We say that area under the curve gives the work done. So for an irreversible iso thermal expansion, we have work done= -P external( V final- V initial) signifying a rectangular graph. But we know it's isothermal so it would be along an isotherm? Or no. I believe if we have external pressure at the y-axis, graph would be rectangular and if we take pressure of gas at y-axis we would get a curve, same as isothermal reversible but would end at a lower volume than the former. So what do we take for the p-v diagram.


2 Answers 2


If you have constant external pressure $P_{ext}$, the work on the PV diagram will be minus the area of a rectangle that is $P_{ext}$ high and $V_{final} - V_{initial}$ long.

enter image description here

When you want to show work as an area, you must plot external pressure on the y-axis of a PV diagram, not gas pressure. In the special case of a reversible expansion or compression, the gas pressure $P$ is equal to the external pressure $P_{ext}$ at every point during the process, so you are still plotting external pressure if you're relating work to the area under the curve.

Why does the external pressure matter in the calculation of expansion work in irreversible processes, and not the gas pressure?

The reasoning goes like this:

expansion work = opposing force x distance moved

opposing force = external pressure x area

volume change = area x distance moved


expansion work = -(external pressure) x (volume change)

The minus sign comes from our decision to make work negative when energy leaves the system in an expansion, and positive when energy enters the system from the surroundings in a compression. This is the convention used in chemistry; you'll find the opposite convention used in physics!

  • $\begingroup$ That would be plotting the external pressure with volume of the gas. What if we plotted pressure and volume of the gas? $\endgroup$
    – jatin
    Mar 19, 2016 at 14:00
  • $\begingroup$ You could do that, but the area under the curve would NOT correspond to the work in that case. $\endgroup$ Mar 19, 2016 at 14:03
  • $\begingroup$ What would that graph look like? $\endgroup$
    – jatin
    Mar 19, 2016 at 16:19
  • $\begingroup$ Is external pressure greater than internal pressure in irreversible expansion $\endgroup$
    – Scáthach
    Aug 12, 2018 at 6:33

The most extreme form of an irreversible isothermal expansion is the adiabatic isothermal expansion of an ideal gas into vacuum. In this case no pressure-volume work is done, so in the PV diagram pressure first drops to zero, then volume increases to the final volume, and the pressure increases to the final value. The area under the curve is zero.

            enter image description here

A more common irreversible isothermal process can be seen as a combination of this totally irreversible process with the reversible process.

  • $\begingroup$ Why does the pressure drop to 0? There is some pressure inside the gas, right? Pressure would drop to 0 if we are plotting external pressure and volume of gas. If we plot pressure of gas, then? Which one is more commonly used? $\endgroup$
    – jatin
    Mar 19, 2016 at 13:59
  • $\begingroup$ When expanding into vacuum there is no pressure to work against. The pressure in the diagram is external pressure since the work we are interested in is energy transferred by the system to its surroundings. $\endgroup$
    – aventurin
    Mar 19, 2016 at 14:22
  • $\begingroup$ @aventurin The pressure of the gas won’t suddenly drop to zero. And the external pressure is zero throughout. $\endgroup$
    – Atom
    Dec 8, 2019 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.