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The most probable speed of $\mathrm{O}_2$ molecules at $T(\mathrm{~K})$ is
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2889 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\frac{R T}{16}}$
$$
\begin{aligned}
& \text { Most probable speed, } v_{\mathrm{mp}}=\sqrt{\frac{2 R T}{M}} \\
& =\sqrt{\frac{2 R T}{32}} \quad \text { (for oxygen, } M=32 \text { ) } \\
& =\sqrt{\frac{R T}{16}} \\
&
\end{aligned}
$$
\begin{aligned}
& \text { Most probable speed, } v_{\mathrm{mp}}=\sqrt{\frac{2 R T}{M}} \\
& =\sqrt{\frac{2 R T}{32}} \quad \text { (for oxygen, } M=32 \text { ) } \\
& =\sqrt{\frac{R T}{16}} \\
&
\end{aligned}
$$
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