contestada

An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules

Respuesta :

Answer:

6.624 x 10^-21 J

Explanation:

The temperature of the ideal gas = 320 K

The average translational energy of an ideal gas is gotten as

[tex]K_{ave}[/tex] = [tex]\frac{3}{2}K_{b}T[/tex]

where

[tex]K_{ave}[/tex]  is the average translational energy of the molecules

[tex]K_{b}[/tex] = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1

T is the temperature of the gas = 320 K

substituting value, we have

[tex]K_{ave}[/tex] = [tex]\frac{3}{2} * 1.38*10^{-23} * 320[/tex] = 6.624 x 10^-21 J