• # question_answer X ml of ${{H}_{2}}$ 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 [IIT 1996] A)                 10 seconds : He B)                 20 seconds : ${{O}_{2}}$ C)                 25 seconds : CO D)                 55 seconds : $C{{O}_{2}}$

Correct Answer: B

Solution :

$r\propto \frac{1}{\sqrt{M}}$      $\because \,\,r=\frac{Volume\,\,effused}{time\,\,taken}=\frac{V}{t}$                    $\frac{V}{t}\propto \frac{1}{\sqrt{M}}$ \for same volumes (V constant)                    $t\propto \sqrt{M}\,\therefore \frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{M}_{1}}}{{{M}_{2}}}}$                    ${{t}_{He}}={{t}_{{{H}_{2}}}}\sqrt{\frac{{{M}_{He}}}{{{M}_{{{H}_{2}}}}}}=5\sqrt[{}]{\frac{4}{2}}=5\sqrt{2}s.$                    ${{t}_{{{O}_{2}}}}=t$$=5\sqrt[{}]{\frac{32}{2}}=20s$                                 ${{t}_{CO}}=$$5\sqrt[{}]{\frac{28}{2}}=5\sqrt{14}s$ ;  ${{t}_{C{{O}_{2}}}}=$$5\sqrt[{}]{\frac{44}{2}}=5\sqrt{22}s$

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