What's the problem to show that $\Delta S > 0$ in this process?

In a book on Thermodynamics I found one exercise that asked the following:

What are the difficulties in showing explicitly that the entropy of an ideal gas must increase during an irreversible adiabatic compression?

Now, I really didn't understand the point with this. First of all, I thought that $\Delta S > 0$ was the very definition of irreversible.

Second, if the process is adiabatic $dQ = 0$ and so $dS=dQ/T$ seems to be zero also. So there is something really wrong here.

I thought for some time on this now and couldn't get any conclusion.

What are those difficulties, and how would anyone actually show that $\Delta S > 0$ in such process?

• An irreversible adiabatic process is not isentropic. Hence as far as the process is not reversible, $dS\neq 0$ – UKH Oct 17 '16 at 3:47

2 Answers

Entropy change of a system has two causes: 1) Flux of entropy due to exchange of mass or heat with surroundings, $\Delta S_{flux}$, and 2) Entropy generated internally, $\Delta S_{generated}$. Total change in entropy of a system is therefore, $\Delta S_{total}=\Delta S_{flux}+\Delta S_{generated}$. While $\Delta S_{flux}$ can be either positive or negative, $\Delta S_{generated}$ is strictly non-negative, i.e. $\Delta S_{generated}\geq 0$. In fact this may be taken as an alternative way of stating the second law (see: Modern Thermodynamics, by Kondepudi and Prigogine). $\Delta S_{generated}\geq 0$ has the status of a postulate, and cannot be proved (unless you get into statistical mechanics, where again a new postulate such as ergodic hypothesis is required).

In a closed system undergoing adiabatic process, flux of entropy is zero (there is neither mass nor heat exchange with surroundings). But entropy can be generated internally due to dissipative processes, such as due to viscous action within flowing gas when it is being compressed. Since $\Delta S_{flux}=0$ and $\Delta S_{generated}\geq 0$, we have $\Delta S_{total}\geq 0$, where in this case $\Delta S_{total}$ also equals change of entropy of the universe. If you can compress the gas quasi-statically such that $\Delta S_{generated}=0$, then and only then do you have a reversible process.

P.S. You are right in saying that $\Delta S>0$ implies irreversible process, with the caveat that entropy change of the entire universe must be considered.

$dS=\frac{dQ_\textrm{rev}}{T}$....(heat supplied in the reversible process) ...in other words, it means that the formula is valid only for a reversible process where $\frac{dQ_\textrm{rev}}{T}$ is a state function. The formula you have to use for say an irreversible adiabatic expansion would be $\Delta S = nC_{v}\ln\frac{T_2}{T_1} + n\mathcal R\ln\frac{V_2}{V_1} > 0$. The reasoning is as follows....

You cannot use $T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ because the final temperature reached by the same amount of expansion in a reversible and an irreversible process are different ( $T_\textrm{2irrev}$ > $T_\textrm{2rev}$ )