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Question: Answered & Verified by Expert
$0.5$ mole of each of $\mathrm{H}_{2}, \mathrm{SO}_{2}$ and $\mathrm{CH}_{4}$ are kept in a container. A hole was made in the container. After $3 \mathrm{~h}$, the order of partial pressures in the container will be
ChemistryStates of MatterKCETKCET 2009
Options:
  • A $\mathrm{pSO}_{2}>\mathrm{pH}_{2}>\mathrm{pCH}_{4}$
  • B $\mathrm{pSO}_{2}>\mathrm{pCH}_{4}>\mathrm{pH}_{2}$
  • C $\mathrm{pH}_{2}>\mathrm{pSO}_{2}>\mathrm{pCH}_{4}$
  • D $\mathrm{pH}_{2}>\mathrm{pCH}_{4}>\mathrm{pSO}_{2}$
Solution:
1299 Upvotes Verified Answer
The correct answer is: $\mathrm{pSO}_{2}>\mathrm{pCH}_{4}>\mathrm{pH}_{2}$
Rate of diffusion $\propto \frac{1}{\sqrt{\text { molecular mass }}}$
$\therefore$ Order of diffusion $: \mathrm{H}_{2}>\mathrm{CH}_{4}>\mathrm{SO}_{2}$ and amount left is in the order $\mathrm{SO}_{2}>\mathrm{CH}_{4}>\mathrm{H}_{2}$ Hence, order of partial pressure is
$$
\mathrm{pSO}_{2}>\mathrm{pCH}_{4}>\mathrm{pH}_{2}
$$

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