(A) The average translational kinetic energy of a gas molecule is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The average translational kinetic energy of a gas molecule is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.The kinetic energy of an electron accelerated through a potential difference $V$ is $E_2 = eV$.Given the condition $E_1 = E_2$, we have:$\frac{3}{2} k_B T = eV$We know that the Boltzmann constant $k_B = \frac{R}{N}$, where $R$ is the universal gas constant and $N$ is Avogadro's number.Substituting $k_B$ into the equation:$\frac{3}{2} (\frac{R}{N}) T = eV$Solving for $T$:$T = \frac{2eVN}{3R}$Thus, the correct option is $A$.

Class 11 Physics Kinetic Theory of Gases Mix Examples-Kinetic Theory of Gases

Difficult

The average translational kinetic energy of a molecule in a gas is $E_1$. The kinetic energy of an electron $(e)$ accelerated from rest through a potential difference of $V$ volts is $E_2$. The temperature at which $E_1 = E_2$ is possible (assuming the mass of the molecule and electron are the same) is: (where $N$ is Avogadro's number, $e$ is the elementary charge, $R$ is the gas constant)

MHT CET 2026 All MHT CET

A
$\frac{2VNe}{3R}$
B
$\frac{VNe}{2R}$
C
$\frac{3NeV}{2R}$
D
$\frac{5NeV}{3R}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Kinetic Theory of Gases

Subtopic

Mix Examples-Kinetic Theory of Gases

Previous Year

MHT CET 2026 Paper All MHT CET years →

For a diatomic gas,the change in internal energy for a unit change in temperature at constant pressure and constant volume is $U_1$ and $U_2$ respectively. The ratio $U_1 : U_2$ is

Easy

View Solution

The density of water is $1000 \; kg \; m^{-3}$. The density of water vapour at $100 \; ^{\circ}C$ and $1 \; atm$ pressure is $0.6 \; kg \; m^{-3}$. The volume of a molecule multiplied by the total number gives what is called molecular volume. What is the average distance (in $\mathring{A}$) between atoms (interatomic distance) in water? (Radius of a water molecule $= 2 \; \mathring{A}$)

Medium

View Solution

$A$ vessel contains $14 \, g$ ($7$ moles) of hydrogen and $96 \, g$ ($3$ moles) of oxygen at $STP$. $A$ chemical reaction is induced by passing an electric spark in the vessel until one of the gases is consumed. The temperature is brought back to its starting value of $273 \, K$. The pressure in the vessel is ...... $atm$.

Medium

View Solution

If the heat required to increase the rms speed of $4$ moles of a diatomic gas from $v$ to $\sqrt{3} v$ is $83.1 \ kJ$, then the initial temperature of the gas is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $^{\circ} C$)

MediumTS EAMCET 2025

View Solution

$A$ box of $1 \ m^{3}$ is filled with nitrogen at $1.5 \ atm$ at $300 \ K$. The box has a hole of an area $0.010 \ mm^{2}$. How much time is required for the pressure to reduce by $0.10 \ atm$,if the pressure outside is $1 \ atm$?

Difficult

View Solution

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

Similar Questions

For a diatomic gas,the change in internal energy for a unit change in temperature at constant pressure and constant volume is $U_1$ and $U_2$ respectively. The ratio $U_1 : U_2$ is

If the heat required to increase the rms speed of $4$ moles of a diatomic gas from $v$ to $\sqrt{3} v$ is $83.1 \ kJ$, then the initial temperature of the gas is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $^{\circ} C$)

$A$ box of $1 \ m^{3}$ is filled with nitrogen at $1.5 \ atm$ at $300 \ K$. The box has a hole of an area $0.010 \ mm^{2}$. How much time is required for the pressure to reduce by $0.10 \ atm$,if the pressure outside is $1 \ atm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module