H8) = 3 * 12.01 g/mol (carbon) + 8 * 1.01 g/mol (hydrogen) = 44.11 g/mol
moles of propane (C3H8) = mass / molecular weight = x g / 44.11 g/mol
Now that we have the number of moles for each gas, we can calculate the partial pressure using the ideal gas law equation.
For methane (CH4):
P_methane = n_methane * R * T / V
P_methane = 0.498 mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L
For ethane (C2H6):
P_ethane = n_ethane * R * T / V
P_ethane = 0.598 mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L
For propane (C3H8):
P_propane = n_propane * R * T / V
P_propane = (x g / 44.11 g/mol) * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L
Since the total pressure in the container is given as 4.50 atm, we can use this information to find x.
Total pressure = P_methane + P_ethane + P_propane
4.50 atm = P_methane + P_ethane + P_propane
Now, we can substitute the calculated pressure expressions and solve for x.
Substituting the expressions for P_methane, P_ethane, and P_propane, and solving for x:
4.50 atm = (0.498 mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L) + (0.598 mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L) + (x g / 44.11 g/mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L)
Simplifying the equation and solving for x:
4.50 atm = 0.445 atm + 0.534 atm + (x * 0.0821 L·atm/mol·K * 296 K / 10.0 L)
3.521 atm = (x * 2.448896 L·atm/mol)
x = 3.521 atm / 2.448896 L·atm/mol
x = 1.435 mol
Now that we have found the number of moles of propane (C3H8), we can calculate its partial pressure:
P_propane = (1.435 mol * 0.0821 L·atm/(mol·K) * (23.0 + 273) K / 10.0 L)
Therefore, the partial pressures are as follows:
Methane (CH4): 0.445 atm
Ethane (C2H6): 0.534 atm
Propane (C3H8): 1.435 atm
To calculate the partial pressure of each gas in the container, we can use the equation:
Partial Pressure = (n/M) * R * T
Where:
Partial Pressure is the pressure exerted by each gas,
n is the number of moles of each gas,
M is the molar mass of each gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin.
Step 1: Calculate the number of moles for each gas.
For methane (CH4):
n(CH4) = mass(CH4) / molar mass(CH4)
Given that the mass of methane (CH4) is 8.00 g and its molar mass is 16.04 g/mol:
n(CH4) = 8.00 g / 16.04 g/mol = 0.498 mol
For ethane (C2H6):
n(C2H6) = mass(C2H6) / molar mass(C2H6)
Given that the mass of ethane (C2H6) is 18.0 g and its molar mass is 30.07 g/mol:
n(C2H6) = 18.0 g / 30.07 g/mol = 0.598 mol
For propane (C3H8):
n(C3H8) = mass(C3H8) / molar mass(C3H8)
The mass of propane (C3H8) is not given. So we need to use the given information that the total pressure is 4.50 atm to calculate the partial pressure of propane (C3H8). We will come back to this calculation in Step 3.
Step 2: Convert the given temperature to Kelvin.
The given temperature is 23.0 ∘C. To convert it to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
T(K) = 23.0 ∘C + 273.15 = 296.15 K
Step 3: Calculate the partial pressure of propane (C3H8).
To find the partial pressure of propane (C3H8), we will subtract the sum of the partial pressures of methane (CH4) and ethane (C2H6) from the total pressure of the container, which is 4.50 atm.
Partial Pressure(C3H8) = Total Pressure - (Partial Pressure(CH4) + Partial Pressure(C2H6))
Partial Pressure(C3H8) = 4.50 atm - (Partial Pressure(CH4) + Partial Pressure(C2H6))
Step 4: Calculate the partial pressures.
Now we can calculate the partial pressures of methane (CH4), ethane (C2H6), and propane (C3H8) using the equation:
Partial Pressure = (n/M) * R * T
Partial Pressure(CH4) = (0.498 mol / 16.04 g/mol) * 0.0821 L·atm/(mol·K) * 296.15 K
Partial Pressure(C2H6) = (0.598 mol / 30.07 g/mol) * 0.0821 L·atm/(mol·K) * 296.15 K
Partial Pressure(C3H8) = 4.50 atm - (Partial Pressure(CH4) + Partial Pressure(C2H6))
Now we can plug in the values and calculate the partial pressures of each gas.
Partial Pressure(CH4) = (0.498 mol / 16.04 g/mol) * 0.0821 L·atm/(mol·K) * 296.15 K = 0.972 atm
Partial Pressure(C2H6) = (0.598 mol / 30.07 g/mol) * 0.0821 L·atm/(mol·K) * 296.15 K = 0.976 atm
Partial Pressure(C3H8) = 4.50 atm - (0.972 atm + 0.976 atm) = 2.552 atm
Therefore, the partial pressures of each gas in the container are:
Methane (CH4): 0.972 atm
Ethane (C2H6): 0.976 atm
Propane (C3H8): 2.552 atm