Question

Work done by system on the piston should be Fsys⋅(ds)

and work done by the surrounding on the piston should be Fsur⋅(−ds)
, thus the net work is stored in the piston as its kinetic energy. The piston exerts equal and opposite force on the system and surrounding, therefore dEsystem
should be equal to dq−(psys not surr)dV
. But is it not so? If no then how?
A. Yes, that's correct.
B. No, dEsystem should not be equal to dq−(psys not surr)dV.
C. The relationship between work, forces, and energy changes is more complex than indicated.
D. The piston's kinetic energy does not necessarily account for the net work in this scenario.
A) Yes, Tμν includes stress-energy due to gravity.
B) No, Tμν only includes stress-energy density from non-gravitational sources.
C) Tμν includes stress-energy from both gravitational and non-gravitational sources.
D) Tμν doesn't relate to stress-energy in the field equations.

Answer 1

The equation dEsystem = dq - (psys - psurr)dV is not the correct expression for the change in system's energy. The piston's kinetic energy does not necessarily account for the net work in this scenario.

Step-by-step explanation:

The work-energy theorem states that the net work done on a system is equal to the change in its kinetic energy. In the given scenario, the work done by the system on the piston is Fsys ⋅ (ds), while the work done by the surroundings on the piston is Fsur ⋅ (-ds). The net work is stored in the piston as its kinetic energy.

The equation dEsystem = dq - (psys - psurr)dV is not the correct expression for the change in system's energy. It seems to be a misconception. The relationship between work, forces, and energy changes is more complex than indicated, as mentioned in option C.

Therefore, the correct option to choose is C. The piston's kinetic energy does not necessarily account for the net work in this scenario.