- #1
DreadyPhysics
- 21
- 0
When using the equipartition theorem to derive the heat capacity of an ideal gas, you have
[tex]\left\langle H \right\rangle=\left\langle \frac{1}{2}m\left(v^{2}_{x}+v^{2}_{y}+v^{2}_{z} \right) \right\rangle[/tex]
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.
[tex]\left\langle H \right\rangle=\left\langle \frac{1}{2}m\left(v^{2}_{x}+v^{2}_{y}+v^{2}_{z} \right) \right\rangle[/tex]
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.