A certain gas possessess 3 degree of freedom corresponding to the translational motion and 2 degrees of freedom corresponding to rotational motion. (i) What is the kinetic energy of translational motion of one such molecule of gas at 300 K ? (ii) If the temperature is raised by `1^(@)C`, what energy must be supplied to the one molecule of the gas. Given Avogadro's number N = `6.023 xx 10^(23) mol^(-1)` and Boltzmann's constant `k_(B)=1.38xx10^(-23)J mo^(-1)K^(-1)`.

A certain gas possessess 3 degree of freedom corresponding to the translational motion and 2 degrees of freedom corresponding to rotational motion. (i) What is the kinetic energy of translational motion of one such molecule of gas at 300 K ? (ii) If the temperature is raised by 1^(@)C , what energy must be supplied to the one molecule of the gas. Given Avogadro's number N = 6.023 xx 10^(23) mol^(-1) and Boltzmann'a constant k_(B)=1.38xx10^(-23)J molecule ^(-1)K^(-1) .

Degrees of freedom of a monatomic gas due to its rotational motion will be

Calculate (a) the average kinetic energy of translation of an oxygen molecule at 27^(@) C (b) the total kinetic energy of an oxygen molecule at 27^(@)C (c ) the total kinetic energy in joule of one mole of oxygen at 27^(@) . Given Avogadro's number = 6.02 xx 10^(23) and Boltzmann's constant = 1.38 xx 10^(-23) J//(mol-K) .

The number of translational degree of freedom for a diatomic gas is

A gas have 3 translation and 2 Rotational degree of freedom. Find (C_p)/(C_v) ?

If k is the Boltzmann constant, the average translational kinetic energy of a gas molecules at absolute temperature T is:

Total degrees of freedom of one molecule of a diatomic gas at normal temperature is

The average translational kinetic energy of the molecules of a gas will be doubled if at constant