## heat capacity of an ideal gas

When using the equipartition theorem to derive the heat capacity of an ideal gas, you have
$$\left\langle H \right\rangle=\left\langle \frac{1}{2}m\left(v^{2}_{x}+v^{2}_{y}+v^{2}_{z} \right) \right\rangle$$

and each degree of freedom contributes 1/2 kT to the total energy and 1/2 k to the total heat capacity, hence the total heat capacity is 3/2 k N .

My question is, why doesn't the Hamiltonian include the rotational energy? Is this what we mean by "ideal gas"? But even if it is an idealized approximation, why then should it give applicable real-world results when applied to real gases that presumably have rotational energies? There must be a deeper physical reason why we can discard rotational energy when dealing with gases.
 PhysOrg.com physics news on PhysOrg.com >> Atomic-scale investigations solve key puzzle of LED efficiency>> Study provides better understanding of water's freezing behavior at nanoscale>> Iron-platinum alloys could be new-generation hard drives
 This is considering just a monoatomic gas, such as any of the noble gases. For a diatomic gas you would include rotational kinetic energy, and it would end up adding an additional Nk to the heat capacity since there are two internal angles to specify the orientation of a diatomic molecule.
 But cannot a single atom rotate on its own axis like a top?

## heat capacity of an ideal gas

Nope. Though the reasons are pretty deep in quantum.
 Well it would be nice to have some reference to those reasons :)

 Quote by DreadyPhysics When using the equipartition theorem to derive the heat capacity of an ideal gas, you have $$\left\langle H \right\rangle=\left\langle \frac{1}{2}m\left(v^{2}_{x}+v^{2}_{y}+v^{2}_{z} \right) \right\rangle$$ and each degree of freedom contributes 1/2 kT to the total energy and 1/2 k to the total heat capacity, hence the total heat capacity is 3/2 k N . My question is, why doesn't the Hamiltonian include the rotational energy? Is this what we mean by "ideal gas"? But even if it is an idealized approximation, why then should it give applicable real-world results when applied to real gases that presumably have rotational energies? There must be a deeper physical reason why we can discard rotational energy when dealing with gases.
One can discard rotational and vibrational energy only when dealing with monatomic gases. A single atoms can't be mechanically rotated.
Classically, one can rotate an atom by rotating the position of its electrons. However, quantum mechanics dominates on the atomic scale. The position of an electron in a single atoms is indeterminate according to the uncertainty principle. Therefore, you can't rotate the electron around the atom. So you can't really rotate the atom.
 Recognitions: Science Advisor As the nucleus is nearly a point particle, the low lying rotational excitations of an atom are rotations of the electrons around the nucleus and typically correspond to electronic transitions in the range of several eV. Hence they cannot be excited at ordinary temperatures. E.g. in the case of a hydrogen atom, the first rotational excited state is an electron in the 2p orbital. The 1s - 2p energy difference is 3/4 * 13.6 eV=10.2 eV. This corresponds a temperature of about 100000 K.
 Can I ask how you calculated that 1s to 2p energy difference?

 Quote by CAF123 Can I ask how you calculated that 1s to 2p energy difference?
The energy of a Hydrogen state with main quantum number n is -13.6 eV/n² (as taught in introductionary Quantum Mechanics or Physical Chemistry courses, and also readily found on the Internet). Therefore, the energy difference between n=1 and n=2 is 13.6 eV - 13.6 eV/4 = 3/4*13.6 eV.