Medium
The $rms$ velocity of molecules of a given mass of gas at $27^{\circ}C$ and $1.0 \times 10^5 \, Nm^{-2}$ pressure is $200 \, m \, s^{-1}$. What will be the $rms$ velocity if the temperature and pressure are $127^{\circ}C$ and $0.5 \times 10^5 \, Nm^{-2}$ respectively?
- A$\frac{400}{\sqrt{3}} \, m \, s^{-1}$
- B$100\sqrt{2} \, m \, s^{-1}$
- C$100 \frac{\sqrt{2}}{3} \, m \, s^{-1}$
- D$50 \frac{\sqrt{2}}{3} \, m \, s^{-1}$
Explore More
Similar Questions
The root mean square speed of molecules of a given mass of a gas at $27^{\circ} C$ and $1$ atmosphere pressure is $200\, ms^{-1}$. The root mean square speed of molecules of the gas at $127^{\circ} C$ and $2$ atmosphere pressure is $\frac{x}{\sqrt{3}}\, ms^{-1}$. The value of $x$ will be ......$ms^{-1}$.
DifficultJEE MAIN 2021
View SolutionThe temperature at which oxygen molecules will have the same r.m.s. speed as helium molecules at $57^{\circ} C$ is (molecular masses of oxygen and helium are $32$ and $4$ respectively.) (in $K$)
The rms speed of an oxygen molecule at a certain temperature is $600 \,ms^{-1}$. If the temperature is doubled and the oxygen molecule dissociates into atomic oxygen atoms,the new rms speed is (in $\,ms^{-1}$)
The number of molecules in a closed container is doubled. What will be the effect on $v_{rms}$?
Easy
View SolutionTwo molecules of a gas have speeds of $9 \times 10^{6} \ m/s$ and $1 \times 10^{6} \ m/s$ respectively. What is the root mean square speed of these molecules?
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo