Q. The $r.m.s.$ speed of the molecules of a gas at a pressure $10^{5}Pa$ and temperature $0℃$ is $0.5km\ \sec^{- 1}$. If the pressure is kept constant but temperature is raised to $819℃,$ the velocity will become
A) $1.5\ kms^{- 1}$
B) $2\ kms^{- 1}$
C) $5\ kms^{- 1}$
D) $1\ kms^{- 1}$
Explanation
(d) $\frac{\left( v_{rms} \right)_{1}}{\left( v_{rms} \right)_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{500}{\left( v_{rms} \right)_{2}} = \sqrt{\frac{0 + 273}{819 + 273}} = \sqrt{\frac{273}{1092}}$
$\left( v_{rms} \right)_{2} = 500\sqrt{\frac{1092}{273}} = 500\sqrt{4} = 1000\frac{m}{s} = 1\frac{km}{s}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.