The translational kinetic energy of the molecules of a gas at absolute temperature $T$ can be doubled by

  • A
    decreasing $T$ to $\frac{T}{2}$
  • B
    increasing $T$ to $4 T$
  • C
    increasing $T$ to $\sqrt{2} T$
  • D
    increasing $T$ to $2 T$

Explore More

Similar Questions

$A$ flask contains argon and oxygen in the ratio of $3: 2$ in mass and the mixture is kept at $27^{\circ} C$. The ratio of their average kinetic energy per molecule respectively will be ...........

At what temperature in $^\circ C$ is the average molecular kinetic energy of any gas double that of its value at $20^\circ C$?

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 $^{\circ} C$)

$A$ gas is contained in a vessel at pressure $P_0$. If the mass of all molecules is halved and their speed is doubled,what will be the final pressure?

Difficult
View Solution

$A$ gas has volume $V$ and pressure $P$. The total translational kinetic energy of all the molecules of the gas is :-

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