Question

State the law of equipartition of energy and hence calculate molar specific heat of mono- and di-atomic gases at constant volume and constant pressure.

Answer 1

Law of equipartition of energy states that for a dynamical system in thermal equilibrium the total energy of the system is shared equally by all the degrees of freedom. The energy associated with each degree of freedom per molecule is \(\frac 12\)k_BT, where k_B is the Boltzmann's constant.

Monatomic gas: For a monatomic gas, each atom has only three degrees of freedom as there can be only translational motion. Hence, the average energy per atom is \(\frac 32\) k_BT

The total internal energy per mole of the gas is E= \(\frac 32\) N_Ak_BT … where N_A is the Avogadro number.

Therefore, the molar specific heat of the gas at constant volume is

C_V =\(\frac{dE}{dT}\) = \(\frac 32\) N_Ak_B = \(\frac 32\) R

where R is the universal gas constant.

Now, by Mayer’s relation, C_P — C_V = R, where

C_P is the specific heat of the gas at constant pressure.

C_P = C_V +R = \(\frac 32\) R +R = \(\frac 52\) R

Diatomic gas : Treating the molecules of a diatomic gas as rigid rotators, each molecule has three translational degrees of freedom and two rotational degrees of freedom. Hence, the average energy per molecule is

3(\(\frac 12\) k_BT) + 2(\(\frac 12\) k_BT)= \(\frac 52\) k_BT

The total internal energy of mole is \(\frac 52\) N_Ak_BT

The molar specific heat at a constant volume C_V is

For an ideal gas 

C_V (monoatomic gas) = \(\frac{dE}{dT}\) = \(\frac 52\) N_Ak_B = \(\frac 52\) R

For an ideal gas  C_P — C_V = R

Where C_P is the molar specific heat at constant pressure

thus C_P = R +  \(\frac 52\) R =  \(\frac 72\) R

A soft or non-rigid diatomic molecule has, in addition, one frequency of vibration which contributes two quadratic terms to the energy.

Hence, the energy per molecule of a soft diatomic molecule is

E = 3(\(\frac 12\) k_BT) + 2(\(\frac 12\) k_BT) + 2(\(\frac 12\) k_BT)= \(\frac 72\) k_BT

Therefore energy per mole of a soft diatomic module is

E = \(\frac 72\) k_BT × N_A = \(\frac 72\) RT

In this case C_V  = \(\frac{dE}{dT}\) = \(\frac 72\) R and 

C_P = C_V + R = \(\frac 72\) R + R = \(\frac 92\) R

[Note :for monatomic gas adiabatic constant, γ =  C_P/C_V = \(\frac 53\), for a diatomic gas γ = \(\frac 75\) or \(\frac 97\)]