Velocity distribution of evaporating water molecules

AI Thread Summary
The velocity distribution of evaporating water molecules is not Maxwellian due to the nature of the evaporation process. Evaporation involves complex interactions that are not purely elastic, challenging the assumption of a Maxwellian distribution. The discussion references several academic sources that explore specific aspects of this topic. It raises fundamental questions about the interactions involved in evaporation and whether they can be classified as two-particle interactions. Overall, the evaporation of H2O molecules requires a reevaluation of traditional kinetic theory models.
Kazys
Messages
10
Reaction score
1
What is the velocity distribution of evaporating H2O molecules? It can not be Maxwellian.
 
Physics news on Phys.org
Thank you for the references. They deal with rather specific issues of this general question. I would like to
reframe it in more simple terms. A Maxwellian distribution results when freely moving particles interact
elastically. The question can be posed as follows - is the evaporation of an H2O molecule elastic?
Is it a two particle interaction? If not, can the distribution of evaporating molecules be Maxwellian?
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top