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
    $-39$
  • B
    $117$
  • C
    $127$
  • D
    $-78$

Explore More

Similar Questions

The average translational kinetic energy of the oxygen molecules at a temperature of $127^{\circ} C$ is (Boltzmann constant $= 1.38 \times 10^{-23} \,J \,K^{-1}$)

The kinetic energy of $20 \, L$ of ${H_2}$ gas is $1.5 \times 10^5 \, J$. Find the pressure exerted on the container.

Difficult
View Solution

According to the Kinetic Theory of Gases,at a given temperature:

Number of molecules in a volume of $4\, cm^{3}$ of a perfect monoatomic gas at some temperature $T$ and at a pressure of $2\, cm$ of mercury is close to $?$
(Given,mean kinetic energy of a molecule (at $T$) is $4 \times 10^{-14}\, erg$,$g=980\, cm/s^{2}$,density of mercury $=13.6\, g/cm^{3}$)

The temperature at which the kinetic energy of oxygen molecules becomes double its value at $27^{\circ}\,C$ is $............^{\circ}\,C$.

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