Question

1. What is the average translational kinetic energy of molecules in a gas at temperature
27°C?
a) 6.21x10-21;
(b) 3x10-21;
(c) 5x10-10
(d) none of these
SAN
Formula :K.E= - KT
2​

Answer 1

Answer:

Average translational kinetic energy of a gas molecule

The average translational energy of a molecule is given by the equipartition theorem as, E=3kT2 where k is the Boltzmann constant and T is the absolute temperature. For T=(37+273)K we are easily in the required high-temperature limi

Explanation:

The average molecular kinetic energy of gas at a temperature T is: Kavg=32kBT K a v g = 3 2 k B T with k_B being the Boltzmann constant