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
In a vessel,an ideal gas is at a pressure $P$. If the mass of all the molecules is halved and their speed is doubled,then the resultant pressure of the gas will be
Absolute zero temperature is the temperature at which:
Medium
View SolutionAt absolute zero temperature,the kinetic energy of the molecules
Easy
View SolutionAccording to the kinetic theory of gases,the total energy of an ideal gas is equal to:
Easy
View Solution$1\, kg$ of a diatomic gas is at a pressure of $8 \times 10^4\, N/m^2$. The density of the gas is $4\, kg/m^3$. What is the energy of the gas due to its thermal motion?
Difficult
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