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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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In the figure below, $N_2$ and $H_2$ are sent into the metre pipe through the two ends. At what distance $x$ will Ammonia be formed.

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Rate of diffusion depends upon the mass of the gas as $r \propto \large\frac{1}{\sqrt m}$
Also, $x = $ Rate of Diffusion of $H_2 \times t$ and $1-x =$ Rate of Diffusion of $N_2 \times t$
$\Rightarrow \large\frac{x}{r_{H_2}}$ $= \large\frac{1-x}{r_{N_2}}$
$\Rightarrow \large\frac{x}{1-x}$ $= \large\frac{r_{H_2}}{r_{N_2}}$
$\Rightarrow \large\frac{x}{1-x}$ $= \large\frac{\sqrt 28}{\sqrt 2}$
$\Rightarrow x = 0.79$
answered Mar 11, 2014 by balaji.thirumalai

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