According to Wikipedia
In physics, work is the energy transferred to or from an object via the application of force along a displacement.
But for a gas in a closed vessel with movable massless piston, what would they do work on? Since the piston is massless, the energy will not be transferred to it, where would the energy obtained from the work go?
If there is an atmosphere on the outside face of the piston, and the piston is massless, then the force that the gas exerts on the piston is equal to the outside atmosphere pressure times the area of the piston. So the gas does work pushing the outside atmosphere back. The work the gas does is equal to the outside atmosphere pressure times the change in volume of the gas. This is the same as the force of the gas times the displacement of the piston.
If the piston were truly massless, at the slightest touch it would move away at the speed of light.
But that's not what they mean when they say "massless"; what they really want is for you to ignore any effects that the piston's mass would introduce. Otherwise it would unnecessarily complicate the situation without adding to your understanding of the simple principles that they are trying to teach.
You will similarly encounter frictionless surfaces, massless pulleys and ropes, perfectly incompressible solids, etc. None of them exist in reality, they are just a way of simplifying the situation so that you can concentrate on the details that are important.
This particular question doesn't supply any additional details, but presumably the piston causes some other object to move, and that object has the mass that you would use in your calculations.
Here work is done against an external force and if the external forces are zero no work is done. For instance consider Joule's free expansion wherein work done is zero even though the gas expands and there is a drop in pressure simply because external forces are absent and the change in the internal energy (for real gases) is credited to dissipative effects or irreversibility. Answering your question, for a reversible process the energy lost by the system in doing work is stored in the surrounding (in any form, be it pressure energy or potential energy etc.) by changing its state.
For equilibrium, we say that: $$p_{in} = p_{ext} + dp$$
Which means that internal pressure is the external pressure plus or minus some small differential. So, to understand the scenario, simply think of what might be causing the external pressure. If the gas was in a vacuum and expanding $p_{ext}$ is zero and since that is the pressure that comes to the work term, the work is zero.
For the case where the outside environment is the atmosphere, i.e caused by the air outside, then it is easy to understand that the atmosphere is what causes the external pressure. So, when we expand, work is done on the atmosphere.
