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?

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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?

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Answer 1 · 2017-09-01T05:29:08

71 g/ mol

Explanation:

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 $E_{k} = \frac{1}{2} m v^{2}$ $E_k= 1/2 m v^2$ you have that v^2 is proportional to the inverse of the molecular mass.
For that, the Graham's law says $\frac{M b}{M a} = \frac{V_{a}^{2}}{V_{b}^{2}}$ $(Mb)/(Ma) =V_a^2/V_b^2$
but speed is proportional to the inverse of the time occurred to diffuse so
$M b = M a \times \frac{t_{b}^{2}}{t_{a}^{2}} = 40 \frac{g}{m o l} \times \frac{60^{2}}{45^{2}} = 71 \frac{g}{m o l}$ $Mb = Ma xx t_b^2/t_a^2 =40 g/(mol) xx 60^2/45^2= 71 g/(mol)$

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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

Explanation:

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