Question

calculate the average translational kinetic energy of a gas at 54centigrade

Answer 1

What you need here is something called the equipartition theorem.

This states that the thermal average energy of each degree of freedom of a system in equilibrium at (absolute) temperature T is equal to \[\frac{ 1 }{ 2 }k _{B}T\]
where KB is Boltzmann's constant

Now your problem becomes one of deciding how many degrees of freedom to include for the translational energy of a gas.

If you consider a single atom or molecule, this has 3 translational degrees of freedom, one for each of the x, y and z directions.

So, for one mole of the gas the average translational energy will be\[\frac{ 3 }{ 2 }N _{A}k _{B}T=\frac{ 3 }{ 2 }RT\] where NA is Avogadro's constant and R is the molar gas constant

Remember that T has to be in Kelvin.