When I have a gas in one vessel which is expanding into another vessel in vacuum, I know the work done will be $0$.

My question is,

  • what if that gas is not ideal, and it's a superheated vapor,

  • is this assumption valid that the work of the expansion will be $0$?

  • $\begingroup$ Yes. Of course you can still assume that. The walls of the vessel are rigid, so no work is done by the gas on its surroundings. But the internal energy for a non-ideal gas is not just a function of temperature. $\endgroup$ Commented Dec 9, 2020 at 13:03
  • $\begingroup$ Are the two vessels thermally insulated? $\endgroup$
    – Bob D
    Commented Dec 9, 2020 at 15:50

1 Answer 1


There is potential work that can be extracted from 2 tanks before pressure is equalized.

After valve is open and pressure becomes equal, work cant be extracted anymore.

This lost opportunity of getting the work out is likely what you are interested in.

To avoid the loss of opportunity to get the work out, gas has to keep the speed-pressure oscillations between the chambers, which is very unlikely to sustain for any significant amount of time. When speed-pressure oscillation stops, energy is dissipated as heat, this is somewhat balanced out with the gas cooling on expansion.

If you ask about work being extracted, then no. You have no device that extracts the work, so you get no work.

If you ask about the energy, then no. Energy isnt created or destroyed, only transformed.

Closest property to what you are likely want to know is entropy, that is closest to estimating how much work can be extracted from a system, if some ideal machine is used.


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