At $27^{\circ} C$ temperature,the mean kinetic energy of the atoms of an ideal gas is $E_1$. If the temperature is increased to $327^{\circ} C$,then the mean kinetic energy of the atoms will be

  • A
    $\frac{E_1}{\sqrt{2}}$
  • B
    $\sqrt{2} E_1$
  • C
    $2 E_1$
  • D
    $\frac{E_1}{2}$

Explore More

Similar Questions

The average kinetic energy of a gas molecule at $27^oC$ is $6.21 \times 10^{-21} \, J$. Its average kinetic energy at $227^oC$ will be

If the mass of each molecule of a gas kept at a constant pressure $P$ in a container is halved and their speed is doubled,what will be the resulting pressure of the gas?

Difficult
View Solution

The kinetic energy of one mole of a gas at $300 \, K$ is $E$. If the kinetic energy at $400 \, K$ is $E'$,then what is the ratio $E'/E$?

The pressure exerted by the gas on the walls of the container is because:

What is the pressure of hydrogen in a cylinder of volume $10 \text{ litre}$ if its total energy of translation is $7.5 \times 10^3 \text{ J}$?

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