https://www.quora.com/profile/Johnny-Pa ... liams-1067
What kind of magic is this? It's like you're watching sleight of hand knowing it's a magic trick but you still can't see how it's done. In the case of a gas expanding in a vacuum though surely any work done is a temperature change? And then there's that killer last sentence!Peter Williams
Ph.D. in Physics, The University of Texas at Austin (Graduated 2000)
Several ppl have responded that the answer is “zero.”
Actually, it’s a bit more complex than that. First, note that the question is a bit vague: what is the antecedent of “it?”
If by “it” you mean the vacuum, then yes, the answer is zero (neglecting fancy arcane stuff like the Casimir effect).
If by “it” you mean the gas, then the answer is not necessarily zero. True, if we imagine a box with a partition, gas on one side and vacuum on the other, and we remove the partition - poof! - then the work done on (and by) the gas is zero, after we wait around for the gas to come back into thermal equilibrium. For an ideal gas, all we have done is change its entropy, not its energy. Neither heat nor work has been exchanged between the gas and anything else.
If, however, we imagine that on the vacuum side of the partition, the box extends to infinity, then after we remove the partition, the gas expands indefinitely. It cools adiabatically and gains bulk momentum , approaching Mach 1 or so. The gain of momentum comes from the unbalanced pressure force on the walls, and the mechanical energy comes from the thermal energy of the molecules. In accordance with the 2nd law, you can’t get it all out, of course, but you can definitely get a good chunk of it, especially if you build a nice de Laval nozzle.
This is not just an academic exercise. You can now imagine, say, floating out in space, and having a can filled with high-pressure gas, and you open one side of the can. You will experience a force; work is being done on you, and work is being done on the gas that is expanding in the opposite direction (accelerating it), the energy coming from the adiabatic expansion of the gas.
So in that case, no, the answer isn’t “zero.”
There’s no paradox here; it’s just that usually questions like this about the thermodynamics of ideal gases, until you move on to the field of gas dynamics, are framed in very idealistic conditions. The field should really be called “thermostatics,” not “thermodynamics.” One of the idealizations is usually that you consider the bulk state of the gas to be characterized by a very small number of thermodynamic state variables: density, temperature, pressure. That’s it. That’s fine if the gas is in complete thermodynamic equilibrium, but when the gas is in motion, and in particular when the gas is moving different speeds at different points (I don’t mean the gas molecules themselves, I mean the gas viewed as a continuum), then pretty much by definition it’s not in thermodynamic equilibrium. Instead, in that case, ppl adopt the approximate view that in small enough little volumes (a “fluid particle”), the gas is in some kind of local thermodynamic equilibrium (LTE), but not when considered as a bulk. And of course, at some point, even that approximation breaks down, and the entire concept of thermodynamic quantities like “temperature” and “pressure” start to lose their meaning, even in that restricted LTE sense.
Just had to point this all out b/c it can be a bit confusing when you consider the thought experiment of the astronaut with the aerosol can (the example I gave above), and trying to make that jive with basic ideal gas thermodynamics.
And again, in all cases, the work done on the vacuum is always zero.