The diameter of a gas molecule is `2.4 xx 10^(-10) m`. Calculate the mean free path at NTP. Given Boltzmann constant `k = 1.38 xx 10^(-23) J molecule^(-1) K^(-1)`.

The average translational kinetic energy of nitrogen gas molecule is 0.02eV (1eV - 1.6 xx 10^(-19)J) . Calculate the temperatuire of the gas. Boltzmann constant k = 1.38 xx 10^(-23) J//K .

The average translational kinetic energy of nitrogen gas molecule is 0.02eV (1eV - 1.6 xx 10^(-19)J) . Calculate the temperatuire of the gas. Boltzmann constant k = 1.38 xx 10^(-23) J//K .

The average translational kinetic energy of air molecules is 0.040 eV (1eV=1.6xx10^(-19)J). Calculate the temperature of the air. Blozmann constant K=1.38xx10^(-23) J K^(-1).

The average translational kinetic energy of air molecules is 0.040 eV (1eV=1.6xx10^(-19)J). Calculate the temperature of the air. Blozmann constant K=1.38xx10^(-23) J K^(-1).

The rms speed of particle of mass 5 xx 10^(-17) kg . In their random motion in air at NTP will be (Boltzmann's constant) K = 1.38 xx 10^(-23) J//K

At what temperature the kinetic energy of a molecule will be equal to 2.8 xx 10^(-20) J ? Boltzmann constant (k_(B)) = 1.4 xx 10^(-23) J "molecule"^(-1)K^(-1)

At what temperature would the kinetic energy of a gas molecule be equal to the energy of photon of wavelength 6000Å ? Given, Boltzmann's constant, k = 1.38 xx 10^(-23) J .K^(-1) , Planck's constant, h = 6.625 xx 10^(-34) J.s

Calculate the temperature at which the average K.E. of a molecule of a gas will be the same as that of an electron accelerated through 1 volt. Boltzmann constant = 1.4 xx 10^(-23) J "molecule"^(-1) K^(-1) , charge of an electron = 1.6 xx 10^(-19)C .

The kinetic energy of translation of an oxygen molecule at a particular temperature of 6.27 xx 10^(-31) J . Calculate the temperature . Boltzman's constant = 1.38 xx 10^(-23) J//K .