Physics Asked on April 27, 2021
The polytropic process is defined as such that $pV^m=A$, where $A$ is a constant. Generally, the change in entropy is
$$Delta S=nR ln frac{V}{V_0}+nC_Vln frac{T}{T_0}.$$
Using $pV=nRT$ and $pV^m=A$ we get $T=frac{AV^{1-m}}{nR}$. Substituting into previous equation we obtain:
$$Delta S=(gamma -m)C_Vnln frac{V}{V_0},$$
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…
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.
Answered by MaxL on April 27, 2021
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