I would argue that if you are computing the work done by an expanding gas in a frictionless piston, in an irreversible expansion where the inner pressure is greater at all times than the (constant) outer pressure, that you should integrate the internal pressure over volume change, not multiply the volume change by the constant outer pressure. However, there are many sources which say that to compute work by the gas, you should just multiply volume change by external pressure. I believe that the work done by the gas is computed by integrating pressure over volume change, but that the work done by the gas plus the piston is computed by multiplying external pressure by volume change, and that the two are different. So I am wondering if all of those sources are subtly wrong, or if I'm missing something. Suppose that you have a frictionless piston lying on its side (so that the piston weight doesn't add any extra pressure), held in place by a latch. Inside is gas at 4 atm; external pressure is 1 atm. You release the latch and the piston expands to double its volume (so the new pressure is 2 atm), where it is caught in place by another latch. What is the work done by the gas in this process? Here is my argument: 1) To find work done by a force, you integrate force over distance moved -- one of the first things learned in physics -- so the work done by the gas is found by integrating the (changing) internal pressure over the volume change, because the force exerted is proportional to the internal pressure. 2) If you consider "the system" to be the gas plus the piston, then the work done by that system is equal to the external pressure times the volume change, because that's the work required to make room for the enlarged system. 3) The difference between #1 and #2 can be accounted for by the fact that the expanding gas accelerates the piston and gives it kinetic energy. Once the piston locks into place, that kinetic energy is dissipated as heat energy. However, if the gas imparted that energy to the piston, that still counts as work done by the gas, even if it gets dissipated as heat once the piston stops moving. It's similar to the question about how if you lift an object 1 meter that weights 10N, you've done 1 Nm of work -- but, a student asks, what if you lift it 1 meter using a force of 50N, haven't you done 50Nm of work? Yes, but most of that that 50 Nm lifting work gets converted into the kinetic energy of the object, and if the object locks into place at the end, that kinetic energy will be dissipated as heat energy. (Now, I had assumed that the internal pressure of the gas in the piston could be computed using the ideal gas law. However, if it can't be, I think the internal pressure is obviously greater than the external pressure -- otherwise the piston wouldn't move -- and integrating it will give you a larger value for work done, than you would get by multiplying volume change by external pressure.) So first things first, is this correct so far? Because there are many sources saying that under the conditions that I've specified, the work done by the gas equals external pressure times volume change, not internal pressure integrated over volume change. (I'm talking about irreversible processes, where the two pressures differ; in reversible processes where they're the same, it doesn't matter.) If I'm right, then I think these sources are in fact wrong: http://www.columbia.edu/itc/chemistry/f1403/lectures_ppt/Wk_10_Mon.ppt https://facweb.northseattle.edu/hprice/CHEM161/Silberberg PPT to link/ch06_lecture_7e.ppt https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map:_Physical_Chemistry_(Atkins_et_al.)/Chapter_2:_The_First_Law/Topic_2A:_Internal_Energy http://www.quantumstudy.com/chemistry/thermodynamics-4/ https://quizlet.com/77523508/chemistry-133-chapter-6-thermochemistry-flash-cards/ http://docfish.com/wp-content/uploa...Energy_First_Law_of_Thermodynamics_HW_KEY.pdf (In each of those, you can find the part that I'm disputing, by searching for the phrase "work done by the gas". I found these, and many others, just by googling "pressure volume work" "work done by the gas".) On the other hand there are some sources which agree with me: https://www.grc.nasa.gov/WWW/k-12/airplane/work2.html which says that compute the work done by the gas, you integrate the changing internal pressure, which is clearly less than the external. So, which is it? To avoid horrible confusion, please be clear what position you're taking: 1) Is the work done by the gas equal to (A) the external pressure times volume change (and I'm wrong), or is it (B) some larger number that you would get by integrating internal pressure over volume change (and I'm right)? 2) If your answer to #1 is A, what exactly is wrong with my argument that the internal pressure is greater than the external pressure, and to get work done by the gas, you integrate the force over the distance moved? 3) If your answer to #1 is A, is the NASA page in fact wrong (the one I think backs me up)? 4) If your answer to #1 is B, then do you agree that the other sources listed above -- which say to use pressure-times-volume-change -- are in fact wrong?