Q. The molecules of a given mass of a gas have a rms velocity of 200 m/s at 27$℃$ and $1.0 \times 10^{5}\ N/m^{2}$ pressure. When the temperature is 127$℃$ and pressure is $0.5 \times 10^{5}N/m^{2}$, the rms velocity in m/s will be
A) $\frac{100\sqrt{2}}{3}$
B) $100\sqrt{2}$
C) $\frac{400}{\sqrt{3}}$
D) None of these
Explanation
(c) RMS velocity doesn’t depend upon pressure, it depends upon temperature only,
$ie.,\ \ \ v_{rms} \propto \sqrt{T}$.
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}\ \Rightarrow \ \frac{200}{v_{2}} = \sqrt{\frac{(273 + 27)}{(273 + 127)}} = \sqrt{\frac{300}{400}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{2} = \frac{400}{\sqrt{3}}$ m/s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.