Question

Dust particles are$10\mathrm{\mu m}$in diameter. They are pulverized rock, with $\mathrm{\rho }\approx 2500\mathrm{kg}/{\mathrm{m}}^3$. If you treat dust as an ideal gas, what is the rms speed of a dust particle at$20°\mathrm{C}$$?$

Short answer

In ideal gas state the rms speed of a dust particle at${20}^{°}\mathrm{C}$is$9.6·{10}^{-5}\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Formula for root mean square 

Root mean square speed can be given as,

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}$

where is the mass of one particle.

The volume and density of a particle can be used to compute its mass,

$\mathrm{m}=\mathrm{\rho V}$

Calculation for root square mean 

Root mean square of speed,

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{\rho V}}}$

substitute the volume of a sphere with diameter$\mathrm{d}$.

We have,

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2=\frac{1}{6}{\mathrm{\pi d}}^3$

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{\rho }\frac{1}{6}{\mathrm{\pi d}}^3}}=\sqrt{\frac{18{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{\pi \rho d}}^3}}$

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{18·1.38·{10}^{-23}·293}{\mathrm{\pi }·2500·{\left(1·{10}^{-5}\right)}^3}}$

${\mathrm{v}}_{\mathrm{rms}}=9.6·{10}^{-5}\mathrm{m}/\mathrm{s}$