Basically, I need to conceptually understand why the work a gas does is the integral $\int p_\mathrm{external}dv$ and is 0 when pressure external is 0. I understand why $\mathrm{d}w = - p_\mathrm{externa } \mathrm{d}v$ and so obviously I understand why the math says the work is 0; I need to conceptually understand it.
When you have isothermal expansion in a cylinder where the external volume outside the gas has pressure 0, a.k.a. vacuum, must the movable top of the cylinder be massless? The equations obviously say the work is 0, but I think of the stopper itself as being resistance. Could I think of isothermal expansion in a vacuum as the top basically "disappearing" when the gas is released?
This is difficult to phrase for me, so please ask if you don't understand. I also am only a sophomore in high school, so I have a limited understanding of calculus. Like I said, conceptually for the pressure against the gas to be 0, the lid must be frictionless and massless, right? Thank you.