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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,
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Verified Answer
The correct answer is:
kinetic energy of $\mathrm{O}_{2} < $ kinetic energy of $\mathrm{SO}_{2}$
$\mathrm{KE}=\frac{3}{2} R T$ $\mathrm{KE} \propto T$
$$
\begin{aligned}
& \frac{\mathrm{KE}_{\mathrm{O}_{2}}}{\mathrm{KE}_{\mathrm{SO}_{2}}}=\frac{T_{\mathrm{O}_{2}}}{T_{\mathrm{SO}_{2}}}=\frac{273}{546}=\frac{1}{2} \\
\mathrm{KE}_{\mathrm{SO}_{2}} &=2 \mathrm{KE}_{\mathrm{O}_{2}} \\
\therefore \quad \mathrm{KE}_{\mathrm{SO}_{2}} &>2 \mathrm{KE}_{\mathrm{O}_{2}}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{KE}_{\mathrm{O}_{2}}}{\mathrm{KE}_{\mathrm{SO}_{2}}}=\frac{T_{\mathrm{O}_{2}}}{T_{\mathrm{SO}_{2}}}=\frac{273}{546}=\frac{1}{2} \\
\mathrm{KE}_{\mathrm{SO}_{2}} &=2 \mathrm{KE}_{\mathrm{O}_{2}} \\
\therefore \quad \mathrm{KE}_{\mathrm{SO}_{2}} &>2 \mathrm{KE}_{\mathrm{O}_{2}}
\end{aligned}
$$
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