Question

Find average translational kinetic energy of individual nitrogen molecules at 1600K (a)in Joules

and (b) electron volts.​

Answer 1

Question:

Find average translational kinetic energy of individual nitrogen molecules at 1600K

(a)in Joules

and (b) electron volts.

Solution:

Average Translational Kinetic Energy , $K = \frac{5}{2} k_{B} T$

where,

$k_{B} = Boltzmann \ constant = 1.38 * 10^{-23}$

$T = Temperature$

(a) $K = \frac{5}{2} * 1.38 * 10^{-23} * 1600$

$K = 5 * 1.38 * 10^{-23} * 800$

$K = 5520 * 10^{-23}$

$K = 5.52 * 10^{-20} \ J$

(b) To convert in electron volts(eV)

1 Joule(J) = 6.24 * 10¹⁸ eV

$K = 5.52 * 10^{-20} * 6.24 * 10^{18}$

$K = 34.4448 * 10^{-20} * 10^{18}$

$K = 34.4448 * 10^{-2} \ eV$

The Average Translational Kinetic Energy of individual nitrogen molecules at 1600K is 5.52 * 10⁻²⁰ J or 34.4448 * 10⁻² eV