Irreversible Processes in Thermodynamics

Question

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Question: Understanding Work in Irreversible Processes in Thermodynamics

I am currently studying irreversible processes in thermodynamics and have encountered a conceptual challenge. I would appreciate any clarification on the following:

In a reversible process, where the external pressure $P_{\text{ext}}$ differs from the gas pressure $P_{\text{gas}}$ by an infinitesimal amount ($P_{\text{ext}} = P_{\text{gas}} + \delta P$), the work done by the gas and the surroundings on the piston are equal and opposite. This is because the piston, with area $A$ and mass $M$, does not gain any kinetic energy due to the equilibrium of forces during a small displacement $dx$.

However, in an irreversible process, when a gas is compressed against a significantly higher external pressure $P_{\text{ext}}$:

1. The net external force on the piston causes it to accelerate and gain kinetic energy.
2. During the course of compression, the internal gas pressure $P_{\text{gas}}$ increases, which would retard the piston’s motion.
3. This interaction could lead to a to-and-fro motion of the piston (which I assume can be neglected), eventually bringing the piston back to a state of no net kinetic energy when equilibrium is achieved.

Given that the net displacement of the piston results in a change in volume $(V_2 - V_1)$, the work done by the surroundings on the piston is given as:
$ W_{\text{surroundings}} = P_{\text{ext}} (V_1 - V_2). $

Applying the work-energy theorem to the piston:
$ W_{\text{surroundings}} + W_{\text{gas}} = \Delta K_{\text{piston}}. $

Since the piston’s final kinetic energy is zero $\Delta K_{\text{piston}} = 0$, this implies:
$ W_{\text{gas}} = -W_{\text{surroundings}} = P_{\text{ext}} (V_2 - V_1). $

Is this interpretation correct, or is there any flaw in my understandings?

Your insights and explanations will be greatly appreciated!

Answer 1

Your interpretation is correct. However, in an irreversible rapid compression, the force that the gas exerts on the piton is not determined by the ideal gas law, which is valid only at thermodynamic equilibrium. The presence of viscous compressive stresses within the gas cause the behavior to deviate from the ideal gas law. These are also what causes the piston motion to damp out.