The polytropic process is defined as such that $pV^m=A$, where $A$ is a constant. Generally, the change in entropy is
$$\Delta S=nR \ln \frac{V}{V_0}+nC_V\ln \frac{T}{T_0}.$$
Using $pV=nRT$ and $pV^m=A$ we get $T=\frac{AV^{1-m}}{nR}$. Substituting into previous equation we obtain:
$$\Delta S=(\gamma -m)C_Vn\ln \frac{V}{V_0},$$
which is the formula for $\Delta S$ in polytropic process and $\gamma=C_p/C_V$. According to Wikipedia, for $m < 0$ II law of thermodynamics would be violated. The problem is, I can't really see why it would be – the entropy is positive and everything seems fine…
[Physics] II law of thermodynamics & polytropic process
entropyideal-gasthermodynamics
Best Answer
If you decrease the pressure p in the polytropic process, the volume V of the system will be reduced too to fulfill the equality. Therefore the relation between the final and the initial volumes will be less than 1, and the change of the entropy will be negative.