KCET · Chemistry · States of Matter
One mole of oxygen at \(273 \mathrm{~K}\) and one mole of sulphur dioxide at \(546 \mathrm{~K}\) are taken in two separate containers, then,
- A kinetic energy of \(\mathrm{O}_{2}>\) kinetic energy of \(\mathrm{SO}_{2}\)
- B kinetic energy of \(\mathrm{O}_{2} < \) kinetic energy of \(\mathrm{SO}_{2}\)
- C kinetic energy of both are equal
- D None of the above
Answer & Solution
Correct Answer
(B) kinetic energy of \(\mathrm{O}_{2} < \) kinetic energy of \(\mathrm{SO}_{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{KE}=\frac{3}{2} R T\) \(\mathrm{KE} \propto T\)
\[
\begin{aligned}
& \frac{\mathrm{KE}_{\mathrm{O}_{2}}}{\mathrm{KE}_{\mathrm{SO}_{2}}}=\frac{T_{\mathrm{O}_{2}}}{T_{\mathrm{SO}_{2}}}=\frac{273}{546}=\frac{1}{2} \\
\mathrm{KE}_{\mathrm{SO}_{2}} &=2 \mathrm{KE}_{\mathrm{O}_{2}} \\
\therefore \quad \mathrm{KE}_{\mathrm{SO}_{2}} &>2 \mathrm{KE}_{\mathrm{O}_{2}}
\end{aligned}
\]
\[
\begin{aligned}
& \frac{\mathrm{KE}_{\mathrm{O}_{2}}}{\mathrm{KE}_{\mathrm{SO}_{2}}}=\frac{T_{\mathrm{O}_{2}}}{T_{\mathrm{SO}_{2}}}=\frac{273}{546}=\frac{1}{2} \\
\mathrm{KE}_{\mathrm{SO}_{2}} &=2 \mathrm{KE}_{\mathrm{O}_{2}} \\
\therefore \quad \mathrm{KE}_{\mathrm{SO}_{2}} &>2 \mathrm{KE}_{\mathrm{O}_{2}}
\end{aligned}
\]
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