# Temperature of a gas: Is it velocity only?

I've been going through Feynman's lectures. One thing he said, unless I misunderstood, is that the temperature of a gas molecule is determined by its velocity, but not its vibration or rotation.

Did I get that wrong? It seems that a molecule hitting the wall of a vessel would impart more energy if it was spinning.

• Please give a reference and/or an exact quote. How else could we know if you misunderstood?
– user137289
Oct 24, 2020 at 10:03
• You can't really talk about "the temperature of a molecule" on its own. Temperature describes an energy distribution, which involves a whole collection of molecules. Any two of these molecules, with the same temperature, may have very different energies. Oct 24, 2020 at 10:28
• "the only thing that counts is how fast they are moving." found near the end in link Oct 24, 2020 at 14:06
• I join my voice to @Pieter's. Can you please quote? Also, keep in mind that the definition is sensitive to whether you're considering classical vs quantum mechanics. Degrees of freedom get frozen at low temperatures in the quantum formalism. More context is necessary. Finally, as RogerJBarlow says, temperature is not a property of a molecule, but of the whole ensemble. Feb 14, 2021 at 1:44

In diatomic gases (such as oxygen or nitrogen) where each molecule contains two atoms, energy is stored in the vibration and rotation of these atoms (in between and about each other), but temperature is the average translational kinetic energy of the molecules. And translational kinetic energy is $$\frac{1}{2} m v^2$$ where $$m$$, $$v$$ are the mass and velocity of each molecule respectively and since mass stays the same it is the velocity of each molecule that determines kinetic energy and therefore temperature.

This would obviously be the same for molecules with more than two atoms, and as explained above we can apply the same rationale for monoatomic gases. Note that here we are talking about temperature as defined in classical physics.

• Incorrect, every degree of freedom contributes to the temperature en.wikipedia.org/wiki/Equipartition_theorem
– user65081
Oct 24, 2020 at 4:02
• True. I made a correction. Oct 24, 2020 at 6:06
• At room temperature, there is almost no thermal energy in the vibrational modes of nitrogen or oxygen.
– user137289
Oct 24, 2020 at 10:04
• @Pieter That should be kT/2 per degree of freedom, just as for translation. Oct 24, 2020 at 23:00
• @my2cts No, because there are quantum effects. At room temperature, almost all nitrogen and oxygen molecules are in their vibrational ground state. The spacing of the vibrational energy levels is larger than $kT$.
– user137289
Oct 24, 2020 at 23:21

It is the velocity only that gives pressure on the walls, that determines the equation of state, the general gas law.

Imagine a diatomic colliding head-on with a wall. The impact would depend on the phase of the vibrational motion, but on average vibrations would not make a difference. Sometimes the molecule would be stretching, sometimes it would be contracting, but on average the vibrational velocity is zero. It is the same with rotations. The internal motions do not affect pressure.

Internal energy can be transferred, so when the walls are warmer, the gas would become warmer and also rotations would speed up.

As shown in University Physics, the energy of internal modes is not proportional to temperature. Absolute temperature is almost equivalent to gas temperature, which depends on the equation of state of the classical ideal gas.

Temperature is related to the average energy of the molecule, and each degree of freedom contributes 1/2kT to the average internal energy. Thus, if you look at the translation degrees of freedom, you get that $$=3/2kT$$, but, if for instance, you have also 3 vibrational degrees of freedom, you will also have $$=3/2kT$$. Thus, the kinetic energy is a linear function of the temperature, but also is the vibrational and other forms of energy, and viceversa.

• Not true at low temperature. For N$_2$ and O$_2$ not true at room temperature. The vibrational mode (there is only one in diatomics) is frozen.
– user137289
Oct 24, 2020 at 23:26
• @Pieter I am not talking at room temperature, but in general, and of course, only if those degrees of freedom are available
– user65081
Oct 24, 2020 at 23:45
• @Wolphramjonny If it is not even true at room temperature, it is not true in general. It is quite common for vibrational energies (stretching modes) to be large compared to $kT$. Your answer was wrong.
– user137289
Oct 24, 2020 at 23:51
• – user65081
Oct 25, 2020 at 1:27
• @Pieter You're right. Oct 25, 2020 at 20:09

One thing he said, unless I misunderstood, is that the temperature of a gas molecule is determined by its velocity, but not its vibration or rotation.

You did not misunderstand, although rather than velocity I would call it translational kinetic energy (half the mass times the square of the velocity). Up to a constant, the average translational kinetic energy per molecule is the temperature, regardless of the mass of each molecule or its rotational energy (which monatomic gases don’t have but others do).

Did I get that wrong? It seems that a molecule hitting the wall of a vessel would impart more energy if it was spinning.

It does. Your intuition is correct, but that does not mean your understanding of temperature is not also correct. These two things are compatible. Molecules that spin are capable of giving more energy in a collision, but also of taking more. If you start with equivalent quantities of two equally warm gasses, one monatomic and the other diatomic, and you put them in two separate identical cool containers, the final temperature of the diatomic gas will be higher than that of the monatomic gas (the energy given up by the diatomic gas is greater). If you then bring the two containers into thermal equilibrium with each other, the average translational kinetic energy per molecule will then be the same for both. The average energy transfer when a molecule hits the wall is also the same for both (it’s zero—because equilibrium) but the RMS energy transfer per collision (assuming equal molecular weights and therefore equal RMS center-of-mass velocities and equal intervals between collisions) is greater for the diatomic gas.

The subtler point is why it is the translational kinetic energy per molecule that equalizes in thermal equilibrium (that is, that defines the temperature) rather than the total energy or some other property. If it wasn’t so, the ideal gas law would not hold, but that’s not a satisfying explanation. For a nice presentation of the reason, I would refer you to Feynman, the very chapter linked to in your comment.