# Isothermal Gas Expansion, a Reversible or Irreversible process?

Consider the following situation. A certain quantity of ideal monatomic gas (say one mole) is confined in a cylinder by a piston and is maintained at constant temperature (say $T_0$) by thermal contact with a heat reservoir. Then the gas slowly expands from $V_1$ to $V_2$ while being held at the same temperature $T_0$.

Question: Is this process reversible or irreversible?

Attempt: When the gas expands, the temperature must decrease, so the heat reservoir gives energy to the gas so the gas is maintained to the same temperature, right? Then If we do work on the gas so that it returns to the initial volume $V_1$, we know that due to $\Delta T=0$, then $\Delta U=0$, right? So, the work done on the gas is going to transform itself to heat that would be absorbed by the heat reservoir. My question is: how do we know if the heat given by the heat reservoir is the same that the heat absorbed by itself? If this is true, then I guess the process will be reversible, but if it's not true, does the process would be irreversible? Or it does not matter due that we have a heat reservoir?

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I edited my question @Manishearth –  Anuar May 20 '13 at 15:34

Your process will be reversible only if it is a) quasi-static and b) non-dissipative. It will be quasi-static if it is carried out infinitely slowly in such a manner that the pressure on either sides of the piston varies only infinitesimally. It will be non-dissipative if the piston is frictionless and there is no viscous heating of the gas as it expands.

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So, for practical results, can we consider that the process is reversible? –  Anuar May 21 '13 at 12:18
@Anuar. You can approximate the process to be reversible if you are carrying it out sufficiently slowly. From the definition of reversible process, you would have guessed that real world processes are far from that ideal. –  Amey Joshi May 21 '13 at 14:10

Check the definition of "reversible". The process is only reversible if the external pressure and the internal pressure are the "same", where "same" means that for an expansion the internal pressure is infinitesimally greater than the external pressure.

Put another way, work $W$ is given by $W = \int_{V_1}^{V_2} p_{ext} dV$, where $p_{ext}$ is the external pressure. So you cannot determine the work or the temperature change unless you can do this integral, which you can't, because you do not know if the internal and external pressures are even related.

Thus the question is ill-posed and cannot be answered as stated.

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I don't understand why you say that. According to Blundell (Thermal Physics), the expression $đW=-pdV$ can be used only in a reversible process. Also I understand that "in a reversible cycle, the system and its surroundings will be exactly the same after each cycle". that's why I'm asking for the $Q$'s. Because they are equal then I'll conclude that the process is reversible. –  Anuar May 20 '13 at 18:29
I have an error in the above answer. There should be a minus sign in front of the integral. That said, the physical and thermodynamic definition of work is as I have it in the equation. IF the process is reversible, then and only then can the external pressure be replaced by the internal pressure since the two are the same. As for your point about the Q's, please note that your process is not a cycle. –  Paul J. Gans May 21 '13 at 3:20
I know that the process is not a cycle. But in order to find out if the process is reversible or not, I must know what would happen if I return to the initial state. Finally, can you make me a favor? Which book can you suggest me to read about that fact involving the pressures (external & internal)? I've never listened that. Thanks. –  Anuar May 21 '13 at 6:46