Internal energy during the expansion of a gas?

[Pericles98]

I have question regarding the (W = ∫Pdv) formula for the work done during the expansion of an ideal gas and the change in internal energy during the process. If we were to have a gas enclosed inside an insulated cylinder with a movable piston at one end with cross sectional area "a", I understand that the work done by the gas on its surroundings is derived from:

W = Fd
W = PAd
W = PΔV

Where P is the internal pressure of the gas. If the pressure varies as a function of volume,

W = ∫Pdv

Because the container is insulated, there is no exchange of heat between the gas and its surroundings, and its change in internal energy is simply given by ΔU = -W = -∫Pdv. However, I do not quite understand why the change in internal energy does not take into consideration the work done by the atmospheric pressure onto the gas.

It would be analogous to having a block being pushed by two opposite forces F1 to the right and F2 to the left. If the block moves a distance x to the right, then F1 performed a work of F1x on the object. However, F2 also performed a work of -F2x, and so the total energy change in the object would equal F1x - F2x.

Why then, is the change in internal energy of the gas not ΔU = -∫Pdv + P_atmΔV? Where P_atmΔV is the work done by the atmospheric pressure onto the gas?

 

[DoItForYourself]

You assume that in the initial condition the piston stays at a constant place and does not move. Consequently, ΣF=0 (on the piston) and this means inner pressure=outer pressure=P_atmospheric (since surface area in both sides is the same).

In other words, the total pressure that is applied to the piston (initially) is zero.

 

[Chestermiller]

In the example you gave involving two forces acting on a mass, you are not dealing with an action-reaction pair, but rather with two external forces, and both must be included in calculating the work done on the mass. If you were dealing with the work done on the piston by the gas and by the external atmosphere, the same principle would apply. But if you are dealing with the work done by the gas on the piston, you would include only the force of the gas on the piston, and not the force of the piston on the gas, since this is not an external force acting on the piston (it is part of the action-reaction pair).

 

[Pericles98]

Thank you!