First slide

Graham's law

Question

X mL of H₂ gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical condition is:

Moderate

A

10 sec : He
B

20 sec : O₂
C

25 sec : CO
D

55 sec : CO₂

Solution

$\begin{array}{l} \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \\ \frac{V_{1}}{t_{1}} \times \frac{t_{2}}{V_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \\ \frac{t_{2}}{t_{1}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{t_{2}}{5} = \sqrt{\frac{M_{2}}{2}} \end{array}$
From options, (2) is suitable.

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