-1
$\begingroup$

Are all collisions between gaseous molecules and boundary in a system at Equilibrium, elastic? Why?

$\endgroup$
1
$\begingroup$

One has to be careful when considering dissipation in the microscopic scale. Collisions being inelastic means, in a way analogous to the elementary physics, that there are energy lost from motion. Here are several scenarios which may affect the elasticity of the collision.

  1. Energy transferred to the internal degrees of freedom of the particles. When two atoms collide, one or both may be excited, transferring energy from its motion to the internal energy levels. In fact, the Franck–Hertz experiment, one of the earliest experiments which provided evidence of energy quantization, exactly demonstrates this point. In a box of gas under room temperature, however, this excitation is usually negligible, as can be estimated using statistical mechanics.

  2. Energy dissipated in forms of radiation. Yes. This happens all the time, in fact, more often than two atoms or molecules collide, in a dilute gas at least. A completely rigorous treatment should take both the particle and the ambient EM field into account. If the system is in the equilibrium state, that means not only the particle system, but the combined system of the particles and the EM field is in equilibrium. In this case, the average photon energy absorbed by the particles is equal to the average photon energy emitted. If the temperature is not very high, or if the particles are relatively heavy, then up to a considerable accuracy we can forget about the ambient EM radiation field, and assume there is no energy gain or lost in terms of radiations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.