Question

The kinetic molecular theory attributes an average kinetic energy of $3/2KT$ to each particle. What rms speed would a mist particle of mass ${10}^{12}gm$ have at room temperture $\left({27}^0\circ C\right)$ according to the kinetic molecular theory.

Answer 1

Ans: $V_{rms}=3.525\times {10}^{-15}m{s}^{-1}$

Root mean square velocity

$V_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ pen mole

$\therefore RT=KT$ per molecule / particle

$K=1.380\times {10}^{-23}{m}^{2}k{g}^{5^{-2}}{K}^{-1}$

$M={10}^{12}gm={10}^{9}kg$

$T={27}^{o}C=(20+273)K=300K$

$\therefore V_{rms}={\left(\frac{3KT}{M}\right)}^{1/2}$

$={\left(\frac{3\times 138064\times {10}^{-23}{m}^{2}kg{s}^{-2}\times 300K}{{10}^{9}kg}\right)}^{1/2}$

$={(\frac{12.4258}{10}\times {10}^{-21})}^{1/2}m{s}^{-1}$

$=3.525\times {10}^{-15}m{s}^{-1}$