Why hot molecules have higher velocities

Question

I'm looking for a something different than the "T is average KE, heat is energy, so adding heat increases the KE and therefore T" explanation. Let's call that the "thermo" side of the argument, which I agree with, but I'm looking at the other side. I'm talking about putting the "dynamics" back in thermodynamics.

That's right - good old fashioned, classical, high school, Newtonian dynamics - the kind grandma used to make: F * d = 1/2 mv^2. My thought is that both F and d increase at higher T, and below is my line of thinking.

Just to keep it as simple as possible, consider a monoatomic gas (say He) at constant V. No vibrations or rotations, just translation. No attractive forces, just electronic repulsion. The number of molecules (atoms) can either be two, or 10^23, as long as they all act the same. I'll assume two molecules, A and B. Here is my line of reasoning:

• Heat is added to the gas in the form of IR radiation.
• Both A and B absorb a photon.
• Both electron clouds increase in size.
• @ constant V, this forces the electrons of the two molecules closer together.
• This increases the (Coulomb) force of repulsion.
• Thus F (at least Fmax) between the two is higher than before the heat addition.
• Now for the distance d through which F acts. F starts at a separation distance closer than before heat addition.
• F falls off as 1/d^2 until reaching "0" at some dfinal. This dfinal is the same as before heat addition.
• Thus, the d through which F acts is also larger than before heat addition.

Therefore, F * d has increased, so has 1/2 mv^2 and therefore T.

Does the above line of reasoning seem valid?

Answer 1

• Heat is added to the gas in the form of IR radiation.

• Both A and B absorb a photon.

That's not how most IR radiation interacts with the molecules. (And by picking a monoatomic gas, you've picked one that does not interact strongly in the IR anyway. It's almost transparent). Without lower-energy vibrational modes, the interactions are rare. But there will still be some scattering. (see also: Difference in Raman, Rayleigh, and Compton scattering)

• Both electron clouds increase in size.

• @ constant V, this forces the electrons of the two molecules closer together.

This change in size isn't relevant. For an ideal gas, we presume the size of particles is small compared to the distance between them. Even if absorption happens (and in this case, it really doesn't), the change in distance between the electron shells is insignificant and the coulomb forces are basically zero.

Instead, you can think of the photons as "kicking" the particles. The particle goes one way, and the photon goes another way. The momentum transfer in this process is tiny, and that helps to make monoatomic gases poor IR absorbers. The gas is more likely to interact thermally with the container and equilibrate via those collisions rather than through radiation.

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Why there aren’t many photons kicking the molecules? Or, in other words, why there is a poor momentum transfer?

The first problem is why is the interaction rate so low? In the interaction, you have to conserve momentum and energy. For a monoatomic molecule, there's not many places for the energy to go, just into KE. This heavily restricts the interactions that are possible with "low-energy" photons. Instead, the tendency is for the two to not interact.

Secondly, when there is an interaction, the amount of energy transferred isn't large. Momentum transfer depends on the ratio of the momenta. It's most efficient when the momenta are similar in size. Bouncing a flea off a freight train doesn't change the train much. Further, the flea (or photon in this case), still has some of the incoming energy as it departs.

In an absorption interaction, then it's possible for all of the photon's energy to be retained and available as thermal energy later.