Degrees of freedom in a diatomic gas in 2-dimensions Question: What is the specific heat capacity at constant volume of a two-dimensional diatomic ideal gas of N particles at room temperature?
My answer: A diatomic gas can move in both directions, can vibrate, and can spin. This is 4 degrees of freedom and by the Equipartition theorem I know that each of these degrees of freedom have energy $k_bT/2$. Heat capacity at constant volume is defined as the change in energy per unit temperature, so my total comes to be:
$$C = \left (\frac{\partial U}{\partial T} \right )_V = \frac{\partial}{\partial T} \frac{4Nk_bT}{2} = 2Nk_b.$$
The actual answer is $(5/2)Nk_b$. I'm not sure where I'm missing the extra degree of freedom. 
 A: For each vibrational degree of freedom, the energy contained is $k_bT$, not $k_bT/2$. 
See also: http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html
A: The equipartition theorem says that each quadratic degree of freedom that appears in the energy function contributes $\frac{1}{2}k T$ to the internal energy. So all we have to do is to count the degrees of freedom:


*

*The translational kinetic energy of the diatomic molecule has two degrees of freedom because the molecule can move independently in the $x$ and $y$ direction.

*In two dimensions there is only one rotation axis, so there is only one degree of freedom in this case.

*The vibration is slightly more subtle. Firstly the vibrational motion has one translational degree of freedom because each atom can oscillate along the axis that binds the molecule together. Secondly (this is the part I suspect you have not counted) the atom also has potential energy. The potential energy only depends on the distance between the atoms, so there is one degree of freedom here as well. In total there is thus two degrees of freedom for the vibrational motion.


Adding all of these together gives us the $\frac{5}{2}kT$ as expected from your solution manual. You obtain the heat capacity by differentiating this w.r.t temperature.
