The molecular weight of an unknown gas is found by measuring the time required for a known volume of the gas to effuse through a small hole, under constant pressure. The apparatus is calibrated by measuring the time needed for the same volume of O2 (Molecular weight =32g/mol) to effuse through the same pinhole, under the same conditions. The time found for O2 is 60s and that for the unknown gas is 120s compute the molecular weight of the unknown gas.
rate = volume/time and mm stands for molar mass.
(rateO2/rate unk) = sqrt[(mm unk)/mm O2)]
To compute the molecular weight of the unknown gas, we can use Graham's law of effusion, which states that the rates of effusion of two gases are inversely proportional to the square roots of their molecular weights.
Given:
Time for O2 to effuse (t1) = 60 s
Time for unknown gas to effuse (t2) = 120 s
Molecular weight of O2 (M1) = 32 g/mol
According to Graham's law, the ratio of the effusion rates is equal to the inverse ratio of the square roots of the molecular weights:
Rate of O2 / Rate of unknown gas = √(Molecular weight of unknown gas / Molecular weight of O2)
We can rearrange this equation to solve for the molecular weight of the unknown gas:
Molecular weight of unknown gas = (Rate of O2 / Rate of unknown gas)² * Molecular weight of O2
Now, let's calculate the rate of O2 (Rate1) and the rate of the unknown gas (Rate2):
Rate1 = Volume / Time1
Rate2 = Volume / Time2
Since the volumes remain the same (as mentioned in the question), we can simplify the equation to:
Rate1 / Rate2 = Time2 / Time1
Substituting the given values:
Rate1 / Rate2 = 120 s / 60 s = 2
Now, substitute the values into the equation for the molecular weight of the unknown gas:
Molecular weight of unknown gas = (2)² * 32 g/mol
Molecular weight of unknown gas = 4 * 32 g/mol
Molecular weight of unknown gas = 128 g/mol
Therefore, the molecular weight of the unknown gas is 128 g/mol.