Question

An ideal gas is at a temperature of 422 K. What is the average translational kinetic energy (in x10^-21 J) of one of its molecules? Round your answer to the nearest decimal place.

Answer 1

Average kinetic energy of a molecule of ideal gas=3kT/2

where k(Boltzmann constant)=1.38X10^-23J/k

T=422k

so using these value we get-

average K.E. of one molecule

=3X1.38X10^-23X422/2

=873.54X10^-23J

=8.7354X10^-21J

⁼8.7X10^-21J(approx.)