How can specific heat capacity of monoatomic, diatomic and polyatomic gases be explained on the basis of Law of equipartition of Energy?

How can specific heat capacity of monoatomic, diatomic and polyatomic

1.	How can specific heat capacity of monoatomic, diatomic and polyatomic gases be explained on the basis of Law of equipartition of Energy?
Answer» Solution :From LAW of equipartition of energy, energy per each degree of freedom is `1/2 k_(B)T` per atom or molecule. 1. Monoatomic GAS has three degrees of freedom. `:. U_(1) = 3xx1/2K_(B).T or (dU)/(dT) = 3/2K_(B).T` But molar specific heat at constant volume `C_(V) = (dU)/(dT)` `:. C_(V) = 3/2 K_(B)TN_(A) , But K_(B)N_(A)=R` `:. C_(V) = 3/2RT, But C_(P) = C_(V) +R :. C_(P) = 5/2 R` (2) A diatomic gas has 3 TRANSLATIONAL and two rational degrees of freedom `:.` Kinetic energy per molecule `U_(1) = 5.1/2 K_(B).T` For one gram MOLE total energy `U = 5/2K_(B).T.N_(A)` Molar specific heat at constant volume `C_(V) = (dU)/(dT) =5/2 K_(B) . N_(A)` `:. C_(V) = 5/2 R & C_(P) = 5/2 R + R= 7/2 R` (3) A polyatomic gas has three translational three ratitional and at least one vibrational degrees of freedom. `:.` Kinetic energy per molecule `U_(1) = 3.1/2K_(B) . T + 3 .1/2 + f = (3+f)K_(B)T` f = Number of vibrational degree of freedom kinetic energy = per gram mole of MOLECULES `= U_(1)N_(A) = U = (3+f)K_(B) . N_(A) . T = (3+f)RT` Molar specific heat `C_(V) = (dU)/(dT) = (3+f)R` `:. C_(P) = (u+f)R` `:.` For polyatomic gases `C_(V) = (3+f)R & C_(P) = (4+f)R`.