We mix 146 grams of oxygen gas with
117 grams of argon gas in a volume of 436 mL
at 58�C. What will be the final pressure of
the gas mixture?
Answer in units of atm
Convert grams to mol, find total mols then use PV = nRT and solve for P.
To find the final pressure of the gas mixture, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K) )
T is the temperature of the gas in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 58°C + 273.15 = 331.15 K
Next, we need to calculate the number of moles of oxygen gas (O2) and argon gas (Ar) using their respective molar masses.
The molar mass of oxygen gas (O2) is 32 g/mol.
The molar mass of argon gas (Ar) is 40 g/mol.
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 146 g / 32 g/mol
Moles of O2 = 4.5625 mol
Moles of Ar = Mass of Ar / Molar mass of Ar
Moles of Ar = 117 g / 40 g/mol
Moles of Ar = 2.925 mol
Now that we have the number of moles for each gas, we can calculate the total number of moles:
Total moles = Moles of O2 + Moles of Ar
Total moles = 4.5625 mol + 2.925 mol
Total moles = 7.4875 mol
Next, let's convert the volume from mL to L:
Volume (V) = 436 mL / 1000 mL/L
Volume (V) = 0.436 L
Now we have all the required values to calculate the final pressure using the ideal gas law equation:
PV = nRT
P * 0.436 L = (7.4875 mol) * (0.0821 L·atm/(mol·K)) * 331.15 K
We can rearrange the equation to solve for P:
P = (nRT) / V
P = (7.4875 * 0.0821 * 331.15) / 0.436
P ≈ 34.587 atm
Therefore, the final pressure of the gas mixture is approximately 34.587 atm.