Question

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

Answer 1

6.624 x 10^-21 J

Step-by-step explanation:

The temperature of the ideal gas = 320 K

The average translational energy of an ideal gas is gotten as

`K_(ave)` =
`(3)/(2)K_(b)T`

where

`K_(ave)` is the average translational energy of the molecules

`K_(b)` = 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

`K_(ave)` =
`(3)/(2) * 1.38*10^(-23) * 320` = 6.624 x 10^-21 J