Answer to Question #4769 in Molecular Physics | Thermodynamics for diwash
If you manage to pump all the air out of a steel can, for example, you will have a vacuum in there, but there will be photons constantly radiated off of the walls and re-absorbed by them. This soup of photons will be in thermal equilibrium with the walls, and therefore will have a defined "temperature". In fact, even the deepest of deep space (outside the galaxy, for example), is in a radiation bath of temperature 3K, left over from the Big Bang. There may be other stuff, like the neutrinos, for example, which are not in thermal equilibrium with the 3K radiation because they don't interact with it, and so space may have two or more "temperatures".
But we said a vacuum is a region of space with nothing in it, and that means those photons have to go. Cooling the walls down to as close to absolute zero as you can get (and the limit here is that photons of energies that would be radiated by a wall of a cold temperature would have wavelengths longer than the size of the can -- that'll let you freeze out all of the photons) will give you a vacuum. You have to also shield it from outside sources of energy. There's little you can do about the neutrinos and dark matter -- they penetrate ordinary matter, but also don't really interact with it so to a good approximation you can neglect them.
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