I have just started learning thermodynamics and the concept of entropy confuses me.

Suppose I have a gas in a cylindrical container fitted with a piston. I take it through an adiabatic irreversible process to an other state. There is certainly some entropy change. But when I take it through an adiabatic reversible process to the same state as I did in the irreversible process, then the entropy change would be zero, because $Q_\text{reversible}$ in this case is zero. Now since entropy is a state function, $\Delta S$ should be the same in the two cases, but it is not.

Where am I going wrong?

  • $\begingroup$ Mark these words: every time you have a problem reconciling $\Delta S$ with irreversibility, you're probably disregarding the change of entropy in the surroundings. $\endgroup$ – André Chalella Dec 1 '14 at 15:35
  • $\begingroup$ I don't have time to answer right now but look up the proof for the clausius inequality. This will prove everything for you. That inequality is crucial to know! $\endgroup$ – DLV Mar 15 '15 at 7:34

What you are saying is correct: a reversible and adiabatic process between two states, $A$ and $B$, does not change the entropy either of the system or of its environment (surroundings). An irreversible and adiabatic process between two states, $A'$ and $B'$, increases the system's entropy. The two statements are reconciled by noting that if $A=A'$ then $B\ne B'$ and vice versa.

Another way of stating the same, is that an irreversible and adiabatic cycle is impossible. In fact, this statement is almost equivalent to Caratheodory's axiom (a standard formulation of the 2nd law), namely, that in any neighborhood of any state there are states inaccessible via a purely adiabatic process.

The apparent one-sidedness of this, is a verbalization of the increase of entropy function, whose existence is a mathematical consequence of the same. You can sense the physical intuition of an entropy increase as a manifestation of the excess work expended to compensate for the irreversibility of the process to reach a certain state.

  • $\begingroup$ "Another way of stating the same is that an irreversible and adiabatic cycle is impossible." What do you mean by irreversible cycle? I am not sure this exists either. $\endgroup$ – Ján Lalinský Feb 7 '15 at 11:32
  • $\begingroup$ You may wish to read this really excellent article by Pau-Chang Lu from which I paraphrased my comment:"Didactic remarks on the Sears-Kestin statement of the second law of thermodynamics" Am. J. Phys. 50(3), March 1982, pp247-251 $\endgroup$ – hyportnex Feb 7 '15 at 23:30

There is a general formula to get the entropy change between two states, $A$ and $B$, of a system, which is:

\begin{equation} \Delta S_{A \rightarrow B} = \int_{\Gamma^{rev}(A \rightarrow B)} \: \frac{\delta Q_{rev}(\Gamma)}{T} \end{equation}

This formula states that the variation of entropy of a system between the states $A$ and $B$ can always (even for irreversible "real" transformations) be gotten by the integral of $\delta Q_{rev}/T$ along any reversible path that goes from $A$ to $B$.

For instance, the path can be a combination of isothermal, isobaric and isothermal transformations; it does not matter as long as the path goes from $A$ to $B$.

If you consider an initial state $A$ that undergoes either:

  • a reversible adiabatic transformation to a state $B_r$

  • an irreversible adiabatic transformation to a state $B_i$

Then, by definition of the terms reversible and irreversible, and the fact that the entropy is a state function, you know (as emphasized by user31748) that $B_r$ has to be a different state from $B_i$.

That's the actual nature of the real transformation (in your example, it is often a monobaric and adiabatic transformation), that you can find out the state $B_i$ and see how different it is from $B_r$.

In particular, in the above formula, you can see that the variation of entropy between an isobaric and adiabatic transformation, compared to a monobaric and adiabatic transformation, is exactly:

\begin{equation} \Delta S_{A \rightarrow B_i} = \Delta S_{B_r \rightarrow B_i} \int_{\Gamma^{rev}(B_r \rightarrow B_i)} \: \frac{\delta Q_{rev}(\Gamma)}{T} \end{equation}

for any reversible path that goes from $B_r$ to $B_i$.


But when I take it through an adiabatic reversible process to the same state as I did in the irreversible process,

You assume this is possible, but nobody has succeeded doing so. If someone did, it would contradict the second law of thermodynamics for the reason you mentioned.

  • $\begingroup$ What about the adiabatic free expansion of a gas? $\endgroup$ – user106570 Sep 10 '16 at 0:07
  • $\begingroup$ @KaumudiHarikumar, what do you mean? Free adiabatic expansion results in state that has higher entropy. It is an irreversible process. There is no reversible adiabatic process that would end up in the same final state. $\endgroup$ – Ján Lalinský Sep 10 '16 at 8:39
  • $\begingroup$ Oh, OK, sorry; I may have been mistaken. $\endgroup$ – user106570 Sep 10 '16 at 8:52

Entropy is a state function, correct ! But entropy itself depends upon how you got to that state (final state).

In thermodynamics Entropy is considered the quality of heat (hotness/coolness) whereas temperature is considered quantity/degree of heat (hotness/coolness).

The reversible processes (and adiabatic i.e. no heat exchange) do not mess with the internal structure and probability of predictability of the states of each molecules of the system.

Whereas the irreversible processes definitely modify the internal structure of the states/positions of the molecules of the system (or you may want to say that the molecules after having being irreversibly processed not remain so very predictable anymore). This is the reason behind such processes are named irreversible in the first place.

Concept of Entropy is not intuitive. Although should you, for example, consider an air-conditioner, what you will find in the evaporator fins of the air-conditioner is that mainly entropy is exchanged not temperature. Because entropy is exchanged (entropy is like sucked by the refrigerant from the room air), temperature of the room-air automatically goes down and of the refrigerant (working fluid) goes up. Refrigerant from being in a cool liquid form becomes superheated (because it kind of sucked in entropy from the room). This single branch of process (out of the whole refrigeration cycle) is isothermal i.e. temperature constant. But at the same time entropy does its thing and cools down the room air.

Although this example might not be the perfect example it has worked well for me to understand the concept of entropy. I can now understand that there is not only the temperature (quantity/degree of hotness/coolness) but entropy (quality of hotness/coolness) as well.


Let me explain it in simple words without equations.

When you have a closed system at state A, you have to use energy on it to convert it into state B. If the process of shifting from state A to B required no energy, the system would already be in state B.

Now when you have the system at state B and want to return it to state A, you again have to use energy for this purpose.

This use of energy for changing the system state is what increases the entropy. The lesser energy applied, the smaller the change in entropy, but since you must always spend energy, even if very little, the entropy will change with it.

  • $\begingroup$ "When you have a closed system at state A, you have to use energy on it to convert it into state B." Expansion of gas in a cylinder requires no expenditure of energy. It happens by itself if the motion of the piston is allowed. $\endgroup$ – Ján Lalinský Feb 7 '15 at 11:22

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