# Why is adiabatic process isentropic?

I have read that adiabatic process is isentropic because there is no heat exchange in an adiabatic process and thus no change in entropy.

But my question is - Even in adiabatic process, work can be done. Let's take an example of an adiabatic vessel with a piston attached. That vessel does not exchange heat but work can be done by pulling in or out the piston.

If the volume of the system changes, isn't entropy also changed, even in a reversible adiabatic process?

EDIT : I know that change in entropy is defined as change in heat divided by temperature. Since there is no change in heat in an adiabatic process, the entropy is zero. My question is different - What I see entropy as - It is a measure of the different microstates in which a system can be. So, even if there is no exchange of heat, can't number of thermally achievable microstates increase if we increase the volume by letting the system do volume work? Why is only heat exchange considered as a mechanism for entropy change?

If I understand the question, you are wondering how to justify the statement that a (reverible) adiabatic process is isentropic from the point of view of statistical mechanics (the classical thermodynamics definition makes sense to you). Let us then start with the entropic fundamental relationship, S = S (U, V, N), where U stands for energy, V for volume, N for number of particles. In many a statistical mechanics texts you will find the explicit definition of S for a system of particles (under usual simplifying assumptions). Inyour example N is constant, but U and V are not: I would be glad to help further if needed, but if you looked at the expression for S as a function of V and N this would answer your question alone. I believe the discussion at Isentropic processes be useful.

• Hi ! Actually I have no knowledge of stastical thermodynamics. Is it possible to answer my question without aany use of it ? – biogirl Apr 17 '14 at 18:36
• just to clarify my question a bit more - I simply want to know why we do not consider that a change in entropy is happening in adiabatic reversible processes even if we can change the volume of the system. Why has only heat been considered a measure of entropy over here ? Doesnt volume also increase the number of microstates ? – biogirl Apr 17 '14 at 18:43
• It does indeed!, but energy is lowered, analogously to a gas cooling while expanding. Did the link I attached at the end of my answer make sense? I think that is a very nice answer, if not I will be glad to try to contribute further – Smerdjakov Apr 17 '14 at 19:36

By definition a reversible adiabatic system has $dQ = 0$.

We also know the following from the Clausius Theorem :

$dS = \frac{dQ}{T}$

Then it is easy to see that there can be no change in entropy.

Note that irreversible adiabatic systems CAN see a change in entropy because in that case the above equation is no longer an equality but an inequality :

$dS < \frac{dQ}{T}$

• hi! thank you but I already know the things you say in the answer . My question is a bit different. – biogirl Apr 15 '14 at 15:58
• Yikes, wish I could be of greater help but my physical-chemistry knowledge is pretty limited ! – Ari Ben Canaan Apr 15 '14 at 16:10
• AriBenCanaan No worries ! – biogirl Apr 15 '14 at 16:39
• AriBenCanaan, from your comment one might deduce dS is zero or negative even in the irreversible case. I believe one should not forget that in an irreversible process entropy is being created within the system, without need for heat supply at the boundaries: it is easyto imagine an adiabatic system in which entrioy is growing. – Smerdjakov Apr 17 '14 at 10:49
• "irreversible adiabatic systems" makes sense only if your meaning of adiabatic is "isolated from heat exchange". Another, quite popular is some circles, meaning of "adiabatic" = "slow enough to be reversible". – Slaviks Apr 17 '14 at 11:16

Consider an isentropic (idealistic adiabatic process, with no friction) process, in which the volume of a system is increased. For that to happen, the internal energy of the system would decrease because the system would be performing work. So any gain in the number of available microstates from an increase in volume is negated by the loss of internal energy by the system, which decreases the number of microstates (lower temperature means smaller molecular speeds means fewer microstates).

Similarly, isentropically decreasing the volume of a system would seem to decrease the number of microstates, but work must be done on the system for this change to occur, causing an increase in the internal energy of the system, which counteracts the decrease in volume and keeps the number of microstates the same.

Others answered the question but I wanted to correct something that you said:

I have read that adiabatic process is isentropic because there is no heat exchange in an adiabatic process and thus no change in entropy.

This is not true. Two comments in particular:

1. An adiabatic process is not necessarily isentropic. It is isentropic only if it is reversible. In fact this is a good rule to memorize:

$$\text{reversible+adiabatic} = \text{isentropic}$$

1. Entropy can change even if heat is not exchanged. That's what we call irreversible adiabatic process.