I've been looking at some sample problems involving work done by gas in an irreversible expansion in a piston. They all say $W = P\Delta V$, but some of them say to use internal pressure for $P$, some of them say to use external pressure, and I'm wondering if a large portion of the sources are simply wrong. My argument is that you should use (changing) internal pressure to compute work done by the gas, and (usually constant) external pressure to compute work done by the gas plus the piston. But many sources say to use external pressure to compute work done by the gas, and this sounds incorrect.
Suppose you have gas compressed in a piston at 4 atm, currently volume 1 liter, and the piston is frictionless but held in place by a catch (with normal 1 atm pressure outside). (Edited to add: let's assume the piston is lying on its side so the mass of the piston does not contribute to the internal pressure.) You release the catch and the piston expands to twice its volume (and pressure decreases to 2 atm) where it's caught by another catch. What is the work done by the gas?
Here is my reasoning:
- If you consider "the system" to be the gas, then the work done by the gas as it expands is the integral of the internal pressure over the change in volume, because the first thing I learned in physics is that work is the integral of force over distance moved.
- If you consider "the system" to be the piston as a whole, then the work done by the piston is the external pressure times the change in volume. This is the work required to make room in the surroundings for the expanded system.
- The difference between #1 and #2 is because when the gas expands, it imparts kinetic energy to the piston, and when the piston hits the second latch and locks into place, the kinetic energy gets diffused as heat energy. But that should still count as work that the gas did to get the piston moving, even if the piston lost the kinetic energy when it hit the latch.
(This is similar to the question posted at:
where someone asked: if I'm lifting a 10 N weight 1 meter by using a 50 N force, why is the work done only 10 Nm and not 50 Nm? Because if you lift it using a 50 N force, the rest of that work goes into the kinetic energy of the object. Presumably if the object hit something and got locked into place then the kinetic energy would be diffused as heat energy.)
Is my reasoning right so far?
Because then, here are several supposedly reliable sources which all say that the work done by the gas when it expands, is the external pressure times the volume change. My position is that unless it's specified that it's reversible (i.e. external pressure = internal pressure), this is (subtly) incorrect, and you would use external pressure to compute work done by the gas plus the piston, but not the work done by the gas by itself. I found these just by googling "pressure volume work" "work done by the gas", and about half of the results say to use external pressure, and that's what I'm arguing is incorrect. Here are the sources saying to use external pressure; for each document, you can search for the phrase "work done by the gas":
why do we use the external pressure to calculate the work done by gas
(the "accepted" answer says to compute work done by the gas, you use the external pressure; the other answer says that you use the internal pressure) http://www.quantumstudy.com/chemistry/thermodynamics-4/
and many more which come up just googling "pressure volume work" "work done by the gas".
As long as work is the integral of force over distance moved, I'm saying that to compute work done by an expanding gas, you use internal pressure (if it's different from external), and these sources are incorrect. Am I missing something?
(To avoid confusion, please specify: (1) do you agree that to compute work done by an irreversibly expanding gas, you use internal pressure, or am I wrong; and (2) if I am right about #1, isn't it also the case that the sources above which say otherwise, are incorrect? If not, why not?)