The equipartion theorem and degree of freedom in case of vibration

Question

I have been taught in chemistry that, the energy of a vibrational freedom is $RT$ (ie, twice that of rotational/translational) The degree of freedom which I found in chemistry, for the vibrational degree of freedom was given by this - $$f_{vib}=3N-f_{rot}-f_{trans}$$ (where N is the number of atoms in the molecule) Now I was taught in Physics that...- Energy of ALL (including vibrational) degrees of freedom is $(1/2)RT$, BUT that the number of degrees of freedom FOR vibration to be $$f_{vib}=2*(3N-f_{rot}-f_{trans})$$ Now this is shocking and confusiong- Both seem to be implying the same when it comes to Internal energy calculation, In the case of calculations of $Cv$, $Cp$, and $gamma$ I am now really confused as I was taught that $$ C_{v}=\frac{fR}{2}, C_{p}=\frac{(f+2)R}{2}$$ (They give different ans in the "phy"/"chem " interpretations.) Can anyone help me with this? Also, my thought on this was that it is energy that is twice and not degree of freedom, Though I still find plenty of sites with agreement on both these ideas Any help will appreciated.

Answer 1

tldr:

Chemists and physicists mean slightly different things by the term "vibrational degrees of freedom". What chemists call a vibrational degree of freedom, physicists call a vibrational mode. Each mode is a possible vibration of a molecule. A diatomic molecule has one vibrational mode associated with the relative motion of the two atoms along the axis of the molecule. A water molecule (a nonlinear, triatomic molecule) has three vibrational modes: the bending mode and the symmetric and anti-symmetric stretch modes. For each vibrational mode there are two physical degrees of freedom, corresponding to the two "places" where energy can be stored: the elastic/potential energy and the kinetic energy of the relative motion of the molecules.

Thus, for physicists, the correct expression is the second one the OP has written down, and it's the one that allows you to compute thermodynamic properties of the ideal gas. The first one counts the number of vibrational modes and can't directly be used to compute internal energies and specific heats until you include that factor of 2.

More details below.

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The equipartition theorem states that

Every quadratic degree of freedom contributes $\frac{1}{2}k_{\textrm{B}}T$ to the equilibrium internal energy.

A quadratic degree of freedom is one for which there is an associated energy that can be written as a quadratic function of the coordinates and associated momenta of the particles making up the system. For a one-dimensional vibrational mode, the total energy is $$ E_{\textrm{vib.}} = \frac{1}{2}mv_x^2 + \frac{1}{2}m\omega^2 x^2\,, $$ i.e., there are two "places" where thermal energy is "stored": the elastic potential energy and the relative kinetic energy of the vibration.

For rotation about a single (principal) axis of a molecule, there is only one physical degree of freedom, whose energy is given by $$ E_{\textrm{rot.}} = \frac{1}{2}I \Omega^2\,, $$ where $\Omega$ is the angular velocity of rotation and $I$ is the moment of inertia about this principal axis.

Finally, for translational motion (of the center of mass) of the molecule, there are three degrees of freedom corresponding to the three terms in the translational kinetic energy, i.e., $$ \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2+\frac{1}{2}mv_z^2\,. $$

Then, the internal energy of such an ideal gas is $$ U=\frac{Nf}{2}k_{\textrm{B}}T\,, $$ where $f$ is the total number of quadratic degrees of freedom per molecule, and $N$ is the number of molecules, making the molar specific heat at constant volume equal to $$ \mathcal{C}_V = \frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V = \frac{1}{n}\frac{Nf}{2}k_{\textrm{B}}T =\frac{f}{2}RT\,. $$ You have to count every quadratic degree of freedom. Otherwise, you'll get the internal energy wrong.

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If, instead, we count only the number of vibrational modes (e.g., bending mode and two stretch modes of a water molecule), we have to double this number to get the total number of (quadratic) degrees of freedom, since to each mode is associated a potential energy and a kinetic energy. Then, we write the number of degrees of freedom as $$ N(3 + 2N_{\textrm{vib.}} + N_{\textrm{rot.}})\,, $$ where $N$ is the number of particles, $N_{\textrm{vib.}}$ is the number of vibrational modes, and $N_{\textrm{rot.}}$ is the number of rotational degrees of freedom, which is 0 for a single atom, 2 for a linear molecule, and 3 for any non-linear molecule.

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As an example, consider a gas of heteronuclear diatomic molecules (like nitrogen). Such a molecule has three translational degrees of freedom associated with the movement of the center of mass, corresponding to the three terms in the translational kinetic energy, i.e., $$ \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2+\frac{1}{2}mv_z^2\,. $$ It also has two$^1$ rotational degrees of freedom, corresponding to rotation about two perpendicular axes that are perpendicular to the axis of the molecule; the associated energies are $$ \frac{1}{2}I_x \Omega_x^2+ \frac{1}{2}I_y \Omega_y^2\,. $$ Finally, there are two internal vibrational degrees of freedom, as discussed above, associated with the motion of the two atoms in the molecule relative to each other along the axis of the molecule; the associated energy is $$ \frac{1}{2}mv_z^2 +\frac{1}{2}m\omega^2 z^2\,. $$ This corresponds to seven degrees of freedom per molecule, resulting in an internal equilibrium energy of $$ E = N\times7\times\frac{1}{2}k_{\textrm{B}}T\,, $$ where $N$ is the number of molecules.

_{$^1$ The rotation about the $axis$ of the molecule is neglected under the approximation that the atoms are point-like, and hence the corresponding moment of inertia is zero.}

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One of the (possible) major reasons for the discrepancy between chemistry and physics here is that chemists tend to think in terms of spectroscopy, and there are single lines associated with one vibrational mode, even though there are two physical degrees of freedom associated with a single vibrational mode, and so they count vibrational modes rather than vibrational degrees of freedom. Nonetheless, they might still call them degrees of freedom, so you have to be careful about the terminology.