Prove that average molecular energy for an ideal gas is 3/2kt
Answers
We have PV =1/3 [mNc2] but c2 is proportional to the absolute temperature T.
For a given mass of gas at constant temperature 1/3 [mNc2] is constant and therefore PV is constant and this is Boyle's law.
The Ideal Gas equation for n moles of gas is PV = nRT and so for one mole of gas we have PV = RT, where R is the gas constant. But in one mole of gas there are NA molecules where NA is the Avogadro constant (6.02x1023) and therefore:
PV = (1/3)[mNAc2] = RT
Therefore: 1/3 [mc2] = RT/NA = kT
and the quantity k (= R/NA) is known as Boltzmann's constant (k) .
The equation for one molecule may therefore be rewritten as:
(1/3)[mc2] = kT
However the mean kinetic energy of a molecule in the gas is ½mc2 and so:
Mean kinetic energy of a molecule in an ideal gas = (1/2)mc2 = (3/2)kT = (3/2)RT/NA
You should be able to see from this formula that the mean kinetic energy of a molecule in an ideal gas is proportional to the absolute temperature (T) of the gas.