# Sudden volume increase piston

I am wondering what happens in the following situation. I have a piston filled with an ideal gas for which I suddenly/instantaneaouly iscrease the volume. In particular I want to know what happens to the temperature and entropy of the gas.

After some thinking I reckon that the temperature will remain constant since no work is done by the gas and no heat flow is present.

For the entropy I use the relation: $$d \Omega = TdS + VdP = 0$$ Then I would get: $$dS = - \frac{V}{T}dP = - \frac{R}{P}dP \Rightarrow S_f = S_i -R \ln{\left(\frac{P_f}{P_i}\right)}$$ Where $P_f,S_f$ are the final pressure and entropy, and $P_i, S_i$ the initial pressure and entropy. ($R$ the gas constant).

As a bonus question that springs to mind, why isn't this a quasi-static process?

• I reckon you're right, and so do these guys: learnthermo.com/examples/ch05/p-5e-3.php – Jasen Sep 15 '15 at 10:45
• The temperature would still drop for an expanding gas, just not as much as if the energy had been transferred to the piston by slow expansion. – Stuart Van Horne Sep 27 '18 at 0:53
• Also important, perhaps, is that one knows a priori that the energy of an ideal gas depends only on the temperature (and not the volume.) The original process is not quasistatic, but the virtual process (that is used to compute/define the entropy change) is. Sounds like a miscommunication. – TotallyRhombus Nov 28 '18 at 21:15
• Quasi-static and reversible are different things. All reversible processes are quasi-static, but not vice-versa. – Alexander Dec 29 '18 at 14:07