This issue was also addressed by Miller in his earlier response. The piston is treated as a rigid body, and a rigid body can have no change in its internal energy (e.g., a rigid body cannot be compressed.) The bottom of the piston is exposed to the gas and the top is exposed to the atmosphere. Calculate the work done on the system in the following reversible processes: (i) Isothermal compression of 1 mole of an ideal gas at 273K from 1 to 2atm. Let the system be defined as the gas in a container, a closed thermodynamic system of constant mass, with a moveable boundary, that being the interface of the gas with a moveable piston on top of the gas. The answers to both these questions follow, using a detailed evaluation of the work done on the piston by a single force, from the pressure of the gas, contrasted with the net work done by the total force on the piston, the vector sum of all forces on the piston: the force from the gas pressure, the weight of the piston, the external atmospheric pressure on the piston, and other forces on the piston due to external loads driven by the piston. And, if there is also an external pressure on the piston (e.g., from the atmosphere), and from other forces on the piston due to external loads driven by the piston, why are they also not included in the work done by the gas? The answer is: we are interested in the work done by a system on its surroundings for the piston example, this is the work done by the gas (the system) on the piston (the surroundings), but this is not the total work done on the piston. Specifically, why is the weight of the piston not included in the work done by the gas? Based on comments and answers by others, there is confusion about the answer to this question. The detailed evaluation also answers a question by in an earlier comment for a gas expanding and moving a piston. Today in our chemistry class we derived the pressure-volume work done on an ideal gas. For a quasi-equilibrium process this work can also be evaluated considering changes in the internal pressure and volume of the gas but in general for a non quasi-equilibrium, irreversible, process the work cannot be evaluated using changes in the internal pressure and volume in the gas, because the gas is not in a definite state.Ī detailed evaluation of work for a gas expanding and pushing a piston follows. During an isothermal compression of an ideal gas, 335 J of heat must be removed from the gas to maintain constant temperature. The answer to the question by the OP is that the work done by a system can always be evaluated considering movement against the external force (pressure).
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