- If the temperature doesn't change, why is the expansion irreversible? Doesn't reversibility require no net change?
- Irreversibility is not about changes in state functions like temperature, but about the path and the entropy of the universe. While the gas itself returns to its initial temperature, its entropy has permanently increased (ΔS > 0). To compress the gas back to its original volume reversibly, we would have to do work on it and transfer heat out, increasing the entropy of the surroundings. The combined entropy of the system and surroundings would still be greater than zero, making the original free expansion irreversible.
- Is any real-world process a true Joule expansion?
- A perfect Joule expansion with absolutely no heat transfer or work is an idealization. However, a very rapid expansion of a gas into a well-evacuated chamber through a porous plug (in the Joule-Thomson experiment) is a close approximation where enthalpy, not internal energy, is constant. The simple free expansion model is crucial for understanding the conceptual foundations of entropy and irreversibility.
- Why is the work done equal to zero? The gas is expanding, so isn't it pushing on something?
- Work in thermodynamics is defined as energy transfer due to organized, macroscopic motion against an opposing force. In free expansion, the gas expands into a vacuum where the opposing pressure is zero. While individual molecules collide with the moving boundary, these collisions are disordered and not against a net opposing force. Therefore, there is no macroscopic, pressure-volume work (W = ∫ P_external dV = 0).
- How does the entropy increase if no heat is transferred (since dS = dQ_rev/T)?
- The formula dS = dQ_rev / T applies only to reversible heat transfers. For an irreversible process like free expansion, we must calculate entropy change using a reversible path connecting the same initial and final states. For an ideal gas at constant temperature, that reversible path is an isothermal expansion, which does involve heat transfer. The calculated ΔS = nR ln(V2/V1) is the same for both the reversible and irreversible paths between those states, highlighting that entropy is a state function.