Register to reply

Time constant for gas to reach equipartion equilibrium

by Calvadosser
Tags: constant, equilibrium, equipartion, reach, time
Share this thread:
Calvadosser
#1
Feb18-12, 05:12 PM
P: 36
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
Phys.Org News Partner Physics news on Phys.org
Mapping the optimal route between two quantum states
Spin-based electronics: New material successfully tested
Verifying the future of quantum computing
Rap
#2
Feb18-12, 08:38 PM
P: 789
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.
Calvadosser
#3
Feb19-12, 01:20 PM
P: 36
Thank you for the reply.

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

Martin


Register to reply

Related Discussions
Time for Hydro System to Reach Static Equilibrium Mechanical Engineering 0
Time to reach thermal equilibrium and/or a steady state Classical Physics 5
Time taken for two substances to reach equilibrium (equation?) Chemistry 2
Time to reach thermal equilibrium with radiation Astronomy & Astrophysics 2
Time it takes to reach equilibrium Classical Physics 7