Question

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

• A. $10^6$ times the average speed of an air molecule
• B. $10^{-12}$ times the average speed of an air molecule
• C. $10^{12}$ times the average speed of an air molecule
• D. $10^{-6}$ times the average speed of an air molecule
• E. $10^{-10}$ times the average speed of an air molecule

Answer 1

Answer: (D)

Let there is elastic collision and both act like two ideal gases then.
$v_{\text{avg }}=\sqrt{\frac{8kT}{\pi m}}$
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.