# Ideal gas with vibrational degrees of freedom

Consider the model of an ideal gas, where every molecule has an additional vibrational degree of freedom, which can be described by an oscillator potential. Since there is no interaction between the molecules (apart from a very weak interaction which allows a thermal equilibrium to arise) the oscillators can be regarded as decoupled. The total energy of the system is therefore given by the sum of the kinetic energies and the oscillatory energies of the molecules:

E = E(kin) + E(osc) =  (E(kin)i + E(osc)i) where i ranges from 1 to N.

The entropy of an ensemble of decoupled harmonic oscillators is given by

S(osc) = kBN [ln (2E(osc)/h) + 1]

a) Use this result to express the total entropy as a function of E(kin) and E(osc)

Due to scattering, kinetic energy can be transferred to vibrational degrees of freedom and vice versa, so that the total energy E is conserved. Through this exchange a thermal equilibrium between the various degrees of freedom arises.

b) Express the total entropy S as a function of E(kin) and maximize S, to find the connection between E(kin) and E(osc)

c) Use the result from b) to write the entropy as a function which depends only on the total energy E. Derive from this the relationship between total energy and temperature. Compare this result with the relationship between energy and temperature of an ideal gas with no internal degrees of freedom.

d) We generalize now to a situation where every particle of the ideal gas has M vibrational degrees of freedom, which are in thermal equilibrium with each other and with the kinetic degrees of freedom. This means that only the total energy E = E(kin) + E(osc)m is constant, where m ranges from 1 to M. Derive for this case also the relationship between E and the temperature T.

See attached file for full problem description.

#### Solution Preview

The formula for the entropy is only valid in the high temperature limit. The exact formula is:

S_{osc} = k [q Log(1+N/q) + N Log(1+q/N)] (1)

where

q = E_{osc}/(h-bar omega)

I'm going to use this formula first instead of the formula given in the text, because in that case this problem becomes very simple. You regain the formula given in the text by considering the case q>>N. Then Log(1+N/q) ---> q/N and Log(1+q/N) ---> Log(q/N) and (1) reduces to:

S_{osc, HT} = Nk[Log(q/N) + 1]

I'll use this formula in the end of the problem when some explicit results are needed.

The entropy for the translational degrees of freedom is:

S_{trans} = Nk [log[V/N (E_{kin}/N)^(3/2)]+5/2 + 3/2 Log((4 pi m)/(3 h^2))] (2)

The total entropy is:

S = ...

#### Solution Summary

A detailed solution is given.