Question

What rms velocity would a dust particle of mass ${10}^{-13}g$ have at ${27}^{0}C$ according to molecular kinetic theory?

• A. $1.11cm{s}^{-1}$
• B. $3.11cm{s}^{-1}$
• C. $5.0cm{s}^{-1}$
• D. $6.5cm{s}^{-1}$

Answer 1

The correct option is A $1.11cm{s}^{-1}$

K. E. per molecule $=\frac{3}{2}kT$

$=\frac{3}{2}.\frac{R}{N}.T$

lf the mass of the particle is m, the Kinetic Energy of this molecule $=\frac{1}{2}m{c}^2$, where c is the rms speed.

$\frac{1}{2}m{c}^2=\frac{3}{2}.\frac{R}{N}.T$

$c=\sqrt{3\times \frac{R}{N}\times \frac{T}{m}}$

$=\sqrt{\frac{3\times 8.314\times {10}^7\times 300}{6.02\times {10}^{23}\times {10}^{-13}}}$

$=\sqrt{1.24}=1.11cm{s}^{-1}$