Q. For a monoatomic gas kinetic energy $= E.$ The relation with rms velocity is
A) $u = \left( \frac{2E}{m} \right)^{1/2}$
B) $u = \left( \frac{3E}{2m} \right)^{1/2}$
C) $u = \left( \frac{E}{2m} \right)^{1/2}$
D) $u = \left( \frac{E}{3m} \right)^{1/2}$
Explanation
(a)
RMS velocity $u_{rms} = \sqrt{\frac{3pV}{M}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
and $pV = nkT\ \ \ \ \ \left( k \rightarrow {Boltzmann}^{'}s\ constant \right)$
For a molecule $n = 1$
$pV = kT$
So, $u_{rms} = \sqrt{\frac{3kT}{m}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Kinetic energy $(E) = \frac{3}{2}\ kT\ or\ kT = \frac{2}{3}\ E$
$u_{rms} = \sqrt{\frac{3 \times \frac{2}{3}\ E}{m}} = \sqrt{\frac{2E}{m}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.