The average translational kinetic energy of O | vedclass

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

Question

(C) The average translational kinetic energy of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann con

Vedclass · Accepted Answer

$0.048$

Vedclass · Answer

(C) The average translational kinetic energy of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Since the average translational kinetic energy depends only on the temperature $T$ and not on the nature or mass of the gas molecules,it is the same for all ideal gases at the same temperature.Given that the average translational kinetic energy of $O_2$ molecules is $0.048 \ eV$ at temperature $T$,the average translational kinetic energy of $N_2$ molecules at the same temperature $T$ will also be $0.048 \ eV$.

The average translational kinetic energy of $O_2$ molecules at a particular temperature is $0.048 \ eV$. The average translational kinetic energy of an equal number of $N_2$ molecules at the same temperature is (in $eV$):

Similar Questions

At absolute zero temperature,the kinetic energy of the molecules

Helium and argon are placed in a flask at the same room temperature $(300 \ K)$. What is the ratio of the average kinetic energies (per molecule) of helium and argon?

Write the equation of pressure of an ideal gas in terms of its density.

At constant volume,the temperature of a gas is increased. Then:

The average force applied on the walls of a closed container depends as $T^{x}$ where $T$ is the temperature of an ideal gas. The value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module