At what temperature does the average translational kinetic energy of a molecule in a gas become equal to the kinetic energy of an electron accelerated from rest through a potential difference of $V$ volt? ($N=$ Avogadro number,$R=$ gas constant,$e=$ electronic charge)

  • A
    $\frac{2 eVN}{3 R}$
  • B
    $\frac{eVN}{R}$
  • C
    $\frac{eVN}{4 R}$
  • D
    $\frac{3 eVN}{2 R}$

Explore More

Similar Questions

At what temperature in $^\circ C$ will the kinetic energy of gas molecules be double the kinetic energy at $27^\circ C$?

The kinetic energy,due to translational motion,of most of the molecules of an ideal gas at absolute temperature $T$ is proportional to:

If the kinetic energy of a helium atom at temperature $T$ is $E$,then the Avogadro constant $N_A$ is given by:

The average kinetic energy of hydrogen molecules at $300\, K$ is $E$. At the same temperature,the average kinetic energy of oxygen molecules will be

The temperature of a gas is $-78^{\circ} C$ and the average translational kinetic energy of its molecules is $K$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is: (in $^{\circ} C$)

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