Doubts regarding work and pressure during adiabatic compression of a gas in a cylinder

Question

So, I have doubts regarding the work done on the gas in this situation:

Let's say I have an ideal gas inside an adiabatic cylinder fitted with a piston of area $A$.
The piston has mass $M$.
Outside the cylinder there's atmospheric pressure $p_{atm}$.
Suppose the gas and the piston are at equilibrium and the piston is at a distance $h$ from the bottom of the cylinder. Then, a mass $m$ is put on the piston, which compresses the gas until equilibrium is reached again.

In calculating the work done on the gas by the enviroment, which is: $$ W =\int p \,dV $$ I would assume that I should use the external pressure that can be written as: $$p_{ext} = p_{atm} + \frac{mg}{A} + \frac{Mg}{A}$$
Therefore, the work becomes: $$W = \big(p_{atm} + \frac{mg}{A} + \frac{Mg}{A}\big) \Delta V$$ Is this right? The fact that there's a contribution to the work given by atmospheric pressure and the weight of the piston bugs me a little, because before the mass $m$ was put on the cylinder, everything was at rest. To me, it doesn't feel right that something other than the mass $m$ is doing work on the cylinder.
Should I replace the pressure in the work integral with just the pressure given by the mass $m$, instead? Because that's the difference in pressure that breaks the equilibrium? In that case the work would become: $$ W_? = \frac{mg}{A} \Delta V$$ Now, if I call the linear displacement of the piston $x$. The final volume of the gas would be $V_1 = (h - x) A$.
Since the initial volume is $ V_0 = h A$, $\Delta V$ becomes:$$\Delta V = V_1 - V_0 = - A x$$ Therefore the work, following this reasoning, would be:$$ W_? =\frac{mg}{A} (-Ax) = -mgx$$ To me, this makes more sense because it means that the work done on the gas is equal to the potential energy that the mass $m$ loses going from $h$ to $h-x$.
Could you explain to me which is correct, $W$ or $W_?$ and why?
Thanks for taking the time

Answer 1

The first version is correct. The atmospheric pressure is accompanied by a volume change in going from the initial state to the final state, and the potential energy of both masses change when they do work on the gas.