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I'm having some trouble trying to find the irreversible work done by a perfect gas when expanding. I know that the reversible work is greater than the irreversible one, but how can I find the irreversible work for an isothermal process?

The exercise asks me to find the irreversible work for the expansion of a perfect gas from 10000 to 2000 bar at 315 K with $T_{cte}$. I also found that the irreversible work is simply $P\Delta V$, and even though the volume does vary, the exercise also tells me the pressure does too. I found the volumes for each state and got $V_1=2.619*10^{-3}L$ and $V_2=0.013095 L$. But now I'm not sure what pressure I should use? Or why should I just use one pressure. Thanks in advance.

Also, is the irreversible work for a gas the same independently of the process? Is it the same for an isothermal, an adiabatic and isobaric process? Or just like the reversible work, does it change?

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The expansion takes place when the pressure is changed suddenly and drastically modifying the initial equilibrium state, remember that at difference of the reversible case the pressure is not changed infinitesimally. So in the formula $P \Delta V$ use the final pressure.

The other questions you set up are very interesting, I recommend you this paper that might help you.

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  • $\begingroup$ It looks like I've got to pay to read that paper. $\endgroup$
    – Nicholas J.
    Aug 24, 2014 at 3:49
  • $\begingroup$ I updated the link $\endgroup$
    – dapias
    Aug 24, 2014 at 3:54
  • $\begingroup$ Thank you! Would you mind telling me how is it possible for a gas expansion to do PV work at constant pressure and temperature? I thought if V changed then either P or T must change. $\endgroup$
    – Nicholas J.
    Aug 24, 2014 at 4:07
  • $\begingroup$ Well, a gas at constant number of particles has two degrees of freedom, it means that knowing the values of two variables it is enough to determine the remaining one. Applying this to your question, it implies that if pressure and temperature are constant, then volume also is constant and there is no process. $\endgroup$
    – dapias
    Aug 24, 2014 at 4:29
  • $\begingroup$ But if we take a look at a phase change? Say water from liquid to gas held at 100°C and 1 atm, the V does change, but T and P remain constant, and there's still PV work, is that right? $\endgroup$
    – Nicholas J.
    Aug 24, 2014 at 4:41

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