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The kinetic molecular theory attributes an average translational kinetic energy of $\large\frac{3}{2}\large\frac{RT}{N}$ to each particle. What rms speed woul a mist particle of mass $10^{-12}g$ have a root temperature $(27^{\large\circ}C)$ according to kinetic theory of gases.

$(a)\;0.35 \;cm\; sec^{-1}\qquad(b)\;35 \;cm\; sec^{-1}\qquad(c)\;3.5 \;cm\; sec^{-1}\qquad(d)\;0.035 \;cm\; sec^{-1}$

1 Answer

Let mass of mist particle be m , then K.E of this particle = $\large\frac{1}{2}mu^2$
Where u is the its rms velocity
Also K.E per molecule = $\large\frac{3RT}{2\times N}$
$\therefore \large\frac{1}{2}mu^2 = \large\frac{3RT}{2\times N}$
$u = \sqrt{[3\large\frac{R}{N}\times \large\frac{T}{m}]}$
$m = 10^{-12}g R = 8.314\times10^7 \;erg \;T = 300K$
$u = \sqrt{\large\frac{3\times8.314\times10^7\times300}{6.023\times10^{23}\times10^{-12}}}$
Hence answer is (A)
answered Apr 7, 2014 by sharmaaparna1

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