# Gas molecules & energy exchange

1. Dec 3, 2011

### Nuklear99

My question is about the energy exchange between gas particles and the walls of their container...

If you consider a collection of gas molecules enclosed in a container, if the whole system is cooled (ie. like a balloon dipped in liquid nitrogen) as the gas particles collide with the inner wall of the enclosure they lose kinetic energy & therefore velocity, Thus the gas is cooled...

My question is, do the gas particles only lose velocity in a vector direction perpendicular to the wall?? Or is the velocity lost in the direction of the particle's movement????

In other words, if you think of a box enclosure, and a molecule hits the wall at 30 degrees to perpendicular, would the velocity lost be directly in that direction or would it be lost solely in the perpendicular component??

2. Dec 3, 2011

### Andrew Mason

What do you think?

Keep in mind that the walls consist of molecules. The gas molecules do not hit a flat surface - they hit other molecules. As the temperature decreases the kinetic energy of those molecules in the wall (appearing as random vibrational motion at the molecular level) decreases.

AM

3. Dec 3, 2011

### Jedi_Sawyer

The collisions would be inelastic and the coeffient of restitution would be less than one. That is the walls are cold and absorb energy, and after the collision the molecule will have less velocity than before. The velocity vector of the molecule has a perpendicular and parallel component relative to the plane of a surfaceit contacts, the coeffient of restitution applies to the perpendicular component ordinarily, there might be some value of COR for the parallel component also because of friction, but it would be less.

4. Dec 3, 2011

### Jedi_Sawyer

Another way that an exhaust gas molecule can lose directional momentum is to have collisions that are not elastic, after this type of collision the gas molecule loses some of its momentum or Kinetic Energy. If the collision is to a wall that prevents the escape of the gas from the system the molecule will transfer to the wall less momentum than it would have transferred had the collision been elastic, so it the particle is exactly turned around it will transfer less than twice the momentum that it had to this wall. If the wall becomes warmer or gains heat from this collision with a gas molecule in an inelastic collision we would expect that the gas molecule lost an equal amount of heat energy from the collision so that both energy and momentum could be conserved.

Perhaps the amount of heat energy gained by the wall exactly equals the amount of Kinetic Energy that was lost by the molecule in the inelastic collision. If that is the case then the amount of KE lost by the molecule in the inelastic collision is equal to the amount of lost energy that is stored in the internal vibrations of the molecule. That is the molecule has become cooler and slower after it has imparted momentum and energy to the wall.

5. Dec 4, 2011

### klimatos

The energy loss is normal to the wall only.

6. Dec 4, 2011

### klimatos

If the wall molecules are bonded in place, then the only way in which they can gain kinetic energy from molecular collisions with the gas is through an increase in their internal kinetic energies of vibration and libration. Here, however, quantum constraints rule. Not every collision offers the exact required energy jump; and some collisions therefore result in no net energy transfer. This is all the domain of statistical thermodynamics.

7. Dec 4, 2011

### Andrew Mason

I think I would disagree with Jedi. Would the collisions at the molecular level not be perfectly elastic. There is no "friction" at the molecular level. Would the total kinetic energy of molecules (those of the gas and of the wall) involved in a collision, not be the same before and after a collision.

I think I would also disagree with klimatos. The molecules in the wall are vibrating in random directions. Could it not be that a gas molecule colliding with the molecules in the wall might strike at an oblique angle and lose energy in a direction other than in the direction normal to the (macroscopic) surface?

AM

8. Dec 4, 2011

### klimatos

The impulse transfer is normal to the real (molecular) surface, not to the imaginary (macroscopic) one. At the molecular level, this surface is highly irregular. I would suspect that virtually all impulse transfers would not be normal to the macroscopic surface. Statistical mechanics, however, mandates that the average impulse transfer would be normal to the macroscopic surface under conditions of equilibrium.

I would like to reemphasize that most collisions would be elastic and would result in no net transfer of kinetic energy. Only those collisions whose impulse had a magnitude equal to the allowable changes in the internal energy levels of both the impacting molecule and the target molecule would result in a net transfer of energy. In other words, only inelastic collisions result in heat (KE) transfer. This is probably less than one in a million collisions. However, since the average molecule undergoes some five billion collisions per second (25°C and 1000 hPa), this is more than enough to maintain equipartition.

9. Dec 4, 2011

### Andrew Mason

Why wouldn't elastic collisions result in the transfer of kinetic energy to the wall molecules? Isn't heat transfer simply a statistical process: a loss of average kinetic energy of the gas molecules and an average increase of the molecules of the wall?

