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Suppose there's a spaceship that has a hole in it. Gas is expanding through the hole from the spaceship to outer space (vacuum). According to pressure volume work ($P\Delta V$) the work done by the gas is infinity. According to my homework assignment, it's $0$. Can someone please explain why?

Edit:

According to Wikipedia:

$P$ denotes the pressure inside the system and outside the system, against which the system expands; the two pressures are practically equal for a reversible process;

How can $P$ denote two different quantities? Can't this rule be applied for situations where the pressures inside and outside the system are different?

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  • $\begingroup$ In this case external pressure is 0 (since it expands against vacuum). Therefore work done Is 0.What makes you think work done is infinity? $\endgroup$ – Karan Singh Mar 10 '15 at 21:20
  • $\begingroup$ The example of a spaceship is not very good, since gas expanding through a hole in a spaceship will do work by accelerating the spaceship, unless the spaceship is fixed. This is the principle by which rockets fly and by which inflated baloons move after you let them go. $\endgroup$ – Curt F. Mar 10 '15 at 21:43
  • $\begingroup$ @CurtF. Is work done? Or is the acceleration of the craft just a consequence of conservation of momentum, without a thermodynamic work interaction? $\endgroup$ – hBy2Py Aug 26 '15 at 17:49
  • $\begingroup$ I don't think its controversial to say that all energy changes involve transfers of either heat or work. Take the system as the spacecraft (or balloon) only, i.e. it doesn't include the gas. The transfer of energy that makes it accelerate sure isn't heat. So it must be work. $\endgroup$ – Curt F. Aug 26 '15 at 18:13
  • $\begingroup$ @CurtF. Mm, of course -- to tell whether work is done you have to choose your control volume such that its boundary includes the surface at which the putative work interaction occurs. Gah, this is why thermo always drove me batty! $\endgroup$ – hBy2Py Aug 26 '15 at 19:01
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The external pressure in space is 0 (well, depends where and not quite, but for the discussion let's say that it is). So $P \cdot \Delta V$ is also 0. Gas expanding from a hole in a spaceship is not an irreversible process. Such processes are very slow and gradual, such that the external pressure has time to equilibrate with the internal pressure.

On the contrary - you have the most irreversible process in here. Gas leaking from a hole in a spaceship into a medium (space) that will never equilibrate with the internal pressure.

Think about it from an intuitive point of view. Work done by an expanding system is there because it moves other things around. In space there is nothing to move, so no work done.

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