Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Actually, this process is reversible, but it is not cyclic. In reversible processes, the internal pressure is equal to the external pressure. That is the reason equilibrium remains constant.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.