A New Thermodynamics

Blog: Thermal Equilibrium: "A new understanding"

By Kent W. Mayhew


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Thermal Equilibrium

Walls & Thermal Equilibrium in Dilute Gaseous Closed Systems 

Consider a closed gaseous system in thermal equilibrium: This now means

1)      The walls absorb as much blackbody/thermal radiation, as they radiate.

2)      The gas molecules absorb as much blackbody/thermal radiation, as they radiate, hence maintain a constant vibrational energy (assuming they have it).

3)      The gas molecules receive a given amount of kinetic energy (translational plus rotational) from the wallís vibrational energy: Elastic or inelastic? Collisions are inelastic!

4)     The gas molecules equally (give vs take) exchange vibrational energy with the walls.In general net flow of vibrational energy in either direction will be zero. 

The above provides us with a new understanding of thermal equilibrium inside of closed system that takes into account a systemís thermal radiation which was never traditionally considered. See thermal equilibrium atmosphere, In the above description there will be two cases. In the above, the gas is sufficiently dilute that the intermolecular gas collisions are considered as being insignicant, when compared to the gas-wall collisions, hence the gas adheres (or approximately so) toKinetic Theory

It must be understood that inelastic collisions as desrcibed in my peer reviewed kinetic theory (webpage, book,paper) just means that intermolecular collisions result in conservation of momentum but not kinetic energy, hence in order to abide by the principle of energy conservation, radiation is given off during such collisions. More that radiation is no part of the system's blackbody/thermal radiation, which is continually absorbed and re-emitted by both condensed matter and polyatomic molecules.

Even so, it should be stated that the total energy associated with thermal radiation is generally considerably less than the total energy associated with matter (condensed matter and/or gas molecules). However the fact that it travels at the speed of light, means that it can contribute significantly to thermal energy exchanges. Moreover it all becomes part of a systemís temperature, as measured with a thermometer. 

Next consider that the surroundings are either the atmosphere or the heat bath as illustrated in Fig 6.1. During expansion, the gas can remain isothermal   (Tatm=Tsys ) if it expands quasi-statically, thus allowing blackbody/thermal radiation to infinitesimally enter the system through its walls. Thus the blackbody/thermal radiation density within the system remains constant.

As the ideal gas isothermally expands its pressure decreases while its volume increase such that d(PV)=0. The only energy change to the system is the addition of blackbody/thermal radiation into the expanding freespace. Therefore, in terms of radiation density (@), the change in energy within our expanding isothermal system becomes:

dEsystem=dEblackbody=@dV     (1)


   Since the energy associated with blackbody/thermal radiation generally is extremely small, when compared to the kinematic energies associated with gas molecules, then the energy change as defined by eqn 1, generally can be approximated as zero during the quasi-static expansion. Remember, too often this energy is freely given into the expanding systems from the surroundings hence its transfer goes unnoticed.

Rather than quasi-static, consider what happens if the system experiences a rapid large volume increase? If the blackbody/ thermal radiation cannot be transferred from/through the walls quickly enough, then the blackbody energy density will not remain constant. If a thermometer is placed into the above rapidly expanding system, then the thermometer will still be giving as much kinematic energy onto its surrounding gas molecules, as it receives. But, the same can no longer be said of the blackbody radiation and its accompanying thermal radiation!

One could rightfully argue that given enough time then thermal equilibrium will again be attained. But this is missing the point. Furthermore, some might argue that the expanding gas must do work? Perhaps, but unless both the expanding systemís surroundings, and expanding force, are clearly defined   then any work cannot be determined. I.e., if an external force is expanding the system, then it is the external force, which is displacing the surroundings and hence is doing the work.



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This website is copyright of Kent W. Mayhew who in 2018 resides in Ottawa Ontario Canada
   This website is full of new ideas, which are the property of Kent W. Mayhew.  
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