What volume will be required to contain a mixture of 0.547 mol of N2 and 0.547 mol of O2 at 1.11 atm and 22.1 degrees celsius?
To find the volume required to contain the mixture of N2 and O2, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L*atm/K*mol)
T = temperature (in Kelvin)
First, we need to convert the given temperature from degrees Celsius to Kelvin:
T(Kelvin) = T(Celsius) + 273.15
T(K) = 22.1 + 273.15
T(K) = 295.25 K
Now, we can plug in the known values into the ideal gas law equation:
P * V = n * R * T
We are given:
P = 1.11 atm
n(N2) = 0.547 mol
n(O2) = 0.547 mol
R = 0.0821 L*atm/K*mol
T = 295.25 K
Since we have equal moles of N2 and O2, we can add the number of moles together:
n(total) = n(N2) + n(O2)
n(total) = 0.547 mol + 0.547 mol
n(total) = 1.094 mol
Now we can rearrange the ideal gas law equation to solve for V:
V = (n * R * T) / P
V = (1.094 mol * 0.0821 L*atm/K*mol * 295.25 K) / 1.11 atm
Evaluating this expression:
V = 23.38 L
Therefore, the volume required to contain the mixture of N2 and O2 is approximately 23.38 liters.