$p V^{\frac{5}{3}} \cdot\left(e^{\frac{-p V}{E_{0}}}\right)=C_{1}$
$\Rightarrow \quad p V^{\frac{5}{3}}=C_{1} e^{\frac{p V}{E_{0}}}$
Taking $\log$ to $e$ base, we have
$\log (p)+\frac{5}{3} \log (V)=\log \left(C_{1}\right)+\frac{p V}{E_{0}}$
Differentiating with respect to volume, we have
$\frac{1}{p} \cdot \frac{d p}{d V}+\frac{5 l}{3 V}=0+\frac{1}{E_{0}}\left(p+\frac{V d p}{d V}\right)$
$\Rightarrow \frac{d p}{p}+\frac{5}{3} \frac{d V}{V}=\frac{1}{E_{0}}(p d V+V d p)$
$\Rightarrow \frac{-d V}{d p}=\frac{\left(\frac{1}{p}-\frac{V}{E_{0}}\right)}{\left(\frac{p}{E_{0}}-\frac{5}{3 V}\right)}$
$\text { As, } p V=R T$
$\Rightarrow \quad-\frac{d V}{d p}=\left(\frac{R T}{\frac{R}{E_{0}}-\frac{5 p}{E_{0}}}\right)$
$\Rightarrow \quad-\frac{1}{V} \frac{d V}{d p} =\frac{(R T}{p\left(\frac{1}{E_{0}}-\frac{5}{3 R T}\right)}$
But this is compressibility of gas, so compressibility,
$k=-\frac{1}{V} \frac{d V}{d p}$
$=\frac{\left(\frac{1}{RT}-\frac{1}{E_{0}}\right)}{p\left(\frac{1}{E_{0}}-\frac{5}{3 R T}\right)}$
As $T \rightarrow \infty, \frac{1}{R T} \rightarrow 0$
So, $k$ approaches a constant value at higher temperatures.
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($C$ and $R$ represent capacitance and resistance respectively)