3 posts • Page 1 of 1
I am a bit confused on the piston problem. Does the external pressure on the piston cause work to be done, which then causes displacement? I don't understand how applying downward force causes the distance part of the equation (w=PAD).
The system has internal energy, ∆U, which is q + w. Therefore, the system has the capacity to do work. In the particular problem I believe you are referring to, we are letting the system do work on surroundings, meaning it is pushing the piston out. Were the piston to be compressed, the volume would decrease, meaning ∆V would be negative and therefore w would increase (because w = -P∆V) because work is being done ON the system. This would increase internal energy. However, in this problem, we seen in the third picture than internal energy decreases. The downward force, F = PexA, in this problem is not actually forcing the piston down any further at this moment. Instead, imagine that the system has already been compressed and we are looking at its expansion. The system is expanding by doing work against the external pressure, represented by P, and is changing the volume of the system by the value of the Area pressure is being exerted on (A) times the distance the piston moves out (D). ∆V is therefore positive in this problem because volume of the system is increasing. Because w = -P∆V, the internal energy of the system is decreasing, which is expected because work is being done on the surroundings.
The external pressure in the equation is really just the normal pressure outside of the system. When work is done by the system and pushes the piston up, this displaces the particles on the other side of the piston and thus results in the energy change. The external pressure changes so little when this happens that we don't even consider it.
Who is online
Users browsing this forum: No registered users and 3 guests