A certain volume of argon gas ( Mol.wt = 40) requires 45 s to diffuse through a hole at a certain pressure and temperature. The same volume of another gas of unknown molecular weight requires 60 s to pass through the same hole under same conditions?

1 Answer
Sep 1, 2017

71 g/ mol


since kinetic energy of every gases dipends only from the absolute temperature (Ek = KT) and then at a certain temerature is equal for every gasses, and kinetic energy is also given from # E_k= 1/2 m v^2# you have that v^2 is proportional to the inverse of the molecular mass.
For that, the Graham's law says # (Mb)/(Ma) =V_a^2/V_b^2 #
but speed is proportional to the inverse of the time occurred to diffuse so
# Mb = Ma xx t_b^2/t_a^2 =40 g/(mol) xx 60^2/45^2= 71 g/(mol)#