If oxygen effuses through an opening at a rate of 67 particles per second, how fast would effuses Hydrogen gas effuse through the same opening?

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Answer 1 · 2016-05-01T10:23:27

$268$ particles per second.

Graham's law of effusion says that

$R_1/R_2=sqrt(M_2/M_1)$ ,

where $R$ is the rate of effusion and $M$ is the mass. The subscript number refers to either oxygen or hydrogen - it's a relative equation.

If we say that substance $1$ is hydrogen and $2$ is oxygen, then

$R_1=?$
$R_2=67$

$M_1=1$
$M_2=16$

( $M$ values from the periodic table).

Putting these values into a rearranged equation,

$R_1=R_2*sqrt(M_2/M_1)$
$R_1=67*sqrt(16/1)=67*4=268$