Why the sunset is red and a possible experiment? Before downvoting: I have read some of the answers and will cite them as well.
Not trying to debunk physics or create conspiracies. I love physics but I don't understand it.
Background
This is more or less the model in my mind:

*

*We see the sky blue because blue light is scattered more relative to the rest of the wavelengths composing light, due to its shorter wavelength. Blue light ends up traveling in the atmosphere rather than going through it, and so we get blue all around. The phenomenon is called Rayleigh Scattering.


*Heuristically we can accept that, once you subtract blue from white light, you get yellow. The sun then has to be white, and we just see it yellow.
Now, for the redness of the sunset, I understand this:

When the sun is near the horizon, it the path of the light through the
atmosphere (and in particular through layers which have a higher
concentration of dust particles) is longer, thus the scattering away
of the non-red components is more pronounced.

That is from the current accepted answer.
I also took a look at Rayleigh Scattering to Hyperphysics.
And I remember a lecture (I think it is by Walter Lewin) where a dense smoke is shine with white light and you see it blue. And the light source should look yellow (through the smoke).
Question
I wonder the following: Suppose we have a massive room with movable walls (maybe also mirrors) full of air or smoke.
Would we see different colors, using white light as a source, as we increase the length of the room (same density of the gas), and the path of the light through it? From blue to red?
 A: For particle spacing much larger than wavelength, one can treat each individual particle interaction as a discrete, isolated scattering event. The probability of the scattering event varies according to $\lambda ^{-4}$, where $\lambda$ is wavelength (larger wavelengths are redder, smaller wavelengths are bluer).
For any given photon of a particular wavelength, it is equally likely to scatter on its first close encounter with a particle as on its trillionth. However, given the very strong wavelength dependence of the probability, the smaller-wavelength photons have a much higher probability of scattering on every interaction and will tend to scatter early. For something that scatters only very weakly, like the atmosphere, this means that the small wavelengths constitute most of the scattered light across the whole sky (hence the sky being blue from horizon to horizon). For something like dense smoke, the effect can be pronounced over short distances.
Head on, the color of a white light through a particulate scattering medium will dim and redden with increasing distance. For the scattered visible beam, it will become bluer with increasing angle (0 degrees being head on, 90 degrees being side on), but redder and dimmer with increasing distance.
An example of this can be seen in the video on this news article from about 2:05 to 2:20 - although cameras are even more unreliable than the naked eye for recording the true color and brightness of lights, so take the following with due skepticism.
The camera approaches the white-light headlights of two emergency vehicles at a fairly consistent 30-degree angle as it gets closer while going around a curve. At first the lights appear to be dim orange and the beams, also dim orange, are barely visible. As the camera is getting closer, still at about 30 degrees, the color of the lights is shifting to yellow-white while the beams are clearly visible and are white in color. As the vehicle with the camera passes the emergency vehicles, the angle between the camera and the beam increases, and the color of the lights shifts to their true white color while the color of the beams is blue (although in real life it was likely not as vivid as what the camera recorded). Meanwhile, the red-tinted light-bar on the fire truck doesn't shift noticeably in color during the clip.



A: Not sure I follow, if you filled a room with an atmosphere, not just oxygen. And had a white light shone from one end to the other, assuming the light is not reflected from the walls and is absorbed mostly. Say a black wall.
The colour on the other end of the wall, will get less and less blue .[ White-blue].
Not only will it get less and less blue, as the distance light travels through the atmosphere increases, the scattering of all of the other colours is also more defined.
Red gets scattered less than any other colour so there is a point where the only remaining colour is red
