# Time constant for gas to reach equipartion equilibrium

 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 $\lambda\approx 1/n\sigma$ where $n$ is the number density of unexcited CO2 atoms and $\sigma$ is the cross sectional area for absorption. The equilibration time is something like $\lambda/c$. 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.