Can vapour pressure be used to 'generate' work for free?
I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.
This will then prompt the system (water sample) to take energy, in the form of heat, from the surroundings to reestablish the thermal equilibrium. Hence taking back the work done by system on surrounding via expansion of the bottle through heat to maintain its temperature. Although, as you can see in the Maxwell distribution of speed, the particles with enough kinetic energy to escape from the water surface are very less at low temperatures.
So, if you take a system with just enough temperature and manage to isolate it from the surroundings, I think you should see a visible drop in the temperature of water proportional to the work done in expanding the water bottle.
Mathematical description of the above Explanation:
Let's say the total energy change taking place after the particles escape water surface is $\Delta U$ out of which $W$ is used to do work on the surroundings. The remaining energy ($\Delta U - W$) is used to establish an equilibrium temperature inside the bottle and hence is given back to the air-water system inside the bottle.
This step may be happening simultaneously with the next steps I am writing but I am assuming that the air- water system inside the bottle will reach an equilibrium first before any work is done. This is purely to make the math easier and because thermodynamic is the study of equilibriums rather than processes so it won't affect the results as long as this equilibrium is established even when we are studying the final change taken place.
So the net change in temperature inside the bottle is $$\frac{-W}{C_{system}}$$ where $C_{system}$ is the heat capacity of the system.
Net temperature difference between system and surrounding $$\Delta T=\frac{W}{C_{system}}$$
So the system will absorb energy in the form of heat to bring the temperature back to equilibrium $$Q=C_{system}\Delta T=&gt;Q=C_{system}\biggr[\frac{W}{C_{system}}\biggr]=&gt;Q=W$$
So we can see mathematically that the system will take back all the energy it gave to the surroundings by doing work in the form of heat.