The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given $\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$]
Solution
At NTP, $T=298 \mathrm{~K}$
<br/><br/>$$
\begin{aligned}
v_{\mathrm{rms}} &=\sqrt{\frac{3 R T}{M}} \\
&=\sqrt{\frac{3 k N_{A} \times 298}{5 \times 10^{-17} \times N_{A}}}
\end{aligned}
$$
<br/><br/>$\simeq 15 \mathrm{~mm} / \mathrm{s}$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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