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Q. Some smoke is trapped in a small glass container and is viewed through a microscope. A number of very small smoke particles are seen in continuous random motion as a result of their bombardment by air molecules. If the mass of the smoke particle is about $10^{12}$ times higher than that of an air molecule the average speed of a smoke particle is

KEAMKEAM 2019Kinetic Theory

Solution:

Let there is elastic collision and both act like two ideal gases then.
$v_{\text {avg }}=\sqrt{\frac{8 k T}{\pi m}}$
where temperature is constant, so $v \propto \sqrt{\frac{1}{m}}$
Hence the ratio $\frac{v_{\text {smoke }}}{v_{\text {air }}}=\sqrt{\frac{m_{\text {air }}}{m_{\text {smoke }}}}$
Given, $m_{\text {smoke }}=10^{12} m_{\text {air }}$
$\Rightarrow \frac{v_{\text {smoke }}}{v_{\text {air }}}=\sqrt{\frac{m_{\text {air }}}{10^{12} m _{\text {air }}}}=10^{-6}$
So, $v_{\text {smoke }}$ is $10^{-6}$ times the average speed of an air molecule.