Time constant for gas to reach equipartion equilibrium

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SUMMARY

The discussion focuses on the dynamics of gas, specifically CO2, reaching equipartition equilibrium after being subjected to electromagnetic radiation at a specific frequency, F1. It establishes that the gas initially enters a non-equilibrium state with excess energy at F1, and the return to equilibrium can be approximated by an exponential decay. Key factors influencing the time constant for this process include the lifetime of the excited state, mean free path, and optical depth, with the equilibration time being proportional to the mean free path divided by the speed of light. The discussion also highlights the importance of understanding re-radiation and energy absorption in this context.

PREREQUISITES
  • Understanding of molecular vibrational and rotational modes
  • Knowledge of the law of equipartition of energy
  • Familiarity with concepts of mean free path and optical depth
  • Basic principles of electromagnetic radiation interaction with gases
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  • Learn about the mean free path and its calculation in gases
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Physicists, chemists, and students studying thermodynamics or gas dynamics, particularly those interested in the interaction of gases with electromagnetic radiation and the principles of energy distribution in molecular systems.

Calvadosser
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My understanding is that a gas, such as CO2 for example, that interacts with electromagnetic radiation, can only absorb energy at specific frequencies, corresponding to the resonant frequencies of its molecule - equivalently the quantized energy associated with each molecular vibrational/rotational mode.

I also understand that such as gas reaches equilibrium, with equal energy in each possible mode (including kinetic energy) of its molecules [law of equipartition].

My question: If a volume of the gas is subjected to an impulse of electromagnetic energy at just one of the frequencies, F1 say, at which it absorbs radiation, then presumably immediately after it will be in a non-equilibrium state, with an excess of molecules energized at that F1.

MY QUESTION: I'd like to understand the dynamics of the gas once again reaching equipartition equilibrium. Presumably, it is actually very complicated but perhaps it can be approximated by a simple exponential curve by which the excess of molecules that have been energized at frequency F1 diminish until equilibrium is finally reached. Presumably some of the newly excited molecules will re-radiate the absorbed energy at F1 before the gas has reached equilibrium.

- Is my description correct? (Or are there misconceptions in what I have said?)

- If so, what is the approximate time constant involved?

- Does the time constant depend greatly on the temperature or the pressure of the gas? If so, how?

- What proportion of the energy at F1 absorbed from the impulse is re-radiated prior to the gas reaching equilibrium?

Thank you,
Martin
 
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Just some rough ideas - first there is the lifetime of the excited state. Equilibrium has to be longer than that. Then, upon de-exictation, the photon emitted will have a certain "mean free path" or average distance it travels until absorption. The optical depth of the medium. The time for a photon to travel a mean free path will be on the order of the equilibration time, as long as the lifetime of the excited state is much smaller. The mean free path should be something on the order of [itex]\lambda\approx 1/n\sigma[/itex] where [itex]n[/itex] is the number density of unexcited CO2 atoms and [itex]\sigma[/itex] is the cross sectional area for absorption. The equilibration time is something like [itex]\lambda/c[/itex]. That ignores excitation and re-radiation.

Its the optical depth that is the number you are looking for, really. The distance a photon or its offspring go before being converted to heat.
 
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Thank you for the reply.

Could you recommend an book where I can read up on all this? [preferably inexpensive]

Martin
 

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