Question

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 10 − 21 J ?

Answer 1

404K

Step-by-step explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

`K.E=(3)/(2)KT`

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

`T=(2/3(8.37*10^(-21)))/(1.38*10^-23)\\ T=404K`

converting to degree we have
`131^(0)C`