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The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes
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Verified Answer
The correct answer is:
$2v$
As the rms speed is given by
$$
\begin{aligned}
& v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}{V_{\text {ms }}} \propto \sqrt{\frac{T}{M}}
\end{aligned}
$$
When temperature is doubled and molecules dissociates into atoms, then
$$
\begin{aligned}
\frac{v_{\mathrm{rms}_{1}}}{v_{\mathrm{rms}_{2}}} &=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{2 T}{M / 2}}}=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{4 T}{M}}} \\
&=\frac{\sqrt{T}}{\sqrt{4 T}}=\frac{1}{2}
\end{aligned}
$$
If $v_{\mathrm{rms}},$ is $v,$ then $v_{\mathrm{rms}_{2}}$ will be $2 v$.
$$
\begin{aligned}
& v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}{V_{\text {ms }}} \propto \sqrt{\frac{T}{M}}
\end{aligned}
$$
When temperature is doubled and molecules dissociates into atoms, then
$$
\begin{aligned}
\frac{v_{\mathrm{rms}_{1}}}{v_{\mathrm{rms}_{2}}} &=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{2 T}{M / 2}}}=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{4 T}{M}}} \\
&=\frac{\sqrt{T}}{\sqrt{4 T}}=\frac{1}{2}
\end{aligned}
$$
If $v_{\mathrm{rms}},$ is $v,$ then $v_{\mathrm{rms}_{2}}$ will be $2 v$.
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