If two gases CH_4 and SO_2 are allowed to ent | vedclass

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(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \f

Vedclass · Accepted Answer

$667$

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(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Let $d_1$ be the distance traveled by $CH_4$ and $d_2$ be the distance traveled by $SO_2$.Since they start at the same time and meet at time $t$,the ratio of distances is equal to the ratio of their rates of diffusion: $\frac{d_1}{d_2} = \frac{r_{CH_4}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{CH_4}}}$.Given $M_{CH_4} = 16 \ g/mol$ and $M_{SO_2} = 64 \ g/mol$.$\frac{d_1}{d_2} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2$.So,$d_1 = 2d_2$.Since the total length is $1 \ km = 1000 \ m$,$d_1 + d_2 = 1000$.Substituting $d_2 = \frac{d_1}{2}$,we get $d_1 + \frac{d_1}{2} = 1000$.$\frac{3d_1}{2} = 1000 \implies d_1 = \frac{2000}{3} \approx 667 \ m$.

If two gases $CH_4$ and $SO_2$ are allowed to enter from the two ends of a $1 \ km$ long vacuum tube at the same time,where will the gases meet from the $CH_4$ end (in $m$)?

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