# Does the temperature of a non-ideal gas only depend on the translational kinetic energy of the particles?

I know this is true for ideal gases, but is it true for non-ideal gases?

I can imagine for example a case of perfectly plastic collisions between atoms which could transfer all the translational kinetic energy into other degrees of freedom (eg vibrational kinetic energy). Would the temperature of the gas be zero in that case?

## 1 Answer

I would not say that "temperature depends on the translational kinetic energy" but rather that "the mean translational energy at equilibrium depends of $$T$$". This is indeed true in general for all matter, whether gas (ideal or not), liquid or solid, pure or multicomponent.

The distribution of kinetic energies of particles at equilibrium is given by $$f(E_K) = \frac{2}{\sqrt\pi (kT)^{3/2}}~~ E_K^{1/2} e^{-E_K/kT}$$ and is independent of the potential interaction between molecules. This result follows from the facts that (i) kinetic and potential energy are additive, and (ii) kinetic energy is additive with respect to the kinetic energy of the constituent particles. The mean kinetic energy is $$\bar E_K = \int E_K f(E_K) dE_K = \frac{3}{2} k T$$ and this answers the question: A equilibrium the mean translational kinetic energy is proportional to temperature and the proportionality constant is the same for all substances.