The gases $A$ and $B$ are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by gas $A$ is four times the time taken by gas $B$ for diffusion of equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular weights?

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Solution

$\frac{γ_{A}}{γ_{B}} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}} ⟹ \frac{t_{B}}{t_{A}} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}}$ (since equal volume of both gases is taken)
$⟹ \frac{1}{4} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}} ⟹ \frac{1}{16} = \frac{M_{B}}{M_{A}} ⟹ M_{A} : M_{B} = 16 : 1$

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The gases A and B are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by gas A is four times the time taken by gas B for diffusion of equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular weights?

γAγB=√MB√MA⟹tBtA=√MB√MA (since equal volume of both gases is taken)⟹14=√MB√MA⟹116=MBMA⟹MA:MB=16:1

$\frac{γ_{A}}{γ_{B}} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}} ⟹ \frac{t_{B}}{t_{A}} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}}$ (since equal volume of both gases is taken)
$⟹ \frac{1}{4} = \frac{\sqrt{M_{B}}}{\sqrt{M_{A}}} ⟹ \frac{1}{16} = \frac{M_{B}}{M_{A}} ⟹ M_{A} : M_{B} = 16 : 1$