- #1
Pericles98
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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 + PatmΔV? Where PatmΔV is the work done by the atmospheric pressure onto the gas?
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 + PatmΔV? Where PatmΔV is the work done by the atmospheric pressure onto the gas?