AM

10. Dec 5, 2011

### klimatos

By definition, elastic collisions result in neither a loss nor a gain in kinetic energy. Both sides have exactly the same KE after the collision as before. With no net transfer of KE there is no net transfer of conductive heat. Radiative heat transfer, of course, is another matter; but one in which quantum constraints once again play the dominant role.

With just one inelastic collision in a million elastic collisions, the average still favors one side or the other, and conductive heat will flow.

11. Dec 5, 2011

### Andrew Mason

Elastic simply means that the kinetic energy is conserved overall. It does not mean there is no transfer of kinetic energy in the collision.

What causes a collision to be inelastic?

AM

12. Dec 5, 2011

### klimatos

You are certainly correct when speaking of collisions between molecules of a gas. As long as the total kinetic energy is conserved, the collision remains elastic. However, in collisions with a container wall, if the wall doesn't move, no deformation takes place, and no change in internal energies takes place, then the molecule leaves with the same kinetic energy it had when it arrived. In order for the internal energies of the wall molecules to be increased (heat exchange) the collision has to be inelastic.

It will only be inelastic if a photon is emitted or if the KE available for exchange exactly matches one of the allowable quantum jumps. Partial matches are not allowable. This doesn't happen very often.

"In an inelastic collision, the total momentum of the two bodies remains the same, but some of the initial kinetic energy is transformed into heat energy internal to the bodies, used up in deforming the bodies, or radiated away in some other fashion."

http://www.thefreedictionary.com/inelastic+collision

13. Dec 5, 2011

### Andrew Mason

But the molecules in the wall can move even if the average position of all the molecules doesn't. Why wouldn't a collision between a gas molecule and a molecule in the wall transfer kinetic energy from one to the other in just about every collision?

Why would a photon be emitted due to energy transfer in a collision between a gas molecule and a container molecules any more or less frequently, on average, than a collision between two gas molecules or between two container molecules (as they vibrate, they continuously transfer energy to one another)?

A quantum jump would cause the molecule to increase its energy of vibration or rotation (ie kinetic energy), would it not? Why do you say that is inelastic? That quantum jump is not creating waste heat within the molecule.

AM

14. Dec 5, 2011

### cjl

The collisions of gas molecules with a wall are most certainly not elastic, nor are they dependent on a photon (nor in fact are they specular, in which the normal momentum vector is reflected and the parallel momentum vector is unchanged). The only exception to this is extremely carefully prepared surfaces in high vacuum, where specular reflection can be achieved. Normally, gas molecules interacting with a wall will briefly stick to the wall when they hit, transferring all of their momentum and kinetic energy. They will then be emitted from the surface in a somewhat random direction, with their speed determined almost exclusively by the temperature of the wall. As a result, the outgoing molecules have, on average, zero slip velocity (they have no parallel velocity), and they will have the same temperature as the wall temperature. Heat transfer is determined by the incoming velocity as compared to the outgoing velocity - if on average, the incoming speed distribution of the gas is higher than the outgoing one, there is heat transfer to the wall (which of course will be the case if the gas has a higher temperature on average than the wall temperature).

15. Dec 6, 2011

### klimatos

This is a very common simplifying assumption, but there is no evidence that it is actually true. What force overcomes the natural inclination to rebound (as in gas molecule collisions)? I find it hard to believe that a water molecule will automatically stick to a paraffin surface. Again, what force or combination of forces accounts for this adhesion?

No, this is merely a simplifying assumption, and many texts admit as much.

Rebound directions are largely random, but this is easily accounted for by the irregularity of wall surfaces at the molecular scale.

16. Dec 6, 2011

### klimatos

1. Quantum thermodynamics. The translatory KE of the gas molecule is quantized. The vibratory KE of the wall is quantized. There has to be an exact match before KE can be transferred.

2. I don't know that it would, unless the wall were extremely hot. I don't think molecules "know" what it is that they are colliding with. Solids vibrate at low temperatures. Gases do not. Some gases (O2 for instance) have no vibratory mode at all.

3. See definition of inelastic collision is post #12 in this thread. I don't make the rules.

Last edited: Dec 6, 2011
17. Dec 12, 2011

### Nuklear99

Thanks everyone for your input.. I understand all the points presented, but I still feel inclined to assume the gas molecules lose velocity in a direction other than just normal to the surface of the container.. Doesn't viscosity have some effect on the tangential component of the gas molecules' velocity loss??? Doesn't viscosity have an effect on the pressure??