Using the Kinetic gas equation show that the average kinetic energy per mole of gas is given by 3/2 RT.

Question

Using the Kinetic gas equation show that the average kinetic energy per mole of gas is given by \(\frac{3}{2}\) RT.

Answer 1

From the Kinetic model for gases

PV = \(\frac{1}{3}\)mNc^-2

For 1 mole of the gas

PVm = \(\frac{1}{3}\)mNAc^-2

⁼ \(\frac{1}{3}\)mc^-2

For ideal gas PVm = RT

\(\frac{1}{3}\)mc^-2 = RT

\(\frac{2}{3}.\frac{1}{2}\)mc^-2 = RT

\(\frac{1}{2}\)mc^-2 = \(\frac{3}{2}\)RT

But \(\frac{1}{2}\)mc^-2 Average kinetic energy per mole gas (K.E).

Hence K.E. = \(\frac{3}{2}\)RT