At $133.33 \ K$,the $RMS$ velocity of an ideal gas is $(M = 0.083 \ kg \ mol^{-1}; R = 8.3 \ J \ mol^{-1} \ K^{-1})$ (in $m \ s^{-1}$)

  • A
    $200$
  • B
    $150$
  • C
    $2000$
  • D
    $400$

Explore More

Similar Questions

The $RMS$ velocity of $CO$ gas molecules at $27^{\circ} C$ is approximately $1000 \ m / s$. For molecules at $600 \ K$, the $RMS$ velocity is approximately (in $m / s$)

At $400 \ K$,the root mean square $(rms)$ speed of a gas $X$ (molecular weight $= 40$) is equal to the most probable speed of gas $Y$ at $60 \ K$. The molecular weight of the gas $Y$ is:

Predict the correct order of rate of diffusion of the following molecules.

If the average velocity of $N_2$ molecule is $0.3 \, m/s$ at $27 \, ^\circ C$,then the velocity will be $0.6 \, m/s$ at ........... $K$.

At what temperature $K$ will the $rms$ speed of $SO_2$ gas be equal to the $rms$ speed of $O_2$ gas at $303 \ K$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo