Question
An ideal gas $\Big(\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma\Big)$ is taken through a process in which the pressure and the volume vary as $p = aV^b$. Find the value of b for which the specific heat capacity in the process is zero.
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Answer
$\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma,\ \text{C}_\text{p}-\text{C}_\text{V}=\text{R},$
$\text{C}_\text{V}=\frac{\text{r}}{\gamma-1},\text{C}_\text{P}=\frac{\gamma\text{R}}{\gamma-1}$
$\text{Pdv}=\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow0=\text{C}_\text{V}\text{dT}+\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow\frac{1}{\text{b}+1}=\frac{-\text{C}_\text{V}}{\text{R}}$
$\Rightarrow\text{b}+1=\frac{-\text{R}}{\text{C}_\text{V}}=\frac{-(\text{C}_\text{P}-\text{C}_\text{V})}{\text{C}_\text{V}}=-\gamma+1$
$\Rightarrow\text{b}=-\gamma$
$\text{C}_\text{V}=\frac{\text{r}}{\gamma-1},\text{C}_\text{P}=\frac{\gamma\text{R}}{\gamma-1}$
$\text{Pdv}=\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow0=\text{C}_\text{V}\text{dT}+\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow\frac{1}{\text{b}+1}=\frac{-\text{C}_\text{V}}{\text{R}}$
$\Rightarrow\text{b}+1=\frac{-\text{R}}{\text{C}_\text{V}}=\frac{-(\text{C}_\text{P}-\text{C}_\text{V})}{\text{C}_\text{V}}=-\gamma+1$
$\Rightarrow\text{b}=-\gamma$
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