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From $$Pv^n = \mathrm{constant}$$ When $n = - \infty$ What kind of process is it.... Eg... When $n= \infty$, It is constant volume....

Therefore when $n =- \infty$... What is constant?

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    $\begingroup$ Is this a serious question or is it an "I was just wondering" question? $\endgroup$ – Chet Miller Mar 16 at 11:37
  • $\begingroup$ Hi Chimaobi, welcome to the Physics SE. Please use the MathJax syntax for your math expressions to make them better readable. $\endgroup$ – flaudemus Mar 16 at 14:20
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\begin{align} P v^{-\infty} &= K \\ \frac{P}{v^\infty} &= K \\ \frac{v^\infty}{P} &= \frac{1}{K} \\ v^\infty\frac{1}{P} &= \underbrace{\frac{1}{K}}_{K'} \end{align} Variations in $v$ get scaled by an exponent of $\infty$ and are therefore infinitely more important in determining whether the result is $K'$ than variations in $1/P$ are. The process is still constant volume.

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