Question

A diatomic molecule is rotating about its center of mass with an angular speed of 3.90 ✕ 10^12 rad/s. Determine the rotational kinetic energy (in J) of one molecule, if the gas is nitrogen which has a bond length of 1.10 Å and a molecular molar mass of 28.0 g/mole.

Answer 1

Answer:

$E_s = 7.35\times 10^{-22}\ J$

Explanation:

given,

angular speed, ω = 3.90 x 10¹² rad/s

bond length, L = 1.10 Å

molar mass = 28.0 g/mole

Rotational Kinetic energy

$E_s = \dfrac{1}{2} I \omega^2$

now,

moment of inertia ,$I = \dfrac{1}{2}MR^2$......(1)

mass of O₂ = m

mass of O = M = m/2

now, mass

$M = \dfrac{28\times 1.67\times 10^{-27}}{2}$

form equation(1)

$I = \dfrac{1}{2}\times \dfrac{28\times 1.67\times 10^{-27}}{2}\ R^2$

$I = \dfrac{1}{2}\times \dfrac{28\times 1.67\times 10^{-27}}{2}\ (1.10\times 10^{-10})^2$

 I = 9.66 x 10⁻⁴⁷ kg.m²

putting value in Rotational kinetic energy equation

$E_s = \dfrac{1}{2}\times 9.66\times 10^{-47}(3.9\times 10^{12})^2$

$E_s = 7.35\times 10^{-22}\ J$

Hence, rotational kinetic energy of the molecule is equal to $E_s = 7.35\times 10^{-22}\ J$