# Particle in kinetic gas theory without collision

Hello I have a (Gedankenexperiment) question about the kinetic gas theory. As far as I understand temperature corresponds to kinetic energy. Assume now there would be one particle that does not collide with any other particle in a period of time. Would such a particle contribute to heat capacity in this period of time?

Interesting question!

Like you say, temperature is related to molecular kinetic energy as

$$e_t = \frac{3}{2} \frac{k T}{m} ,$$

where $$m$$ is the molecular mass, $$k$$ is Boltzmann's constant, and $$e_t$$ the specific translational internal energy (in J/kg). The latter comes from the kinetic energy from random thermal translational motion of the gas molecules. In other words, it depends on their speed and mass.

Depending on the type of the molecules in the gas, we can also have a specific inner energy $$e_i$$, representing the vibrational and rotational energy of the molecules. The total specific energy is then $$e = e_t + e_i$$, and the heat capacity (at constant volume) is

$$c_V = \left( \frac{\partial (e_t+e_i)}{\partial T} \right)_V = \frac{d_f}{2} \frac{k}{m},$$

where $$d_f$$ is the number of degrees of freedom of the gas molecules. All gas molecules have at least 3 degrees of freedom from movement in the $$x$$, $$y$$, and $$z$$ directions, and any possible modes of rotation and vibration can give them additional inner degrees of freedom.

As a thought experiment, consider if we suddenly randomly increase the speed of some molecules throughout our gas. This suddenly increases $$e_t$$ in the gas. Collisions between molecules will gradually spread this extra energy among the rest of the molecules of the gas.

In a noble gas (single-atom molecules) in equilibrium with its surroundings, $$e_t$$ will remain constant after its increase, as there is nowhere for that extra translational energy to go: With no internal molecular structure, $$d_f = 3$$ and $$e_i = 0$$.

If we have a gas of molecules that do have an internal structure (i.e. having two or more atoms), there is somewhere for that extra translational energy to go, namely inner energy: molecular vibration and rotation. These represent additional degrees of freedom, so that $$d_f > 3$$. Gradually, collisions between molecules will equilibrate translational and inner energy. Thus, in our thought experiment, $$e_t$$ will increase suddenly, before gradually $$e_t$$ decreases and $$e_i$$ increases until they reach an equilibrium.

So, what happens if we have one molecule that doesn't interact with the others?

• In a noble gas, it does not matter whether it does or not. Whether that molecule has a chance to collide with other molecules or not, we still have $$d_f = 3$$, which means that the heat capacity is unchanged.

• However, in a gas of molecules of two or more atoms, it does matter! As it does not collide with other molecules, its inner energy will not absorb any of the translational energy in the gas. In other words, its inner degrees of freedom are not available. Thus, the "average" number of degrees of freedom for molecules in the gas is ever so slightly reduced, thus reducing the heat capacity ever so slightly.