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Why is the work done in an irreversible isothermal expansion against vacuum equal to zero?

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A system does its maximum amount of expansion work when the external pressure is equal to the pressure of the system all the way through the expansion process.

At constant pressure work done is given by: $$w=-p_\mathrm{ex}\Delta V$$ Where $p_\mathrm{ex}$ is the external pressure. Therefore the minimum work done is when the external pressure is $0$, which is the case under vacuum. This would mean that the work done under vacuum is $0$.

This is called free expansion. The greater the opposing force then the greater the amount of work done is (as work is equal to force applied over a distance). Therefore when the external pressure is $0$ there is nothing for the piston to push against, because there is no opposing force, no work is done by the system.


If you are interested to see where $w=-p_\mathrm{ex} \Delta V$ comes from then note that $$w=Fd$$

Because pressure can be given by $P=F/A$ and volume can be given by $V=Ad$ we can see that $$w=\frac{F}{A} Ad = p\Delta V$$

This is work done on the gas. For work done by the gas, we put a negative sign in front of the pressure term.

*Note that this is at constant pressure.

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  • $\begingroup$ Good answer. It really is intuitive as well, work is defined by a transfer of energy from one system to another - if the other system has vacuum, then no change in its volume will transfer any energy to it. Trying to pump something with vacuum on the low pressure side of the pump will not produce any gainful results. $\endgroup$ – Stian Yttervik Jan 11 '17 at 9:09
  • $\begingroup$ @PiyushKumar you wouldn't mind marking my answer as correct then would you? ;) $\endgroup$ – Georgeos Hardo Jan 22 '17 at 17:20
  • $\begingroup$ Yeah sure george!! $\endgroup$ – Piyush Kumar Jan 22 '17 at 17:52

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