# II law of thermodynamics & polytropic process

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...

• I don't see any reason either. Commented Oct 17, 2019 at 15:50
• @ChetMiller. On the Wikipedia page, it says "Negative exponents reflect a process where work and heat flow simultaneously in or out of the system. In the absence of forces except pressure, such a spontaneous process is not allowed by the second law of thermodynamics..." So there, it is talking about spontaneous processes. Clearly, this system cannot be isolated from its environment, because volume and pressure changing together necessarily implies that heat and work flows are in opposite directions.... Commented Oct 25, 2019 at 20:17
• ...Perhaps the answer, then, is that it's possible to show that the change in entropy of the environment forces the total entropy to decrease, meaning that the entropy goes down in an isolate system. Commented Oct 25, 2019 at 20:17
• My understanding is that a polytropic process, by definition, is not spontaneous (i.e., it is reversible). Commented Oct 25, 2019 at 22:33