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Why does $d(p^{1-\gamma}T^\gamma$) give $\frac{dp}{p} = \frac{\gamma}{\gamma - 1} \frac{dT}{T}$? If $(p^{1-\gamma}T^\gamma) = \text{constant}$?

where $p$ is pressure, $T$ is temperature and $\gamma = C_p/C_V$ is the ratio of heat capacities. I've seen examples of this kind of treatment where $d(pV) = p\,dV+V\,dp$, but I'm not sure enough of how it works to reproduce that on the expression above.

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    $\begingroup$ Can you define the symbols you're using? $\endgroup$ – Alex Provost Feb 17 '16 at 22:14
  • $\begingroup$ @AlexProvost Sorry, added to the question! $\endgroup$ – user13948 Feb 17 '16 at 22:23
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Given the notation you are using, I think you are talking about thermodynamics, specifically some kind of adiabatic law.

I think you mean that if $p^{1-\gamma} T^\gamma$ is constant then $\frac{dp}{p} = \frac{\gamma}{\gamma-1} \frac{dT}{T}$. To see that, just calculate $d(p^{1-\gamma} T^\gamma)=(1-\gamma) p^{-\gamma} T^\gamma dp + \gamma p^{1-\gamma} T^{\gamma-1} dT$. Then set that to zero and do algebra. It will probably help to divide both sides by $p^{-\gamma} T^\gamma$.

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  • $\begingroup$ That's exactly what I mean! But why would $d(p^{1-\gamma}t^{\gamma}) = $ the RHS in your answer? Where does that come from? $\endgroup$ – user13948 Feb 17 '16 at 22:22
  • $\begingroup$ @Karacoreable It doesn't. $p^{1-\gamma} T^\gamma$ being constant means its differential is zero, and then you can solve that equation for $\frac{dp}{p}$. $\endgroup$ – Ian Feb 17 '16 at 22:23

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