Well, look at the piston. The force downwards is: (Atmospheric pressure)*(piston area) plus (sum of the masses of the piston and the weights)*g. These are not temperature dependent. If the piston is to stay where it is, the force upwards from the air below the piston (which is (air pressure)*(piston area)) must equal the force downward. Hence: The air pressure must stay constant.
Ok . Thank you very much.
I don't think that's correct. The example shown days nothing about the outside air. The pressure of the air inside the piston should absolutely change.
I think they mean the pressure after a new equilibrium was found.
Not as I read the problem... The volume will change in such as a way as to maintain a constant pressure.
Okay. That makes sense. I wasn't thinking about after equilibrium.
Pressure in the piston will change if the outside pressure changes, but temperature changes won't change anything.
Even far away from a temperature equilibrium the piston will still be in mechanical equilibrium (unless you change the temperature with an explosion...).
Can you elaborate on this?
Force equilibrium, as Svein explained. If the (frictionless) piston does not move, pressure from below multiplied by the area is equal to pressure above multiplied by the area plus the weight of piston and the added weights. If we assume that pressure outside does not change, and we do not change the weights, pressure inside won't change.
I think there's some confusion. I was under the impression that we weren't taking the outside air into account since the 'air' is labeled as being inside the piston.
It is not given that we have a vacuum outside so air is a natural assumption - but even that does not matter because the answer "the pressure does not change" is independent of the absolute pressure.
Oh oh oh, I get it now. I was confused by 'temperature equilibrium' vs 'mechanical equilibrium'.
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