# Contribution of vibrational degrees of freedom

The ratio $$\displaystyle\frac{C_p}{C_v}$$ for linear triatomic molecules including vibration contribution is given by: ?

Attempt:

We are aware that $$C_v = \frac{fR}{2}$$ and $$C_p = C_v+ R$$

Hence, $$\gamma = 1 + \frac 2f$$

Now, for $$f$$ we have:

• $$2$$ rotational degrees of freedom
• $$3$$ translational degrees of freedom
• $$3N - 5 = 3\times 3-5 = 4$$ vibrational degrees of freedom (formula available here)

Thus, $$f = 2+3+4 = 9$$

And $$\gamma = 1+ \frac 29 = 1.22$$

But answer given is $$\gamma = 1.154$$

Someone in chat room told me that each degree of freedom of vibrational actually has two times contribution so we would have $$2\times 4 = 8$$ vibrational degres of freedom and a total of $$f = 13$$, which gives $$\gamma = 1.154$$. However then I see a contradiction here. If vibrational indeed contributed $$2$$ per degree then degree of freedom of methane would be $$3+3+2\times(3\times 5-6) = 24$$ not $$3+3+ (3\times 5 - 6) = 15$$.

Can someone please clear the confusion here?

• Each degree of freedom contributes $R/2$; so each vibration has $R/2$ from potential energy and the same from the kinetic energy making $R$ in total. This is based on the equipartition of energy and we add $R/2$ for each 'squared term' in the energy meaning the squared velocity in kinetic energy as $mv^2/2$ and $x^2$ in the harmonic potential, $kx^2/2$. – porphyrin Apr 18 '19 at 13:54
• VibrationAL KE is a reasonable quantity at high temperature, usually ignored at stp – Chemist Apr 18 '19 at 15:14
• @Chemist, yes this is true, but what is a 'high temperature' depends on the molecule, one with heavy atoms will have some low frequency vibrations, also bending vibrations are usually of low frequency. The heat capacity then depends smoothly on temperature, see chemistry.stackexchange.com/questions/110944/…. (The partition function is needed and then calculate $(\partial U/\partial T)_V$ vs $T$. ) – porphyrin Apr 19 '19 at 8:23

You have to be careful to make a distinction between the coordinates needed to describe a molecule, and the degrees of freedom which the molecule can actually access. For this purpose, one might choose to define the number of degrees of freedom as the minimum number of spatial coordinates needed to fully describe the positions of atoms relative to a fixed reference frame. This will be $$3N$$ total coordinates which corroborates the other answer you link. We then recognize that motion for non-linear (linear) systems, motion in six (five) of these coordinates correspond to translation and rotation. Therefore, the other $$3N-6$$ ($$3N-5$$) coordinates describe the vibrations of the system.

Now, if one is concerned with the total energy of a system, then we should keep track of the Hamiltonian. In a classical context, the Hamiltonian is a function of all coordinates which are capable of holding energy in one form or another, $$H(\mathbf{q},\mathbf{p})$$. For molecules this means we must keep track of $$3N$$ spatial coordinates, $$\mathbf{q}$$, and $$3N$$ momentum coordinates, $$\mathbf{p}$$.

Now, if we want to know about the average kinetic energy of atom 1 in the $$x$$ direction, we perform an ensemble average

$$\langle T_{x,1}\rangle=\frac{1}{2m}\frac{\int p_{x,1}^2e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}{\int e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}$$

It is a nice exercise to show that if all the degrees of freedom in the Hamiltonian are quadratic, then this separates into lots of integrals over gaussians and the top and bottom will cancel in all coordinates except for the $$x$$ direction motion of atom 1. This integral will give you the average energy as $$1/2kT$$.

But, since the integral over the quadratic momentum and the quadratic position as in the harmonic potential are exactly the same, we have just shown that every type of energy which depends quadratically on a coordinate has an average energy of $$1/2kT$$.

Therefore, if we have a particle in 1-D subject to a harmonic potential, the Hamiltonian is, $$H=\frac{p^2}{2m}+\frac{1}{2}kx^2$$

which means the average energy is $$kT$$ since there are two coordinates which the Hamiltonian depends on quadratically.

So, hopefully now it is clear why we double to count of vibrations and arrive at the answer, $$\gamma=1+\frac{2}{13}=1.154$$

I know this was very long-winded and overly detailed, but I hope this gives you a little better picture into where the equipartition theorem comes from and how this convenient counting of coordinates we do is really a coincidence of the fact a harmonic potential and kinetic energy are each quadratic in their respective coordinates.

As an aside, the distinction I have made between coordinates and degrees of freedom is completely arbitrary. Indeed, people often refer to both a harmonic potential and the kinetic energy as quadratic degrees of freedom. I prefer this language, but then one must be careful to clarify between degrees of freedom of position and degrees of freedom of momentum because this is more coordinates than are needed to describe the system completely, so it seems there are more degrees of freedom than there really are.