KCET · Chemistry · States of Matter
\(0.5\) 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
- 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 \(\mathrm{pH}_{2}>\mathrm{pCH}_{4}>\mathrm{pSO}_{2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{pSO}_{2}>\mathrm{pCH}_{4}>\mathrm{pH}_{2}\)
Step-by-step Solution
Detailed explanation
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
\[
\mathrm{pSO}_{2}>\mathrm{pCH}_{4}>\mathrm{pH}_{2}
\]
\(\therefore\) 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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